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# Posts Tagged ‘separation of variables’

## PHY450H1S. Relativistic Electrodynamics Tutorial 4 (TA: Simon Freedman). Waveguides: confined EM waves.

Posted by peeterjoot on March 14, 2011

# Motivation

While this isn’t part of the course, the topic of waveguides is one of so many applications that it is worth a mention, and that will be done in this tutorial.

We will setup our system with a waveguide (conducting surface that confines the radiation) oriented in the $\hat{\mathbf{z}}$ direction. The shape can be arbitrary

PICTURE: cross section of wacky shape.

## At the surface of a conductor.

At the surface of the conductor (I presume this means the interior surface where there is no charge or current enclosed) we have

\begin{aligned}\boldsymbol{\nabla} \times \mathbf{E} &= - \frac{1}{{c}} \frac{\partial {\mathbf{B}}}{\partial {t}} \\ \boldsymbol{\nabla} \times \mathbf{B} &= \frac{1}{{c}} \frac{\partial {\mathbf{E}}}{\partial {t}} \\ \boldsymbol{\nabla} \cdot \mathbf{B} &= 0 \\ \boldsymbol{\nabla} \cdot \mathbf{E} &= 0\end{aligned} \hspace{\stretch{1}}(1.1)

If we are talking about the exterior surface, do we need to make any other assumptions (perfect conductors, or constant potentials)?

## Wave equations.

For electric and magnetic fields in vacuum, we can show easily that these, like the potentials, separately satisfy the wave equation

Taking curls of the Maxwell curl equations above we have

\begin{aligned}\boldsymbol{\nabla} \times (\boldsymbol{\nabla} \times \mathbf{E}) &= - \frac{1}{{c^2}} \frac{\partial^2 {\mathbf{E}}}{\partial {{t}}^2} \\ \boldsymbol{\nabla} \times (\boldsymbol{\nabla} \times \mathbf{B}) &= - \frac{1}{{c^2}} \frac{\partial^2 {\mathbf{B}}}{\partial {{t}}^2},\end{aligned} \hspace{\stretch{1}}(1.5)

but we have for vector $\mathbf{M}$

\begin{aligned}\boldsymbol{\nabla} \times (\boldsymbol{\nabla} \times \mathbf{M})=\boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{M}) - \Delta \mathbf{M},\end{aligned} \hspace{\stretch{1}}(1.7)

which gives us a pair of wave equations

\begin{aligned}\square \mathbf{E} &= 0 \\ \square \mathbf{B} &= 0.\end{aligned} \hspace{\stretch{1}}(1.8)

We still have the original constraints of Maxwell’s equations to deal with, but we are free now to pick the complex exponentials as fundamental solutions, as our starting point

\begin{aligned}\mathbf{E} &= \mathbf{E}_0 e^{i k^a x_a} = \mathbf{E}_0 e^{ i (k^0 x_0 - \mathbf{k} \cdot \mathbf{x}) } \\ \mathbf{B} &= \mathbf{B}_0 e^{i k^a x_a} = \mathbf{B}_0 e^{ i (k^0 x_0 - \mathbf{k} \cdot \mathbf{x}) },\end{aligned} \hspace{\stretch{1}}(1.10)

With $k_0 = \omega/c$ and $x_0 = c t$ this is

\begin{aligned}\mathbf{E} &= \mathbf{E}_0 e^{ i (\omega t - \mathbf{k} \cdot \mathbf{x}) } \\ \mathbf{B} &= \mathbf{B}_0 e^{ i (\omega t - \mathbf{k} \cdot \mathbf{x}) }.\end{aligned} \hspace{\stretch{1}}(1.12)

For the vacuum case, with monochromatic light, we treated the amplitudes as constants. Let’s see what happens if we relax this assumption, and allow for spatial dependence (but no time dependence) of $\mathbf{E}_0$ and $\mathbf{B}_0$. For the LHS of the electric field curl equation we have

\begin{aligned}0 &= \boldsymbol{\nabla} \times \mathbf{E}_0 e^{i k_a x^a} \\ &= (\boldsymbol{\nabla} \times \mathbf{E}_0 - \mathbf{E}_0 \times \boldsymbol{\nabla}) e^{i k_a x^a} \\ &= (\boldsymbol{\nabla} \times \mathbf{E}_0 - \mathbf{E}_0 \times \mathbf{e}^\alpha i k_a \partial_\alpha x^a) e^{i k_a x^a} \\ &= (\boldsymbol{\nabla} \times \mathbf{E}_0 + \mathbf{E}_0 \times \mathbf{e}^\alpha i k^a {\delta_\alpha}^a ) e^{i k_a x^a} \\ &= (\boldsymbol{\nabla} \times \mathbf{E}_0 + i \mathbf{E}_0 \times \mathbf{k} ) e^{i k_a x^a}.\end{aligned}

Similarly for the divergence we have

\begin{aligned}0 &= \boldsymbol{\nabla} \cdot \mathbf{E}_0 e^{i k_a x^a} \\ &= (\boldsymbol{\nabla} \cdot \mathbf{E}_0 + \mathbf{E}_0 \cdot \boldsymbol{\nabla}) e^{i k_a x^a} \\ &= (\boldsymbol{\nabla} \cdot \mathbf{E}_0 + \mathbf{E}_0 \cdot \mathbf{e}^\alpha i k_a \partial_\alpha x^a) e^{i k_a x^a} \\ &= (\boldsymbol{\nabla} \cdot \mathbf{E}_0 - \mathbf{E}_0 \cdot \mathbf{e}^\alpha i k^a {\delta_\alpha}^a ) e^{i k_a x^a} \\ &= (\boldsymbol{\nabla} \cdot \mathbf{E}_0 - i \mathbf{k} \cdot \mathbf{E}_0 ) e^{i k_a x^a}.\end{aligned}

This provides constraints on the amplitudes

\begin{aligned}\boldsymbol{\nabla} \times \mathbf{E}_0 - i \mathbf{k} \times \mathbf{E}_0 &= -i \frac{\omega}{c} \mathbf{B}_0 \\ \boldsymbol{\nabla} \times \mathbf{B}_0 - i \mathbf{k} \times \mathbf{B}_0 &= i \frac{\omega}{c} \mathbf{E}_0 \\ \boldsymbol{\nabla} \cdot \mathbf{E}_0 - i \mathbf{k} \cdot \mathbf{E}_0 &= 0 \\ \boldsymbol{\nabla} \cdot \mathbf{B}_0 - i \mathbf{k} \cdot \mathbf{B}_0 &= 0\end{aligned} \hspace{\stretch{1}}(1.14)

Applying the wave equation operator to our phasor we get

\begin{aligned}0 &=\left(\frac{1}{{c^2}} \partial_{tt} - \boldsymbol{\nabla}^2 \right) \mathbf{E}_0 e^{i (\omega t - \mathbf{k} \cdot \mathbf{x})} \\ &=\left(-\frac{\omega^2}{c^2} - \boldsymbol{\nabla}^2 + \mathbf{k}^2 \right) \mathbf{E}_0 e^{i (\omega t - \mathbf{k} \cdot \mathbf{x})}\end{aligned}

So the momentum space equivalents of the wave equations are

\begin{aligned}\left( \boldsymbol{\nabla}^2 +\frac{\omega^2}{c^2} - \mathbf{k}^2 \right) \mathbf{E}_0 &= 0 \\ \left( \boldsymbol{\nabla}^2 +\frac{\omega^2}{c^2} - \mathbf{k}^2 \right) \mathbf{B}_0 &= 0.\end{aligned} \hspace{\stretch{1}}(1.18)

Observe that if $c^2 \mathbf{k}^2 = \omega^2$, then these amplitudes are harmonic functions (solutions to the Laplacian equation). However, it doesn’t appear that we require such a light like relation for the four vector $k^a = (\omega/c, \mathbf{k})$.

# Back to the tutorial notes.

In class we went straight to an assumed solution of the form

\begin{aligned}\mathbf{E} &= \mathbf{E}_0(x, y) e^{ i(\omega t - k z) } \\ \mathbf{B} &= \mathbf{B}_0(x, y) e^{ i(\omega t - k z) },\end{aligned} \hspace{\stretch{1}}(2.20)

where $\mathbf{k} = k \hat{\mathbf{z}}$. Our Laplacian was also written as the sum of components in the propagation and perpendicular directions

\begin{aligned}\boldsymbol{\nabla}^2 = \frac{\partial^2 {{}}}{\partial {{x_\perp}}^2} + \frac{\partial^2 {{}}}{\partial {{z}}^2}.\end{aligned} \hspace{\stretch{1}}(2.22)

With no $z$ dependence in the amplitudes we have

\begin{aligned}\left( \frac{\partial^2 {{}}}{\partial {{x_\perp}}^2} +\frac{\omega^2}{c^2} - \mathbf{k}^2 \right) \mathbf{E}_0 &= 0 \\ \left( \frac{\partial^2 {{}}}{\partial {{x_\perp}}^2} +\frac{\omega^2}{c^2} - \mathbf{k}^2 \right) \mathbf{B}_0 &= 0.\end{aligned} \hspace{\stretch{1}}(2.23)

# Separation into components.

It was left as an exercise to separate out our Maxwell equations, so that our field components $\mathbf{E}_0 = \mathbf{E}_\perp + \mathbf{E}_z$ and $\mathbf{B}_0 = \mathbf{B}_\perp + \mathbf{B}_z$ in the propagation direction, and components in the perpendicular direction are separated

\begin{aligned}\boldsymbol{\nabla} \times \mathbf{E}_0 &=(\boldsymbol{\nabla}_\perp + \hat{\mathbf{z}}\partial_z) \times \mathbf{E}_0 \\ &=\boldsymbol{\nabla}_\perp \times \mathbf{E}_0 \\ &=\boldsymbol{\nabla}_\perp \times (\mathbf{E}_\perp + \mathbf{E}_z) \\ &=\boldsymbol{\nabla}_\perp \times \mathbf{E}_\perp +\boldsymbol{\nabla}_\perp \times \mathbf{E}_z \\ &=( \hat{\mathbf{x}} \partial_x +\hat{\mathbf{y}} \partial_y ) \times ( \hat{\mathbf{x}} E_x +\hat{\mathbf{y}} E_y ) +\boldsymbol{\nabla}_\perp \times \mathbf{E}_z \\ &=\hat{\mathbf{z}} (\partial_x E_y - \partial_z E_z) +\boldsymbol{\nabla}_\perp \times \mathbf{E}_z.\end{aligned}

We can do something similar for $\mathbf{B}_0$. This allows for a split of 1.14 into $\hat{\mathbf{z}}$ and perpendicular components

\begin{aligned}\boldsymbol{\nabla}_\perp \times \mathbf{E}_\perp &= -i \frac{\omega}{c} \mathbf{B}_z \\ \boldsymbol{\nabla}_\perp \times \mathbf{B}_\perp &= i \frac{\omega}{c} \mathbf{E}_z \\ \boldsymbol{\nabla}_\perp \times \mathbf{E}_z - i \mathbf{k} \times \mathbf{E}_\perp &= -i \frac{\omega}{c} \mathbf{B}_\perp \\ \boldsymbol{\nabla}_\perp \times \mathbf{B}_z - i \mathbf{k} \times \mathbf{B}_\perp &= i \frac{\omega}{c} \mathbf{E}_\perp \\ \boldsymbol{\nabla}_\perp \cdot \mathbf{E}_\perp &= i k E_z - \partial_z E_z \\ \boldsymbol{\nabla}_\perp \cdot \mathbf{B}_\perp &= i k B_z - \partial_z B_z.\end{aligned} \hspace{\stretch{1}}(3.25)

So we see that once we have a solution for $\mathbf{E}_z$ and $\mathbf{B}_z$ (by solving the wave equation above for those components), the components for the fields in terms of those components can be found. Alternately, if one solves for the perpendicular components of the fields, these propagation components are available immediately with only differentiation.

In the case where the perpendicular components are taken as given

\begin{aligned}\mathbf{B}_z &= i \frac{ c }{\omega} \boldsymbol{\nabla}_\perp \times \mathbf{E}_\perp \\ \mathbf{E}_z &= -i \frac{ c }{\omega} \boldsymbol{\nabla}_\perp \times \mathbf{B}_\perp,\end{aligned} \hspace{\stretch{1}}(3.31)

we can express the remaining ones strictly in terms of the perpendicular fields

\begin{aligned}\frac{\omega}{c} \mathbf{B}_\perp &= \frac{c}{\omega} \boldsymbol{\nabla}_\perp \times (\boldsymbol{\nabla}_\perp \times \mathbf{B}_\perp) + \mathbf{k} \times \mathbf{E}_\perp \\ \frac{\omega}{c} \mathbf{E}_\perp &= \frac{c}{\omega} \boldsymbol{\nabla}_\perp \times (\boldsymbol{\nabla}_\perp \times \mathbf{E}_\perp) - \mathbf{k} \times \mathbf{B}_\perp \\ \boldsymbol{\nabla}_\perp \cdot \mathbf{E}_\perp &= -i \frac{c}{\omega} (i k - \partial_z) \hat{\mathbf{z}} \cdot (\boldsymbol{\nabla}_\perp \times \mathbf{B}_\perp) \\ \boldsymbol{\nabla}_\perp \cdot \mathbf{B}_\perp &= i \frac{c}{\omega} (i k - \partial_z) \hat{\mathbf{z}} \cdot (\boldsymbol{\nabla}_\perp \times \mathbf{E}_\perp).\end{aligned} \hspace{\stretch{1}}(3.33)

Is it at all helpful to expand the double cross products?

\begin{aligned}\frac{\omega^2}{c^2} \mathbf{B}_\perp &= \boldsymbol{\nabla}_\perp (\boldsymbol{\nabla}_\perp \cdot \mathbf{B}_\perp) -{\boldsymbol{\nabla}_\perp}^2 \mathbf{B}_\perp + \frac{\omega}{c} \mathbf{k} \times \mathbf{E}_\perp \\ &= i \frac{c}{\omega}(i k - \partial_z)\boldsymbol{\nabla}_\perp \hat{\mathbf{z}} \cdot (\boldsymbol{\nabla}_\perp \times \mathbf{E}_\perp)-{\boldsymbol{\nabla}_\perp}^2 \mathbf{B}_\perp + \frac{\omega}{c} \mathbf{k} \times \mathbf{E}_\perp \end{aligned}

This gives us

\begin{aligned}\left( {\boldsymbol{\nabla}_\perp}^2 + \frac{\omega^2}{c^2} \right) \mathbf{B}_\perp &= - \frac{c}{\omega} (k + i\partial_z) \boldsymbol{\nabla}_\perp \hat{\mathbf{z}} \cdot (\boldsymbol{\nabla}_\perp \times \mathbf{E}_\perp) + \frac{\omega}{c} \mathbf{k} \times \mathbf{E}_\perp \\ \left( {\boldsymbol{\nabla}_\perp}^2 + \frac{\omega^2}{c^2} \right) \mathbf{E}_\perp &= -\frac{c}{\omega} (k + i\partial_z) \boldsymbol{\nabla}_\perp \hat{\mathbf{z}} \cdot (\boldsymbol{\nabla}_\perp \times \mathbf{B}_\perp) - \frac{\omega}{c} \mathbf{k} \times \mathbf{B}_\perp,\end{aligned} \hspace{\stretch{1}}(3.37)

but that doesn’t seem particularly useful for completely solving the system? It appears fairly messy to try to solve for $\mathbf{E}_\perp$ and $\mathbf{B}_\perp$ given the propagation direction fields. I wonder if there is a simplification available that I am missing?

# Solving the momentum space wave equations.

Back to the class notes. We proceeded to solve for $\mathbf{E}_z$ and $\mathbf{B}_z$ from the wave equations by separation of variables. We wish to solve equations of the form

\begin{aligned}\left( \frac{\partial^2 {{}}}{\partial {{x}}^2} + \frac{\partial^2 {{}}}{\partial {{y}}^2} + \frac{\omega^2}{c^2} - \mathbf{k}^2 \right) \phi(x,y) = 0\end{aligned} \hspace{\stretch{1}}(4.39)

Write $\phi(x,y) = X(x) Y(y)$, so that we have

\begin{aligned}\frac{X''}{X} + \frac{Y''}{Y} = \mathbf{k}^2 - \frac{\omega^2}{c^2}\end{aligned} \hspace{\stretch{1}}(4.40)

One solution is sinusoidal

\begin{aligned}\frac{X''}{X} &= -k_1^2 \\ \frac{Y''}{Y} &= -k_2^2 \\ -k_1^2 - k_2^2&= \mathbf{k}^2 - \frac{\omega^2}{c^2}.\end{aligned} \hspace{\stretch{1}}(4.41)

The example in the tutorial now switched to a rectangular waveguide, still oriented with the propagation direction down the z-axis, but with lengths $a$ and $b$ along the $x$ and $y$ axis respectively.

Writing $k_1 = 2\pi m/a$, and $k_2 = 2 \pi n/ b$, we have

\begin{aligned}\phi(x, y) = \sum_{m n} a_{m n} \exp\left( \frac{2 \pi i m}{a} x \right)\exp\left( \frac{2 \pi i n}{b} y \right)\end{aligned} \hspace{\stretch{1}}(4.44)

We were also provided with some definitions

\begin{definition}TE (Transverse Electric)

$\mathbf{E}_3 = 0$.
\end{definition}
\begin{definition}
TM (Transverse Magnetic)

$\mathbf{B}_3 = 0$.
\end{definition}
\begin{definition}
TM (Transverse Electromagnetic)

$\mathbf{E}_3 = \mathbf{B}_3 = 0$.
\end{definition}

\begin{claim}TEM do not existing in a hollow waveguide.
\end{claim}

Why: I had in my notes

\begin{aligned}\boldsymbol{\nabla} \times \mathbf{E} = 0 & \implies \frac{\partial {E_2}}{\partial {x^1}} -\frac{\partial {E_1}}{\partial {x^2}} = 0 \\ \boldsymbol{\nabla} \cdot \mathbf{E} = 0 & \implies \frac{\partial {E_1}}{\partial {x^1}} +\frac{\partial {E_2}}{\partial {x^2}} = 0\end{aligned}

and then

\begin{aligned}\boldsymbol{\nabla}^2 \phi &= 0 \\ \phi &= \text{const}\end{aligned}

In retrospect I fail to see how these are connected? What happened to the $\partial_t \mathbf{B}$ term in the curl equation above?

It was argued that we have $\mathbf{E}_\parallel = \mathbf{B}_\perp = 0$ on the boundary.

So for the TE case, where $\mathbf{E}_3 = 0$, we have from the separation of variables argument

\begin{aligned}\hat{\mathbf{z}} \cdot \mathbf{B}_0(x, y) =\sum_{m n} a_{m n} \cos\left( \frac{2 \pi i m}{a} x \right)\cos\left( \frac{2 \pi i n}{b} y \right).\end{aligned} \hspace{\stretch{1}}(4.45)

No sines because

\begin{aligned}B_1 \propto \frac{\partial {B_3}}{\partial {x_a}} \rightarrow \cos(k_1 x^1).\end{aligned} \hspace{\stretch{1}}(4.46)

The quantity

\begin{aligned}a_{m n}\cos\left( \frac{2 \pi i m}{a} x \right)\cos\left( \frac{2 \pi i n}{b} y \right).\end{aligned} \hspace{\stretch{1}}(4.47)

is called the $TE_{m n}$ mode. Note that since $B = \text{const}$ an ampere loop requires $\mathbf{B} = 0$ since there is no current.

Writing

\begin{aligned}k &= \frac{\omega}{c} \sqrt{ 1 - \left(\frac{\omega_{m n}}{\omega}\right)^2 } \\ \omega_{m n} &= 2 \pi c \sqrt{ \left(\frac{m}{a} \right)^2 + \left(\frac{n}{b} \right)^2 }\end{aligned} \hspace{\stretch{1}}(4.48)

Since $\omega < \omega_{m n}$ we have $k$ purely imaginary, and the term

\begin{aligned}e^{-i k z} = e^{- {\left\lvert{k}\right\rvert} z}\end{aligned} \hspace{\stretch{1}}(4.50)

represents the die off.

$\omega_{10}$ is the smallest.

Note that the convention is that the $m$ in $TE_{m n}$ is the bigger of the two indexes, so $\omega > \omega_{10}$.

The phase velocity

\begin{aligned}V_\phi = \frac{\omega}{k} = \frac{c}{\sqrt{ 1 - \left(\frac{\omega_{m n}}{\omega}\right)^2 }} \ge c\end{aligned} \hspace{\stretch{1}}(4.51)

However, energy is transmitted with the group velocity, the ratio of the Poynting vector and energy density

\begin{aligned}\frac{\left\langle{\mathbf{S}}\right\rangle}{\left\langle{{U}}\right\rangle} = V_g = \frac{\partial {\omega}}{\partial {k}} = 1/\frac{\partial {k}}{\partial {\omega}}\end{aligned} \hspace{\stretch{1}}(4.52)

(This can be shown).

Since

\begin{aligned}\left(\frac{\partial {k}}{\partial {\omega}}\right)^{-1} = \left(\frac{\partial {}}{\partial {\omega}}\sqrt{ (\omega/c)^2 - (\omega_{m n}/c)^2 }\right)^{-1} = c \sqrt{ 1 - (\omega_{m n}/\omega)^2 } \le c\end{aligned} \hspace{\stretch{1}}(4.53)

We see that the energy is transmitted at less than the speed of light as expected.

# Final remarks.

I’d started converting my handwritten scrawl for this tutorial into an attempt at working through these ideas with enough detail that they self contained, but gave up part way. This appears to me to be too big of a sub-discipline to give it justice in one hours class. As is, it is enough to at least get an concept of some of the ideas involved. I think were I to learn this for real, I’d need a good text as a reference (or the time to attempt to blunder through the ideas in much much more detail).

## Oct 12, PHY356F lecture notes.

Posted by peeterjoot on October 12, 2010

Today as an experiment I tried taking live notes in latex in class (previously I’d not taken any notes since the lectures followed the text so closely).

# Oct 12.

Review. What have we learned?

## Chapter 1.

Information about systems comes from vectors and operators. Express the vector ${\lvert {\phi} \rangle}$ describing the system in terms of eigenvectors ${\lvert {a_n} \rangle}$. $n \in 1,2,3,\cdots$.

of some operator $A$. What are the coefficients $c_n$? Act on both sides by ${\langle {a_m} \rvert}$ to find

\begin{aligned}\left\langle{{a_m}} \vert {{\phi}}\right\rangle &= \sum_n c_n \underbrace{\left\langle{{a_m}} \vert {{a_n}}\right\rangle}_{\text{Kronicker delta}} \\ &= \sum c_n \delta_{mn} \\ &= c_m\end{aligned}

\begin{aligned}c_m = \left\langle{{a_m}} \vert {{\phi}}\right\rangle\end{aligned}

Analogy

\begin{aligned}\mathbf{v} = \sum_i v_i \mathbf{e}_i \end{aligned}

\begin{aligned}\mathbf{e}_1 \cdot \mathbf{v} = \sum_i v_i \mathbf{e}_1 \cdot \mathbf{e}_i = v_1\end{aligned}

Physical information comes from the probability for obtaining a measurement of the physical entity associated with operator $A$. The probability of obtaining outcome $a_m$, an eigenvalue of $A$, is ${\left\lvert{c_n}\right\rvert}^2$

## Chapter 2.

Deal with operators that have continuous eigenvalues and eigenvectors.

We now express

\begin{aligned}{\lvert {\phi} \rangle} = \int dk f(k) {\lvert {k} \rangle}\end{aligned}

Here the coeffecients $f(k)$ are analogous to $c_n$.

Now if we project onto $k'$

\begin{aligned}\left\langle{{k'}} \vert {{\phi}}\right\rangle &= \int dk f(k) \underbrace{\left\langle{{k'}} \vert {{k}}\right\rangle}_{\text{Dirac delta}} \\ &= \int dk f(k) \delta(k' -k) \\ &= f(k') \end{aligned}

Unlike the discrete case, this is not a probability. Probability density for obtaining outcome $k'$ is ${\left\lvert{f(k')}\right\rvert}^2$.

Example 2.

\begin{aligned}{\lvert {\phi} \rangle} = \int dk f(k) {\lvert {k} \rangle}\end{aligned}

Now if we project x onto both sides

\begin{aligned}\left\langle{{x}} \vert {{\phi}}\right\rangle &= \int dk f(k) \left\langle{{x}} \vert {{k}}\right\rangle \\ \end{aligned}

With $\left\langle{{x}} \vert {{k}}\right\rangle = u_k(x)$

\begin{aligned}\phi(x) &\equiv \left\langle{{x}} \vert {{\phi}}\right\rangle \\ &= \int dk f(k) u_k(x) \\ &= \int dk f(k) \frac{1}{{\sqrt{L}}} e^{ikx}\end{aligned}

This is with periodic boundary value conditions for the normalization. The infinite normalization is also possible.

\begin{aligned}\phi(x) &= \frac{1}{{\sqrt{L}}} \int dk f(k) e^{ikx}\end{aligned}

Multiply both sides by $e^{-ik'x}/\sqrt{L}$ and integrate. This is analogous to multiplying ${\lvert {\phi} \rangle} = \int f(k) {\lvert {k} \rangle} dk$ by ${\langle {k'} \rvert}$. We get

\begin{aligned}\int \phi(x) \frac{1}{{\sqrt{L}}} e^{-ik'x} dx&= \frac{1}{{L}} \iint dk f(k) e^{i(k-k')x} dx \\ &= \int dk f(k) \Bigl( \frac{1}{{L}} \int e^{i(k-k')x} \Bigr) \\ &= \int dk f(k) \delta(k-k') \\ &= f(k')\end{aligned}

\begin{aligned}f(k') &=\int \phi(x) \frac{1}{{\sqrt{L}}} e^{-ik'x} dx\end{aligned}

We can talk about the state vector in terms of its position basis $\phi(x)$ or in the momentum space via Fourier transformation. This is the equivalent thing, but just expressed different. The question of interpretation in terms of probabilities works out the same. Either way we look at the probability density.

The quantity

\begin{aligned}{\lvert {\phi} \rangle} = \int dk f(k) {\lvert {k} \rangle}\end{aligned}

is also called a wave packet state since it involves a superposition of many stats ${\lvert {k} \rangle}$. Example: See Fig 4.1 (Gaussian wave packet, with ${\left\lvert{\phi}\right\rvert}^2$ as the height). This wave packet is a snapshot of the wave function amplitude at one specific time instant. The evolution of this wave packet is governed by the Hamiltonian, which brings us to chapter 3.

## Chapter 3.

For

\begin{aligned}{\lvert {\phi} \rangle} = \int dk f(k) {\lvert {k} \rangle}\end{aligned}

How do we find ${\lvert {\phi(t)} \rangle}$, the time evolved state? Here we have the option of choosing which of the pictures (Schr\”{o}dinger, Heisenberg, interaction) we deal with. Since the Heisenberg picture deals with time evolved operators, and the interaction picture with evolving Hamiltonian’s, neither of these is required to answer this question. Consider the Schr\”{o}dinger picture which gives

\begin{aligned}{\lvert {\phi(t)} \rangle} = \int dk f(k) {\lvert {k} \rangle} e^{-i E_k t/\hbar}\end{aligned}

where $E_k$ is the eigenvalue of the Hamiltonian operator $H$.

STRONG SEEMING HINT: If looking for additional problems and homework, consider in detail the time evolution of the Gaussian wave packet state.

## Chapter 4.

For three dimensions with $V(x,y,z) = 0$

\begin{aligned}H &= \frac{1}{{2m}} \mathbf{p}^2 \\ \mathbf{p} &= \sum_i p_i \mathbf{e}_i \\ \end{aligned}

In the position representation, where

\begin{aligned}p_i &= -i \hbar \frac{d}{dx_i}\end{aligned}

the Sch equation is

\begin{aligned}H u(x,y,z) &= E u(x,y,z) \\ H &= -\frac{\hbar^2}{2m} \boldsymbol{\nabla}^2 \\ = -\frac{\hbar^2}{2m} \left( \frac{\partial^2}{\partial {x}^2}+\frac{\partial^2}{\partial {y}^2}+\frac{\partial^2}{\partial {z}^2}\right) \end{aligned}

Separation of variables assumes it is possible to let

\begin{aligned}u(x,y,z) = X(x) Y(y) Z(z)\end{aligned}

(these capital letters are functions, not operators).

\begin{aligned}-\frac{\hbar^2}{2m} \left( YZ \frac{\partial^2 X}{\partial {x}^2}+ XZ \frac{\partial^2 Y}{\partial {y}^2}+ YZ \frac{\partial^2 Z}{\partial {z}^2}\right)&= E X Y Z\end{aligned}

Dividing as usual by $XYZ$ we have

\begin{aligned}-\frac{\hbar^2}{2m} \left( \frac{1}{{X}} \frac{\partial^2 X}{\partial {x}^2}+ \frac{1}{{Y}} \frac{\partial^2 Y}{\partial {y}^2}+ \frac{1}{{Z}} \frac{\partial^2 Z}{\partial {z}^2} \right)&= E \end{aligned}

The curious thing is that we have these three derivatives, which is supposed to be related to an Energy, which is independent of any $x,y,z$, so it must be that each of these is separately constant. We can separate these into three individual equations

\begin{aligned}-\frac{\hbar^2}{2m} \frac{1}{{X}} \frac{\partial^2 X}{\partial {x}^2} &= E_1 \\ -\frac{\hbar^2}{2m} \frac{1}{{Y}} \frac{\partial^2 Y}{\partial {x}^2} &= E_2 \\ -\frac{\hbar^2}{2m} \frac{1}{{Z}} \frac{\partial^2 Z}{\partial {x}^2} &= E_3\end{aligned}

or

\begin{aligned}\frac{\partial^2 X}{\partial {x}^2} &= \left( - \frac{2m E_1}{\hbar^2} \right) X \\ \frac{\partial^2 Y}{\partial {x}^2} &= \left( - \frac{2m E_2}{\hbar^2} \right) Y \\ \frac{\partial^2 Z}{\partial {x}^2} &= \left( - \frac{2m E_3}{\hbar^2} \right) Z \end{aligned}

We have then

\begin{aligned}X(x) = C_1 e^{i k x}\end{aligned}

with

\begin{aligned}E_1 &= \frac{\hbar^2 k_1^2 }{2m} = \frac{p_1^2}{2m} \\ E_2 &= \frac{\hbar^2 k_2^2 }{2m} = \frac{p_2^2}{2m} \\ E_3 &= \frac{\hbar^2 k_3^2 }{2m} = \frac{p_3^2}{2m} \end{aligned}

We are free to use any sort of normalization procedure we wish (periodic boundary conditions, infinite Dirac, …)

## Angular momentum.

HOMEWORK: go through the steps to understand how to formulate $\boldsymbol{\nabla}^2$ in spherical polar coordinates. This is a lot of work, but is good practice and background for dealing with the Hydrogen atom, something with spherical symmetry that is most naturally analyzed in the spherical polar coordinates.

In spherical coordinates (We won’t go through this here, but it is good practice) with

\begin{aligned}x &= r \sin\theta \cos\phi \\ y &= r \sin\theta \sin\phi \\ z &= r \cos\theta\end{aligned}

we have with $u = u(r,\theta, \phi)$

\begin{aligned}-\frac{\hbar^2}{2m} \left( \frac{1}{{r}} \partial_{rr} (r u) + \frac{1}{{r^2 \sin\theta}} \partial_\theta (\sin\theta \partial_\theta u) + \frac{1}{{r^2 \sin^2\theta}} \partial_{\phi\phi} u \right)&= E u\end{aligned}

We see the start of a separation of variables attack with $u = R(r) Y(\theta, \phi)$. We end up with

\begin{aligned}-\frac{\hbar^2}{2m} &\left( \frac{r}{R} (r R')' + \frac{1}{{Y \sin\theta}} \partial_\theta (\sin\theta \partial_\theta Y) + \frac{1}{{Y \sin^2\theta}} \partial_{\phi\phi} Y \right) \\ \end{aligned}

\begin{aligned}r (r R')' + \left( \frac{2m E}{\hbar^2} r^2 - \lambda \right) R &= 0\end{aligned}

\begin{aligned}\frac{1}{{Y \sin\theta}} \partial_\theta (\sin\theta \partial_\theta Y) + \frac{1}{{Y \sin^2\theta}} \partial_{\phi\phi} Y &= -\lambda\end{aligned}

Application of separation of variables again, with $Y = P(\theta) Q(\phi)$ gives us

\begin{aligned}\frac{1}{{P \sin\theta}} \partial_\theta (\sin\theta \partial_\theta P) + \frac{1}{{Q \sin^2\theta}} \partial_{\phi\phi} Q &= -\lambda \end{aligned}

\begin{aligned}\frac{\sin\theta}{P } \partial_\theta (\sin\theta \partial_\theta P) +\lambda \sin^2\theta+ \frac{1}{{Q }} \partial_{\phi\phi} Q &= 0\end{aligned}

\begin{aligned}\frac{\sin\theta}{P } \partial_\theta (\sin\theta \partial_\theta P) + \lambda \sin^2\theta - \mu = 0\frac{1}{{Q }} \partial_{\phi\phi} Q &= -\mu\end{aligned}

or

\begin{aligned}\frac{1}{P \sin\theta} \partial_\theta (\sin\theta \partial_\theta P) +\lambda -\frac{\mu}{\sin^2\theta} &= 0\end{aligned} \hspace{\stretch{1}}(1.1)

\begin{aligned}\partial_{\phi\phi} Q &= -\mu Q\end{aligned} \hspace{\stretch{1}}(1.2)

The equation for $P$ can be solved using the Legendre function $P_l^m(\cos\theta)$ where $\lambda = l(l+1)$ and $l$ is an integer

Replacing $\mu$ with $m^2$, where $m$ is an integer

\begin{aligned}\frac{d^2 Q}{d\phi^2} &= -m^2 Q\end{aligned}

Imposing a periodic boundary condition $Q(\phi) = Q(\phi + 2\pi)$, where ($m = 0, \pm 1, \pm 2, \cdots$) we have

\begin{aligned}Q &= \frac{1}{{\sqrt{2\pi}}} e^{im\phi}\end{aligned}

There is the overall solution $r(r,\theta,\phi) = R(r) Y(\theta, \phi)$ for a free particle. The functions $Y(\theta, \phi)$ are

\begin{aligned}Y_{lm}(\theta, \phi) &= N \left( \frac{1}{{\sqrt{2\pi}}} e^{im\phi} \right) \underbrace{ P_l^m(\cos\theta) }_{ -l \le m \le l }\end{aligned}

where $N$ is a normalization constant, and $m = 0, \pm 1, \pm 2, \cdots$. $Y_{lm}$ is an eigenstate of the $\mathbf{L}^2$ operator and $L_z$ (two for the price of one). There’s no specific reason for the direction $z$, but it is the direction picked out of convention.

Angular momentum is given by

\begin{aligned}\mathbf{L} = \mathbf{r} \times \mathbf{p}\end{aligned}

where

\begin{aligned}\mathbf{R} = x \hat{\mathbf{x}} + y\hat{\mathbf{y}} + z\hat{\mathbf{z}}\end{aligned}

and

\begin{aligned}\mathbf{p} = p_x \hat{\mathbf{x}} + p_y\hat{\mathbf{y}} + p_z\hat{\mathbf{z}}\end{aligned}

The important thing to remember is that the aim of following all the math is to show that

\begin{aligned}\mathbf{L}^2 Y_{lm} = \hbar^2 l (l+1) Y_{lm}\end{aligned}

and simultaneously

\begin{aligned}\mathbf{L}_z Y_{lm} = \hbar m Y_{lm}\end{aligned}

Part of the solution involves working with $\left[{L_z},{L_{+}}\right]$, and $\left[{L_z},{L_{-}}\right]$, where

\begin{aligned}L_{+} &= L_x + i L_y \\ L_{-} &= L_x - i L_y\end{aligned}

An exercise (not in the book) is to evaluate

\begin{aligned}\left[{L_z},{L_{+}}\right] &= L_z L_x + i L_z L_y - L_x L_z - i L_y L_z \end{aligned} \hspace{\stretch{1}}(1.3)

where

\begin{aligned}\left[{L_x},{L_y}\right] &= i \hbar L_z \\ \left[{L_y},{L_z}\right] &= i \hbar L_x \\ \left[{L_z},{L_x}\right] &= i \hbar L_y\end{aligned} \hspace{\stretch{1}}(1.4)

Substitution back in 1.3 we have

\begin{aligned}\left[{L_z},{L_{+}}\right] &=\left[{L_z},{L_x}\right] + i \left[{L_z},{L_y}\right] \\ &=i \hbar ( L_y - i L_x ) \\ &=\hbar ( i L_y + L_x ) \\ &=\hbar L_{+}\end{aligned}

## Notes and problems for Desai chapter IV.

Posted by peeterjoot on October 12, 2010

# Notes.

Chapter IV notes and problems for [1].

There’s a lot of magic related to the spherical Harmonics in this chapter, with identities pulled out of the Author’s butt. It would be nice to work through that, but need a better reference to work from (or skip ahead to chapter 26 where some of this is apparently derived).

Other stuff pending background derivation and verification are

\begin{itemize}
\item Antisymmetric tensor summation identity.

\begin{aligned}\sum_i \epsilon_{ijk} \epsilon_{iab} = \delta_{ja} \delta_{kb} - \delta_{jb}\delta_{ka}\end{aligned} \hspace{\stretch{1}}(1.1)

This is obviously the coordinate equivalent of the dot product of two bivectors

\begin{aligned}(\mathbf{e}_j \wedge \mathbf{e}_k) \cdot (\mathbf{e}_a \wedge \mathbf{e}_b) &=( (\mathbf{e}_j \wedge \mathbf{e}_k) \cdot \mathbf{e}_a ) \cdot \mathbf{e}_b) =\delta_{ka}\delta_{jb} - \delta_{ja}\delta_{kb}\end{aligned} \hspace{\stretch{1}}(1.2)

We can prove 1.1 by expanding the LHS of 1.2 in coordinates

\begin{aligned}(\mathbf{e}_j \wedge \mathbf{e}_k) \cdot (\mathbf{e}_a \wedge \mathbf{e}_b)&= \sum_{ie} \left\langle{{\epsilon_{ijk} \mathbf{e}_j \mathbf{e}_k \epsilon_{eab} \mathbf{e}_a \mathbf{e}_b}}\right\rangle \\ &=\sum_{ie}\epsilon_{ijk} \epsilon_{eab}\left\langle{{(\mathbf{e}_i \mathbf{e}_i) \mathbf{e}_j \mathbf{e}_k (\mathbf{e}_e \mathbf{e}_e) \mathbf{e}_a \mathbf{e}_b}}\right\rangle \\ &=\sum_{ie}\epsilon_{ijk} \epsilon_{eab}\left\langle{{\mathbf{e}_i \mathbf{e}_e I^2}}\right\rangle \\ &=-\sum_{ie} \epsilon_{ijk} \epsilon_{eab} \delta_{ie} \\ &=-\sum_i\epsilon_{ijk} \epsilon_{iab}\qquad\square\end{aligned}

\item Question on raising and lowering arguments.

How equation (4.240) was arrived at is not clear. In (4.239) he writes

\begin{aligned}\int_0^{2\pi} \int_0^{\pi} d\theta d\phi(L_{-} Y_{lm})^\dagger L_{-} Y_{lm} \sin\theta\end{aligned}

Shouldn’t that Hermitian conjugation be just complex conjugation? if so one would have

\begin{aligned}\int_0^{2\pi} \int_0^{\pi} d\theta d\phi L_{-}^{*} Y_{lm}^{*}L_{-} Y_{lm} \sin\theta\end{aligned}

How does he end up with the $L_{-}$ and the $Y_{lm}^{*}$ interchanged. What justifies this commutation?

A much clearer discussion of this can be found in The operators $L_{\pm}$, where Dirac notation is used for the normalization discussion.

\item Another question on raising and lowering arguments.

The reasoning leading to (4.238) isn’t clear to me. I fail to see how the $L_{-}$ commutation with $\mathbf{L}^2$ implies this?

\end{itemize}

# Problems

## Problem 1.

### Statement.

Write down the free particle Schr\”{o}dinger equation for two dimensions in (i) Cartesian and (ii) polar coordinates. Obtain the corresponding wavefunction.

### Cartesian case.

For the Cartesian coordinates case we have

\begin{aligned}H = -\frac{\hbar^2}{2m} (\partial_{xx} + \partial_{yy}) = i \hbar \partial_t\end{aligned} \hspace{\stretch{1}}(2.3)

Application of separation of variables with $\Psi = XYT$ gives

\begin{aligned}-\frac{\hbar^2}{2m} \left( \frac{X''}{X} +\frac{Y''}{Y} \right) = i \hbar \frac{T'}{T} = E .\end{aligned} \hspace{\stretch{1}}(2.4)

Immediately, we have the time dependence

\begin{aligned}T \propto e^{-i E t/\hbar},\end{aligned} \hspace{\stretch{1}}(2.5)

with the PDE reduced to

\begin{aligned}\frac{X''}{X} +\frac{Y''}{Y} = - \frac{2m E}{\hbar^2}.\end{aligned} \hspace{\stretch{1}}(2.6)

Introducing separate independent constants

\begin{aligned}\frac{X''}{X} &= a^2 \\ \frac{Y''}{Y} &= b^2 \end{aligned} \hspace{\stretch{1}}(2.7)

provides the pre-normalized wave function and the constraints on the constants

\begin{aligned}\Psi &= C e^{ax}e^{by}e^{-iE t/\hbar} \\ a^2 + b^2 &= -\frac{2 m E}{\hbar^2}.\end{aligned} \hspace{\stretch{1}}(2.9)

### Rectangular normalization.

We are now ready to apply normalization constraints. One possibility is a rectangular periodicity requirement.

\begin{aligned}e^{ax} &= e^{a(x + \lambda_x)} \\ e^{ay} &= e^{a(y + \lambda_y)} ,\end{aligned} \hspace{\stretch{1}}(2.11)

or

\begin{aligned}a\lambda_x &= 2 \pi i m \\ a\lambda_y &= 2 \pi i n.\end{aligned} \hspace{\stretch{1}}(2.13)

This provides a more explicit form for the energy expression

\begin{aligned}E_{mn} &= \frac{1}{{2m}} 4 \pi^2 \hbar^2 \left( \frac{m^2}{{\lambda_x}^2}+\frac{n^2}{{\lambda_y}^2}\right).\end{aligned} \hspace{\stretch{1}}(2.15)

We can also add in the area normalization using

\begin{aligned}\left\langle{{\psi}} \vert {{\phi}}\right\rangle &= \int_{x=0}^{\lambda_x} dx\int_{y=0}^{\lambda_x} dy \psi^{*}(x,y) \phi(x,y).\end{aligned} \hspace{\stretch{1}}(2.16)

Our eigenfunctions are now completely specified

\begin{aligned}u_{mn}(x,y,t) &= \frac{1}{{\sqrt{\lambda_x \lambda_y}}}e^{2 \pi i x/\lambda_x}e^{2 \pi i y/\lambda_y}e^{-iE t/\hbar}.\end{aligned} \hspace{\stretch{1}}(2.17)

\begin{aligned}f(x,y) = a_{mn} u_{mn}\end{aligned} \hspace{\stretch{1}}(2.18)

and then “solve” for $a_{mn}$, for an arbitrary $f(x,y)$ by taking inner products

\begin{aligned}a_{mn} = \left\langle{{u_mn}} \vert {{f}}\right\rangle =\int_{x=0}^{\lambda_x} dx \int_{y=0}^{\lambda_x} dy f(x,y) u_mn^{*}(x,y).\end{aligned} \hspace{\stretch{1}}(2.19)

This gives the appearance that any function $f(x,y)$ is a solution, but the equality of 2.18 only applies for functions in the span of this function vector space. The procedure works for arbitrary square integrable functions $f(x,y)$, but the equality really means that the RHS will be the periodic extension of $f(x,y)$.

### Infinite space normalization.

An alternate normalization is possible by using the Fourier transform normalization, in which we substitute

\begin{aligned}\frac{2 \pi m }{\lambda_x} &= k_x \\ \frac{2 \pi n }{\lambda_y} &= k_y \end{aligned} \hspace{\stretch{1}}(2.20)

Our inner product is now

\begin{aligned}\left\langle{{\psi}} \vert {{\phi}}\right\rangle &= \int_{-\infty}^{\infty} dx\int_{\infty}^{\infty} dy \psi^{*}(x,y) \phi(x,y).\end{aligned} \hspace{\stretch{1}}(2.22)

And the corresponding normalized wavefunction and associated energy constant $E$ are

\begin{aligned}u_{\mathbf{k}}(x,y,t) &= \frac{1}{{2\pi}}e^{i k_x x}e^{i k_y y}e^{-iE t/\hbar} = \frac{1}{{2\pi}}e^{i \mathbf{k} \cdot \mathbf{x}}e^{-iE t/\hbar} \\ E &= \frac{\hbar^2 \mathbf{k}^2 }{2m}\end{aligned} \hspace{\stretch{1}}(2.23)

Now via this Fourier inner product we are able to construct a solution from any square integrable function. Again, this will not be
an exact equality since the Fourier transform has the effect of averaging across discontinuities.

### Polar case.

In polar coordinates our gradient is

\begin{aligned}\boldsymbol{\nabla} &= \hat{\mathbf{r}} \partial_r + \frac{\hat{\boldsymbol{\theta}}}{r} \partial_\theta.\end{aligned} \hspace{\stretch{1}}(2.25)

with

\begin{aligned}\hat{\mathbf{r}} &= \mathbf{e}_1 e^{\mathbf{e}_1 \mathbf{e}_2 \theta} \\ \hat{\boldsymbol{\theta}} &= \mathbf{e}_2 e^{\mathbf{e}_1 \mathbf{e}_2 \theta} .\end{aligned} \hspace{\stretch{1}}(2.26)

Squaring the gradient for the Laplacian we’ll need the partials, which are

\begin{aligned}\partial_r \hat{\mathbf{r}} &= 0 \\ \partial_r \hat{\boldsymbol{\theta}} &= 0 \\ \partial_\theta \hat{\mathbf{r}} &= \hat{\boldsymbol{\theta}} \\ \partial_\theta \hat{\boldsymbol{\theta}} &= -\hat{\mathbf{r}}.\end{aligned}

The Laplacian is therefore

\begin{aligned}\boldsymbol{\nabla}^2 &= (\hat{\mathbf{r}} \partial_r + \frac{\hat{\boldsymbol{\theta}}}{r} \partial_\theta) \cdot(\hat{\mathbf{r}} \partial_r + \frac{\hat{\boldsymbol{\theta}}}{r} \partial_\theta) \\ &= \partial_{rr} + \frac{\hat{\boldsymbol{\theta}}}{r} \cdot \partial_\theta \hat{\mathbf{r}} \partial_r \frac{\hat{\boldsymbol{\theta}}}{r} \cdot \partial_\theta \frac{\hat{\boldsymbol{\theta}}}{r} \partial_\theta \\ &= \partial_{rr} + \frac{\hat{\boldsymbol{\theta}}}{r} \cdot (\partial_\theta \hat{\mathbf{r}}) \partial_r + \frac{\hat{\boldsymbol{\theta}}}{r} \cdot \frac{\hat{\boldsymbol{\theta}}}{r} \partial_{\theta\theta} + \frac{\hat{\boldsymbol{\theta}}}{r} \cdot (\partial_\theta \hat{\boldsymbol{\theta}}) \frac{1}{{r}} \partial_\theta .\end{aligned}

Evalating the derivatives we have

\begin{aligned}\boldsymbol{\nabla}^2 = \partial_{rr} + \frac{1}{{r}} \partial_r + \frac{1}{r^2} \partial_{\theta\theta},\end{aligned} \hspace{\stretch{1}}(2.28)

and are now prepared to move on to the solution of the Hamiltonian $H = -(\hbar^2/2m) \boldsymbol{\nabla}^2$. With separation of variables again using $\Psi = R(r) \Theta(\theta) T(t)$ we have

\begin{aligned}-\frac{\hbar^2}{2m} \left( \frac{R''}{R} + \frac{R'}{rR} + \frac{1}{{r^2}} \frac{\Theta''}{\Theta} \right) = i \hbar \frac{T'}{T} = E.\end{aligned} \hspace{\stretch{1}}(2.29)

Rearranging to separate the $\Theta$ term we have

\begin{aligned}\frac{r^2 R''}{R} + \frac{r R'}{R} + \frac{2 m E}{\hbar^2} r^2 E = -\frac{\Theta''}{\Theta} = \lambda^2.\end{aligned} \hspace{\stretch{1}}(2.30)

The angular solutions are given by

\begin{aligned}\Theta = \frac{1}{{\sqrt{2\pi}}} e^{i \lambda \theta}\end{aligned} \hspace{\stretch{1}}(2.31)

Where the normalization is given by

\begin{aligned}\left\langle{{\psi}} \vert {{\phi}}\right\rangle &= \int_{0}^{2 \pi} d\theta \psi^{*}(\theta) \phi(\theta).\end{aligned} \hspace{\stretch{1}}(2.32)

And the radial by the solution of the PDE

\begin{aligned}r^2 R'' + r R' + \left( \frac{2 m E}{\hbar^2} r^2 E - \lambda^2 \right) R = 0\end{aligned} \hspace{\stretch{1}}(2.33)

## Problem 2.

### Statement.

Use the orthogonality property of $P_l(\cos\theta)$

\begin{aligned}\int_{-1}^1 dx P_l(x) P_{l'}(x) = \frac{2}{2l+1} \delta_{l l'},\end{aligned} \hspace{\stretch{1}}(2.34)

confirm that at least the first two terms of (4.171)

\begin{aligned}e^{i k r \cos\theta} = \sum_{l=0}^\infty (2l + 1) i^l j_l(kr) P_l(\cos\theta)\end{aligned} \hspace{\stretch{1}}(2.35)

are correct.

### Solution.

Taking the inner product using the integral of 2.34 we have

\begin{aligned}\int_{-1}^1 dx e^{i k r x} P_l'(x) = 2 i^l j_l(kr) \end{aligned} \hspace{\stretch{1}}(2.36)

To confirm the first two terms we need

\begin{aligned}P_0(x) &= 1 \\ P_1(x) &= x \\ j_0(\rho) &= \frac{\sin\rho}{\rho} \\ j_1(\rho) &= \frac{\sin\rho}{\rho^2} - \frac{\cos\rho}{\rho}.\end{aligned} \hspace{\stretch{1}}(2.37)

On the LHS for $l'=0$ we have

\begin{aligned}\int_{-1}^1 dx e^{i k r x} = 2 \frac{\sin{kr}}{kr}\end{aligned} \hspace{\stretch{1}}(2.41)

On the LHS for $l'=1$ note that

\begin{aligned}\int dx x e^{i k r x} &= \int dx x \frac{d}{dx} \frac{e^{i k r x}}{ikr} \\ &= x \frac{e^{i k r x}}{ikr} - \frac{e^{i k r x}}{(ikr)^2}.\end{aligned}

So, integration in $[-1,1]$ gives us

\begin{aligned}\int_{-1}^1 dx e^{i k r x} = -2i \frac{\cos{kr}}{kr} + 2i \frac{1}{{(kr)^2}} \sin{kr}.\end{aligned} \hspace{\stretch{1}}(2.42)

Now compare to the RHS for $l'=0$, which is

\begin{aligned}2 j_0(kr) = 2 \frac{\sin{kr}}{kr},\end{aligned} \hspace{\stretch{1}}(2.43)

which matches 2.41. For $l'=1$ we have

\begin{aligned}2 i j_1(kr) = 2i \frac{1}{{kr}} \left( \frac{\sin{kr}}{kr} - \cos{kr} \right),\end{aligned} \hspace{\stretch{1}}(2.44)

which in turn matches 2.42, completing the exersize.

## Problem 3.

### Statement.

Obtain the commutation relations $\left[{L_i},{L_j}\right]$ by calculating the vector $\mathbf{L} \times \mathbf{L}$ using the definition $\mathbf{L} = \mathbf{r} \times \mathbf{p}$ directly instead of introducing a differential operator.

### Solution.

Expressing the product $\mathbf{L} \times \mathbf{L}$ in determinant form sheds some light on this question. That is

\begin{aligned}\begin{vmatrix} \mathbf{e}_1 & \mathbf{e}_2 & \mathbf{e}_3 \\ L_1 & L_2 & L_3 \\ L_1 & L_2 & L_3\end{vmatrix}&= \mathbf{e}_1 \left[{L_2},{L_3}\right] +\mathbf{e}_2 \left[{L_3},{L_1}\right] +\mathbf{e}_3 \left[{L_1},{L_2}\right]= \mathbf{e}_i \epsilon_{ijk} \left[{L_j},{L_k}\right]\end{aligned} \hspace{\stretch{1}}(2.45)

We see that evaluating this cross product in turn requires evaluation of the set of commutators. We can do that with the canonical commutator relationships directly using $L_i = \epsilon_{ijk} r_j p_k$ like so

\begin{aligned}\left[{L_i},{L_j}\right]&=\epsilon_{imn} r_m p_n \epsilon_{jab} r_a p_b- \epsilon_{jab} r_a p_b \epsilon_{imn} r_m p_n \\ &=\epsilon_{imn} \epsilon_{jab} r_m (p_n r_a) p_b- \epsilon_{jab} \epsilon_{imn} r_a (p_b r_m) p_n \\ &=\epsilon_{imn} \epsilon_{jab} r_m (r_a p_n -i \hbar \delta_{an}) p_b- \epsilon_{jab} \epsilon_{imn} r_a (r_m p_b - i \hbar \delta{mb}) p_n \\ &=\epsilon_{imn} \epsilon_{jab} (r_m r_a p_n p_b - r_a r_m p_b p_n )- i \hbar ( \epsilon_{imn} \epsilon_{jnb} r_m p_b - \epsilon_{jam} \epsilon_{imn} r_a p_n ).\end{aligned}

The first two terms cancel, and we can employ (4.179) to eliminate the antisymmetric tensors from the last two terms

\begin{aligned}\left[{L_i},{L_j}\right]&=i \hbar ( \epsilon_{nim} \epsilon_{njb} r_m p_b - \epsilon_{mja} \epsilon_{min} r_a p_n ) \\ &=i \hbar ( (\delta_{ij} \delta_{mb} -\delta_{ib} \delta_{mj}) r_m p_b - (\delta_{ji} \delta_{an} -\delta_{jn} \delta_{ai}) r_a p_n ) \\ &=i \hbar (\delta_{ij} \delta_{mb} r_m p_b - \delta_{ji} \delta_{an} r_a p_n - \delta_{ib} \delta_{mj} r_m p_b + \delta_{jn} \delta_{ai} r_a p_n ) \\ &=i \hbar (\delta_{ij} r_m p_m- \delta_{ji} r_a p_a- r_j p_i+ r_i p_j ) \\ \end{aligned}

For $k \ne i,j$, this is $i\hbar (\mathbf{r} \times \mathbf{p})_k$, so we can write

\begin{aligned}\mathbf{L} \times \mathbf{L} &= i\hbar \mathbf{e}_k \epsilon_{kij} ( r_i p_j - r_j p_i ) = i\hbar \mathbf{L} = i\hbar \mathbf{e}_k L_k = i\hbar \mathbf{L}.\end{aligned} \hspace{\stretch{1}}(2.46)

In [2], the commutator relationships are summarized this way, instead of using the antisymmetric tensor (4.224)

\begin{aligned}\left[{L_i},{L_j}\right] &= i \hbar \epsilon_{ijk} L_k\end{aligned} \hspace{\stretch{1}}(2.47)

as here in Desai. Both say the same thing.

TODO.

## Problem 5.

### Statement.

A free particle is moving along a path of radius $R$. Express the Hamiltonian in terms of the derivatives involving the polar angle of the particle and write down the Schr\”{o}dinger equation. Determine the wavefunction and the energy eigenvalues of the particle.

### Solution.

In classical mechanics our Lagrangian for this system is

\begin{aligned}\mathcal{L} = \frac{1}{{2}} m R^2 \dot{\theta}^2,\end{aligned} \hspace{\stretch{1}}(2.48)

with the canonical momentum

\begin{aligned}p_\theta = \frac{\partial {\mathcal{L}}}{\partial {\dot{\theta}}} = m R^2 \dot{\theta}.\end{aligned} \hspace{\stretch{1}}(2.49)

Thus the classical Hamiltonian is

\begin{aligned}H = \frac{1}{{2m R^2}} {p_\theta}^2.\end{aligned} \hspace{\stretch{1}}(2.50)

By analogy the QM Hamiltonian operator will therefore be

\begin{aligned}H = -\frac{\hbar^2}{2m R^2} \partial_{\theta\theta}.\end{aligned} \hspace{\stretch{1}}(2.51)

For $\Psi = \Theta(\theta) T(t)$, separation of variables gives us

\begin{aligned}-\frac{\hbar^2}{2m R^2} \frac{\Theta''}{\Theta} = i \hbar \frac{T'}{T} = E,\end{aligned} \hspace{\stretch{1}}(2.52)

from which we have

\begin{aligned}T &\propto e^{-i E t/\hbar} \\ \Theta &\propto e^{ \pm i \sqrt{2m E} R \theta/\hbar }.\end{aligned} \hspace{\stretch{1}}(2.53)

Requiring single valued $\Theta$, equal at any multiples of $2\pi$, we have

\begin{aligned}e^{ \pm i \sqrt{2m E} R (\theta + 2\pi)/\hbar } = e^{ \pm i \sqrt{2m E} R \theta/\hbar },\end{aligned}

or

\begin{aligned}\pm \sqrt{2m E} \frac{R}{\hbar} 2\pi = 2 \pi n,\end{aligned}

Suffixing the energy values with this index we have

\begin{aligned}E_n = \frac{n^2 \hbar^2}{2 m R^2}.\end{aligned} \hspace{\stretch{1}}(2.55)

Allowing both positive and negative integer values for $n$ we have

\begin{aligned}\Psi = \frac{1}{{\sqrt{2\pi}}} e^{i n \theta} e^{-i E_n t/\hbar},\end{aligned} \hspace{\stretch{1}}(2.56)

where the normalization was a result of the use of a $[0,2\pi]$ inner product over the angles

\begin{aligned}\left\langle{{\psi}} \vert {{\phi}}\right\rangle \equiv \int_0^{2\pi} \psi^{*}(\theta) \phi(\theta) d\theta.\end{aligned} \hspace{\stretch{1}}(2.57)

## Problem 6.

### Statement.

Determine $\left[{L_i},{r}\right]$ and $\left[{L_i},{\mathbf{r}}\right]$.

### Solution.

Since $L_i$ contain only $\theta$ and $\phi$ partials, $\left[{L_i},{r}\right] = 0$. For the position vector, however, we have an angular dependence, and are left to evaluate $\left[{L_i},{\mathbf{r}}\right] = r \left[{L_i},{\hat{\mathbf{r}}}\right]$. We’ll need the partials for $\hat{\mathbf{r}}$. We have

\begin{aligned}\hat{\mathbf{r}} &= \mathbf{e}_3 e^{I \hat{\boldsymbol{\phi}} \theta} \\ \hat{\boldsymbol{\phi}} &= \mathbf{e}_2 e^{\mathbf{e}_1 \mathbf{e}_2 \phi} \\ I &= \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3\end{aligned} \hspace{\stretch{1}}(2.58)

Evaluating the partials we have

\begin{aligned}\partial_\theta \hat{\mathbf{r}} = \hat{\mathbf{r}} I \hat{\boldsymbol{\phi}}\end{aligned}

With

\begin{aligned}\hat{\boldsymbol{\theta}} &= \tilde{R} \mathbf{e}_1 R \\ \hat{\boldsymbol{\phi}} &= \tilde{R} \mathbf{e}_2 R \\ \hat{\mathbf{r}} &= \tilde{R} \mathbf{e}_3 R\end{aligned} \hspace{\stretch{1}}(2.61)

where $\tilde{R} R = 1$, and $\hat{\boldsymbol{\theta}} \hat{\boldsymbol{\phi}} \hat{\mathbf{r}} = \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3$, we have

\begin{aligned}\partial_\theta \hat{\mathbf{r}} &= \tilde{R} \mathbf{e}_3 \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3 \mathbf{e}_2 R = \tilde{R} \mathbf{e}_1 R = \hat{\boldsymbol{\theta}}\end{aligned} \hspace{\stretch{1}}(2.64)

For the $\phi$ partial we have

\begin{aligned}\partial_\phi \hat{\mathbf{r}}&= \mathbf{e}_3 \sin\theta I \hat{\boldsymbol{\phi}} \mathbf{e}_1 \mathbf{e}_2 \\ &= \sin\theta \hat{\boldsymbol{\phi}}\end{aligned}

We are now prepared to evaluate the commutators. Starting with the easiest we have

\begin{aligned}\left[{L_z},{\hat{\mathbf{r}}}\right] \Psi&=-i \hbar (\partial_\phi \hat{\mathbf{r}} \Psi - \hat{\mathbf{r}} \partial_\phi \Psi ) \\ &=-i \hbar (\partial_\phi \hat{\mathbf{r}}) \Psi \\ \end{aligned}

So we have

\begin{aligned}\left[{L_z},{\hat{\mathbf{r}}}\right]&=-i \hbar \sin\theta \hat{\boldsymbol{\phi}}\end{aligned} \hspace{\stretch{1}}(2.65)

Observe that by virtue of chain rule, only the action of the partials on $\hat{\mathbf{r}}$ itself contributes, and all the partials applied to $\Psi$ cancel out due to the commutator differences. That simplifies the remaining commutator evaluations. For reference the polar form of $L_x$, and $L_y$ are

\begin{aligned}L_x &= -i \hbar (-S_\phi \partial_\theta - C_\phi \cot\theta \partial_\phi) \\ L_y &= -i \hbar (C_\phi \partial_\theta - S_\phi \cot\theta \partial_\phi),\end{aligned} \hspace{\stretch{1}}(2.66)

where the sines and cosines are written with $S$, and $C$ respectively for short.

We therefore have

\begin{aligned}\left[{L_x},{\hat{\mathbf{r}}}\right]&= -i \hbar (-S_\phi (\partial_\theta \hat{\mathbf{r}}) - C_\phi \cot\theta (\partial_\phi \hat{\mathbf{r}}) ) \\ &= -i \hbar (-S_\phi \hat{\boldsymbol{\theta}} - C_\phi \cot\theta S_\theta \hat{\boldsymbol{\phi}} ) \\ &= -i \hbar (-S_\phi \hat{\boldsymbol{\theta}} - C_\phi C_\theta \hat{\boldsymbol{\phi}} ) \\ \end{aligned}

and

\begin{aligned}\left[{L_y},{\hat{\mathbf{r}}}\right]&= -i \hbar (C_\phi (\partial_\theta \hat{\mathbf{r}}) - S_\phi \cot\theta (\partial_\phi \hat{\mathbf{r}})) \\ &= -i \hbar (C_\phi \hat{\boldsymbol{\theta}} - S_\phi C_\theta \hat{\boldsymbol{\phi}} ).\end{aligned}

Adding back in the factor of $r$, and summarizing we have

\begin{aligned}\left[{L_i},{r}\right] &= 0 \\ \left[{L_x},{\mathbf{r}}\right] &= -i \hbar r (-\sin\phi \hat{\boldsymbol{\theta}} - \cos\phi \cos\theta \hat{\boldsymbol{\phi}} ) \\ \left[{L_y},{\mathbf{r}}\right] &= -i \hbar r (\cos\phi \hat{\boldsymbol{\theta}} - \sin\phi \cos\theta \hat{\boldsymbol{\phi}} ) \\ \left[{L_z},{\mathbf{r}}\right] &= -i \hbar r \sin\theta \hat{\boldsymbol{\phi}}\end{aligned} \hspace{\stretch{1}}(2.68)

## Problem 7.

### Statement.

Show that

\begin{aligned}e^{-i\pi L_x /\hbar } {\lvert {l,m} \rangle} = {\lvert {l,m-1} \rangle}\end{aligned} \hspace{\stretch{1}}(2.72)

TODO.

# References

[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.

[2] R. Liboff. Introductory quantum mechanics. 2003.