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# Posts Tagged ‘determinant’

## phy456 Problem set 4, problem 2 notes

Posted by peeterjoot on October 12, 2011

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# Problem 2.

I was deceived by an incorrect result in Mathematica, which led me to believe that the second order energy perturbation was zero (whereas part (c) of the problem asked if it was greater or lesser than zero). I started starting writing this up to show my reasoning, but our Professor quickly provided an example after class showing how this zero must be wrong, and I didn’t have to show him any of this.

## Setup

Recall first the one dimensional particle in a box. Within the box we have to solve

\begin{aligned}\frac{P^2}{2m} \psi = E\psi\end{aligned} \hspace{\stretch{1}}(1.1)

and find

\begin{aligned}\psi \sim e^{\frac{i}{\hbar} \sqrt{2 m E} x} \end{aligned} \hspace{\stretch{1}}(1.2)

With

\begin{aligned}k = \frac{\sqrt{2 m E}}{\hbar}\end{aligned} \hspace{\stretch{1}}(1.3)

our general state, involving terms of each sign, takes the form

\begin{aligned}\psi = A e^{ i k x } +B e^{ -i k x }\end{aligned} \hspace{\stretch{1}}(1.4)

Inserting boundary conditions gives us

\begin{aligned}\begin{bmatrix}\psi(-L/2) \\ \psi(L/2)\end{bmatrix}\begin{bmatrix}e^{ -i k \frac{L}{2} } +e^{ i k \frac{L}{2} } \\ e^{ i k \frac{L}{2} } +e^{ -i k \frac{L}{2} }\end{bmatrix}\begin{bmatrix}A \\ B\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(1.5)

The determinant is zero

\begin{aligned}e^{-i k L} - e^{i k L} = 0,\end{aligned} \hspace{\stretch{1}}(1.6)

which provides our constraint on $k$

\begin{aligned}e^{2 i k L} = 1.\end{aligned} \hspace{\stretch{1}}(1.7)

We require $2 k L = 2 \pi n$ for any integer $n$, or

\begin{aligned}k = \frac{\pi n}{L}.\end{aligned} \hspace{\stretch{1}}(1.8)

This quantizes the energy, and inverting 1.3 gives us

\begin{aligned}E = \frac{1}{{2m}} \left( \frac{\hbar \pi n }{L} \right)^2.\end{aligned} \hspace{\stretch{1}}(1.9)

To complete the task of matching boundary value conditions we cheat and recall that the particular linear combinations that we need to match the boundary constraint of zero at $\pm L/2$ were sums and differences yielding cosines and sines respectively. Since

\begin{aligned}{\left.{{\sin\left( \frac{\pi n x }{L} \right) }}\right\vert}_{{x = \pm L/2}} = \pm \sin\left(\frac{\pi n}{2}\right)\end{aligned} \hspace{\stretch{1}}(1.10)

So sines are the wave functions for $n = 2, 4, ...$ since $\sin(n \pi) = 0$ for integer $n$. Similarly

\begin{aligned}{\left.{{\cos\left( \frac{\pi n x }{L} \right) }}\right\vert}_{{x = \pm L/2}} = \cos\left(\frac{\pi n}{2}\right).\end{aligned} \hspace{\stretch{1}}(1.11)

Cosine becomes zero at $\pi/2, 3\pi/2, \cdots$, so our wave function is the cosine for $n = 1, 3, 5, \cdots$.

Normalizing gives us

\begin{aligned}\psi_n(x) = \sqrt{\frac{2}{L}}\left\{\begin{array}{l l}\cos\left(\frac{\pi n x}{L}\right) & \quad n = 1, 3, 5, \cdots \\ \sin\left(\frac{\pi n x}{L}\right) & \quad n = 2, 4, 6, \cdots \end{array}\right.\end{aligned} \hspace{\stretch{1}}(1.12)

## Two non-interacting particles. Three lowest energy levels and degeneracies

Forming the Hamiltonian for two particles in the box without interaction, we have within the box

\begin{aligned}H = \frac{P_1^2}{2m} +\frac{P_2^2}{2m} \end{aligned} \hspace{\stretch{1}}(1.13)

we can apply separation of variables, and it becomes clear that our wave functions have the form

\begin{aligned}\psi_{nm}(x_1, x_2) = \psi_n(x_1) \psi_m(x_2)\end{aligned} \hspace{\stretch{1}}(1.14)

Plugging in

\begin{aligned}H \psi = E \psi,\end{aligned} \hspace{\stretch{1}}(1.15)

supplies the energy levels for the two particle wavefunction, giving

\begin{aligned}\begin{aligned}H \psi_{nm} &= \frac{\hbar^2}{2m}\left(\left(\frac{\pi n}{L}\right)^2+\left(\frac{\pi m}{L}\right)^2\right)\psi_{nm} \\ &= \frac{1}{2m} \left(\frac{\hbar \pi}{L}\right)^2 ( n^2 + m^2 ) \psi_{nm}\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.16)

Letting $n, m$ each range over $[1,3]$ for example we find

\begin{aligned}\begin{array}{l l l}n & m & n^2 + m^2 \\ 1 & 1 & 2 \\ 1 & 2 & 5 \\ 1 & 3 & 10 \\ 2 & 1 & 5 \\ 2 & 2 & 8 \\ 2 & 3 & 13 \\ 3 & 1 & 10 \\ 3 & 2 & 13 \\ 3 & 3 & 18\end{array}\end{aligned} \hspace{\stretch{1}}(1.17)

It’s clear that our lowest energy levels are

\begin{aligned}\frac{1}{m} \left(\frac{\hbar \pi}{L}\right)^2 \\ \frac{5}{2m} \left(\frac{\hbar \pi}{L}\right)^2 \\ \frac{4}{m} \left(\frac{\hbar \pi}{L}\right)^2 \end{aligned}

with degeneracies $1, 2, 1$ respectively.

## Ground state energy with interaction perturbation to first order.

With $c_0$ positive and an interaction potential of the form

\begin{aligned}U(X_1, X_2) = - c_0 \delta(X_1 - X_2)\end{aligned} \hspace{\stretch{1}}(1.18)

The second order perturbation of the ground state energy is

\begin{aligned}E = E_{11}^{(0)} + H_{11;11}' + \sum_{nm \ne 11} \frac{{\left\lvert{H_{11;11}' }\right\rvert}^2}{E_{11} - E_{nm}}\end{aligned} \hspace{\stretch{1}}(1.19)

where

\begin{aligned}E_{11}^{(0)} = \frac{1}{m} \left(\frac{\hbar \pi}{L}\right)^2,\end{aligned} \hspace{\stretch{1}}(1.20)

and

\begin{aligned}H_{nm;ab}' = -c_0 {\langle {\psi_{nm}} \rvert} \delta(X_1 - X_2) {\lvert {\psi_{ab}} \rangle}\end{aligned} \hspace{\stretch{1}}(1.21)

to proceed, we need to expand the matrix element

\begin{aligned}{\langle {\psi_{nm}} \rvert} \delta(X_1 - X_2) {\lvert {\psi_{ab}} \rangle}&=\int dx_1 dx_2 dy_1 dy_2\left\langle{{\psi_{nm}}} \vert {{x_1 x_2}}\right\rangle {\langle {x_1 x_2} \rvert} \delta(X_1 - X_2) {\lvert {y_1 y_2 } \rangle} \left\langle{{y_1 y_2}} \vert {{\psi_{ab}}}\right\rangle \\ &=\int dx_1 dx_2 dy_1 dy_2\left\langle{{\psi_{nm}}} \vert {{x_1 x_2}}\right\rangle \delta(x_1 - x_2) \delta^2(\mathbf{x} - \mathbf{y}) \left\langle{{y_1 y_2}} \vert {{\psi_{ab}}}\right\rangle \\ &=\int dx_1 dx_2 \left\langle{{\psi_{nm}}} \vert {{x_1 x_2}}\right\rangle \delta(x_1 - x_2) \left\langle{{x_1 x_2}} \vert {{\psi_{ab}}}\right\rangle \\ &=\int_{-L/2}^{L/2} dx\psi_{nm}(x, x)\psi_{ab}(x, x)\end{aligned}

So, for our first order calculation we need

\begin{aligned}H_{11; 11}' &= - c_0\int_{-L/2}^{L/2} dx\psi_{11}(x, x)\psi_{11}(x, x) \\ &=\frac{4}{L^2}\int_{-L/2}^{L/2} dx\cos^4( \pi x /L ) \\ &=- \frac{3 c_0}{2 L}\end{aligned}

For the second order perturbation of the energy, it is clear that this will reduce the first order approximation for each matrix element that is non-zero.

Attempting that calculation with \href{https://github.com/peeterjoot/physicsplay/blob/796c8e3739ae1a9ca26270a0e91384afba45661d/notes/phy456/problem\

This worksheet can be seen to be giving misleading results, by evaluating

\begin{aligned}\int_{-\frac{L}{2}}^{\frac{L}{2}} \left(\frac{2}{L}\right)^2 \cos ^2\left(\frac{\pi x}{L}\right) \cos ^2\left(\frac{3 \pi x}{L}\right) \, dx = \frac{1}{L}\end{aligned} \hspace{\stretch{1}}(1.22)

Yet, the FullSimplify gives

\begin{aligned}\text{FullSimplify}\left[\int_{-\frac{L}{2}}^{\frac{L}{2}} \text{Cos}\left[\frac{\pi x}{L}\right]^2 \left(\frac{2}{L}\right)^2 \text{Cos}\left[\frac{(2 n+1) \pi x}{L}\right] \text{Cos}\left[\frac{(2 m+1) \pi x}{L}\right] \, dx,\{m,n\}\in \text{Integers}\right] = 0\end{aligned} \hspace{\stretch{1}}(1.23)

## PHY450H1S (relativistic electrodynamics) Problem Set 3.

Posted by peeterjoot on March 2, 2011

# Disclaimer.

This problem set is as yet ungraded (although only the second question will be graded).

# Problem 1. Fun with $\epsilon_{\alpha\beta\gamma}$, $\epsilon^{ijkl}$, $F_{ij}$, and the duality of Maxwell’s equations in vacuum.

## 1. Statement. rank 3 spatial antisymmetric tensor identities.

Prove that

\begin{aligned}\epsilon_{\alpha \beta \gamma}\epsilon_{\mu \nu \gamma}=\delta_{\alpha\mu} \delta_{\beta\nu}-\delta_{\alpha\nu} \delta_{\beta\mu}\end{aligned} \hspace{\stretch{1}}(2.1)

and use it to find the familiar relation for

\begin{aligned}(\mathbf{A} \times \mathbf{B}) \cdot (\mathbf{C} \times \mathbf{D})\end{aligned} \hspace{\stretch{1}}(2.2)

Also show that

\begin{aligned}\epsilon_{\alpha \beta \gamma}\epsilon_{\mu \beta \gamma}=2 \delta_{\alpha\mu}.\end{aligned} \hspace{\stretch{1}}(2.3)

(Einstein summation implied all throughout this problem).

## 1. Solution

We can explicitly expand the (implied) sum over indexes $\gamma$. This is

\begin{aligned}\epsilon_{\alpha \beta \gamma}\epsilon_{\mu \nu \gamma}=\epsilon_{\alpha \beta 1} \epsilon_{\mu \nu 1}+\epsilon_{\alpha \beta 2} \epsilon_{\mu \nu 2}+\epsilon_{\alpha \beta 3} \epsilon_{\mu \nu 3}\end{aligned} \hspace{\stretch{1}}(2.4)

For any $\alpha \ne \beta$ only one term is non-zero. For example with $\alpha,\beta = 2,3$, we have just a contribution from the $\gamma = 1$ part of the sum

\begin{aligned}\epsilon_{2 3 1} \epsilon_{\mu \nu 1}.\end{aligned} \hspace{\stretch{1}}(2.5)

The value of this for $(\mu,\nu) = (\alpha,\beta)$ is

\begin{aligned}(\epsilon_{2 3 1})^2\end{aligned} \hspace{\stretch{1}}(2.6)

whereas for $(\mu,\nu) = (\beta,\alpha)$ we have

\begin{aligned}-(\epsilon_{2 3 1})^2\end{aligned} \hspace{\stretch{1}}(2.7)

Our sum has value one when $(\alpha, \beta)$ matches $(\mu, \nu)$, and value minus one for when $(\mu, \nu)$ are permuted. We can summarize this, by saying that when $\alpha \ne \beta$ we have

\begin{aligned}\boxed{\epsilon_{\alpha \beta \gamma}\epsilon_{\mu \nu \gamma}=\delta_{\alpha\mu} \delta_{\beta\nu}-\delta_{\alpha\nu} \delta_{\beta\mu}.}\end{aligned} \hspace{\stretch{1}}(2.8)

However, observe that when $\alpha = \beta$ the RHS is

\begin{aligned}\delta_{\alpha\mu} \delta_{\alpha\nu}-\delta_{\alpha\nu} \delta_{\alpha\mu} = 0,\end{aligned} \hspace{\stretch{1}}(2.9)

as desired, so this form works in general without any $\alpha \ne \beta$ qualifier, completing this part of the problem.

\begin{aligned}(\mathbf{A} \times \mathbf{B}) \cdot (\mathbf{C} \times \mathbf{D})&=(\epsilon_{\alpha \beta \gamma} \mathbf{e}^\alpha A^\beta B^\gamma ) \cdot(\epsilon_{\mu \nu \sigma} \mathbf{e}^\mu C^\nu D^\sigma ) \\ &=\epsilon_{\alpha \beta \gamma} A^\beta B^\gamma\epsilon_{\alpha \nu \sigma} C^\nu D^\sigma \\ &=(\delta_{\beta \nu} \delta_{\gamma\sigma}-\delta_{\beta \sigma} \delta_{\gamma\nu} )A^\beta B^\gammaC^\nu D^\sigma \\ &=A^\nu B^\sigmaC^\nu D^\sigma-A^\sigma B^\nuC^\nu D^\sigma.\end{aligned}

This gives us

\begin{aligned}\boxed{(\mathbf{A} \times \mathbf{B}) \cdot (\mathbf{C} \times \mathbf{D})=(\mathbf{A} \cdot \mathbf{C})(\mathbf{B} \cdot \mathbf{D})-(\mathbf{A} \cdot \mathbf{D})(\mathbf{B} \cdot \mathbf{C}).}\end{aligned} \hspace{\stretch{1}}(2.10)

We have one more identity to deal with.

\begin{aligned}\epsilon_{\alpha \beta \gamma}\epsilon_{\mu \beta \gamma}\end{aligned} \hspace{\stretch{1}}(2.11)

We can expand out this (implied) sum slow and dumb as well

\begin{aligned}\epsilon_{\alpha \beta \gamma}\epsilon_{\mu \beta \gamma}&=\epsilon_{\alpha 1 2} \epsilon_{\mu 1 2}+\epsilon_{\alpha 2 1} \epsilon_{\mu 2 1} \\ &+\epsilon_{\alpha 1 3} \epsilon_{\mu 1 3}+\epsilon_{\alpha 3 1} \epsilon_{\mu 3 1} \\ &+\epsilon_{\alpha 2 3} \epsilon_{\mu 2 3}+\epsilon_{\alpha 3 2} \epsilon_{\mu 3 2} \\ &=2 \epsilon_{\alpha 1 2} \epsilon_{\mu 1 2}+ 2 \epsilon_{\alpha 1 3} \epsilon_{\mu 1 3}+ 2 \epsilon_{\alpha 2 3} \epsilon_{\mu 2 3}\end{aligned}

Now, observe that for any $\alpha \in (1,2,3)$ only one term of this sum is picked up. For example, with no loss of generality, pick $\alpha = 1$. We are left with only

\begin{aligned}2 \epsilon_{1 2 3} \epsilon_{\mu 2 3}\end{aligned} \hspace{\stretch{1}}(2.12)

This has the value

\begin{aligned}2 (\epsilon_{1 2 3})^2 = 2\end{aligned} \hspace{\stretch{1}}(2.13)

when $\mu = \alpha$ and is zero otherwise. We can therefore summarize the evaluation of this sum as

\begin{aligned}\boxed{\epsilon_{\alpha \beta \gamma}\epsilon_{\mu \beta \gamma}= 2\delta_{\alpha\mu},}\end{aligned} \hspace{\stretch{1}}(2.14)

completing this problem.

## 2. Statement. Determinant of three by three matrix.

Prove that for any $3 \times 3$ matrix ${\left\lVert{A_{\alpha\beta}}\right\rVert}$: $\epsilon_{\mu\nu\lambda} A_{\alpha \mu} A_{\beta\nu} A_{\gamma\lambda} = \epsilon_{\alpha \beta \gamma} \text{Det} A$ and that $\epsilon_{\alpha\beta\gamma} \epsilon_{\mu\nu\lambda} A_{\alpha \mu} A_{\beta\nu} A_{\gamma\lambda} = 6 \text{Det} A$.

## 2. Solution

In class Simon showed us how the first identity can be arrived at using the triple product $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \text{Det}(\mathbf{a} \mathbf{b} \mathbf{c})$. It occurred to me later that I’d seen the identity to be proven in the context of Geometric Algebra, but hadn’t recognized it in this tensor form. Basically, a wedge product can be expanded in sums of determinants, and when the dimension of the space is the same as the vector, we have a pseudoscalar times the determinant of the components.

For example, in $\mathbb{R}^{2}$, let’s take the wedge product of a pair of vectors. As preparation for the relativistic $\mathbb{R}^{4}$ case We won’t require an orthonormal basis, but express the vector in terms of a reciprocal frame and the associated components

\begin{aligned}a = a^i e_i = a_j e^j\end{aligned} \hspace{\stretch{1}}(2.15)

where

\begin{aligned}e^i \cdot e_j = {\delta^i}_j.\end{aligned} \hspace{\stretch{1}}(2.16)

When we get to the relativistic case, we can pick (but don’t have to) the standard basis

\begin{aligned}e_0 &= (1, 0, 0, 0) \\ e_1 &= (0, 1, 0, 0) \\ e_2 &= (0, 0, 1, 0) \\ e_3 &= (0, 0, 0, 1),\end{aligned} \hspace{\stretch{1}}(2.17)

for which our reciprocal frame is implicitly defined by the metric

\begin{aligned}e^0 &= (1, 0, 0, 0) \\ e^1 &= (0, -1, 0, 0) \\ e^2 &= (0, 0, -1, 0) \\ e^3 &= (0, 0, 0, -1).\end{aligned} \hspace{\stretch{1}}(2.21)

Anyways. Back to the problem. Let’s examine the $\mathbb{R}^{2}$ case. Our wedge product in coordinates is

\begin{aligned}a \wedge b=a^i b^j (e_i \wedge e_j)\end{aligned} \hspace{\stretch{1}}(2.25)

Since there are only two basis vectors we have

\begin{aligned}a \wedge b=(a^1 b^2 - a^2 b^1) e_1 \wedge e_2 = \text{Det} {\left\lVert{a^i b^j}\right\rVert} (e_1 \wedge e_2).\end{aligned} \hspace{\stretch{1}}(2.26)

Our wedge product is a product of the determinant of the vector coordinates, times the $\mathbb{R}^{2}$ pseudoscalar $e_1 \wedge e_2$.

This doesn’t look quite like the $\mathbb{R}^{3}$ relation that we want to prove, which had an antisymmetric tensor factor for the determinant. Observe that we get the determinant by picking off the $e_1 \wedge e_2$ component of the bivector result (the only component in this case), and we can do that by dotting with $e^2 \cdot e^1$. To get an antisymmetric tensor times the determinant, we have only to dot with a different pseudoscalar (one that differs by a possible sign due to permutation of the indexes). That is

\begin{aligned}(e^t \wedge e^s) \cdot (a \wedge b)&=a^i b^j (e^t \wedge e^s) \cdot (e_i \wedge e_j) \\ &=a^i b^j\left( {\delta^{s}}_i {\delta^{t}}_j-{\delta^{t}}_i {\delta^{s}}_j \right) \\ &=a^i b^j{\delta^{[t}}_j {\delta^{s]}}_i \\ &=a^i b^j{\delta^{t}}_{[j} {\delta^{s}}_{i]} \\ &=a^{[i} b^{j]}{\delta^{t}}_{j} {\delta^{s}}_{i} \\ &=a^{[s} b^{t]}\end{aligned}

Now, if we write $a^i = A^{1 i}$ and $b^j = A^{2 j}$ we have

\begin{aligned}(e^t \wedge e^s) \cdot (a \wedge b)=A^{1 s} A^{2 t} -A^{1 t} A^{2 s}\end{aligned} \hspace{\stretch{1}}(2.27)

We can write this in two different ways. One of which is

\begin{aligned}A^{1 s} A^{2 t} -A^{1 t} A^{2 s} =\epsilon^{s t} \text{Det} {\left\lVert{A^{ij}}\right\rVert}\end{aligned} \hspace{\stretch{1}}(2.28)

and the other of which is by introducing free indexes for $1$ and $2$, and summing antisymmetrically over these. That is

\begin{aligned}A^{1 s} A^{2 t} -A^{1 t} A^{2 s}=A^{a s} A^{b t} \epsilon_{a b}\end{aligned} \hspace{\stretch{1}}(2.29)

So, we have

\begin{aligned}\boxed{A^{a s} A^{b t} \epsilon_{a b} =A^{1 i} A^{2 j} {\delta^{[t}}_j {\delta^{s]}}_i =\epsilon^{s t} \text{Det} {\left\lVert{A^{ij}}\right\rVert},}\end{aligned} \hspace{\stretch{1}}(2.30)

This result hold regardless of the metric for the space, and does not require that we were using an orthonormal basis. When the metric is Euclidean and we have an orthonormal basis, then all the indexes can be dropped.

The $\mathbb{R}^{3}$ and $\mathbb{R}^{4}$ cases follow in exactly the same way, we just need more vectors in the wedge products.

For the $\mathbb{R}^{3}$ case we have

\begin{aligned}(e^u \wedge e^t \wedge e^s) \cdot ( a \wedge b \wedge c)&=a^i b^j c^k(e^u \wedge e^t \wedge e^s) \cdot (e_i \wedge e_j \wedge e_k) \\ &=a^i b^j c^k{\delta^{[u}}_k{\delta^{t}}_j{\delta^{s]}}_i \\ &=a^{[s} b^t c^{u]}\end{aligned}

Again, with $a^i = A^{1 i}$ and $b^j = A^{2 j}$, and $c^k = A^{3 k}$ we have

\begin{aligned}(e^u \wedge e^t \wedge e^s) \cdot ( a \wedge b \wedge c)=A^{1 i} A^{2 j} A^{3 k}{\delta^{[u}}_k{\delta^{t}}_j{\delta^{s]}}_i\end{aligned} \hspace{\stretch{1}}(2.31)

and we can choose to write this in either form, resulting in the identity

\begin{aligned}\boxed{\epsilon^{s t u} \text{Det} {\left\lVert{A^{ij}}\right\rVert}=A^{1 i} A^{2 j} A^{3 k}{\delta^{[u}}_k{\delta^{t}}_j{\delta^{s]}}_i=\epsilon_{a b c} A^{a s} A^{b t} A^{c u}.}\end{aligned} \hspace{\stretch{1}}(2.32)

The $\mathbb{R}^{4}$ case follows exactly the same way, and we have

\begin{aligned}(e^v \wedge e^u \wedge e^t \wedge e^s) \cdot ( a \wedge b \wedge c \wedge d)&=a^i b^j c^k d^l(e^v \wedge e^u \wedge e^t \wedge e^s) \cdot (e_i \wedge e_j \wedge e_k \wedge e_l) \\ &=a^i b^j c^k d^l{\delta^{[v}}_l{\delta^{u}}_k{\delta^{t}}_j{\delta^{s]}}_i \\ &=a^{[s} b^t c^{u} d^{v]}.\end{aligned}

This time with $a^i = A^{0 i}$ and $b^j = A^{1 j}$, and $c^k = A^{2 k}$, and $d^l = A^{3 l}$ we have

\begin{aligned}\boxed{\epsilon^{s t u v} \text{Det} {\left\lVert{A^{ij}}\right\rVert}=A^{0 i} A^{1 j} A^{2 k} A^{3 l}{\delta^{[v}}_l{\delta^{u}}_k{\delta^{t}}_j{\delta^{s]}}_i=\epsilon_{a b c d} A^{a s} A^{b t} A^{c u} A^{d v}.}\end{aligned} \hspace{\stretch{1}}(2.33)

This one is almost the identity to be established later in problem 1.4. We have only to raise and lower some indexes to get that one. Note that in the Minkowski standard basis above, because $s, t, u, v$ must be a permutation of $0,1,2,3$ for a non-zero result, we must have

\begin{aligned}\epsilon^{s t u v} = (-1)^3 (+1) \epsilon_{s t u v}.\end{aligned} \hspace{\stretch{1}}(2.34)

So raising and lowering the identity above gives us

\begin{aligned}-\epsilon_{s t u v} \text{Det} {\left\lVert{A_{ij}}\right\rVert}=\epsilon^{a b c d} A_{a s} A_{b t} A_{c u} A_{d u}.\end{aligned} \hspace{\stretch{1}}(2.35)

No sign changes were required for the indexes $a, b, c, d$, since they are paired.

Until we did the raising and lowering operations here, there was no specific metric required, so our first result 2.33 is the more general one.

There’s one more part to this problem, doing the antisymmetric sums over the indexes $s, t, \cdots$. For the $\mathbb{R}^{2}$ case we have

\begin{aligned}\epsilon_{s t} \epsilon_{a b} A^{a s} A^{b t}&=\epsilon_{s t} \epsilon^{s t} \text{Det} {\left\lVert{A^{ij}}\right\rVert} \\ &=\left( \epsilon_{1 2} \epsilon^{1 2} +\epsilon_{2 1} \epsilon^{2 1} \right)\text{Det} {\left\lVert{A^{ij}}\right\rVert} \\ &=\left( 1^2 + (-1)^2\right)\text{Det} {\left\lVert{A^{ij}}\right\rVert}\end{aligned}

We conclude that

\begin{aligned}\boxed{\epsilon_{s t} \epsilon_{a b} A^{a s} A^{b t} = 2! \text{Det} {\left\lVert{A^{ij}}\right\rVert}.}\end{aligned} \hspace{\stretch{1}}(2.36)

For the $\mathbb{R}^{3}$ case we have the same operation

\begin{aligned}\epsilon_{s t u} \epsilon_{a b c} A^{a s} A^{b t} A^{c u}&=\epsilon_{s t u} \epsilon^{s t u} \text{Det} {\left\lVert{A^{ij}}\right\rVert} \\ &=\left( \epsilon_{1 2 3} \epsilon^{1 2 3} +\epsilon_{1 3 2} \epsilon^{1 3 2} + \cdots\right)\text{Det} {\left\lVert{A^{ij}}\right\rVert} \\ &=(\pm 1)^2 (3!)\text{Det} {\left\lVert{A^{ij}}\right\rVert}.\end{aligned}

So we conclude

\begin{aligned}\boxed{\epsilon_{s t u} \epsilon_{a b c} A^{a s} A^{b t} A^{c u}= 3! \text{Det} {\left\lVert{A^{ij}}\right\rVert}.}\end{aligned} \hspace{\stretch{1}}(2.37)

It’s clear what the pattern is, and if we evaluate the sum of the antisymmetric tensor squares in $\mathbb{R}^{4}$ we have

\begin{aligned}\epsilon_{s t u v} \epsilon_{s t u v}&=\epsilon_{0 1 2 3} \epsilon_{0 1 2 3}+\epsilon_{0 1 3 2} \epsilon_{0 1 3 2}+\epsilon_{0 2 1 3} \epsilon_{0 2 1 3}+ \cdots \\ &= (\pm 1)^2 (4!),\end{aligned}

So, for our SR case we have

\begin{aligned}\boxed{\epsilon_{s t u v} \epsilon_{a b c d} A^{a s} A^{b t} A^{c u} A^{d v}= 4! \text{Det} {\left\lVert{A^{ij}}\right\rVert}.}\end{aligned} \hspace{\stretch{1}}(2.38)

This was part of question 1.4, albeit in lower index form. Here since all indexes are matched, we have the same result without major change

\begin{aligned}\boxed{\epsilon^{s t u v} \epsilon^{a b c d} A_{a s} A_{b t} A_{c u} A_{d v}= 4! \text{Det} {\left\lVert{A_{ij}}\right\rVert}.}\end{aligned} \hspace{\stretch{1}}(2.39)

The main difference is that we are now taking the determinant of a lower index tensor.

## 3. Statement. Rotational invariance of 3D antisymmetric tensor

Use the previous results to show that $\epsilon_{\mu\nu\lambda}$ is invariant under rotations.

## 3. Solution

We apply transformations to coordinates (and thus indexes) of the form

\begin{aligned}x_\mu \rightarrow O_{\mu\nu} x_\nu\end{aligned} \hspace{\stretch{1}}(2.40)

With our tensor transforming as its indexes, we have

\begin{aligned}\epsilon_{\mu\nu\lambda} \rightarrow \epsilon_{\alpha\beta\sigma} O_{\mu\alpha} O_{\nu\beta} O_{\lambda\sigma}.\end{aligned} \hspace{\stretch{1}}(2.41)

We’ve got 2.32, which after dropping indexes, because we are in a Euclidean space, we have

\begin{aligned}\epsilon_{\mu \nu \lambda} \text{Det} {\left\lVert{A_{ij}}\right\rVert} = \epsilon_{\alpha \beta \sigma} A_{\alpha \mu} A_{\beta \nu} A_{\sigma \lambda}.\end{aligned} \hspace{\stretch{1}}(2.42)

Let $A_{i j} = O_{j i}$, which gives us

\begin{aligned}\epsilon_{\mu\nu\lambda} \rightarrow \epsilon_{\mu\nu\lambda} \text{Det} A^\text{T}\end{aligned} \hspace{\stretch{1}}(2.43)

but since $\text{Det} O = \text{Det} O^\text{T}$, we have shown that $\epsilon_{\mu\nu\lambda}$ is invariant under rotation.

## 4. Statement. Rotational invariance of 4D antisymmetric tensor

Use the previous results to show that $\epsilon_{i j k l}$ is invariant under Lorentz transformations.

## 4. Solution

This follows the same way. We assume a transformation of coordinates of the following form

\begin{aligned}(x')^i &= {O^i}_j x^j \\ (x')_i &= {O_i}^j x_j,\end{aligned} \hspace{\stretch{1}}(2.44)

where the determinant of ${O^i}_j = 1$ (sanity check of sign: ${O^i}_j = {\delta^i}_j$).

Our antisymmetric tensor transforms as its coordinates individually

\begin{aligned}\epsilon_{i j k l} &\rightarrow \epsilon_{a b c d} {O_i}^a{O_j}^b{O_k}^c{O_l}^d \\ &= \epsilon^{a b c d} O_{i a}O_{j b}O_{k c}O_{l d} \\ \end{aligned}

Let $P_{ij} = O_{ji}$, and raise and lower all the indexes in 2.46 for

\begin{aligned}-\epsilon_{s t u v} \text{Det} {\left\lVert{P_{ij}}\right\rVert}=\epsilon^{a b c d} P_{a s} P_{b t} P_{c u} P_{d v}.\end{aligned} \hspace{\stretch{1}}(2.46)

We have

\begin{aligned}\epsilon_{i j k l} &= \epsilon^{a b c d} P_{a i}P_{a j}P_{a k}P_{a l} \\ &=-\epsilon_{i j k l} \text{Det} {\left\lVert{P_{ij}}\right\rVert} \\ &=-\epsilon_{i j k l} \text{Det} {\left\lVert{O_{ij}}\right\rVert} \\ &=-\epsilon_{i j k l} \text{Det} {\left\lVert{g_{im} {O^m}_j }\right\rVert} \\ &=-\epsilon_{i j k l} (-1)(1) \\ &=\epsilon_{i j k l}\end{aligned}

Since $\epsilon_{i j k l} = -\epsilon^{i j k l}$ both are therefore invariant under Lorentz transformation.

## 5. Statement. Sum of contracting symmetric and antisymmetric rank 2 tensors

Show that $A^{ij} B_{ij} = 0$ if $A$ is symmetric and $B$ is antisymmetric.

## 5. Solution

We swap indexes in $B$, switch dummy indexes, then swap indexes in $A$

\begin{aligned}A^{i j} B_{i j} &= -A^{i j} B_{j i} \\ &= -A^{j i} B_{i j} \\ &= -A^{i j} B_{i j} \\ \end{aligned}

Our result is the negative of itself, so must be zero.

## 6. Statement. Characteristic equation for the electromagnetic strength tensor

Show that $P(\lambda) = \text{Det} {\left\lVert{F_{i j} - \lambda g_{i j}}\right\rVert}$ is invariant under Lorentz transformations. Consider the polynomial of $P(\lambda)$, also called the characteristic polynomial of the matrix ${\left\lVert{F_{i j}}\right\rVert}$. Find the coefficients of the expansion of $P(\lambda)$ in powers of $\lambda$ in terms of the components of ${\left\lVert{F_{i j}}\right\rVert}$. Use the result to argue that $\mathbf{E} \cdot \mathbf{B}$ and $\mathbf{E}^2 - \mathbf{B}^2$ are Lorentz invariant.

## 6. Solution

### The invariance of the determinant

Let’s consider how any lower index rank 2 tensor transforms. Given a transformation of coordinates

\begin{aligned}(x^i)' &= {O^i}_j x^j \\ (x_i)' &= {O_i}^j x^j ,\end{aligned} \hspace{\stretch{1}}(2.47)

where $\text{Det} {\left\lVert{ {O^i}_j }\right\rVert} = 1$, and ${O_i}^j = {O^m}_n g_{i m} g^{j n}$. Let’s reflect briefly on why this determinant is unit valued. We have

\begin{aligned}(x^i)' (x_i)'= {O_i}^a x^a {O^i}_b x^b = x^b x_b,\end{aligned} \hspace{\stretch{1}}(2.49)

which implies that the transformation product is

\begin{aligned}{O_i}^a {O^i}_b = {\delta^a}_b,\end{aligned} \hspace{\stretch{1}}(2.50)

the identity matrix. The identity matrix has unit determinant, so we must have

\begin{aligned}1 = (\text{Det} \hat{G})^2 (\text{Det} {\left\lVert{ {O^i}_j }\right\rVert})^2.\end{aligned} \hspace{\stretch{1}}(2.51)

Since $\text{Det} \hat{G} = -1$ we have

\begin{aligned}\text{Det} {\left\lVert{ {O^i}_j }\right\rVert} = \pm 1,\end{aligned} \hspace{\stretch{1}}(2.52)

which is all that we can say about the determinant of this class of transformations by considering just invariance. If we restrict the transformations of coordinates to those of the same determinant sign as the identity matrix, we rule out reflections in time or space. This seems to be the essence of the $SO(1,3)$ labeling.

Why dwell on this? Well, I wanted to be clear on the conventions I’d chosen, since parts of the course notes used $\hat{O} = {\left\lVert{O^{i j}}\right\rVert}$, and $X' = \hat{O} X$, and gave that matrix unit determinant. That $O^{i j}$ looks like it is equivalent to my ${O^i}_j$, except that the one in the course notes is loose when it comes to lower and upper indexes since it gives $(x')^i = O^{i j} x^j$.

I’ll write

\begin{aligned}\hat{O} = {\left\lVert{{O^i}_j}\right\rVert},\end{aligned} \hspace{\stretch{1}}(2.53)

and require this (not ${\left\lVert{O^{i j}}\right\rVert}$) to be the matrix with unit determinant. Having cleared the index upper and lower confusion I had trying to reconcile the class notes with the rules for index manipulation, let’s now consider the Lorentz transformation of a lower index rank 2 tensor (not necessarily antisymmetric or symmetric)

We have, transforming in the same fashion as a lower index coordinate four vector (but twice, once for each index)

\begin{aligned}A_{i j} \rightarrow A_{k m} {O_i}^k{O_j}^m.\end{aligned} \hspace{\stretch{1}}(2.54)

The determinant of the transformation tensor ${O_i}^j$ is

\begin{aligned}\text{Det} {\left\lVert{ {O_i}^j }\right\rVert} = \text{Det} {\left\lVert{ g^{i m} {O^m}_n g^{n j} }\right\rVert} = (\text{Det} \hat{G}) (1) (\text{Det} \hat{G} ) = (-1)^2 (1) = 1.\end{aligned} \hspace{\stretch{1}}(2.55)

We see that the determinant of a lower index rank 2 tensor is invariant under Lorentz transformation. This would include our characteristic polynomial $P(\lambda)$.

### Expanding the determinant.

Utilizing 2.39 we can now calculate the characteristic polynomial. This is

\begin{aligned}\text{Det} {\left\lVert{F_{ij} - \lambda g_{ij} }\right\rVert}&= \frac{1}{{4!}}\epsilon^{s t u v} \epsilon^{a b c d} (F_{ a s } - \lambda g_{a s}) (F_{ b t } - \lambda g_{b t}) (F_{ c u } - \lambda g_{c u}) (F_{ d v } - \lambda g_{d v}) \\ &=\frac{1}{{24}}\epsilon^{s t u v} \epsilon_{a b c d} ({F^a}_s - \lambda {g^a}_s) ({F^b}_t - \lambda {g^b}_t) ({F^c}_u - \lambda {g^c}_u) ({F^d}_v - \lambda {g^d}_v) \\ \end{aligned}

However, ${g^a}_b = g_{b c} g^{a c}$, or ${\left\lVert{{g^a}_b}\right\rVert} = \hat{G}^2 = I$. This means we have

\begin{aligned}{g^a}_b = {\delta^a}_b,\end{aligned} \hspace{\stretch{1}}(2.56)

and our determinant is reduced to

\begin{aligned}\begin{aligned}P(\lambda) &=\frac{1}{{24}}\epsilon^{s t u v} \epsilon_{a b c d} \Bigl({F^a}_s {F^b}_t - \lambda( {\delta^a}_s {F^b}_t + {\delta^b}_t {F^a}_s ) + \lambda^2 {\delta^a}_s {\delta^b}_t \Bigr) \\ &\times \qquad \qquad \Bigl({F^c}_u {F^d}_v - \lambda( {\delta^c}_u {F^d}_v + {\delta^d}_v {F^c}_u ) + \lambda^2 {\delta^c}_u {\delta^d}_v \Bigr) \end{aligned}\end{aligned} \hspace{\stretch{1}}(2.57)

If we expand this out we have our powers of $\lambda$ coefficients are

\begin{aligned}\lambda^0 &:\frac{1}{{24}} \epsilon^{s t u v} \epsilon_{a b c d} {F^a}_s {F^b}_t {F^c}_u {F^d}_v \\ \lambda^1 &:\frac{1}{{24}} \epsilon^{s t u v} \epsilon_{a b c d} \Bigl(- ({\delta^c}_u {F^d}_v + {\delta^d}_v {F^c}_u ) {F^a}_s {F^b}_t - ({\delta^a}_s {F^b}_t + {\delta^b}_t {F^a}_s ) {F^c}_u {F^d}_v \Bigr) \\ \lambda^2 &:\frac{1}{{24}} \epsilon^{s t u v} \epsilon_{a b c d} \Bigl({\delta^c}_u {\delta^d}_v {F^a}_s {F^b}_t +( {\delta^a}_s {F^b}_t + {\delta^b}_t {F^a}_s ) ( {\delta^c}_u {F^d}_v + {\delta^d}_v {F^c}_u ) + {\delta^a}_s {\delta^b}_t {F^c}_u {F^d}_v \Bigr) \\ \lambda^3 &:\frac{1}{{24}} \epsilon^{s t u v} \epsilon_{a b c d} \Bigl(- ( {\delta^a}_s {F^b}_t + {\delta^b}_t {F^a}_s ) {\delta^c}_u {\delta^d}_v - {\delta^a}_s {\delta^b}_t ( {\delta^c}_u {F^d}_v + {\delta^d}_v {F^c}_u ) \Bigr) \\ \lambda^4 &:\frac{1}{{24}} \epsilon^{s t u v} \epsilon_{a b c d} \Bigl({\delta^a}_s {\delta^b}_t {\delta^c}_u {\delta^d}_v \Bigr) \\ \end{aligned}

By 2.39 the $\lambda^0$ coefficient is just $\text{Det} {\left\lVert{F_{i j}}\right\rVert}$.

The $\lambda^3$ terms can be seen to be zero. For example, the first one is

\begin{aligned}-\frac{1}{{24}} \epsilon^{s t u v} \epsilon_{a b c d} {\delta^a}_s {F^b}_t {\delta^c}_u {\delta^d}_v &=-\frac{1}{{24}} \epsilon^{s t u v} \epsilon_{s b u v} {F^b}_t \\ &=-\frac{1}{{12}} \delta^{t}_b {F^b}_t \\ &=-\frac{1}{{12}} {F^b}_b \\ &=-\frac{1}{{12}} F^{bu} g_{ub} \\ &= 0,\end{aligned}

where the final equality to zero comes from summing a symmetric and antisymmetric product.

Similarly the $\lambda$ coefficients can be shown to be zero. Again the first as a sample is

\begin{aligned}-\frac{1}{{24}} \epsilon^{s t u v} \epsilon_{a b c d} {\delta^c}_u {F^d}_v {F^a}_s {F^b}_t &=-\frac{1}{{24}} \epsilon^{u s t v} \epsilon_{u a b d} {F^d}_v {F^a}_s {F^b}_t \\ &=-\frac{1}{{24}} \delta^{[s}_a\delta^{t}_b\delta^{v]}_d{F^d}_v {F^a}_s {F^b}_t \\ &=-\frac{1}{{24}} {F^a}_{[s}{F^b}_{t}{F^d}_{v]} \\ \end{aligned}

Disregarding the $-1/24$ factor, let’s just expand this antisymmetric sum

\begin{aligned}{F^a}_{[a}{F^b}_{b}{F^d}_{d]}&={F^a}_{a}{F^b}_{b}{F^d}_{d}+{F^a}_{d}{F^b}_{a}{F^d}_{b}+{F^a}_{b}{F^b}_{d}{F^d}_{a}-{F^a}_{a}{F^b}_{d}{F^d}_{b}-{F^a}_{d}{F^b}_{b}{F^d}_{a}-{F^a}_{b}{F^b}_{a}{F^d}_{d} \\ &={F^a}_{d}{F^b}_{a}{F^d}_{b}+{F^a}_{b}{F^b}_{d}{F^d}_{a} \\ \end{aligned}

Of the two terms above that were retained, they are the only ones without a zero ${F^i}_i$ factor. Consider the first part of this remaining part of the sum. Employing the metric tensor, to raise indexes so that the antisymmetry of $F^{ij}$ can be utilized, and then finally relabeling all the dummy indexes we have

\begin{aligned}{F^a}_{d}{F^b}_{a}{F^d}_{b}&=F^{a u}F^{b v}F^{d w}g_{d u}g_{a v}g_{b w} \\ &=(-1)^3F^{u a}F^{v b}F^{w d}g_{d u}g_{a v}g_{b w} \\ &=-(F^{u a}g_{a v})(F^{v b}g_{b w} )(F^{w d}g_{d u})\\ &=-{F^u}_v{F^v}_w{F^w}_u\\ &=-{F^a}_b{F^b}_d{F^d}_a\\ \end{aligned}

This is just the negative of the second term in the sum, leaving us with zero.

Finally, we have for the $\lambda^2$ coefficient ($\times 24$)

\begin{aligned}&\epsilon^{s t u v} \epsilon_{a b c d} \Bigl({\delta^c}_u {\delta^d}_v {F^a}_s {F^b}_t +{\delta^a}_s {F^b}_t {\delta^c}_u {F^d}_v +{\delta^b}_t {F^a}_s {\delta^d}_v {F^c}_u \\ &\qquad +{\delta^b}_t {F^a}_s {\delta^c}_u {F^d}_v +{\delta^a}_s {F^b}_t {\delta^d}_v {F^c}_u + {\delta^a}_s {\delta^b}_t {F^c}_u {F^d}_v \Bigr) \\ &=\epsilon^{s t u v} \epsilon_{a b u v} {F^a}_s {F^b}_t +\epsilon^{s t u v} \epsilon_{s b u d} {F^b}_t {F^d}_v +\epsilon^{s t u v} \epsilon_{a t c v} {F^a}_s {F^c}_u \\ &\qquad +\epsilon^{s t u v} \epsilon_{a t u d} {F^a}_s {F^d}_v +\epsilon^{s t u v} \epsilon_{s b c v} {F^b}_t {F^c}_u + \epsilon^{s t u v} \epsilon_{s t c d} {F^c}_u {F^d}_v \\ &=\epsilon^{s t u v} \epsilon_{a b u v} {F^a}_s {F^b}_t +\epsilon^{t v s u } \epsilon_{b d s u} {F^b}_t {F^d}_v +\epsilon^{s u t v} \epsilon_{a c t v} {F^a}_s {F^c}_u \\ &\qquad +\epsilon^{s v t u} \epsilon_{a d t u} {F^a}_s {F^d}_v +\epsilon^{t u s v} \epsilon_{b c s v} {F^b}_t {F^c}_u + \epsilon^{u v s t} \epsilon_{c d s t} {F^c}_u {F^d}_v \\ &=6\epsilon^{s t u v} \epsilon_{a b u v} {F^a}_s {F^b}_t \\ &=6 (2){\delta^{[s}}_a{\delta^{t]}}_b{F^a}_s {F^b}_t \\ &=12{F^a}_{[a} {F^b}_{b]} \\ &=12( {F^a}_{a} {F^b}_{b} - {F^a}_{b} {F^b}_{a} ) \\ &=-12 {F^a}_{b} {F^b}_{a} \\ &=-12 F^{a b} F_{b a} \\ &=12 F^{a b} F_{a b}\end{aligned}

Therefore, our characteristic polynomial is

\begin{aligned}\boxed{P(\lambda) = \text{Det} {\left\lVert{F_{i j}}\right\rVert} + \frac{\lambda^2}{2} F^{a b} F_{a b} + \lambda^4.}\end{aligned} \hspace{\stretch{1}}(2.58)

Observe that in matrix form our strength tensors are

\begin{aligned}{\left\lVert{ F^{ij} }\right\rVert} &= \begin{bmatrix}0 & -E_x & -E_y & -E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0\end{bmatrix} \\ {\left\lVert{ F_{ij} }\right\rVert} &= \begin{bmatrix}0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_z & -B_y & B_x & 0\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(2.59)

From these we can compute $F^{a b} F_{a b}$ easily by inspection

\begin{aligned}F^{a b} F_{a b} = 2 (\mathbf{B}^2 - \mathbf{E}^2).\end{aligned} \hspace{\stretch{1}}(2.61)

Computing the determinant is not so easy. The dumb and simple way of expanding by cofactors takes two pages, and yields eventually

\begin{aligned}\text{Det} {\left\lVert{ F^{i j} }\right\rVert} = (\mathbf{E} \cdot \mathbf{B})^2.\end{aligned} \hspace{\stretch{1}}(2.62)

That supplies us with a relation for the characteristic polynomial in $\mathbf{E}$ and $\mathbf{B}$

\begin{aligned}\boxed{P(\lambda) = (\mathbf{E} \cdot \mathbf{B})^2 + \lambda^2 (\mathbf{B}^2 - \mathbf{E}^2) + \lambda^4.}\end{aligned} \hspace{\stretch{1}}(2.63)

Observe that we found this for the special case where $\mathbf{E}$ and $\mathbf{B}$ were perpendicular in homework 2. Observe that when we have that perpendicularity, we can solve for the eigenvalues by inspection

\begin{aligned}\lambda \in \{ 0, 0, \pm \sqrt{ \mathbf{E}^2 - \mathbf{B}^2 } \},\end{aligned} \hspace{\stretch{1}}(2.64)

and were able to diagonalize the matrix ${F^{i}}_j$ to solve the Lorentz force equation in parametric form. When ${\left\lvert{\mathbf{E}}\right\rvert} > {\left\lvert{\mathbf{B}}\right\rvert}$ we had real eigenvalues and an orthogonal diagonalization when $\mathbf{B} = 0$. For the ${\left\lvert{\mathbf{B}}\right\rvert} > {\left\lvert{\mathbf{E}}\right\rvert}$, we had a two purely imaginary eigenvalues, and when $\mathbf{E} = 0$ this was a Hermitian diagonalization. For the general case, when one of $\mathbf{E}$, or $\mathbf{B}$ was zero, things didn’t have the same nice closed form solution.

In general our eigenvalues are

\begin{aligned}\lambda = \pm \frac{1}{{\sqrt{2}}} \sqrt{ \mathbf{E}^2 - \mathbf{B}^2 \pm \sqrt{ (\mathbf{E}^2 - \mathbf{B}^2)^2 - 4 (\mathbf{E} \cdot \mathbf{B})^2 }}.\end{aligned} \hspace{\stretch{1}}(2.65)

For the purposes of this problem we really only wish to show that $\mathbf{E} \cdot \mathbf{B}$ and $\mathbf{E}^2 - \mathbf{B}^2$ are Lorentz invariants. When $\lambda = 0$ we have $P(\lambda) = (\mathbf{E} \cdot \mathbf{B})^2$, a Lorentz invariant. This must mean that $\mathbf{E} \cdot \mathbf{B}$ is itself a Lorentz invariant. Since that is invariant, and we require $P(\lambda)$ to be invariant for any other possible values of $\lambda$, the difference $\mathbf{E}^2 - \mathbf{B}^2$ must also be Lorentz invariant.

## 7. Statement. Show that the pseudoscalar invariant has only boundary effects.

Use integration by parts to show that $\int d^4 x \epsilon^{i j k l} F_{ i j } F_{ k l }$ only depends on the values of $A^i(x)$ at the “boundary” of spacetime (e.g. the “surface” depicted on page 105 of the notes) and hence does not affect the equations of motion for the electromagnetic field.

## 7. Solution

This proceeds in a fairly straightforward fashion

\begin{aligned}\int d^4 x \epsilon^{i j k l} F_{ i j } F_{ k l }&=\int d^4 x \epsilon^{i j k l} (\partial_i A_j - \partial_j A_i) F_{ k l } \\ &=\int d^4 x \epsilon^{i j k l} (\partial_i A_j) F_{ k l } -\epsilon^{j i k l} (\partial_i A_j) F_{ k l } \\ &=2 \int d^4 x \epsilon^{i j k l} (\partial_i A_j) F_{ k l } \\ &=2 \int d^4 x \epsilon^{i j k l} \left( \frac{\partial {}}{\partial {x^i}}(A_j F_{ k l }-A_j \frac{\partial { F_{ k l } }}{\partial {x^i}}\right)\\ \end{aligned}

Now, observe that by the Bianchi identity, this second term is zero

\begin{aligned}\epsilon^{i j k l} \frac{\partial { F_{ k l } }}{\partial {x^i}}=-\epsilon^{j i k l} \partial_i F_{ k l } = 0\end{aligned} \hspace{\stretch{1}}(2.66)

Now we have a set of perfect differentials, and can integrate

\begin{aligned}\int d^4 x \epsilon^{i j k l} F_{ i j } F_{ k l }&= 2 \int d^4 x \epsilon^{i j k l} \frac{\partial {}}{\partial {x^i}}(A_j F_{ k l })\\ &= 2 \int dx^j dx^k dx^l\epsilon^{i j k l} {\left.{{(A_j F_{ k l })}}\right\vert}_{{\Delta x^i}}\\ \end{aligned}

We are left with a only contributions to the integral from the boundary terms on the spacetime hypervolume, three-volume normals bounding the four-volume integration in the original integral.

## 8. Statement. Electromagnetic duality transformations.

Show that the Maxwell equations in vacuum are invariant under the transformation: $F_{i j} \rightarrow \tilde{F}_{i j}$, where $\tilde{F}_{i j} = \frac{1}{{2}} \epsilon_{i j k l} F^{k l}$ is the dual electromagnetic stress tensor. Replacing $F$ with $\tilde{F}$ is known as “electric-magnetic duality”. Explain this name by considering the transformation in terms of $\mathbf{E}$ and $\mathbf{B}$. Are the Maxwell equations with sources invariant under electric-magnetic duality transformations?

## 8. Solution

Let’s first consider the explanation of the name. First recall what the expansions are of $F_{i j}$ and $F^{i j}$ in terms of $\mathbf{E}$ and $\mathbf{E}$. These are

\begin{aligned}F_{0 \alpha} &= \partial_0 A_\alpha - \partial_\alpha A_0 \\ &= -\frac{1}{{c}} \frac{\partial {A^\alpha}}{\partial {t}} - \frac{\partial {\phi}}{\partial {x^\alpha}} \\ &= E_\alpha\end{aligned}

with $F^{0 \alpha} = -E^\alpha$, and $E^\alpha = E_\alpha$.

The magnetic field components are

\begin{aligned}F_{\beta \alpha} &= \partial_\beta A_\alpha - \partial_\alpha A_\beta \\ &= -\partial_\beta A^\alpha + \partial_\alpha A^\beta \\ &= \epsilon_{\alpha \beta \sigma} B^\sigma\end{aligned}

with $F^{\beta \alpha} = \epsilon^{\alpha \beta \sigma} B_\sigma$ and $B_\sigma = B^\sigma$.

Now let’s expand the dual tensors. These are

\begin{aligned}\tilde{F}_{0 \alpha} &=\frac{1}{{2}} \epsilon_{0 \alpha i j} F^{i j} \\ &=\frac{1}{{2}} \epsilon_{0 \alpha \beta \sigma} F^{\beta \sigma} \\ &=\frac{1}{{2}} \epsilon_{0 \alpha \beta \sigma} \epsilon^{\sigma \beta \mu} B_\mu \\ &=-\frac{1}{{2}} \epsilon_{0 \alpha \beta \sigma} \epsilon^{\mu \beta \sigma} B_\mu \\ &=-\frac{1}{{2}} (2!) {\delta_\alpha}^\mu B_\mu \\ &=- B_\alpha \\ \end{aligned}

and

\begin{aligned}\tilde{F}_{\beta \alpha} &=\frac{1}{{2}} \epsilon_{\beta \alpha i j} F^{i j} \\ &=\frac{1}{{2}} \left(\epsilon_{\beta \alpha 0 \sigma} F^{0 \sigma} +\epsilon_{\beta \alpha \sigma 0} F^{\sigma 0} \right) \\ &=\epsilon_{0 \beta \alpha \sigma} (-E^\sigma) \\ &=\epsilon_{\alpha \beta \sigma} E^\sigma\end{aligned}

Summarizing we have

\begin{aligned}F_{0 \alpha} &= E^\alpha \\ F^{0 \alpha} &= -E^\alpha \\ F^{\beta \alpha} &= F_{\beta \alpha} = \epsilon_{\alpha \beta \sigma} B^\sigma \\ \tilde{F}_{0 \alpha} &= - B_\alpha \\ \tilde{F}^{0 \alpha} &= B_\alpha \\ \tilde{F}_{\beta \alpha} &= \tilde{F}^{\beta \alpha} = \epsilon_{\alpha \beta \sigma} E^\sigma\end{aligned} \hspace{\stretch{1}}(2.67)

Is there a sign error in the $\tilde{F}_{0 \alpha} = - B_\alpha$ result? Other than that we have the same sort of structure for the tensor with $E$ and $B$ switched around.

Let’s write these in matrix form, to compare

\begin{aligned}\begin{array}{l l l l}{\left\lVert{ \tilde{F}_{i j} }\right\rVert} &= \begin{bmatrix}0 & -B_x & -B_y & -B_z \\ B_x & 0 & -E_z & E_y \\ B_y & E_z & 0 & E_x \\ B_z & -E_y & -E_x & 0 \\ \end{bmatrix} ^{i j} }\right\rVert} &= \begin{bmatrix}0 & B_x & B_y & B_z \\ -B_x & 0 & -E_z & E_y \\ -B_y & E_z & 0 & -E_x \\ -B_z & -E_y & E_x & 0 \\ \end{bmatrix} \\ {\left\lVert{ F^{ij} }\right\rVert} &= \begin{bmatrix}0 & -E_x & -E_y & -E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0\end{bmatrix} }\right\rVert} &= \begin{bmatrix}0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_z & -B_y & B_x & 0\end{bmatrix}.\end{array}\end{aligned} \hspace{\stretch{1}}(2.73)

From these we can see by inspection that we have

\begin{aligned}\tilde{F}^{i j} F_{ij} = \tilde{F}_{i j} F^{ij} = 4 (\mathbf{E} \cdot \mathbf{B})\end{aligned} \hspace{\stretch{1}}(2.74)

This is consistent with the stated result in [1] (except for a factor of $c$ due to units differences), so it appears the signs above are all kosher.

Now, let’s see if the if the dual tensor satisfies the vacuum equations.

\begin{aligned}\partial_j \tilde{F}^{i j}&=\partial_j \frac{1}{{2}} \epsilon^{i j k l} F_{k l} \\ &=\frac{1}{{2}} \epsilon^{i j k l} \partial_j (\partial_k A_l - \partial_l A_k) \\ &=\frac{1}{{2}} \epsilon^{i j k l} \partial_j \partial_k A_l - \frac{1}{{2}} \epsilon^{i j l k} \partial_k A_l \\ &=\frac{1}{{2}} (\epsilon^{i j k l} - \epsilon^{i j k l} \partial_k A_l \\ &= 0 \qquad\square\end{aligned}

So the first checks out, provided we have no sources. If we have sources, then we see here that Maxwell’s equations do not hold since this would imply that the four current density must be zero.

How about the Bianchi identity? That gives us

\begin{aligned}\epsilon^{i j k l} \partial_j \tilde{F}_{k l} &=\epsilon^{i j k l} \partial_j \frac{1}{{2}} \epsilon_{k l a b} F^{a b} \\ &=\frac{1}{{2}} \epsilon^{k l i j} \epsilon_{k l a b} \partial_j F^{a b} \\ &=\frac{1}{{2}} (2!) {\delta^i}_{[a} {\delta^j}_{b]} \partial_j F^{a b} \\ &=\partial_j (F^{i j} - F^{j i} ) \\ &=2 \partial_j F^{i j} .\end{aligned}

The factor of two is slightly curious. Is there a mistake above? If there is a mistake, it doesn’t change the fact that Maxwell’s equation

\begin{aligned}\partial_k F^{k i} = \frac{4 \pi}{c} j^i\end{aligned} \hspace{\stretch{1}}(2.75)

Gives us zero for the Bianchi identity under source free conditions of $j^i = 0$.

# Problem 2. Transformation properties of $\mathbf{E}$ and $\mathbf{B}$, again.

## 1. Statement

Use the form of $F^{i j}$ from page 82 in the class notes, the transformation law for ${\left\lVert{ F^{i j} }\right\rVert}$ given further down that same page, and the explicit form of the $SO(1,3)$ matrix $\hat{O}$ (say, corresponding to motion in the positive $x_1$ direction with speed $v$) to derive the transformation law of the fields $\mathbf{E}$ and $\mathbf{B}$. Use the transformation law to find the electromagnetic field of a charged particle moving with constant speed $v$ in the positive $x_1$ direction and check that the result agrees with the one that you obtained in Homework 2.

## 1. Solution

Given a transformation of coordinates

\begin{aligned}{x'}^i \rightarrow {O^i}_j x^j\end{aligned} \hspace{\stretch{1}}(3.76)

our rank 2 tensor $F^{i j}$ transforms as

\begin{aligned}F^{i j} \rightarrow {O^i}_aF^{a b}{O^j}_b.\end{aligned} \hspace{\stretch{1}}(3.77)

Introducing matrices

\begin{aligned}\hat{O} &= {\left\lVert{{O^i}_j}\right\rVert} \\ \hat{F} &= {\left\lVert{F^{ij}}\right\rVert} = \begin{bmatrix}0 & -E_x & -E_y & -E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0\end{bmatrix} \end{aligned} \hspace{\stretch{1}}(3.78)

and noting that $\hat{O}^\text{T} = {\left\lVert{{O^j}_i}\right\rVert}$, we can express the electromagnetic strength tensor transformation as

\begin{aligned}\hat{F} \rightarrow \hat{O} \hat{F} \hat{O}^\text{T}.\end{aligned} \hspace{\stretch{1}}(3.80)

The class notes use ${x'}^i \rightarrow O^{ij} x^j$, which violates our conventions on mixed upper and lower indexes, but the end result 3.80 is the same.

\begin{aligned}{\left\lVert{{O^i}_j}\right\rVert} =\begin{bmatrix}\cosh\alpha & -\sinh\alpha & 0 & 0 \\ -\sinh\alpha & \cosh\alpha & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(3.81)

Writing

\begin{aligned}C &= \cosh\alpha = \gamma \\ S &= -\sinh\alpha = -\gamma \beta,\end{aligned} \hspace{\stretch{1}}(3.82)

we can compute the transformed field strength tensor

\begin{aligned}\hat{F}' &=\begin{bmatrix}C & S & 0 & 0 \\ S & C & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}\begin{bmatrix}0 & -E_x & -E_y & -E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0\end{bmatrix} \begin{bmatrix}C & S & 0 & 0 \\ S & C & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix} \\ &=\begin{bmatrix}C & S & 0 & 0 \\ S & C & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}\begin{bmatrix}- S E_x & -C E_x & -E_y & -E_z \\ C E_x & S E_x & -B_z & B_y \\ C E_y + S B_z & S E_y + C B_z & 0 & -B_x \\ C E_z - S B_y & S E_z - C B_y & B_x & 0 \end{bmatrix} \\ &=\begin{bmatrix}0 & -E_x & -C E_y - S B_z & - C E_z + S B_y \\ E_x & 0 & -S E_y - C B_z & - S E_z + C B_y \\ C E_y + S B_z & S E_y + C B_z & 0 & -B_x \\ C E_z - S B_y & S E_z - C B_y & B_x & 0\end{bmatrix} \\ &=\begin{bmatrix}0 & -E_x & -\gamma(E_y - \beta B_z) & - \gamma(E_z + \beta B_y) \\ E_x & 0 & - \gamma (-\beta E_y + B_z) & \gamma( \beta E_z + B_y) \\ \gamma (E_y - \beta B_z) & \gamma(-\beta E_y + B_z) & 0 & -B_x \\ \gamma (E_z + \beta B_y) & -\gamma(\beta E_z + B_y) & B_x & 0\end{bmatrix}.\end{aligned}

As a check we have the antisymmetry that is expected. There is also a regularity to the end result that is aesthetically pleasing, hinting that things are hopefully error free. In coordinates for $\mathbf{E}$ and $\mathbf{B}$ this is

\begin{aligned}E_x &\rightarrow E_x \\ E_y &\rightarrow \gamma ( E_y - \beta B_z ) \\ E_z &\rightarrow \gamma ( E_z + \beta B_y ) \\ B_z &\rightarrow B_x \\ B_y &\rightarrow \gamma ( B_y + \beta E_z ) \\ B_z &\rightarrow \gamma ( B_z - \beta E_y ) \end{aligned} \hspace{\stretch{1}}(3.84)

Writing $\boldsymbol{\beta} = \mathbf{e}_1 \beta$, we have

\begin{aligned}\boldsymbol{\beta} \times \mathbf{B} = \begin{vmatrix} \mathbf{e}_1 & \mathbf{e}_2 & \mathbf{e}_3 \\ \beta & 0 & 0 \\ B_x & B_y & B_z\end{vmatrix} = \mathbf{e}_2 (-\beta B_z) + \mathbf{e}_3( \beta B_y ),\end{aligned} \hspace{\stretch{1}}(3.90)

which puts us enroute to a tidier vector form

\begin{aligned}E_x &\rightarrow E_x \\ E_y &\rightarrow \gamma ( E_y + (\boldsymbol{\beta} \times \mathbf{B})_y ) \\ E_z &\rightarrow \gamma ( E_z + (\boldsymbol{\beta} \times \mathbf{B})_z ) \\ B_z &\rightarrow B_x \\ B_y &\rightarrow \gamma ( B_y - (\boldsymbol{\beta} \times \mathbf{E})_y ) \\ B_z &\rightarrow \gamma ( B_z - (\boldsymbol{\beta} \times \mathbf{E})_z ).\end{aligned} \hspace{\stretch{1}}(3.91)

For a vector $\mathbf{A}$, write $\mathbf{A}_\parallel = (\mathbf{A} \cdot \hat{\mathbf{v}})\hat{\mathbf{v}}$, $\mathbf{A}_\perp = \mathbf{A} - \mathbf{A}_\parallel$, allowing a compact description of the field transformation

\begin{aligned}\mathbf{E} &\rightarrow \mathbf{E}_\parallel + \gamma \mathbf{E}_\perp + \gamma (\boldsymbol{\beta} \times \mathbf{B})_\perp \\ \mathbf{B} &\rightarrow \mathbf{B}_\parallel + \gamma \mathbf{B}_\perp - \gamma (\boldsymbol{\beta} \times \mathbf{E})_\perp.\end{aligned} \hspace{\stretch{1}}(3.97)

Now, we want to consider the field of a moving particle. In the particle’s (unprimed) rest frame the field due to its potential $\phi = q/r$ is

\begin{aligned}\mathbf{E} &= \frac{q}{r^2} \hat{\mathbf{r}} \\ \mathbf{B} &= 0.\end{aligned} \hspace{\stretch{1}}(3.99)

Coordinates for a “stationary” observer, who sees this particle moving along the x-axis at speed $v$ are related by a boost in the $-v$ direction

\begin{aligned}\begin{bmatrix}ct' \\ x' \\ y' \\ z'\end{bmatrix}\begin{bmatrix}\gamma & \gamma (v/c) & 0 & 0 \\ \gamma (v/c) & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}\begin{bmatrix}ct \\ x \\ y \\ z\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(3.101)

Therefore the fields in the observer frame will be

\begin{aligned}\mathbf{E}' &= \mathbf{E}_\parallel + \gamma \mathbf{E}_\perp - \gamma \frac{v}{c}(\mathbf{e}_1 \times \mathbf{B})_\perp = \mathbf{E}_\parallel + \gamma \mathbf{E}_\perp \\ \mathbf{B}' &= \mathbf{B}_\parallel + \gamma \mathbf{B}_\perp + \gamma \frac{v}{c}(\mathbf{e}_1 \times \mathbf{E})_\perp = \gamma \frac{v}{c}(\mathbf{e}_1 \times \mathbf{E})_\perp \end{aligned} \hspace{\stretch{1}}(3.102)

More explicitly with $\mathbf{E} = \frac{q}{r^3}(x, y, z)$ this is

\begin{aligned}\mathbf{E}' &= \frac{q}{r^3}(x, \gamma y, \gamma z) \\ \mathbf{B}' &= \gamma \frac{q v}{c r^3} ( 0, -z, y )\end{aligned} \hspace{\stretch{1}}(3.104)

Comparing to Problem 3 in Problem set 2, I see that this matches the result obtained by separately transforming the gradient, the time partial, and the scalar potential. Actually, if I am being honest, I see that I made a sign error in all the coordinates of $\mathbf{E}'$ when I initially did (this ungraded problem) in problem set 2. That sign error should have been obvious by considering the $v=0$ case which would have mysteriously resulted in inversion of all the coordinates of the observed electric field.

## 2. Statement

A particle is moving with velocity $\mathbf{v}$ in perpendicular $\mathbf{E}$ and $\mathbf{B}$ fields, all given in some particular “stationary” frame of reference.

\begin{enumerate}
\item Show that there exists a frame where the problem of finding the particle trajectory can be reduced to having either only an electric or only a magnetic field.
\item Explain what determines which case takes place.
\item Find the velocity $\mathbf{v}_0$ of that frame relative to the “stationary” frame.
\end{enumerate}

## 2. Solution

\paragraph{Part 1 and 2:} Existence of the transformation.

In the single particle Lorentz trajectory problem we wish to solve

\begin{aligned}m c \frac{du^i}{ds} = \frac{e}{c} F^{i j} u_j,\end{aligned} \hspace{\stretch{1}}(3.106)

which in matrix form we can write as

\begin{aligned}\frac{d U}{ds} = \frac{e}{m c^2} \hat{F} \hat{G} U.\end{aligned} \hspace{\stretch{1}}(3.107)

where we write our column vector proper velocity as $U = {\left\lVert{u^i}\right\rVert}$. Under transformation of coordinates ${u'}^i = {O^i}_j x^j$, with $\hat{O} = {\left\lVert{{O^i}_j}\right\rVert}$, this becomes

\begin{aligned}\hat{O} \frac{d U}{ds} = \frac{e}{m c^2} \hat{O} \hat{F} \hat{O}^\text{T} \hat{G} \hat{O} U.\end{aligned} \hspace{\stretch{1}}(3.108)

Suppose we can find eigenvectors for the matrix $\hat{O} \hat{F} \hat{O}^\text{T} \hat{G}$. That is for some eigenvalue $\lambda$, we can find an eigenvector $\Sigma$

\begin{aligned}\hat{O} \hat{F} \hat{O}^\text{T} \hat{G} \Sigma = \lambda \Sigma.\end{aligned} \hspace{\stretch{1}}(3.109)

Rearranging we have

\begin{aligned}(\hat{O} \hat{F} \hat{O}^\text{T} \hat{G} - \lambda I) \Sigma = 0\end{aligned} \hspace{\stretch{1}}(3.110)

and conclude that $\Sigma$ lies in the null space of the matrix $\hat{O} \hat{F} \hat{O}^\text{T} \hat{G} - \lambda I$ and that this difference of matrices must have a zero determinant

\begin{aligned}\text{Det} (\hat{O} \hat{F} \hat{O}^\text{T} \hat{G} - \lambda I) = -\text{Det} (\hat{O} \hat{F} \hat{O}^\text{T} - \lambda \hat{G}) = 0.\end{aligned} \hspace{\stretch{1}}(3.111)

Since $\hat{G} = \hat{O} \hat{G} \hat{O}^\text{T}$ for any Lorentz transformation $\hat{O}$ in $SO(1,3)$, and $\text{Det} ABC = \text{Det} A \text{Det} B \text{Det} C$ we have

\begin{aligned}\text{Det} (\hat{O} \hat{F} \hat{O}^\text{T} - \lambda G)= \text{Det} (\hat{F} - \lambda \hat{G}).\end{aligned} \hspace{\stretch{1}}(3.112)

In problem 1.6, we called this our characteristic equation $P(\lambda) = \text{Det} (\hat{F} - \lambda \hat{G})$. Observe that the characteristic equation is Lorentz invariant for any $\lambda$, which requires that the eigenvalues $\lambda$ are also Lorentz invariants.

In problem 1.6 of this problem set we computed that this characteristic equation expands to

\begin{aligned}P(\lambda) = \text{Det} (\hat{F} - \lambda \hat{G}) = (\mathbf{E} \cdot \mathbf{B})^2 + \lambda^2 (\mathbf{B}^2 - \mathbf{E}^2) + \lambda^4.\end{aligned} \hspace{\stretch{1}}(3.113)

The eigenvalues for the system, also each necessarily Lorentz invariants, are

\begin{aligned}\lambda = \pm \frac{1}{{\sqrt{2}}} \sqrt{ \mathbf{E}^2 - \mathbf{B}^2 \pm \sqrt{ (\mathbf{E}^2 - \mathbf{B}^2)^2 - 4 (\mathbf{E} \cdot \mathbf{B})^2 }}.\end{aligned} \hspace{\stretch{1}}(3.114)

Observe that in the specific case where $\mathbf{E} \cdot \mathbf{B} = 0$, as in this problem, we must have $\mathbf{E}' \cdot \mathbf{B}'$ in all frames, and the two non-zero eigenvalues of our characteristic polynomial are simply

\begin{aligned}\lambda = \pm \sqrt{\mathbf{E}^2 - \mathbf{B}^2}.\end{aligned} \hspace{\stretch{1}}(3.115)

These and $\mathbf{E} \cdot \mathbf{B} = 0$ are the invariants for this system. If we have $\mathbf{E}^2 > \mathbf{B}^2$ in one frame, we must also have ${\mathbf{E}'}^2 > {\mathbf{B}'}^2$ in another frame, still maintaining perpendicular fields. In particular if $\mathbf{B}' = 0$ we maintain real eigenvalues. Similarly if $\mathbf{B}^2 > \mathbf{E}^2$ in some frame, we must always have imaginary eigenvalues, and this is also true in the $\mathbf{E}' = 0$ case.

While the problem can be posed as a pure diagonalization problem (and even solved numerically this way for the general constant fields case), we can also work symbolically, thinking of the trajectories problem as simply seeking a transformation of frames that reduce the scope of the problem to one that is more tractable. That does not have to be the linear transformation that diagonalizes the system. Instead we are free to transform to a frame where one of the two fields $\mathbf{E}'$ or $\mathbf{B}'$ is zero, provided the invariants discussed are maintained.

\paragraph{Part 3:} Finding the boost velocity that wipes out one of the fields.

Let’s now consider a Lorentz boost $\hat{O}$, and seek to solve for the boost velocity that wipes out one of the fields, given the invariants that must be maintained for the system

To make things concrete, suppose that our perpendicular fields are given by $\mathbf{E} = E \mathbf{e}_2$ and $\mathbf{B} = B \mathbf{e}_3$.

Let also assume that we can find the velocity $\mathbf{v}_0$ for which one or more of the transformed fields is zero. Suppose that velocity is

\begin{aligned}\mathbf{v}_0 = v_0 (\alpha_1, \alpha_2, \alpha_3) = v_0 \hat{\mathbf{v}}_0,\end{aligned} \hspace{\stretch{1}}(3.116)

where $\alpha_i$ are the direction cosines of $\mathbf{v}_0$ so that $\sum_i \alpha_i^2 = 1$. We will want to compute the components of $\mathbf{E}$ and $\mathbf{B}$ parallel and perpendicular to this velocity.

Those are

\begin{aligned}\mathbf{E}_\parallel &= E \mathbf{e}_2 \cdot (\alpha_1, \alpha_2, \alpha_3) (\alpha_1, \alpha_2, \alpha_3) \\ &= E \alpha_2 (\alpha_1, \alpha_2, \alpha_3) \\ \end{aligned}

\begin{aligned}\mathbf{E}_\perp &= E \mathbf{e}_2 - \mathbf{E}_\parallel \\ &= E (-\alpha_1 \alpha_2, 1 - \alpha_2^2, -\alpha_2 \alpha_3) \\ &= E (-\alpha_1 \alpha_2, \alpha_1^2 + \alpha_3^2, -\alpha_2 \alpha_3) \\ \end{aligned}

For the magnetic field we have

\begin{aligned}\mathbf{B}_\parallel &= B \alpha_3 (\alpha_1, \alpha_2, \alpha_3),\end{aligned}

and

\begin{aligned}\mathbf{B}_\perp &= B \mathbf{e}_3 - \mathbf{B}_\parallel \\ &= B (-\alpha_1 \alpha_3, -\alpha_2 \alpha_3, \alpha_1^2 + \alpha_2^2) \\ \end{aligned}

Now, observe that $(\boldsymbol{\beta} \times \mathbf{B})_\parallel \propto ((\mathbf{v}_0 \times \mathbf{B}) \cdot \mathbf{v}_0) \mathbf{v}_0$, but this is just zero. So we have $(\boldsymbol{\beta} \times \mathbf{B})_\parallel = \boldsymbol{\beta} \times \mathbf{B}$. So our cross products terms are just

\begin{aligned}\hat{\mathbf{v}}_0 \times \mathbf{B} &= \begin{vmatrix} \mathbf{e}_1 & \mathbf{e}_2 & \mathbf{e}_3 \\ \alpha_1 & \alpha_2 & \alpha_3 \\ 0 & 0 & B \end{vmatrix} = B (\alpha_2, -\alpha_1, 0) \\ \hat{\mathbf{v}}_0 \times \mathbf{E} &= \begin{vmatrix} \mathbf{e}_1 & \mathbf{e}_2 & \mathbf{e}_3 \\ \alpha_1 & \alpha_2 & \alpha_3 \\ 0 & E & 0 \end{vmatrix} = E (-\alpha_3, 0, \alpha_1)\end{aligned}

We can now express how the fields transform, given this arbitrary boost velocity. From 3.97, this is

\begin{aligned}\mathbf{E} &\rightarrow E \alpha_2 (\alpha_1, \alpha_2, \alpha_3) + \gamma E (-\alpha_1 \alpha_2, \alpha_1^2 + \alpha_3^2, -\alpha_2 \alpha_3) + \gamma \frac{v_0^2}{c^2} B (\alpha_2, -\alpha_1, 0) \\ \mathbf{B} &\rightarrowB \alpha_3 (\alpha_1, \alpha_2, \alpha_3)+ \gamma B (-\alpha_1 \alpha_3, -\alpha_2 \alpha_3, \alpha_1^2 + \alpha_2^2) - \gamma \frac{v_0^2}{c^2} E (-\alpha_3, 0, \alpha_1)\end{aligned} \hspace{\stretch{1}}(3.117)

### Zero Electric field case.

Let’s tackle the two cases separately. First when ${\left\lvert{\mathbf{B}}\right\rvert} > {\left\lvert{\mathbf{E}}\right\rvert}$, we can transform to a frame where $\mathbf{E}'=0$. In coordinates from 3.117 this supplies us three sets of equations. These are

\begin{aligned}0 &= E \alpha_2 \alpha_1 (1 - \gamma) + \gamma \frac{v_0^2}{c^2} B \alpha_2 \\ 0 &= E \alpha_2^2 + \gamma E (\alpha_1^2 + \alpha_3^2) - \gamma \frac{v_0^2}{c^2} B \alpha_1 \\ 0 &= E \alpha_2 \alpha_3 (1 - \gamma).\end{aligned} \hspace{\stretch{1}}(3.119)

With an assumed solution the $\mathbf{e}_3$ coordinate equation implies that one of $\alpha_2$ or $\alpha_3$ is zero. Perhaps there are solutions with $\alpha_3 = 0$ too, but inspection shows that $\alpha_2 = 0$ nicely kills off the first equation. Since $\alpha_1^2 + \alpha_2^2 + \alpha_3^2 = 1$, that also implies that we are left with

\begin{aligned}0 = E - \frac{v_0^2}{c^2} B \alpha_1 \end{aligned} \hspace{\stretch{1}}(3.122)

Or

\begin{aligned}\alpha_1 &= \frac{E}{B} \frac{c^2}{v_0^2} \\ \alpha_2 &= 0 \\ \alpha_3 &= \sqrt{1 - \frac{E^2}{B^2} \frac{c^4}{v_0^4} }\end{aligned} \hspace{\stretch{1}}(3.123)

Our velocity was $\mathbf{v}_0 = v_0 (\alpha_1, \alpha_2, \alpha_3)$ solving the problem for the ${\left\lvert{\mathbf{B}}\right\rvert}^2 > {\left\lvert{\mathbf{E}}\right\rvert}^2$ case up to an adjustable constant $v_0$. That constant comes with constraints however, since we must also have our cosine $\alpha_1 \le 1$. Expressed another way, the magnitude of the boost velocity is constrained by the relation

\begin{aligned}\frac{\mathbf{v}_0^2}{c^2} \ge {\left\lvert{\frac{E}{B}}\right\rvert}.\end{aligned} \hspace{\stretch{1}}(3.126)

It appears we may also pick the equality case, so one velocity (not unique) that should transform away the electric field is

\begin{aligned}\boxed{\mathbf{v}_0 = c \sqrt{{\left\lvert{\frac{E}{B}}\right\rvert}} \mathbf{e}_1 = \pm c \sqrt{{\left\lvert{\frac{E}{B}}\right\rvert}} \frac{\mathbf{E} \times \mathbf{B}}{{\left\lvert{\mathbf{E}}\right\rvert} {\left\lvert{\mathbf{B}}\right\rvert}}.}\end{aligned} \hspace{\stretch{1}}(3.127)

This particular boost direction is perpendicular to both fields. Observe that this highlights the invariance condition ${\left\lvert{\frac{E}{B}}\right\rvert} < 1$ since we see this is required for a physically realizable velocity. Boosting in this direction will reduce our problem to one that has only the magnetic field component.

### Zero Magnetic field case.

Now, let’s consider the case where we transform the magnetic field away, the case when our characteristic polynomial has strictly real eigenvalues $\lambda = \pm \sqrt{\mathbf{E}^2 - \mathbf{B}^2}$. In this case, if we write out our equations for the transformed magnetic field and require these to separately equal zero, we have

\begin{aligned}0 &= B \alpha_3 \alpha_1 ( 1 - \gamma ) + \gamma \frac{v_0^2}{c^2} E \alpha_3 \\ 0 &= B \alpha_2 \alpha_3 ( 1 - \gamma ) \\ 0 &= B (\alpha_3^2 + \gamma (\alpha_1^2 + \alpha_2^2)) - \gamma \frac{v_0^2}{c^2} E \alpha_1.\end{aligned} \hspace{\stretch{1}}(3.128)

Similar to before we see that $\alpha_3 = 0$ kills off the first and second equations, leaving just

\begin{aligned}0 = B - \frac{v_0^2}{c^2} E \alpha_1.\end{aligned} \hspace{\stretch{1}}(3.131)

We now have a solution for the family of direction vectors that kill the magnetic field off

\begin{aligned}\alpha_1 &= \frac{B}{E} \frac{c^2}{v_0^2} \\ \alpha_2 &= \sqrt{ 1 - \frac{B^2}{E^2} \frac{c^4}{v_0^4} } \\ \alpha_3 &= 0.\end{aligned} \hspace{\stretch{1}}(3.132)

In addition to the initial constraint that ${\left\lvert{\frac{B}{E}}\right\rvert} < 1$, we have as before, constraints on the allowable values of $v_0$

\begin{aligned}\frac{\mathbf{v}_0^2}{c^2} \ge {\left\lvert{\frac{B}{E}}\right\rvert}.\end{aligned} \hspace{\stretch{1}}(3.135)

Like before we can pick the equality $\alpha_1^2 = 1$, yielding a boost direction of

\begin{aligned}\boxed{\mathbf{v}_0 = c \sqrt{{\left\lvert{\frac{B}{E}}\right\rvert}} \mathbf{e}_1 = \pm c \sqrt{{\left\lvert{\frac{B}{E}}\right\rvert}} \frac{\mathbf{E} \times \mathbf{B}}{{\left\lvert{\mathbf{E}}\right\rvert} {\left\lvert{\mathbf{B}}\right\rvert}}.}\end{aligned} \hspace{\stretch{1}}(3.136)

Again, we see that the invariance condition ${\left\lvert{\mathbf{B}}\right\rvert} < {\left\lvert{\mathbf{E}}\right\rvert}$ is required for a physically realizable velocity if that velocity is entirely perpendicular to the fields.

# Problem 3. Continuity equation for delta function current distributions.

## Statement

Show explicitly that the electromagnetic 4-current $j^i$ for a particle moving with constant velocity (considered in class, p. 100-101 of notes) is conserved $\partial_i j^i = 0$. Give a physical interpretation of this conservation law, for example by integrating $\partial_i j^i$ over some spacetime region and giving an integral form to the conservation law ($\partial_i j^i = 0$ is known as the “continuity equation”).

## Solution

First lets review. Our four current was defined as

\begin{aligned}j^i(x) = \sum_A c e_A \int_{x(\tau)} dx_A^i(\tau) \delta^4(x - x_A(\tau)).\end{aligned} \hspace{\stretch{1}}(4.137)

If each of the trajectories $x_A(\tau)$ represents constant motion we have

\begin{aligned}x_A(\tau) = x_A(0) + \gamma_A \tau ( c, \mathbf{v}_A ).\end{aligned} \hspace{\stretch{1}}(4.138)

The spacetime split of this four vector is

\begin{aligned}x_A^0(\tau) &= x_A^0(0) + \gamma_A \tau c \\ \mathbf{x}_A(\tau) &= \mathbf{x}_A(0) + \gamma_A \tau \mathbf{v},\end{aligned} \hspace{\stretch{1}}(4.139)

with differentials

\begin{aligned}dx_A^0(\tau) &= \gamma_A d\tau c \\ d\mathbf{x}_A(\tau) &= \gamma_A d\tau \mathbf{v}_A.\end{aligned} \hspace{\stretch{1}}(4.141)

Writing out the delta functions explicitly we have

\begin{aligned}\begin{aligned}j^i(x) = \sum_A &c e_A \int_{x(\tau)} dx_A^i(\tau) \delta(x^0 - x_A^0(0) - \gamma_A c \tau) \delta(x^1 - x_A^1(0) - \gamma_A v_A^1 \tau) \\ &\delta(x^2 - x_A^2(0) - \gamma_A v_A^2 \tau) \delta(x^3 - x_A^3(0) - \gamma_A v_A^3 \tau)\end{aligned}\end{aligned} \hspace{\stretch{1}}(4.143)

So our time and space components of the current can be written

\begin{aligned}j^0(x) &= \sum_A c^2 e_A \gamma_A \int_{x(\tau)} d\tau\delta(x^0 - x_A^0(0) - \gamma_A c \tau)\delta^3(\mathbf{x} - \mathbf{x}_A(0) - \gamma_A \mathbf{v}_A \tau) \\ \mathbf{j}(x) &= \sum_A c e_A \mathbf{v}_A \gamma_A \int_{x(\tau)} d\tau\delta(x^0 - x_A^0(0) - \gamma_A c \tau)\delta^3(\mathbf{x} - \mathbf{x}_A(0) - \gamma_A \mathbf{v}_A \tau).\end{aligned} \hspace{\stretch{1}}(4.144)

Each of these integrals can be evaluated with respect to the time coordinate delta function leaving the distribution

\begin{aligned}j^0(x) &= \sum_A c e_A \delta^3(\mathbf{x} - \mathbf{x}_A(0) - \frac{\mathbf{v}_A}{c} (x^0 - x_A^0(0))) \\ \mathbf{j}(x) &= \sum_A e_A \mathbf{v}_A \delta^3(\mathbf{x} - \mathbf{x}_A(0) - \frac{\mathbf{v}_A}{c} (x^0 - x_A^0(0)))\end{aligned} \hspace{\stretch{1}}(4.146)

With this more general expression (multi-particle case) it should be possible to show that the four divergence is zero, however, the problem only asks for one particle. For the one particle case, we can make things really easy by taking the initial point in space and time as the origin, and aligning our velocity with one of the coordinates (say $x$).

Doing so we have the result derived in class

\begin{aligned}j = e \begin{bmatrix}c \\ v \\ 0 \\ 0 \end{bmatrix}\delta(x - v x^0/c)\delta(y)\delta(z).\end{aligned} \hspace{\stretch{1}}(4.148)

Our divergence then has only two portions

\begin{aligned}\frac{\partial {j^0}}{\partial {x^0}} &= e c (-v/c) \delta'(x - v x^0/c) \delta(y) \delta(z) \\ \frac{\partial {j^1}}{\partial {x}} &= e v \delta'(x - v x^0/c) \delta(y) \delta(z).\end{aligned} \hspace{\stretch{1}}(4.149)

and these cancel out when summed. Note that this requires us to be loose with our delta functions, treating them like regular functions that are differentiable.

For the more general multiparticle case, we can treat the sum one particle at a time, and in each case, rotate coordinates so that the four divergence only picks up one term.

As for physical interpretation via integral, we have using the four dimensional divergence theorem

\begin{aligned}\int d^4 x \partial_i j^i = \int j^i dS_i\end{aligned} \hspace{\stretch{1}}(4.151)

where $dS_i$ is the three-volume element perpendicular to a $x^i = \text{constant}$ plane. These volume elements are detailed generally in the text [2], however, they do note that one special case specifically $dS_0 = dx dy dz$, the element of the three-dimensional (spatial) volume “normal” to hyperplanes $ct = \text{constant}$.

Without actually computing the determinants, we have something that is roughly of the form

\begin{aligned}0 = \int j^i dS_i=\int c \rho dx dy dz+\int \mathbf{j} \cdot (\mathbf{n}_x c dt dy dz + \mathbf{n}_y c dt dx dz + \mathbf{n}_z c dt dx dy).\end{aligned} \hspace{\stretch{1}}(4.152)

This is cheating a bit to just write $\mathbf{n}_x, \mathbf{n}_y, \mathbf{n}_z$. Are there specific orientations required by the metric. To be precise we’d have to calculate the determinants detailed in the text, and then do the duality transformations.

Per unit time, we can write instead

\begin{aligned}\frac{\partial {}}{\partial {t}} \int \rho dV= -\int \mathbf{j} \cdot (\mathbf{n}_x dy dz + \mathbf{n}_y dx dz + \mathbf{n}_z dx dy)\end{aligned} \hspace{\stretch{1}}(4.153)

Rather loosely this appears to roughly describe that the rate of change of charge in a volume must be matched with the “flow” of current through the surface within that amount of time.

# References

[1] Wikipedia. Electromagnetic tensor — wikipedia, the free encyclopedia [online]. 2011. [Online; accessed 27-February-2011]. http://en.wikipedia.org/w/index.php?title=Electromagnetic_tensor&oldid=414989505.

[2] L.D. Landau and E.M. Lifshitz. The classical theory of fields. Butterworth-Heinemann, 1980.

## Some worked problems from old PHY356 exams.

Posted by peeterjoot on January 9, 2011

# Motivation.

Some of the old exam questions that I did for preparation for the exam I liked, and thought I’d write up some of them for potential future reference.

# Questions from the Dec 2007 PHY355H1F exam.

## 1b. Parity operator.

\paragraph{Q:} If $\Pi$ is the parity operator, defined by $\Pi {\lvert {x} \rangle} = {\lvert {-x} \rangle}$, where ${\lvert {x} \rangle}$ is the eigenket of the position operator $X$ with eigenvalue $x$), and $P$ is the momentum operator conjugate to $X$, show (carefully) that $\Pi P \Pi = -P$.

\paragraph{A:}

Consider the matrix element ${\langle {-x'} \rvert} \left[{\Pi},{P}\right] {\lvert {x} \rangle}$. This is

\begin{aligned}{\langle {-x'} \rvert} \left[{\Pi},{P}\right] {\lvert {x} \rangle}&={\langle {-x'} \rvert} \Pi P - P \Pi {\lvert {x} \rangle} \\ &={\langle {-x'} \rvert} \Pi P {\lvert {x} \rangle} - {\langle {-x} \rvert} P \Pi {\lvert {x} \rangle} \\ &={\langle {x'} \rvert} P {\lvert {x} \rangle} - {\langle {-x} \rvert} P {\lvert {-x} \rangle} \\ &=- i \hbar \left(\delta(x'-x) \frac{\partial {}}{\partial {x}}-\underbrace{\delta(-x -(-x'))}_{= \delta(x'-x) = \delta(x-x')} \frac{\partial {}}{\partial {-x}}\right) \\ &=- 2 i \hbar \delta(x'-x) \frac{\partial {}}{\partial {x}} \\ &=2 {\langle {x'} \rvert} P {\lvert {x} \rangle} \\ &=2 {\langle {-x'} \rvert} \Pi P {\lvert {x} \rangle} \\ \end{aligned}

We’ve taken advantage of the Hermitian property of $P$ and $\Pi$ here, and can rearrange for

\begin{aligned}{\langle {-x'} \rvert} \Pi P - P \Pi - 2 \Pi P {\lvert {x} \rangle} = 0\end{aligned} \hspace{\stretch{1}}(2.1)

Since this is true for all ${\langle {-x} \rvert}$ and ${\lvert {x} \rangle}$ we have

\begin{aligned}\Pi P + P \Pi = 0.\end{aligned} \hspace{\stretch{1}}(2.2)

Right multiplication by $\Pi$ and rearranging we have

\begin{aligned}\Pi P \Pi = - P \Pi \Pi = - P.\end{aligned} \hspace{\stretch{1}}(2.3)

## 1f. Free particle propagator.

\paragraph{Q:} For a free particle moving in one-dimension, the propagator (i.e. the coordinate representation of the evolution operator),

\begin{aligned}G(x,x';t) = {\langle {x} \rvert} U(t) {\lvert {x'} \rangle}\end{aligned} \hspace{\stretch{1}}(2.4)

is given by

\begin{aligned}G(x,x';t) = \sqrt{\frac{m}{2 \pi i \hbar t}} e^{i m (x-x')^2/ (2 \hbar t)}.\end{aligned} \hspace{\stretch{1}}(2.5)

\paragraph{A:}

This problem is actually fairly straightforward, but it is nice to work it having had a similar problem set question where we were asked about this time evolution operator matrix element (ie: what it’s physical meaning is). Here we have a concrete example of the form of this matrix operator.

Proceeding directly, we have

\begin{aligned}{\langle {x} \rvert} U {\lvert {x'} \rangle}&=\int \left\langle{x} \vert {p'}\right\rangle {\langle {p'} \rvert} U {\lvert {p} \rangle} \left\langle{p} \vert {x'}\right\rangle dp dp' \\ &=\int u_{p'}(x) {\langle {p'} \rvert} e^{-i P^2 t/(2 m \hbar)} {\lvert {p} \rangle} u_p^{*}(x') dp dp' \\ &=\int u_{p'}(x) e^{-i p^2 t/(2 m \hbar)} \delta(p-p') u_p^{*}(x') dp dp' \\ &=\int u_{p}(x) e^{-i p^2 t/(2 m \hbar)} u_p^{*}(x') dp \\ &=\frac{1}{(\sqrt{2 \pi \hbar})^2} \int e^{i p (x-x')/\hbar} e^{-i p^2 t/(2 m \hbar)} dp \\ &=\frac{1}{2 \pi \hbar} \int e^{i p (x-x')/\hbar} e^{-i p^2 t/(2 m \hbar)} dp \\ &=\frac{1}{2 \pi} \int e^{i k (x-x')} e^{-i \hbar k^2 t/(2 m)} dk \\ &=\frac{1}{2 \pi} \int dk e^{- \left(k^2 \frac{ i \hbar t}{2m} - i k (x-x')\right)} \\ &=\frac{1}{2 \pi} \int dk e^{- \frac{ i \hbar t}{2m}\left(k - i \frac{2m}{i \hbar t}\frac{(x-x')}{2} \right)^2- \frac{i^2 2 m (x-x')^2}{4 i \hbar t} } \\ &=\frac{1}{2 \pi} \sqrt{\pi} \sqrt{\frac{2m}{i \hbar t}}e^{\frac{ i m (x-x')^2}{2 \hbar t}},\end{aligned}

which is the desired result. Now, let’s look at how this would be used. We can express our time evolved state using this matrix element by introducing an identity

\begin{aligned}\left\langle{{x}} \vert {{\psi(t)}}\right\rangle &={\langle {x} \rvert} U {\lvert {\psi(0)} \rangle} \\ &=\int dx' {\langle {x} \rvert} U {\lvert {x'} \rangle} \left\langle{{x'}} \vert {{\psi(0)}}\right\rangle \\ &=\sqrt{\frac{m}{2 \pi i \hbar t}} \int dx' e^{i m (x-x')^2/ (2 \hbar t)}\left\langle{{x'}} \vert {{\psi(0)}}\right\rangle \\ \end{aligned}

This gives us

\begin{aligned}\psi(x, t)=\sqrt{\frac{m}{2 \pi i \hbar t}} \int dx' e^{i m (x-x')^2/ (2 \hbar t)} \psi(x', 0)\end{aligned} \hspace{\stretch{1}}(2.6)

However, note that our free particle wave function at time zero is

\begin{aligned}\psi(x, 0) = \frac{e^{i p x/\hbar}}{\sqrt{2 \pi \hbar}}\end{aligned} \hspace{\stretch{1}}(2.7)

So the convolution integral 2.6 does not exist. We likely have to require that the solution be not a pure state, but instead a superposition of a set of continuous states (a wave packet in position or momentum space related by Fourier transforms). That is

\begin{aligned}\psi(x, 0) &= \frac{1}{{\sqrt{2 \pi \hbar}}} \int \hat{\psi}(p, 0) e^{i p x/\hbar} dp \\ \hat{\psi}(p, 0) &= \frac{1}{{\sqrt{2 \pi \hbar}}} \int \psi(x'', 0) e^{-i p x''/\hbar} dx''\end{aligned} \hspace{\stretch{1}}(2.8)

The time evolution of this wave packet is then determined by the propagator, and is

\begin{aligned}\psi(x,t) =\sqrt{\frac{m}{2 \pi i \hbar t}} \frac{1}{{\sqrt{2 \pi \hbar}}} \int dx' dpe^{i m (x-x')^2/ (2 \hbar t)}\hat{\psi}(p, 0) e^{i p x'/\hbar} ,\end{aligned} \hspace{\stretch{1}}(2.10)

or in terms of the position space wave packet evaluated at time zero

\begin{aligned}\psi(x,t) =\sqrt{\frac{m}{2 \pi i \hbar t}}\frac{1}{{2 \pi}}\int dx' dx'' dke^{i m (x-x')^2/ (2 \hbar t)}e^{i k (x' - x'')} \psi(x'', 0)\end{aligned} \hspace{\stretch{1}}(2.11)

We see that the propagator also ends up with a Fourier transform structure, and we have

\begin{aligned}\psi(x,t) &= \int dx' U(x, x' ; t) \psi(x', 0) \\ U(x, x' ; t) &=\sqrt{\frac{m}{2 \pi i \hbar t}}\frac{1}{{2 \pi}}\int du dke^{i m (x - x' - u)^2/ (2 \hbar t)}e^{i k u }\end{aligned} \hspace{\stretch{1}}(2.12)

Does that Fourier transform exist? I’d not be surprised if it ended up with a delta function representation. I’ll hold off attempting to evaluate and reduce it until another day.

## 4. Hydrogen atom.

This problem deals with the hydrogen atom, with an initial ket

\begin{aligned}{\lvert {\psi(0)} \rangle} = \frac{1}{{\sqrt{3}}} {\lvert {100} \rangle}+\frac{1}{{\sqrt{3}}} {\lvert {210} \rangle}+\frac{1}{{\sqrt{3}}} {\lvert {211} \rangle},\end{aligned} \hspace{\stretch{1}}(2.14)

where

\begin{aligned}\left\langle{\mathbf{r}} \vert {{100}}\right\rangle = \Phi_{100}(\mathbf{r}),\end{aligned} \hspace{\stretch{1}}(2.15)

etc.

\paragraph{Q: (a)}

If no measurement is made until time $t = t_0$,

\begin{aligned}t_0 = \frac{\pi \hbar}{ \frac{3}{4} (13.6 \text{eV}) } = \frac{ 4 \pi \hbar }{ 3 E_I},\end{aligned} \hspace{\stretch{1}}(2.16)

what is the ket ${\lvert {\psi(t)} \rangle}$ just before the measurement is made?

\paragraph{A:}

Our time evolved state is

\begin{aligned}{\lvert {\psi{t_0}} \rangle} = \frac{1}{{\sqrt{3}}} e^{-i E_1 t_0 /\hbar } {\lvert {100} \rangle}+\frac{1}{{\sqrt{3}}} e^{- i E_2 t_0/\hbar } ({\lvert {210} \rangle} + {\lvert {211} \rangle}).\end{aligned} \hspace{\stretch{1}}(2.17)

Also observe that this initial time was picked to make the exponential values come out nicely, and we have

\begin{aligned}\frac{E_n t_0 }{\hbar} &= - \frac{E_I \pi \hbar }{\frac{3}{4} E_I n^2 \hbar} \\ &= - \frac{4 \pi }{ 3 n^2 },\end{aligned}

so our time evolved state is just

\begin{aligned}{\lvert {\psi(t_0)} \rangle} = \frac{1}{{\sqrt{3}}} e^{-i 4 \pi / 3} {\lvert {100} \rangle}+\frac{1}{{\sqrt{3}}} e^{- i \pi / 3 } ({\lvert {210} \rangle} + {\lvert {211} \rangle}).\end{aligned} \hspace{\stretch{1}}(2.18)

\paragraph{Q: (b)}

Suppose that at time $t_0$ an $L_z$ measurement is made, and the outcome 0 is recorded. What is the appropriate ket $\psi_{\text{after}}(t_0)$ right after the measurement?

\paragraph{A:}

A measurement with outcome 0, means that the $L_z$ operator measurement found the state at that point to be the eigenstate for $L_z$ eigenvalue 0. Recall that if ${\lvert {\phi} \rangle}$ is an eigenstate of $L_z$ we have

\begin{aligned}L_z {\lvert {\phi} \rangle} = m \hbar {\lvert {\phi} \rangle},\end{aligned} \hspace{\stretch{1}}(2.19)

so a measurement of $L_z$ with outcome zero means that we have $m=0$. Our measurement of $L_z$ at time $t_0$ therefore filters out all but the $m=0$ states and our new state is proportional to the projection over all $m=0$ states as follows

\begin{aligned}{\lvert {\psi_{\text{after}}(t_0)} \rangle}&\propto \left( \sum_{n l} {\lvert {n l 0} \rangle}{\langle {n l 0} \rvert} \right) {\lvert {\psi(t_0)} \rangle} \\ &\propto \left( {\lvert {1 0 0} \rangle}{\langle {1 0 0} \rvert} +{\lvert {2 1 0} \rangle}{\langle {2 1 0} \rvert} \right) {\lvert {\psi(t_0)} \rangle} \\ &= \frac{1}{{\sqrt{3}}} e^{-i 4 \pi / 3} {\lvert {100} \rangle}+\frac{1}{{\sqrt{3}}} e^{- i \pi / 3 } {\lvert {210} \rangle} \end{aligned}

A final normalization yields

\begin{aligned}{\lvert {\psi_{\text{after}}(t_0)} \rangle}= \frac{1}{{\sqrt{2}}} ({\lvert {210} \rangle} - {\lvert {100} \rangle})\end{aligned} \hspace{\stretch{1}}(2.20)

\paragraph{Q: (c)}

Right after this $L_z$ measurement, what is ${\left\lvert{\psi_{\text{after}}(t_0)}\right\rvert}^2$?

\paragraph{A:}

Our amplitude is

\begin{aligned}\left\langle{\mathbf{r}} \vert {{\psi_{\text{after}}(t_0)}}\right\rangle&= \frac{1}{{\sqrt{2}}} (\left\langle{\mathbf{r}} \vert {{210}}\right\rangle - \left\langle{\mathbf{r}} \vert {{100}}\right\rangle) \\ &= \frac{1}{{\sqrt{2 \pi a_0^3}}}\left(\frac{r}{4\sqrt{2} a_0} e^{-r/2a_0} \cos\theta-e^{-r/a_0}\right) \\ &= \frac{1}{{\sqrt{2 \pi a_0^3}}}e^{-r/2 a_0} \left(\frac{r}{4\sqrt{2} a_0} \cos\theta-e^{-r/2 a_0}\right),\end{aligned}

so the probability density is

\begin{aligned}{\left\lvert{\left\langle{\mathbf{r}} \vert {{\psi_{\text{after}}(t_0)}}\right\rangle}\right\rvert}^2= \frac{1}{{2 \pi a_0^3}}e^{-r/a_0} \left(\frac{r}{4\sqrt{2} a_0} \cos\theta-e^{-r/2 a_0}\right)^2 \end{aligned} \hspace{\stretch{1}}(2.21)

\paragraph{Q: (d)}

If then a position measurement is made immediately, which if any components of the expectation value of $\mathbf{R}$ will be nonvanishing? Justify your answer.

\paragraph{A:}

The expectation value of this vector valued operator with respect to a radial state ${\lvert {\psi} \rangle} = \sum_{nlm} a_{nlm} {\lvert {nlm} \rangle}$ can be expressed as

\begin{aligned}\left\langle{\mathbf{R}}\right\rangle = \sum_{i=1}^3 \mathbf{e}_i \sum_{nlm, n'l'm'} a_{nlm}^{*} a_{n'l'm'} {\langle {nlm} \rvert} X_i{\lvert {n'l'm'} \rangle},\end{aligned} \hspace{\stretch{1}}(2.22)

where $X_1 = X = R \sin\Theta \cos\Phi, X_2 = Y = R \sin\Theta \sin\Phi, X_3 = Z = R \cos\Phi$.

Consider one of the matrix elements, and expand this by introducing an identity twice

\begin{aligned}{\langle {nlm} \rvert} X_i {\lvert {n'l'm'} \rangle}&=\int r^2 \sin\theta dr d\theta d\phi{r'}^2 \sin\theta' dr' d\theta' d\phi'\left\langle{{nlm}} \vert {{r \theta \phi}}\right\rangle {\langle {r \theta \phi} \rvert} X_i {\lvert {r' \theta' \phi' } \rangle}\left\langle{{r' \theta' \phi'}} \vert {{n'l'm'}}\right\rangle \\ &=\int r^2 \sin\theta dr d\theta d\phi{r'}^2 \sin\theta' dr' d\theta' d\phi'R_{nl}(r) Y_{lm}^{*}(\theta,\phi)\delta^3(\mathbf{x} - \mathbf{x}') x_iR_{n'l'}(r') Y_{l'm'}(\theta',\phi')\\ &=\int r^2 \sin\theta dr d\theta d\phi{r'}^2 \sin\theta' dr' d\theta' d\phi'R_{nl}(r) Y_{lm}^{*}(\theta,\phi) \\ &\qquad{r'}^2 \sin\theta' \delta(r-r') \delta(\theta - \theta') \delta(\phi-\phi')x_iR_{n'l'}(r') Y_{l'm'}(\theta',\phi')\\ &=\int r^2 \sin\theta dr d\theta d\phidr' d\theta' d\phi'R_{nl}(r) Y_{lm}^{*}(\theta,\phi) \delta(r-r') \delta(\theta - \theta') \delta(\phi-\phi')x_iR_{n'l'}(r') Y_{l'm'}(\theta',\phi')\\ &=\int r^2 \sin\theta dr d\theta d\phiR_{nl}(r) R_{n'l'}(r) Y_{lm}^{*}(\theta,\phi) Y_{l'm'}(\theta,\phi)x_i\\ \end{aligned}

Because our state has only $m=0$ contributions, the only $\phi$ dependence for the $X$ and $Y$ components of $\mathbf{R}$ come from those components themselves. For $X$, we therefore integrate $\int_0^{2\pi} \cos\phi d\phi = 0$, and for $Y$ we integrate $\int_0^{2\pi} \sin\phi d\phi = 0$, and these terms vanish. Our expectation value for $\mathbf{R}$ for this state, therefore lies completely on the $z$ axis.

# Questions from the Dec 2008 PHY355H1F exam.

## 1b. Trace invariance for unitary transformation.

\paragraph{Q:} Show that the trace of an operator is invariant under unitary transforms, i.e. if $A' = U^\dagger A U$, where $U$ is a unitary operator, prove $\text{Tr}(A') = \text{Tr}(A)$.

\paragraph{A:}

The bulk of this question is really to show that commutation of operators leaves the trace invariant (unless this is assumed). To show that we start with the definition of the trace

\begin{aligned}\text{Tr}(AB) &= \sum_n {\langle {n} \rvert} A B {\lvert {n} \rangle} \\ &= \sum_{n m} {\langle {n} \rvert} A {\lvert {m} \rangle} {\langle {m} \rvert} B {\lvert {n} \rangle} \\ &= \sum_{n m} {\langle {m} \rvert} B {\lvert {n} \rangle} {\langle {n} \rvert} A {\lvert {m} \rangle} \\ &= \sum_{m} {\langle {m} \rvert} B A {\lvert {m} \rangle}.\end{aligned}

Thus we have

\begin{aligned}\text{Tr}(A B) = \text{Tr}( B A ).\end{aligned} \hspace{\stretch{1}}(3.23)

For the unitarily transformed operator we have

\begin{aligned}\text{Tr}(A') &= \text{Tr}( U^\dagger A U ) \\ &= \text{Tr}( U^\dagger (A U) ) \\ &= \text{Tr}( (A U) U^\dagger ) \\ &= \text{Tr}( A (U U^\dagger) ) \\ &= \text{Tr}( A ) \qquad \square\end{aligned}

## 1d. Determinant of an exponential operator in terms of trace.

\paragraph{Q:} If $A$ is an Hermitian operator, show that

\begin{aligned}\text{Det}( \exp A ) = \exp ( \text{Tr}(A) )\end{aligned} \hspace{\stretch{1}}(3.24)

where the Determinant ($\text{Det}$) of an operator is the product of all its eigenvectors.

\paragraph{A:}

The eigenvalues clue in the question provides the starting point. We write the exponential in its series form

\begin{aligned}e^A = 1 + \sum_{k=1}^\infty \frac{1}{{k!}} A^k\end{aligned} \hspace{\stretch{1}}(3.25)

Now, suppose that we have the following eigenvalue relationships for $A$

\begin{aligned}A {\lvert {n} \rangle} = \lambda_n {\lvert {n} \rangle}.\end{aligned} \hspace{\stretch{1}}(3.26)

From this the exponential is

\begin{aligned}e^A {\lvert {n} \rangle} &= {\lvert {n} \rangle} + \sum_{k=1}^\infty \frac{1}{{k!}} A^k {\lvert {n} \rangle} \\ &= {\lvert {n} \rangle} + \sum_{k=1}^\infty \frac{1}{{k!}} (\lambda_n)^k {\lvert {n} \rangle} \\ &= e^{\lambda_n} {\lvert {n} \rangle}.\end{aligned}

We see that the eigenstates of $e^A$ are those of $A$, with eigenvalues $e^{\lambda_n}$.

By the definition of the determinant given we have

\begin{aligned}\text{Det}( e^A ) &= \Pi_n e^{\lambda_n} \\ &= e^{\sum_n \lambda_n} \\ &= e^{\text{Tr}ace(A)} \qquad \square\end{aligned}

## 1e. Eigenvectors of the Harmonic oscillator creation operator.

\paragraph{Q:} Prove that the only eigenvector of the Harmonic oscillator creation operator is ${\lvert {\text{null}} \rangle}$.

\paragraph{A:}

Recall that the creation (raising) operator was given by

\begin{aligned}a^\dagger = \sqrt{\frac{m \omega}{2 \hbar}} X - \frac{ i }{\sqrt{2 m \omega \hbar} } P= \frac{1}{{ \alpha \sqrt{2} }} X - \frac{ i \alpha }{\sqrt{2} \hbar } P,\end{aligned} \hspace{\stretch{1}}(3.27)

where $\alpha = \sqrt{\hbar/m \omega}$. Now assume that $a^\dagger {\lvert {\phi} \rangle} = \lambda {\lvert {\phi} \rangle}$ so that

\begin{aligned}{\langle {x} \rvert} a^\dagger {\lvert {\phi} \rangle} = {\langle {x} \rvert} \lambda {\lvert {\phi} \rangle}.\end{aligned} \hspace{\stretch{1}}(3.28)

Write $\left\langle{{x}} \vert {{\phi}}\right\rangle = \phi(x)$, and expand the LHS using 3.27 for

\begin{aligned}\lambda \phi(x) &= {\langle {x} \rvert} a^\dagger {\lvert {\phi} \rangle} \\ &= {\langle {x} \rvert} \left( \frac{1}{{ \alpha \sqrt{2} }} X - \frac{ i \alpha }{\sqrt{2} \hbar } P \right) {\lvert {\phi} \rangle} \\ &= \frac{x \phi(x)}{ \alpha \sqrt{2} } - \frac{ i \alpha }{\sqrt{2} \hbar } (-i\hbar)\frac{\partial {}}{\partial {x}} \phi(x) \\ &= \frac{x \phi(x)}{ \alpha \sqrt{2} } - \frac{ \alpha }{\sqrt{2} } \frac{\partial {\phi(x)}}{\partial {x}}.\end{aligned}

As usual write $\xi = x/\alpha$, and rearrange. This gives us

\begin{aligned}\frac{\partial {\phi}}{\partial {\xi}} +\sqrt{2} \lambda \phi - \xi \phi = 0.\end{aligned} \hspace{\stretch{1}}(3.29)

Observe that this can be viewed as a homogeneous LDE of the form

\begin{aligned}\frac{\partial {\phi}}{\partial {\xi}} - \xi \phi = 0,\end{aligned} \hspace{\stretch{1}}(3.30)

augmented by a forcing term $\sqrt{2}\lambda \phi$. The homogeneous equation has the solution $\phi = A e^{\xi^2/2}$, so for the complete equation we assume a solution

\begin{aligned}\phi(\xi) = A(\xi) e^{\xi^2/2}.\end{aligned} \hspace{\stretch{1}}(3.31)

Since $\phi' = (A' + A \xi) e^{\xi^2/2}$, we produce a LDE of

\begin{aligned}0 &= (A' + A \xi -\xi A + \sqrt{2} \lambda A ) e^{\xi^2/2} \\ &= (A' + \sqrt{2} \lambda A ) e^{\xi^2/2},\end{aligned}

or

\begin{aligned}0 = A' + \sqrt{2} \lambda A.\end{aligned} \hspace{\stretch{1}}(3.32)

This has solution $A = B e^{-\sqrt{2} \lambda \xi}$, so our solution for 3.29 is

\begin{aligned}\phi(\xi) = B e^{\xi^2/2 - \sqrt{2} \lambda \xi} = B' e^{ (\xi - \lambda \sqrt{2} )^2/2}.\end{aligned} \hspace{\stretch{1}}(3.33)

This wave function is an imaginary Gaussian with minimum at $\xi = \lambda\sqrt{2}$. It is also unnormalizable since we require $B' = 0$ for any $\lambda$ if $\int {\left\lvert{\phi}\right\rvert}^2 < \infty$. Since $\left\langle{{\xi}} \vert {{\phi}}\right\rangle = \phi(\xi) = 0$, we must also have ${\lvert {\phi} \rangle} = 0$, completing the exercise.

## 2. Two level quantum system.

Consider a two-level quantum system, with basis states $\{{\lvert {a} \rangle}, {\lvert {b} \rangle}\}$. Suppose that the Hamiltonian for this system is given by

\begin{aligned}H = \frac{\hbar \Delta}{2} ( {\lvert {b} \rangle}{\langle {b} \rvert}- {\lvert {a} \rangle}{\langle {a} \rvert})+ i \frac{\hbar \Omega}{2} ( {\lvert {a} \rangle}{\langle {b} \rvert}- {\lvert {b} \rangle}{\langle {a} \rvert})\end{aligned} \hspace{\stretch{1}}(3.34)

where $\Delta$ and $\Omega$ are real positive constants.

\paragraph{Q: (a)} Find the energy eigenvalues and the normalized energy eigenvectors (expressed in terms of the $\{{\lvert {a} \rangle}, {\lvert {b} \rangle}\}$ basis). Write the time evolution operator $U(t) = e^{-i H t/\hbar}$ using these eigenvectors.

\paragraph{A:}

The eigenvalue part of this problem is probably easier to do in matrix form. Let

\begin{aligned}{\lvert {a} \rangle} &= \begin{bmatrix}1 \\ 0\end{bmatrix} \\ {\lvert {b} \rangle} &= \begin{bmatrix}0 \\ 1\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(3.35)

Our Hamiltonian is then

\begin{aligned}H = \frac{\hbar}{2} \begin{bmatrix}-\Delta & i \Omega \\ -i \Omega & \Delta\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(3.37)

Computing $\det{H - \lambda I} = 0$, we get

\begin{aligned}\lambda = \pm \frac{\hbar}{2} \sqrt{ \Delta^2 + \Omega^2 }.\end{aligned} \hspace{\stretch{1}}(3.38)

Let $\delta = \sqrt{ \Delta^2 + \Omega^2 }$. Our normalized eigenvectors are found to be

\begin{aligned}{\lvert {\pm} \rangle} = \frac{1}{{\sqrt{ 2 \delta (\delta \pm \Delta)} }}\begin{bmatrix}i \Omega \\ \Delta \pm \delta\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(3.39)

In terms of ${\lvert {a} \rangle}$ and ${\lvert {b} \rangle}$, we then have

\begin{aligned}{\lvert {\pm} \rangle} = \frac{1}{{\sqrt{ 2 \delta (\delta \pm \Delta)} }}\left(i \Omega {\lvert {a} \rangle}+ (\Delta \pm \delta) {\lvert {b} \rangle} \right).\end{aligned} \hspace{\stretch{1}}(3.40)

Note that our Hamiltonian has a simple form in this basis. That is

\begin{aligned}H = \frac{\delta \hbar}{2} ({\lvert {+} \rangle}{\langle {+} \rvert} - {\lvert {-} \rangle}{\langle {-} \rvert} )\end{aligned} \hspace{\stretch{1}}(3.41)

Observe that once we do the diagonalization, we have a Hamiltonian that appears to have the form of a scaled projector for an open Stern-Gerlach aparatus.

Observe that the diagonalized Hamiltonian operator makes the time evolution operator’s form also simple, which is, by inspection

\begin{aligned}U(t) = e^{-i t \frac{\delta}{2}} {\lvert {+} \rangle}{\langle {+} \rvert} + e^{i t \frac{\delta}{2}} {\lvert {-} \rangle}{\langle {-} \rvert}.\end{aligned} \hspace{\stretch{1}}(3.42)

Since we are asked for this in terms of ${\lvert {a} \rangle}$, and ${\lvert {b} \rangle}$, the projectors ${\lvert {\pm} \rangle}{\langle {\pm} \rvert}$ are required. These are

\begin{aligned}{\lvert {\pm} \rangle}{\langle {\pm} \rvert} &= \frac{1}{{2 \delta (\delta \pm \Delta)}}\Bigl( i \Omega {\lvert {a} \rangle} + (\Delta \pm \delta) {\lvert {b} \rangle} \Bigr)\Bigl( -i \Omega {\langle {a} \rvert} + (\Delta \pm \delta) {\langle {b} \rvert} \Bigr) \\ \end{aligned}

\begin{aligned}{\lvert {\pm} \rangle}{\langle {\pm} \rvert} = \frac{1}{{2 \delta (\delta \pm \Delta)}}\Bigl(\Omega^2 {\lvert {a} \rangle}{\langle {a} \rvert}+(\delta \pm \delta)^2 {\lvert {b} \rangle}{\langle {b} \rvert}+i \Omega (\Delta \pm \delta) ({\lvert {a} \rangle}{\langle {b} \rvert}-{\lvert {b} \rangle}{\langle {a} \rvert})\Bigr)\end{aligned} \hspace{\stretch{1}}(3.43)

Substitution into 3.42 and a fair amount of algebra leads to

\begin{aligned}U(t) = \cos(\delta t/2) \Bigl( {\lvert {a} \rangle}{\langle {a} \rvert} + {\lvert {b} \rangle}{\langle {b} \rvert} \Bigr)+ i \frac{\Omega}{\delta} \sin(\delta t/2) \Bigl( {\lvert {a} \rangle}{\langle {a} \rvert} - {\lvert {b} \rangle}{\langle {b} \rvert} -i ({\lvert {a} \rangle}{\langle {b} \rvert} - {\lvert {b} \rangle}{\langle {a} \rvert} )\Bigr).\end{aligned} \hspace{\stretch{1}}(3.44)

Note that while a big cumbersome, we can also verify that we can recover the original Hamiltonian from 3.41 and 3.43.

\paragraph{Q: (b)}

Suppose that the initial state of the system at time $t = 0$ is ${\lvert {\phi(0)} \rangle}= {\lvert {b} \rangle}$. Find an expression for the state at some later time $t > 0$, ${\lvert {\phi(t)} \rangle}$.

\paragraph{A:}

Most of the work is already done. Computation of ${\lvert {\phi(t)} \rangle} = U(t) {\lvert {\phi(0)} \rangle}$ follows from 3.44

\begin{aligned}{\lvert {\phi(t)} \rangle} =\cos(\delta t/2) {\lvert {b} \rangle}- i \frac{\Omega}{\delta} \sin(\delta t/2) \Bigl( {\lvert {b} \rangle} +i {\lvert {a} \rangle}\Bigr).\end{aligned} \hspace{\stretch{1}}(3.45)

\paragraph{Q: (c)}

Suppose that an observable, specified by the operator $X = {\lvert {a} \rangle}{\langle {b} \rvert} + {\lvert {b} \rangle}{\langle {a} \rvert}$, is measured for this system. What is the probabilbity that, at time $t$, the result $1$ is obtained? Plot this probability as a function of time, showing the maximum and minimum values of the function, and the corresponding values of $t$.

\paragraph{A:}

The language of questions like these attempt to bring some physics into the mathematics. The phrase “the result $1$ is obtained”, is really a statement that the operator $X$, after measurement is found to have the eigenstate with numeric value 1.

We can calcuate the eigenvectors for this operator easily enough and find them to be $\pm 1$. For the positive eigenvalue we can also compute the eigenstate to be

\begin{aligned}{\lvert {X+} \rangle} = \frac{1}{{\sqrt{2}}} \Bigl( {\lvert {a} \rangle} + {\lvert {b} \rangle} \Bigr).\end{aligned} \hspace{\stretch{1}}(3.46)

The question of what the probability for this measurement is then really a question asking for the computation of the amplitude

\begin{aligned}{\left\lvert{\frac{1}{{\sqrt{2}}}\left\langle{{ (a + b)}} \vert {{\phi(t)}}\right\rangle}\right\rvert}^2\end{aligned} \hspace{\stretch{1}}(3.47)

From 3.45 we find this probability to be

\begin{aligned}{\left\lvert{\frac{1}{{\sqrt{2}}}\left\langle{{ (a + b)}} \vert {{\phi(t)}}\right\rangle}\right\rvert}^2&=\frac{1}{{2}} \left(\left(\cos(\delta t/2) + \frac{\Omega}{\delta} \sin(\delta t/2)\right)^2+ \frac{ \Omega^2 \sin^2(\delta t/2)}{\delta^2}\right) \\ &=\frac{1}{{4}} \left( 1 + 3 \frac{\Omega^2}{\delta^2} + \frac{\Delta^2}{\delta^2} \cos (\delta t) + 2 \frac{ \Omega}{\delta} \sin(\delta t) \right)\end{aligned}

We have a simple superposition of two sinusuiods out of phase, periodic with period $2 \pi/\delta$. I’d attempted a rough sketch of this on paper, but won’t bother scanning it here or describing it further.

\paragraph{Q: (d)}

Suppose an experimenter has control over the values of the parameters $\Delta$ and $\Omega$. Explain how she might prepare the state $({\lvert {a} \rangle} + {\lvert {b} \rangle})/\sqrt{2}$.

\paragraph{A:}

For this part of the question I wasn’t sure what approach to take. I thought perhaps this linear combination of states could be made to equal one of the energy eigenstates, and if one could prepare the system in that state, then for certain values of $\delta$ and $\Delta$ one would then have this desired state.

To get there I note that we can express the states ${\lvert {a} \rangle}$, and ${\lvert {b} \rangle}$ in terms of the eigenstates by inverting

\begin{aligned}\begin{bmatrix}{\lvert {+} \rangle} \\ {\lvert {-} \rangle} \\ \end{bmatrix}=\frac{1}{{\sqrt{2\delta}}}\begin{bmatrix}\frac{i \Omega}{\sqrt{\delta + \Delta}} & \sqrt{\delta + \Delta} \\ \frac{i \Omega}{\sqrt{\delta - \Delta}} & -\sqrt{\delta - \Delta}\end{bmatrix}\begin{bmatrix}{\lvert {a} \rangle} \\ {\lvert {b} \rangle} \\ \end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(3.48)

Skipping all the algebra one finds

\begin{aligned}\begin{bmatrix}{\lvert {a} \rangle} \\ {\lvert {b} \rangle} \\ \end{bmatrix}=\begin{bmatrix}-i\sqrt{\delta - \Delta} & -i\sqrt{\delta + \Delta} \\ \frac{\Omega}{\sqrt{\delta - \Delta}} &-\frac{\Omega}{\sqrt{\delta + \Delta}} \end{bmatrix}\begin{bmatrix}{\lvert {+} \rangle} \\ {\lvert {-} \rangle} \\ \end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(3.49)

Unfortunately, this doesn’t seem helpful. I find

\begin{aligned}\frac{1}{{\sqrt{2}}} ( {\lvert {a} \rangle} + {\lvert {b} \rangle} ) = \frac{{\lvert {+} \rangle}}{\sqrt{\delta - \Delta}}( \Omega - i (\delta - \Delta) )-\frac{{\lvert {-} \rangle}}{\sqrt{\delta + \Delta}}( \Omega + i (\delta + \Delta) )\end{aligned} \hspace{\stretch{1}}(3.50)

There’s no obvious way to pick $\Omega$ and $\Delta$ to leave just ${\lvert {+} \rangle}$ or ${\lvert {-} \rangle}$. When I did this on paper originally I got a different answer for this sum, but looking at it now, I can’t see how I managed to get that answer (it had no factors of $i$ in the result as the one above does).

## 3. One dimensional harmonic oscillator.

Consider a one-dimensional harmonic oscillator with the Hamiltonian

\begin{aligned}H = \frac{1}{{2m}}P^2 + \frac{1}{{2}} m \omega^2 X^2\end{aligned} \hspace{\stretch{1}}(3.51)

Denote the ground state of the system by ${\lvert {0} \rangle}$, the first excited state by ${\lvert {1} \rangle}$ and so on.

\paragraph{Q: (a)}
Evaluate ${\langle {n} \rvert} X {\lvert {n} \rangle}$ and ${\langle {n} \rvert} X^2 {\lvert {n} \rangle}$ for arbitrary ${\lvert {n} \rangle}$.

\paragraph{A:}

Writing $X$ in terms of the raising and lowering operators we have

\begin{aligned}X = \frac{\alpha}{\sqrt{2}} (a^\dagger + a),\end{aligned} \hspace{\stretch{1}}(3.52)

so $\left\langle{{X}}\right\rangle$ is proportional to

\begin{aligned}{\langle {n} \rvert} a^\dagger + a {\lvert {n} \rangle} = \sqrt{n+1} \left\langle{{n}} \vert {{n+1}}\right\rangle + \sqrt{n} \left\langle{{n}} \vert {{n-1}}\right\rangle = 0.\end{aligned} \hspace{\stretch{1}}(3.53)

For $\left\langle{{X^2}}\right\rangle$ we have

\begin{aligned}\left\langle{{X^2}}\right\rangle&=\frac{\alpha^2}{2}{\langle {n} \rvert} (a^\dagger + a)(a^\dagger + a) {\lvert {n} \rangle} \\ &=\frac{\alpha^2}{2}{\langle {n} \rvert} (a^\dagger + a) \left( \sqrt{n+1} {\lvert {n+1} \rangle} + \sqrt{n-1} {\lvert {n-1} \rangle}\right) \\ &=\frac{\alpha^2}{2}{\langle {n} \rvert} \Bigl( (n+1) {\lvert {n} \rangle} + \sqrt{n(n-1)} {\lvert {n-2} \rangle}+ \sqrt{(n+1)(n+2)} {\lvert {n+2} \rangle} + n {\lvert {n} \rangle} \Bigr).\end{aligned}

We are left with just

\begin{aligned}\left\langle{{X^2}}\right\rangle = \frac{\hbar}{2 m \omega} (2n + 1).\end{aligned} \hspace{\stretch{1}}(3.54)

\paragraph{Q: (b)}

Suppose that at $t=0$ the system is prepared in the state

\begin{aligned}{\lvert {\psi(0)} \rangle} = \frac{1}{{\sqrt{2}}} ( {\lvert {0} \rangle} + i {\lvert {1} \rangle} ).\end{aligned} \hspace{\stretch{1}}(3.55)

If a measurement of position $X$ were performaed immediately, sketch the propability distribution $P(x)$ that a particle would be found within $dx$ of $x$. Justify how you construct the sketch.

\paragraph{A:}

The probability that we started in state ${\lvert {\psi(0)} \rangle}$ and ended up in position $x$ is governed by the amplitude $\left\langle{{x}} \vert {{\psi(0)}}\right\rangle$, and the probability of being within an interval $\Delta x$, surrounding the point $x$ is given by

\begin{aligned}\int_{x'=x-\Delta x/2}^{x+\Delta x/2} {\left\lvert{ \left\langle{{x'}} \vert {{\psi(0)}}\right\rangle }\right\rvert}^2 dx'.\end{aligned} \hspace{\stretch{1}}(3.56)

In the limit as $\Delta x \rightarrow 0$, this is just the squared amplitude itself evaluated at the point $x$, so we are interested in the quantity

\begin{aligned}{\left\lvert{ \left\langle{{x}} \vert {{\psi(0)}}\right\rangle }\right\rvert}^2 = \frac{1}{{2}} {\left\lvert{ \left\langle{{x}} \vert {{0}}\right\rangle + i \left\langle{{x}} \vert {{1}}\right\rangle }\right\rvert}^2.\end{aligned} \hspace{\stretch{1}}(3.57)

We are given these wave functions in the supplemental formulas. Namely,

\begin{aligned}\left\langle{{x}} \vert {{0}}\right\rangle &= \psi_0(x) = \frac{e^{-x^2/2\alpha^2}}{ \sqrt{\alpha \sqrt{\pi}}} \\ \left\langle{{x}} \vert {{1}}\right\rangle &= \psi_1(x) = \frac{e^{-x^2/2\alpha^2} 2 x }{ \alpha \sqrt{2 \alpha \sqrt{\pi}}}.\end{aligned} \hspace{\stretch{1}}(3.58)

Substituting these into 3.57 we have

\begin{aligned}{\left\lvert{ \left\langle{{x}} \vert {{\psi(0)}}\right\rangle }\right\rvert}^2 = \frac{1}{{2}} e^{-x^2/\alpha^2}\frac{1}{{ \alpha \sqrt{\pi}}}{\left\lvert{ 1 + \frac{2 i x}{\alpha \sqrt{2} } }\right\rvert}^2=\frac{e^{-x^2/\alpha^2}}{ 2\alpha \sqrt{\pi}}\left( 1 + \frac{2 x^2}{\alpha^2 } \right).\end{aligned} \hspace{\stretch{1}}(3.60)

This \href{http://www.wolframalpha.com/input/?i=graph+e^(-x^2)+(1+

\paragraph{Q: (c)}

Now suppose the state given in (b) above were allowed to evolve for a time $t$, determine the expecation value of $X$ and $\Delta X$ at that time.

\paragraph{A:}

Our time evolved state is

\begin{aligned}U(t) {\lvert {\psi(0)} \rangle} = \frac{1}{{\sqrt{2}}}\left(e^{-i \hbar \omega \left( 0 + \frac{1}{{2}} \right) t/\hbar } {\lvert {0} \rangle}+ i e^{-i \hbar \omega \left( 1 + \frac{1}{{2}} \right) t/\hbar } {\lvert {0} \rangle}\right)=\frac{1}{{\sqrt{2}}}\left(e^{-i \omega t/2 } {\lvert {0} \rangle}+ i e^{- 3 i \omega t/2 } {\lvert {1} \rangle}\right).\end{aligned} \hspace{\stretch{1}}(3.61)

The position expectation is therefore

\begin{aligned}{\langle {\psi(t)} \rvert} X {\lvert {\psi(t)} \rangle}&= \frac{\alpha}{2 \sqrt{2}}\left(e^{i \omega t/2 } {\langle {0} \rvert}- i e^{ 3 i \omega t/2 } {\langle {1} \rvert}\right)(a^\dagger + a)\left(e^{-i \omega t/2 } {\lvert {0} \rangle}+ i e^{- 3 i \omega t/2 } {\lvert {1} \rangle}\right) \\ \end{aligned}

We have already demonstrated that ${\langle {n} \rvert} X {\lvert {n} \rangle} = 0$, so we must only expand the cross terms, but those are just ${\langle {0} \rvert} a^\dagger + a {\lvert {1} \rangle} = 1$. This leaves

\begin{aligned}{\langle {\psi(t)} \rvert} X {\lvert {\psi(t)} \rangle}= \frac{\alpha}{2 \sqrt{2}}\left( -i e^{i \omega t} + i e^{-i \omega t} \right)=\sqrt{\frac{\hbar}{2 m \omega}} \cos(\omega t)\end{aligned} \hspace{\stretch{1}}(3.62)

For the squared position expectation

\begin{aligned}{\langle {\psi(t)} \rvert} X^2 {\lvert {\psi(t)} \rangle}&= \frac{\alpha^2}{4 (2)}\left(e^{i \omega t/2 } {\langle {0} \rvert}- i e^{ 3 i \omega t/2 } {\langle {1} \rvert}\right)(a^\dagger + a)^2\left(e^{-i \omega t/2 } {\lvert {0} \rangle}+ i e^{- 3 i \omega t/2 } {\lvert {1} \rangle}\right) \\ &=\frac{1}{{2}} ( {\langle {0} \rvert} X^2 {\lvert {0} \rangle} + {\langle {1} \rvert} X^2 {\lvert {1} \rangle} )+ i \frac{\alpha^2 }{8} ( - e^{ i \omega t} {\langle {1} \rvert} (a^\dagger + a)^2 {\lvert {0} \rangle}+ e^{ -i \omega t} {\langle {0} \rvert} (a^\dagger + a)^2 {\lvert {1} \rangle})\end{aligned}

Noting that $(a^\dagger + a) {\lvert {0} \rangle} = {\lvert {1} \rangle}$, and $(a^\dagger + a)^2 {\lvert {0} \rangle} = (a^\dagger + a){\lvert {1} \rangle} = \sqrt{2} {\lvert {2} \rangle} + {\lvert {0} \rangle}$, so we see the last two terms are zero. The first two we can evaluate using our previous result 3.54 which was $\left\langle{{X^2}}\right\rangle = \frac{\alpha^2}{2} (2n + 1)$. This leaves

\begin{aligned}{\langle {\psi(t)} \rvert} X^2 {\lvert {\psi(t)} \rangle} = \alpha^2 \end{aligned} \hspace{\stretch{1}}(3.63)

Since $\left\langle{{X}}\right\rangle^2 = \alpha^2 \cos^2(\omega t)/2$, we have

\begin{aligned}(\Delta X)^2 = \left\langle{{X^2}}\right\rangle - \left\langle{{X}}\right\rangle^2 = \alpha^2 \left(1 - \frac{1}{{2}} \cos^2(\omega t) \right)\end{aligned} \hspace{\stretch{1}}(3.64)

\paragraph{Q: (d)}

Now suppose that initially the system were prepared in the ground state ${\lvert {0} \rangle}$, and then the resonance frequency is changed abrubtly from $\omega$ to $\omega'$ so that the Hamiltonian becomes

\begin{aligned}H = \frac{1}{{2m}}P^2 + \frac{1}{{2}} m {\omega'}^2 X^2.\end{aligned} \hspace{\stretch{1}}(3.65)

Immediately, an energy measurement is performed ; what is the probability of obtaining the result $E = \hbar \omega' (3/2)$?

\paragraph{A:}

This energy measurement $E = \hbar \omega' (3/2) = \hbar \omega' (1 + 1/2)$, corresponds to an observation of state ${\lvert {1'} \rangle}$, after an initial observation of ${\lvert {0} \rangle}$. The probability of such a measurement is

\begin{aligned}{\left\lvert{ \left\langle{{1'}} \vert {{0}}\right\rangle }\right\rvert}^2\end{aligned} \hspace{\stretch{1}}(3.66)

Note that

\begin{aligned}\left\langle{{1'}} \vert {{0}}\right\rangle &=\int dx \left\langle{{1'}} \vert {{x}}\right\rangle\left\langle{{x}} \vert {{0}}\right\rangle \\ &=\int dx \psi_{1'}^{*} \psi_0(x) \\ \end{aligned}

The wave functions above are

\begin{aligned}\phi_{1'}(x) &= \frac{ 2 x e^{-x^2/2 {\alpha'}^2 }}{ \alpha' \sqrt{ 2 \alpha' \sqrt{\pi} } } \\ \phi_{0}(x) &= \frac{ e^{-x^2/2 {\alpha}^2 } } { \sqrt{ \alpha \sqrt{\pi} } } \end{aligned} \hspace{\stretch{1}}(3.67)

Putting the pieces together we have

\begin{aligned}\left\langle{{1'}} \vert {{0}}\right\rangle &=\frac{2 }{ \alpha' \sqrt{ 2 \alpha' \alpha \pi } }\int dxx e^{-\frac{x^2}{2}\left( \frac{1}{{{\alpha'}^2}} + \frac{1}{{\alpha^2}} \right) }\end{aligned} \hspace{\stretch{1}}(3.69)

Since this is an odd integral kernel over an even range, this evaluates to zero, and we conclude that the probability of measuring the specified energy is zero when the system is initially prepared in the ground state associated with the original Hamiltonian. Intuitively this makes some sense, if one thinks of the Fourier coefficient problem: one cannot construct an even function from linear combinations of purely odd functions.