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PHY456H1F: Quantum Mechanics II. Lecture 15 (Taught by Prof J.E. Sipe). Rotation operator in spin space

Posted by peeterjoot on October 31, 2011

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

Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

Rotation operator in spin space.

We can formally expand our rotation operator in Taylor series

\begin{aligned}e^{-i \theta \hat{\mathbf{n}} \cdot \mathbf{S}/\hbar}= I +\left(-i \theta \hat{\mathbf{n}} \cdot \mathbf{S}/\hbar\right)+\frac{1}{{2!}}\left(-i \theta \hat{\mathbf{n}} \cdot \mathbf{S}/\hbar\right)^2+\frac{1}{{3!}}\left(-i \theta \hat{\mathbf{n}} \cdot \mathbf{S}/\hbar\right)^3+ \cdots\end{aligned} \hspace{\stretch{1}}(2.1)

or

\begin{aligned}e^{-i \theta \hat{\mathbf{n}} \cdot \boldsymbol{\sigma}/2}&= I +\left(-i \theta \hat{\mathbf{n}} \cdot \boldsymbol{\sigma}/2\right)+\frac{1}{{2!}}\left(-i \theta \hat{\mathbf{n}} \cdot \boldsymbol{\sigma}/2\right)^2+\frac{1}{{3!}}\left(-i \theta \hat{\mathbf{n}} \cdot \boldsymbol{\sigma}/2\right)^3+ \cdots \\ &=\sigma_0 +\left(\frac{-i \theta}{2}\right) (\hat{\mathbf{n}} \cdot \boldsymbol{\sigma})+\frac{1}{{2!}} \left(\frac{-i \theta}{2}\right) (\hat{\mathbf{n}} \cdot \boldsymbol{\sigma})^2+\frac{1}{{3!}} \left(\frac{-i \theta}{2}\right) (\hat{\mathbf{n}} \cdot \boldsymbol{\sigma})^3+ \cdots \\ &=\sigma_0 +\left(\frac{-i \theta}{2}\right) (\hat{\mathbf{n}} \cdot \boldsymbol{\sigma})+\frac{1}{{2!}} \left(\frac{-i \theta}{2}\right) \sigma_0+\frac{1}{{3!}} \left(\frac{-i \theta}{2}\right) (\hat{\mathbf{n}} \cdot \boldsymbol{\sigma}) + \cdots \\ &=\sigma_0 \left( 1 - \frac{1}{{2!}}\left(\frac{\theta}{2}\right)^2 + \cdots \right) +(\hat{\mathbf{n}} \cdot \boldsymbol{\sigma}) \left( \frac{\theta}{2} - \frac{1}{{3!}}\left(\frac{\theta}{2}\right)^3 + \cdots \right) \\ &=\cos(\theta/2) \sigma_0 + \sin(\theta/2) (\hat{\mathbf{n}} \cdot \boldsymbol{\sigma})\end{aligned}

where we’ve used the fact that (\hat{\mathbf{n}} \cdot \boldsymbol{\sigma})^2 = \sigma_0.

So our representation of the spin operator is

\begin{aligned}\begin{aligned}e^{-i \theta \hat{\mathbf{n}} \cdot \mathbf{S}/\hbar} &\rightarrow \cos(\theta/2) \sigma_0 + \sin(\theta/2) (\hat{\mathbf{n}} \cdot \boldsymbol{\sigma}) \\ &=\cos(\theta/2) \sigma_0 + \sin(\theta/2) \left(n_x \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} + n_y \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix} + n_z \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} \right) \\ &=\begin{bmatrix}\cos(\theta/2) -i n_z \sin(\theta/2) & -i (n_x -i n_y) \sin(\theta/2) \\ -i (n_x + i n_y) \sin(\theta/2) & \cos(\theta/2) +i n_z \sin(\theta/2) \end{bmatrix}\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.2)

Note that, in particular,

\begin{aligned}e^{-2 \pi i \hat{\mathbf{n}} \cdot \mathbf{S}/\hbar} \rightarrow \cos\pi \sigma_0 = -\sigma_0\end{aligned} \hspace{\stretch{1}}(2.3)

This “rotates” the ket, but introduces a phase factor.

Can do this in general for other degrees of spin, for s = 1/2, 3/2, 5/2, \cdots.

Unfortunate interjection by me

I mentioned the half angle rotation operator that requires a half angle operator sandwich. Prof. Sipe thought I might be talking about a Heisenberg picture representation, where we have something like this in expectation values

\begin{aligned}{\left\lvert {\psi'} \right\rangle} = e^{-i \theta \hat{\mathbf{n}} \cdot \mathbf{J}/\hbar} {\left\lvert {\psi} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.4)

so that

\begin{aligned}{\left\langle {\psi'} \right\rvert}\mathcal{O}{\left\lvert {\psi'} \right\rangle} = {\left\langle {\psi} \right\rvert} e^{i \theta \hat{\mathbf{n}} \cdot \mathbf{J}/\hbar} \mathcal{O}e^{-i \theta \hat{\mathbf{n}} \cdot \mathbf{J}/\hbar} {\left\lvert {\psi} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.5)

However, what I was referring to, was that a general rotation of a vector in a Pauli matrix basis

\begin{aligned}R(\sum a_k \sigma_k) = R( \mathbf{a} \cdot \boldsymbol{\sigma})\end{aligned} \hspace{\stretch{1}}(2.6)

can be expressed by sandwiching the Pauli vector representation by two half angle rotation operators like our spin 1/2 operators from class today

\begin{aligned}R( \mathbf{a} \cdot \boldsymbol{\sigma}) = e^{-\theta \hat{\mathbf{u}} \cdot \boldsymbol{\sigma} \hat{\mathbf{v}} \cdot \boldsymbol{\sigma}/2} \mathbf{a} \cdot \boldsymbol{\sigma} e^{\theta \hat{\mathbf{u}} \cdot \boldsymbol{\sigma} \hat{\mathbf{v}} \cdot \boldsymbol{\sigma}/2}\end{aligned} \hspace{\stretch{1}}(2.7)

where \hat{\mathbf{u}} and \hat{\mathbf{v}} are two non-colinear orthogonal unit vectors that define the oriented plane that we are rotating in.

For example, rotating in the x-y plane, with \hat{\mathbf{u}} = \hat{\mathbf{x}} and \hat{\mathbf{v}} = \hat{\mathbf{y}}, we have

\begin{aligned}R( \mathbf{a} \cdot \boldsymbol{\sigma}) = e^{-\theta \sigma_1 \sigma_2/2} (a_1 \sigma_1 + a_2 \sigma_2 + a_3 \sigma_3) e^{\theta \sigma_1 \sigma_2/2} \end{aligned} \hspace{\stretch{1}}(2.8)

Observe that these exponentials commute with \sigma_3, leaving

\begin{aligned}R( \mathbf{a} \cdot \boldsymbol{\sigma}) &= (a_1 \sigma_1 + a_2 \sigma_2) e^{\theta \sigma_1 \sigma_2} + a_3 \sigma_3 \\ &= (a_1 \sigma_1 + a_2 \sigma_2) (\cos\theta + \sigma_1 \sigma_2 \sin\theta)+a_3 \sigma_3 \\ &= \sigma_1 (a_1 \cos\theta - a_2 \sin\theta)+ \sigma_2 (a_2 \cos\theta + a_1 \sin\theta)+ \sigma_3 (a_3)\end{aligned}

yielding our usual coordinate rotation matrix. Expressed in terms of a unit normal to that plane, we form the normal by multiplication with the unit spatial volume element I = \sigma_1 \sigma_2 \sigma_3. For example:

\begin{aligned}\sigma_1 \sigma_2 \sigma_3( \sigma_3 )=\sigma_1 \sigma_2 \end{aligned} \hspace{\stretch{1}}(2.9)

and can in general write a spatial rotation in a Pauli basis representation as a sandwich of half angle rotation matrix exponentials

\begin{aligned}R( \mathbf{a} \cdot \boldsymbol{\sigma}) = e^{-I \theta (\hat{\mathbf{n}} \cdot \boldsymbol{\sigma})/2} (\mathbf{a} \cdot \boldsymbol{\sigma})e^{I \theta (\hat{\mathbf{n}} \cdot \boldsymbol{\sigma})/2} \end{aligned} \hspace{\stretch{1}}(2.10)

when \hat{\mathbf{n}} \cdot \mathbf{a} = 0 we get the complex-number like single sided exponential rotation exponentials (since \mathbf{a} \cdot \boldsymbol{\sigma} commutes with \mathbf{n} \cdot \boldsymbol{\sigma} in that case)

\begin{aligned}R( \mathbf{a} \cdot \boldsymbol{\sigma}) = (\mathbf{a} \cdot \boldsymbol{\sigma} )e^{I \theta (\hat{\mathbf{n}} \cdot \boldsymbol{\sigma})} \end{aligned} \hspace{\stretch{1}}(2.11)

I believe it was pointed out in one of [1] or [2] that rotations expressed in terms of half angle Pauli matrices has caused some confusion to students of quantum mechanics, because this 2 \pi “rotation” only generates half of the full spatial rotation. It was argued that this sort of confusion can be avoided if one observes that these half angle rotations exponentials are exactly what we require for general spatial rotations, and that a pair of half angle operators are required to produce a full spatial rotation.

The book [1] takes this a lot further, and produces a formulation of spin operators that is devoid of the normal scalar imaginary i (using the Clifford algebra spatial unit volume element instead), and also does not assume a specific matrix representation of the spin operators. They argue that this leads to some subtleties associated with interpretation, but at the time I was attempting to read that text I did know enough QM to appreciate what they were doing, and haven’t had time to attempt a new study of that content.

Spin dynamics

At least classically, the angular momentum of charged objects is associated with a magnetic moment as illustrated in figure (\ref{fig:qmTwoL15:qmTwoL15fig1})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL15fig1}
\caption{Magnetic moment due to steady state current}
\end{figure}

\begin{aligned}\boldsymbol{\mu} = I A \mathbf{e}_\perp\end{aligned} \hspace{\stretch{1}}(3.12)

In our scheme, following the (cgs?) text conventions of [3], where the \mathbf{E} and \mathbf{B} have the same units, we write

\begin{aligned}\boldsymbol{\mu} = \frac{I A}{c} \mathbf{e}_\perp\end{aligned} \hspace{\stretch{1}}(3.13)

For a charge moving in a circle as in figure (\ref{fig:qmTwoL15:qmTwoL15fig2})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL15fig2}
\caption{Charge moving in circle.}
\end{figure}

\begin{aligned}\begin{aligned}I &= \frac{\text{charge}}{\text{time}} \\ &= \frac{\text{distance}}{\text{time}} \frac{\text{charge}}{\text{distance}} \\ &= \frac{q v}{ 2 \pi r}\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.14)

so the magnetic moment is

\begin{aligned}\begin{aligned}\mu &= \frac{q v}{ 2 \pi r} \frac{\pi r^2}{c} \\ &= \frac{q }{ 2 m c } (m v r) \\ &= \gamma L\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.15)

Here \gamma is the gyromagnetic ratio

Recall that we have a torque, as shown in figure (\ref{fig:qmTwoL15:qmTwoL15fig3})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL15fig3}
\caption{Induced torque in the presence of a magnetic field.}
\end{figure}

\begin{aligned}\mathbf{T} = \boldsymbol{\mu} \times \mathbf{B}\end{aligned} \hspace{\stretch{1}}(3.16)

tending to line up \boldsymbol{\mu} with \mathbf{B}. The energy is then

\begin{aligned}-\boldsymbol{\mu} \cdot \mathbf{B}\end{aligned} \hspace{\stretch{1}}(3.17)

Also recall that this torque leads to precession as shown in figure (\ref{fig:qmTwoL15:qmTwoL15fig4})

\begin{aligned}\frac{d{\mathbf{L}}}{dt} = \mathbf{T} = \gamma \mathbf{L} \times \mathbf{B},\end{aligned} \hspace{\stretch{1}}(3.18)

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL15fig4}
\caption{Precession due to torque.}
\end{figure}

with precession frequency

\begin{aligned}\boldsymbol{\omega} = - \gamma \mathbf{B}.\end{aligned} \hspace{\stretch{1}}(3.19)

For a current due to a moving electron

\begin{aligned}\gamma = -\frac{e}{2 m c} < 0\end{aligned} \hspace{\stretch{1}}(3.20)

where we are, here, writing for charge on the electron -e.

Question: steady state currents only?. Yes, this is only true for steady state currents.

For the translational motion of an electron, even if it is not moving in a steady way, regardless of it’s dynamics

\begin{aligned}\boldsymbol{\mu}_0 = - \frac{e}{2 m c} \mathbf{L}\end{aligned} \hspace{\stretch{1}}(3.21)

Now, back to quantum mechanics, we turn \boldsymbol{\mu}_0 into a dipole moment operator and \mathbf{L} is “promoted” to an angular momentum operator.

\begin{aligned}H_{\text{int}} = - \boldsymbol{\mu}_0 \cdot \mathbf{B}\end{aligned} \hspace{\stretch{1}}(3.22)

What about the “spin”?

Perhaps

\begin{aligned}\boldsymbol{\mu}_s = \gamma_s \mathbf{S}\end{aligned} \hspace{\stretch{1}}(3.23)

we write this as

\begin{aligned}\boldsymbol{\mu}_s = g \left( -\frac{e}{ 2 m c} \right)\mathbf{S}\end{aligned} \hspace{\stretch{1}}(3.24)

so that

\begin{aligned}\gamma_s = - \frac{g e}{ 2 m c} \end{aligned} \hspace{\stretch{1}}(3.25)

Experimentally, one finds to very good approximation

\begin{aligned}g = 2\end{aligned} \hspace{\stretch{1}}(3.26)

There was a lot of trouble with this in early quantum mechanics where people got things wrong, and canceled the wrong factors of 2.

In fact, Dirac’s relativistic theory for the electron predicts g=2.

When this is measured experimentally, one does not get exactly g=2, and a theory that also incorporates photon creation and destruction and the interaction with the electron with such (virtual) photons. We get

\begin{aligned}\begin{aligned}g_{\text{theory}} &= 2 \left(1.001159652140 (\pm 28)\right) \\ g_{\text{experimental}} &= 2 \left(1.0011596521884 (\pm 43)\right)\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.27)

Richard Feynman compared the precision of quantum mechanics, referring to this measurement, “to predicting a distance as great as the width of North America to an accuracy of one human hair’s breadth”.

References

[1] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.

[2] D. Hestenes. New Foundations for Classical Mechanics. Kluwer Academic Publishers, 1999.

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

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PHY456H1F: Quantum Mechanics II. Recitation 3 (Taught by Mr. Federico Duque Gomez). WKB method and Stark shift.

Posted by peeterjoot on October 28, 2011

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

Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

WKB method.

Consider the potential

\begin{aligned}V(x) = \left\{\begin{array}{l l}v(x) & \quad \mbox{if latex x \in [0,a]$} \\ \infty & \quad \mbox{otherwise} \\ \end{array}\right.\end{aligned} \hspace{\stretch{1}}(2.1)$

as illustrated in figure (\ref{fig:qmTwoR3:qmTwoR3fig1})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.4\textheight]{qmTwoR3fig1}
\caption{Arbitrary potential in an infinite well.}
\end{figure}

Inside the well, we have

\begin{aligned}\psi(x) = \frac{1}{{\sqrt{k(x)}}} \left( C_{+} e^{i \int_0^x k(x') dx'}+C_{-} e^{-i \int_0^x k(x') dx'}\right)\end{aligned} \hspace{\stretch{1}}(2.2)

where

\begin{aligned}k(x) = \frac{1}{{\hbar}} \sqrt{ 2m( E - v(x) }\end{aligned} \hspace{\stretch{1}}(2.3)

With

\begin{aligned}\phi(x) = e^{\int_0^x k(x') dx'}\end{aligned} \hspace{\stretch{1}}(2.4)

We have

\begin{aligned}\psi(x) &= \frac{1}{{\sqrt{k(x)}}} \left( C_{+}(\cos \phi + i\sin\phi) + C_{-}(\cos\phi - i \sin\phi)\right) \\ &= \frac{1}{{\sqrt{k(x)}}} \left( (C_{+} + C_{-})\cos \phi + i(C_{+} - C_{-}) \sin\phi\right) \\ &= \frac{1}{{\sqrt{k(x)}}} \left( (C_{+} + C_{-})\cos \phi + i(C_{+} - C_{-}) \sin\phi\right) \\ &\equiv \frac{1}{{\sqrt{k(x)}}} \left( C_2 \cos \phi + C_1 \sin\phi\right),\end{aligned}

Where

\begin{aligned}C_2 &= C_{+} + C_{-} \\ C_1 &= i( C_{+} - C_{-})\end{aligned} \hspace{\stretch{1}}(2.5)

Setting boundary conditions we have

\begin{aligned}\phi(0) = 0\end{aligned} \hspace{\stretch{1}}(2.7)

Noting that we have \phi(0) = 0, we have

\begin{aligned}\frac{1}{{\sqrt{k(0)}}} C_2 = 0\end{aligned} \hspace{\stretch{1}}(2.8)

So

\begin{aligned}\psi(x) \sim\frac{1}{{\sqrt{k(x)}}} \sin\phi\end{aligned} \hspace{\stretch{1}}(2.9)

At the other boundary

\begin{aligned}\psi(a) = 0\end{aligned} \hspace{\stretch{1}}(2.10)

So we require

\begin{aligned}\sin \phi(a) = \sin(n \pi)\end{aligned} \hspace{\stretch{1}}(2.11)

or

\begin{aligned}\frac{1}{{\hbar}} \int_0^a \sqrt{2 m (E - v(x')} dx' = n \pi\end{aligned} \hspace{\stretch{1}}(2.12)

This is called the Bohr-Sommerfeld condition.

Check with v(x) = 0.

We have

\begin{aligned}\frac{1}{{\hbar}} \sqrt{2m E} a = n \pi\end{aligned} \hspace{\stretch{1}}(2.13)

or

\begin{aligned}E = \frac{1}{{2m}} \left(\frac{n \pi \hbar}{a}\right)^2\end{aligned} \hspace{\stretch{1}}(2.14)

Stark Shift

Time independent perturbation theory

\begin{aligned}H = H_0 + \lambda H'\end{aligned} \hspace{\stretch{1}}(3.15)

\begin{aligned}H' = e \mathcal{E}_z \hat{Z}\end{aligned} \hspace{\stretch{1}}(3.16)

where \mathcal{E}_z is the electric field.

To first order we have

\begin{aligned}{\left\lvert {\psi_\alpha^{(1)}} \right\rangle} = {\left\lvert {\psi_\alpha^{(0)}} \right\rangle} + \sum_{\beta \ne \alpha} \frac{ {\left\lvert {\psi_\beta^{(0)}} \right\rangle} {\left\langle {\psi_\beta^{(0)}} \right\rvert} H' {\left\lvert {\psi_\alpha^{(0)}} \right\rangle} }{E_\alpha^{(0)} -E_\beta^{(0)} }\end{aligned} \hspace{\stretch{1}}(3.17)

and

\begin{aligned}E_\alpha^{(1)} = {\left\langle {\psi_\alpha^{(0)}} \right\rvert} H' {\left\lvert {\psi_\alpha^{(0)}} \right\rangle} \end{aligned} \hspace{\stretch{1}}(3.18)

With the default basis \{{\left\lvert {\psi_\beta^{(0)}} \right\rangle}\}, and n=2 we have a 4 fold degeneracy

\begin{aligned}l,m &= 0,0 \\ l,m &= 1,-1 \\ l,m &= 1,0 \\ l,m &= 1,+1\end{aligned}

but can diagonalize as follows

\begin{aligned}\begin{bmatrix}\text{nlm} & 200 & 210 & 211 & 21\,-1 \\ 200 & 0 & \Delta & 0 & 0 \\ 210 & \Delta & 0 & 0 & 0 \\ 211 & 0 & 0 & 0 & 0 \\ 21\,-1 & 0 & 0 & 0 & 0\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.19)

FIXME: show.

where

\begin{aligned}\Delta = -3 e \mathcal{E}_z a_0\end{aligned} \hspace{\stretch{1}}(3.20)

We have a split of energy levels as illustrated in figure (\ref{fig:qmTwoR3:qmTwoR3fig2})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.4\textheight]{qmTwoR3fig2}
\caption{Energy level splitting}
\end{figure}

Observe the embedded Pauli matrix (FIXME: missed the point of this?)

\begin{aligned}\sigma_x = \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.21)

Proper basis for perturbation (FIXME:check) is then

\begin{aligned}\left\{\frac{1}{{\sqrt{2}}}( {\left\lvert {2,0,0} \right\rangle} \pm {\left\lvert {2,1,0} \right\rangle} ), {\left\lvert {2, 1, \pm 1} \right\rangle}\right\}\end{aligned} \hspace{\stretch{1}}(3.22)

and our result is

\begin{aligned}{\left\lvert {\psi_{\alpha, n=2}^{(1)}} \right\rangle} = {\left\lvert {\psi_{\alpha}^{(0)}} \right\rangle} +\sum_{\beta \notin \text{degenerate subspace}} \frac{ {\left\lvert {\psi_\beta^{(0)}} \right\rangle} {\left\langle {\psi_\beta^{(0)}} \right\rvert} H' {\left\lvert {\psi_\alpha^{(0)}} \right\rangle} }{E_\alpha^{(0)} -E_\beta^{(0)} }\end{aligned} \hspace{\stretch{1}}(3.23)

Adiabatic perturbation theory

Utilizing instantaneous eigenstates

\begin{aligned}{\left\lvert {\psi(t)} \right\rangle} = \sum_{\alpha} b_\alpha(t) {\left\lvert {\hat{\psi}_\alpha(t)} \right\rangle}\end{aligned} \hspace{\stretch{1}}(4.24)

where

\begin{aligned}H(t) {\left\lvert {\hat{\psi}_\alpha(t)} \right\rangle}= E_\alpha(t) {\left\lvert {\hat{\psi}_\alpha(t)} \right\rangle}\end{aligned} \hspace{\stretch{1}}(4.25)

We found

\begin{aligned}b_\alpha(t) = \bar{b}_\alpha(t) e^{-\frac{i}{\hbar} \int_0^t (E_\alpha(t') - \hbar \Gamma_\alpha(t')) dt'}\end{aligned} \hspace{\stretch{1}}(4.26)

where

\begin{aligned}\Gamma_\alpha = i{\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}\frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_\alpha(t)} \right\rangle}\end{aligned} \hspace{\stretch{1}}(4.27)

and

\begin{aligned}\frac{d{{}}}{dt}\bar{b}_\alpha(t)=-\sum_{\beta \ne \alpha} \bar{b}_\beta(t)e^{-\frac{i}{\hbar} \int_0^t (E_{\beta\alpha}(t') - \hbar \Gamma_{\beta\alpha}(t')) dt'}{\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}\frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle}\end{aligned} \hspace{\stretch{1}}(4.36)

Suppose we start in a subspace

\begin{aligned}\text{span} \left\{\frac{1}{{\sqrt{2}}}( {\left\lvert {2,0,0} \right\rangle} \pm {\left\lvert {2,1,0} \right\rangle} ), {\left\lvert {2, 1, \pm 1} \right\rangle}\right\}\end{aligned} \hspace{\stretch{1}}(4.29)

Now expand the bra derivative kets

\begin{aligned}{\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}\frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle}=\left({\left\langle {\psi_{\alpha}^{(0)}} \right\rvert} +\sum_{\gamma} \frac{ {\left\langle {\psi_\gamma^{(0)}} \right\rvert} H' {\left\lvert {\psi_\alpha^{(0)}} \right\rangle} {\left\langle {\psi_\gamma^{(0)}} \right\rvert} }{E_\alpha^{(0)} -E_\gamma^{(0)} }\right)\frac{d{{}}}{dt}\left({\left\lvert {\psi_{\beta}^{(0)}} \right\rangle} +\sum_{\gamma'} \frac{ {\left\lvert {\psi_{\gamma'}^{(0)}} \right\rangle} {\left\langle {\psi_{\gamma'}^{(0)}} \right\rvert} H' {\left\lvert {\psi_\beta^{(0)}} \right\rangle} }{E_\beta^{(0)} -E_{\gamma'}^{(0)} }\right)\end{aligned} \hspace{\stretch{1}}(4.30)

To first order we can drop the quadratic terms in \gamma,\gamma' leaving

\begin{aligned}\begin{aligned}{\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}\frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle}&\sim\sum_{\gamma'} \left\langle{{\psi_{\alpha}^{(0)}}} \vert {{\psi_{\gamma'}^{(0)}}}\right\rangle \frac{ {\left\langle {\psi_{\gamma'}^{(0)}} \right\rvert} \frac{d{{H'(t)}}}{dt} {\left\lvert {\psi_\beta^{(0)}} \right\rangle} }{E_\beta^{(0)} -E_{\gamma'}^{(0)} }&=\frac{ {\left\langle {\psi_{\alpha}^{(0)}} \right\rvert} \frac{d{{H'(t)}}}{dt} {\left\lvert {\psi_\beta^{(0)}} \right\rangle} }{E_\beta^{(0)} -E_{\alpha}^{(0)} }\end{aligned}\end{aligned} \hspace{\stretch{1}}(4.31)

so

\begin{aligned}\frac{d{{}}}{dt}\bar{b}_\alpha(t)=-\sum_{\beta \ne \alpha} \bar{b}_\beta(t)e^{-\frac{i}{\hbar} \int_0^t (E_{\beta\alpha}(t') - \hbar \Gamma_{\beta\alpha}(t')) dt'}\frac{ {\left\langle {\psi_{\alpha}^{(0)}} \right\rvert} \frac{d{{H'(t)}}}{dt} {\left\lvert {\psi_\beta^{(0)}} \right\rangle} }{E_\beta^{(0)} -E_{\alpha}^{(0)} }\end{aligned} \hspace{\stretch{1}}(4.32)

A different way to this end result.

A result of this form is also derived in [1] section 20.1, but with a different approach. There he takes derivatives of

\begin{aligned}H(t) {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle} = E_\beta(t) {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle},\end{aligned} \hspace{\stretch{1}}(4.33)

\begin{aligned}\frac{d{{H(t)}}}{dt} {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle} + H(t) \frac{d{{}}}{dt}{\left\lvert {\hat{\psi}_\beta(t)} \right\rangle} = \frac{d{{E_\beta(t)}}}{dt} {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle}+ E_\beta(t) \frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle}\end{aligned} \hspace{\stretch{1}}(4.34)

Bra’ing {\left\langle {\hat{\psi}_\alpha(t)} \right\rvert} into this we have, for \alpha \ne \beta

\begin{aligned}{\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}\frac{d{{H(t)}}}{dt} {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle} + {\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}H(t) \frac{d{{}}}{dt}{\left\lvert {\hat{\psi}_\beta(t)} \right\rangle} &= \not{{{\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}\frac{d{{E_\beta(t)}}}{dt} {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle}}}+ {\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}E_\beta(t) \frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle} \\ {\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}\frac{d{{H(t)}}}{dt} {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle} + E_\alpha(t) {\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}\frac{d{{}}}{dt}{\left\lvert {\hat{\psi}_\beta(t)} \right\rangle} &=\end{aligned}

or

\begin{aligned}{\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}\frac{d{{}}}{dt}{\left\lvert {\hat{\psi}_\beta(t)} \right\rangle} =\frac{{\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}\frac{d{{H(t)}}}{dt} {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle} }{E_\beta(t) - E_\alpha(t)}\end{aligned} \hspace{\stretch{1}}(4.35)

so without the implied \lambda perturbation of {\left\lvert {\hat{\psi}_\alpha(t)} \right\rangle} we can from 4.36 write the exact generalization of 4.32 as

\begin{aligned}\frac{d{{}}}{dt}\bar{b}_\alpha(t)=-\sum_{\beta \ne \alpha} \bar{b}_\beta(t)e^{-\frac{i}{\hbar} \int_0^t (E_{\beta\alpha}(t') - \hbar \Gamma_{\beta\alpha}(t')) dt'}\frac{{\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}\frac{d{{H(t)}}}{dt} {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle} }{E_\beta(t) - E_\alpha(t)}\end{aligned} \hspace{\stretch{1}}(4.36)

References

[1] D. Bohm. Quantum Theory. Courier Dover Publications, 1989.

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A different derivation of the adiabatic perturbation coefficient equation

Posted by peeterjoot on October 27, 2011

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

Motivation.

Professor Sipe’s adiabatic perturbation and that of the text [1] in section 17.5.1 and section 17.5.2 use different notation for \gamma_m and take a slightly different approach. We can find Prof Sipe’s final result with a bit less work, if a hybrid of the two methods is used.

Guts

Our starting point is the same, we have a time dependent slowly varying Hamiltonian

\begin{aligned}H = H(t),\end{aligned} \hspace{\stretch{1}}(2.1)

where our perturbation starts at some specific time from a given initial state

\begin{aligned}H(t) = H_0, \qquad t \le 0.\end{aligned} \hspace{\stretch{1}}(2.2)

We assume that instantaneous eigenkets can be found, satisfying

\begin{aligned}H(t) {\left\lvert {n(t)} \right\rangle} = E_n(t) {\left\lvert {n(t)} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.3)

Here I’ll use {\left\lvert {n} \right\rangle} \equiv {\left\lvert {n(t)} \right\rangle} instead of the {\left\lvert {\hat{\psi}_n(t)} \right\rangle} that we used in class because its easier to write.

Now suppose that we have some arbitrary state, expressed in terms of the instantaneous basis kets {\left\lvert {n} \right\rangle}

\begin{aligned}{\left\lvert {\psi} \right\rangle} = \sum_n \bar{b}_n(t) e^{-i\alpha_n + i \gamma_n} {\left\lvert {n} \right\rangle},\end{aligned} \hspace{\stretch{1}}(2.4)

where

\begin{aligned}\alpha_n(t) = \frac{1}{{\hbar}} \int_0^t dt' E_n(t'),\end{aligned} \hspace{\stretch{1}}(2.5)

and \gamma_n (using the notation in the text, not in class) is to be determined.

For this state, we have at the time just before the perturbation

\begin{aligned}{\left\lvert {\psi(0)} \right\rangle} = \sum_n \bar{b}_n(0) e^{-i\alpha_n(0) + i \gamma_n(0)} {\left\lvert {n(0)} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(2.6)

The question to answer is: How does this particular state evolve?

Another question, for those that don’t like sneaky bastard derivations, is where did that magic factor of e^{-i\alpha_n} come from in our superposition state? We will see after we start taking derivatives that this is what we need to cancel the H(t){\left\lvert {n} \right\rangle} in Schr\”{o}dinger’s equation.

Proceeding to plug into the evolution identity we have

\begin{aligned}0 &={\left\langle {m} \right\rvert} \left( i \hbar \frac{d{{}}}{dt} - H(t) \right) {\left\lvert {\psi} \right\rangle} \\ &={\left\langle {m} \right\rvert} \left(\sum_n e^{-i \alpha_n + i \gamma_n}(i \hbar) \left(\frac{d{{\bar{b}_n}}}{dt}+ \bar{b}_n \left(-i \not{{\frac{E_n}{\hbar}}} + i \dot{\gamma}_m \right)\right) {\left\lvert {n} \right\rangle}+ i \hbar \bar{b}_n \frac{d{{}}}{dt} {\left\lvert {n} \right\rangle}- \not{{E_n \bar{b}_n {\left\lvert {n} \right\rangle}}} \right)\\ &=e^{-i \alpha_m + i \gamma_m}(i \hbar) \frac{d{{\bar{b}_m}}}{dt}+e^{-i \alpha_m + i \gamma_m}(i \hbar) i \dot{\gamma}_m \bar{b}_m+ i \hbar \sum_n \bar{b}_n {\left\langle {m} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {n} \right\rangle}e^{-i \alpha_n + i \gamma_n} \\ &\sim\frac{d{{\bar{b}_m}}}{dt}+i \dot{\gamma}_m \bar{b}_m+ \sum_n e^{-i \alpha_n + i \gamma_n}e^{i \alpha_m - i \gamma_m}\bar{b}_n {\left\langle {m} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {n} \right\rangle} \\ &=\frac{d{{\bar{b}_m}}}{dt}+i \dot{\gamma}_m \bar{b}_m+ \bar{b}_m {\left\langle {m} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {m} \right\rangle}+\sum_{n \ne m} e^{-i \alpha_n + i \gamma_n}e^{i \alpha_m - i \gamma_m}\bar{b}_n {\left\langle {m} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {n} \right\rangle}\end{aligned}

We are free to pick \gamma_m to kill the second and third terms

\begin{aligned}0 =i \dot{\gamma}_m \bar{b}_m+ \bar{b}_m {\left\langle {m} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {m} \right\rangle},\end{aligned} \hspace{\stretch{1}}(2.7)

or

\begin{aligned}\dot{\gamma}_m = i {\left\langle {m} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {m} \right\rangle},\end{aligned} \hspace{\stretch{1}}(2.8)

which after integration is

\begin{aligned}\gamma_m(t)= i \int_0^t dt' {\left\langle {m(t')} \right\rvert} \frac{d}{dt'} {\left\lvert {m(t)} \right\rangle},\end{aligned} \hspace{\stretch{1}}(2.9)

as in class we can observe that this is a purely real function. We are left with

\begin{aligned}\frac{d{{\bar{b}_m}}}{dt}=-\sum_{n \ne m} \bar{b}_n e^{-i \alpha_{nm} + i \gamma_{nm}}{\left\langle {m} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {n} \right\rangle} ,\end{aligned} \hspace{\stretch{1}}(2.10)

where

\begin{aligned}\alpha_{nm} &= \alpha_{n} -\alpha_m \\ \gamma_{nm} &= \gamma_{n} -\gamma_m \end{aligned} \hspace{\stretch{1}}(2.11)

The task is now to find solutions for these \bar{b}_m coefficients, and we can refer to the class notes for that without change.

References

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

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casting away volatile with const_cast: a subtlety.

Posted by peeterjoot on October 27, 2011

I broke our windows platform build yesterday, with a change to a macro that I didn’t realize was only used in our windows code. The macro was of the following form:

void goo(const void *) ;
#define foo(p) goo((const void *)p)

Do you ask why there’s a cast here? Shouldn’t that be redundant since types can be converted without cast to void *? That’s true most of the cases, but this particular macro is used with types that may also have a volatile attribute. I realize now that this must have been the point of the cast in the macro. I’d changed the macro to force a bit of stricter type checking:

#define foo(p) goo( const_cast< T*>(p) )

The const_cast can be a handy beastie, and will force a compilation error if the type is not convertible to T* once const or volatile attributes are stripped off. Example:

volatile int x ;
const int y = 3 ;
short z ;
goo( const_cast<int *>(&x) ) ; // compiles.
goo( const_cast<int *>(&y) ) ; // compiles.
goo( const_cast<int *>(&z) ) ; // compilation error

The last line will produce a nice error message of the form:

error: a const_cast can only adjust type qualifiers; it cannot change the underlying type

From what I’d read I’d thought that const_cast stripped off any const and volatile attribute, but my build break shows that this is not quite true. Here’s an example that demonstrates:

#include <stdio.h>

void foo( void * y )
{
   printf( "%d\n", *(int *)y ) ;
}

typedef volatile int T ;

int main()
{
   T x ;

   foo( const_cast< int* >(&x) ) ; // compiles.
   foo( const_cast< T* >(&x) ) ;   // compilation error

   return 0 ;
}

I end up with an error message like:

t.C(15): error: argument of type "T={volatile int} *" is incompatible with parameter of type "void *"
     foo( const_cast(&x) ) ;   // compilation error
          ^

If your type cast argument in the const_cast also includes a volatile attribute, then the const_cast appears to leave the type unchanged, retaining the volatile attribute. This makes a certain sense. It also appears to be a cross platform behavior. I’d changed a number of macros in this way, and found some coding errors by doing so in my other platform builds. The macro in question that I broke the build with was unluckily only called in windows code, and also the only one where the type happened to have volatile in it (all the rest of the types were structures, and in those structures were members that had volatile on specific parts of them).

In retrospect I wish that I’d coded this one type as a struct with a volatile member, then pointers to this struct type could be passed around, cast to void * in the tracing and problem determination code if desired. That would leave the volatile detail to the innards of the hardware oriented code that required it.

Posted in C/C++ development and debugging. | Tagged: , , , | 4 Comments »

PHY456H1F: Quantum Mechanics II. Lecture 14 (Taught by Prof J.E. Sipe). Representation of two state kets and Pauli spin matrices.

Posted by peeterjoot on October 26, 2011

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

Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

Representation of kets.

Reading: section 5.1 – section 5.9 and section 26 in [1].

We found the representations of the spin operators

\begin{aligned}S_x &\rightarrow \frac{\hbar}{2} \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} \\ S_y &\rightarrow \frac{\hbar}{2} \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix} \\ S_z &\rightarrow \frac{\hbar}{2} \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.1)

How about kets? For example for {\left\lvert {\chi} \right\rangle} \in H_s

\begin{aligned}{\left\lvert {\chi} \right\rangle} \rightarrow \begin{bmatrix}\left\langle{{+}} \vert {{\chi}}\right\rangle \\ \left\langle{{-}} \vert {{\chi}}\right\rangle\end{bmatrix},\end{aligned} \hspace{\stretch{1}}(2.4)

and

\begin{aligned}{\left\lvert {+} \right\rangle} &\rightarrow \begin{bmatrix}1 \\ 0\end{bmatrix} \\ {\left\lvert {0} \right\rangle} &\rightarrow \begin{bmatrix}0 \\ 1\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.5)

So, for example

\begin{aligned}S_y{\left\lvert {+} \right\rangle} \rightarrow \frac{\hbar}{2} \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix}\begin{bmatrix}1 \\ 0\end{bmatrix} =\frac{i\hbar}{2}\begin{bmatrix}0 \\ 1\end{bmatrix} \end{aligned} \hspace{\stretch{1}}(2.7)

Kets in H_o \otimes H_s

\begin{aligned}{\left\lvert {\psi} \right\rangle} \rightarrow \begin{bmatrix}\left\langle{{\mathbf{r}+}} \vert {{\psi}}\right\rangle \\ \left\langle{{\mathbf{r}-}} \vert {{\psi}}\right\rangle\end{bmatrix}=\begin{bmatrix}\psi_{+}(\mathbf{r}) \\ \psi_{-}(\mathbf{r})\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(2.8)

This is a “spinor”

Put

\begin{aligned}\begin{aligned}\left\langle{{\mathbf{r} \pm}} \vert {{\psi}}\right\rangle&= \psi_{\pm}(\mathbf{r}) \\ &= \psi_{+} \begin{bmatrix}1 \\ 0\end{bmatrix}+\psi_{-} \begin{bmatrix}0 \\ 1 \end{bmatrix}\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.9)

with

\begin{aligned}\left\langle{{\psi}} \vert {{\psi}}\right\rangle = 1\end{aligned} \hspace{\stretch{1}}(2.10)

Use

\begin{aligned}\begin{aligned}I &= I_o \otimes I_s \\ &= \int d^3 \mathbf{r} {\left\lvert {\mathbf{r}} \right\rangle}{\left\langle {\mathbf{r}} \right\rvert} \otimes \left( {\left\lvert {{+}} \right\rangle}{\left\langle {{+}} \right\rvert}+{\left\lvert {{-}} \right\rangle}{\left\langle {{-}} \right\rvert}\right) \\ &=\int d^3 \mathbf{r} {\left\lvert {\mathbf{r}} \right\rangle}{\left\langle {\mathbf{r}} \right\rvert} \otimes \sum_{\sigma=\pm} {\left\lvert {{\sigma}} \right\rangle}{\left\langle {{\sigma}} \right\rvert} \\ &=\sum_{\sigma = \pm} \int d^3 \mathbf{r} {\left\lvert {{\mathbf{r} \sigma}} \right\rangle}{\left\langle {{\mathbf{r} \sigma}} \right\rvert} \end{aligned}\end{aligned} \hspace{\stretch{1}}(2.11)

So

\begin{aligned}\begin{aligned}{\left\langle {\psi} \right\rvert} I {\left\lvert {\psi} \right\rangle} &= \sum_{\sigma = \pm} \int d^3 \mathbf{r} \left\langle{{\psi}} \vert {{\mathbf{r} \sigma}}\right\rangle \left\langle{{\mathbf{r} \sigma}} \vert {{\psi}}\right\rangle \\ &= \int d^3 \mathbf{r} \left( {\left\lvert{\psi_{+}(\mathbf{r})}\right\rvert}^2+{\left\lvert{\psi_{-}(\mathbf{r})}\right\rvert}^2\right)\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.12)

Alternatively

\begin{aligned}\begin{aligned}{\left\lvert {\psi} \right\rangle} &= I {\left\lvert {\psi} \right\rangle} \\ &=\int d^3 \mathbf{r} \sum_{\sigma = \pm} {\left\lvert {\mathbf{r} \sigma} \right\rangle}\left\langle{{\mathbf{r} \sigma}} \vert {{\psi}}\right\rangle \\ &=\sum_{\sigma = \pm} \left(\int d^3 \mathbf{r} \psi_\sigma(\mathbf{r})\right){\left\lvert {\mathbf{r} \sigma} \right\rangle} \\ &=\sum_{\sigma = \pm} \left(\int d^3 \mathbf{r} \psi_\sigma(\mathbf{r}) {\left\lvert {\mathbf{r}} \right\rangle}\right)\otimes {\left\lvert {\sigma} \right\rangle} \end{aligned}\end{aligned} \hspace{\stretch{1}}(2.13)

In braces we have a ket in H_o, let’s call it

\begin{aligned}{\left\lvert {\psi_\sigma} \right\rangle} = \int d^3 \mathbf{r} \psi_\sigma(\mathbf{r}) {\left\lvert {\mathbf{r}} \right\rangle},\end{aligned} \hspace{\stretch{1}}(2.14)

then

\begin{aligned}{\left\lvert {\psi} \right\rangle} = {\left\lvert {\psi_{+}} \right\rangle} {\left\lvert {+} \right\rangle} + {\left\lvert {\psi_{-}} \right\rangle} {\left\lvert {-} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.15)

where the direct product \otimes is implied.

We can form a ket in H_s as

\begin{aligned}\left\langle{\mathbf{r}} \vert {{\psi}}\right\rangle = \psi_{+}(\mathbf{r}) {\left\lvert {+} \right\rangle} + \psi_{-}(\mathbf{r}) {\left\lvert {-} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.16)

An operator O_o which acts on H_o alone can be promoted to O_o \otimes I_s, which is now an operator that acts on H_o \otimes H_s. We are sometimes a little cavalier in notation and leave this off, but we should remember this.

\begin{aligned}O_o {\left\lvert {\psi} \right\rangle} = (O_o {\left\lvert {\psi+} \right\rangle}) {\left\lvert {+} \right\rangle}+ (O_o {\left\lvert {\psi+} \right\rangle}) {\left\lvert {+} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.17)

and likewise

\begin{aligned}O_s {\left\lvert {\psi} \right\rangle} = {\left\lvert {\psi+} \right\rangle} (O_s {\left\lvert {+} \right\rangle})+{\left\lvert {\psi-} \right\rangle} (O_s {\left\lvert {-} \right\rangle})\end{aligned} \hspace{\stretch{1}}(2.18)

and

\begin{aligned}O_o O_s {\left\lvert {\psi} \right\rangle} = (O_o {\left\lvert {\psi+} \right\rangle}) (O_s {\left\lvert {+} \right\rangle})+(O_o {\left\lvert {\psi-} \right\rangle}) (O_s {\left\lvert {-} \right\rangle})\end{aligned} \hspace{\stretch{1}}(2.19)

Suppose we want to rotate a ket, we do this with a full angular momentum operator

\begin{aligned}e^{-i \theta \hat{\mathbf{n}} \cdot \mathbf{J}/\hbar} {\left\lvert {\psi} \right\rangle}=e^{-i \theta \hat{\mathbf{n}} \cdot \mathbf{L}/\hbar} e^{-i \theta \hat{\mathbf{n}} \cdot \mathbf{S}/\hbar} {\left\lvert {\psi} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.20)

(recalling that \mathbf{L} and \mathbf{S} commute)

So

\begin{aligned}e^{-i \theta \hat{\mathbf{n}} \cdot \mathbf{J}/\hbar} {\left\lvert {\psi} \right\rangle}=(e^{-i \theta \hat{\mathbf{n}} \cdot \mathbf{L}/\hbar} {\left\lvert {\psi+} \right\rangle}) (e^{-i \theta \hat{\mathbf{n}} \cdot \mathbf{S}/\hbar} {\left\lvert {+} \right\rangle})+(e^{-i \theta \hat{\mathbf{n}} \cdot \mathbf{L}/\hbar} {\left\lvert {\psi-} \right\rangle}) (e^{-i \theta \hat{\mathbf{n}} \cdot \mathbf{S}/\hbar} {\left\lvert {-} \right\rangle})\end{aligned} \hspace{\stretch{1}}(2.21)

A simple example.

\begin{aligned}{\left\lvert {\psi} \right\rangle} = {\left\lvert {\psi_+} \right\rangle} {\left\lvert {+} \right\rangle}+{\left\lvert {\psi_-} \right\rangle} {\left\lvert {-} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.22)

Suppose

\begin{aligned}{\left\lvert {\psi_+} \right\rangle} &= \alpha {\left\lvert {\psi_0} \right\rangle} \\ {\left\lvert {\psi_-} \right\rangle} &= \beta {\left\lvert {\psi_0} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.23)

where

\begin{aligned}{\left\lvert{\alpha}\right\rvert}^2 + {\left\lvert{\beta}\right\rvert}^2 = 1\end{aligned} \hspace{\stretch{1}}(2.25)

Then

\begin{aligned}{\left\lvert {\psi} \right\rangle} = {\left\lvert {\psi_0} \right\rangle} {\left\lvert {\chi} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.26)

where

\begin{aligned}{\left\lvert {\chi} \right\rangle} = \alpha {\left\lvert {+} \right\rangle} + \beta {\left\lvert {-} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.27)

for

\begin{aligned}\left\langle{{\psi}} \vert {{\psi}}\right\rangle = 1,\end{aligned} \hspace{\stretch{1}}(2.28)

\begin{aligned}\left\langle{{\psi_0}} \vert {{\psi_0}}\right\rangle \left\langle{{\chi}} \vert {{\chi}}\right\rangle = 1\end{aligned} \hspace{\stretch{1}}(2.29)

so

\begin{aligned}\left\langle{{\psi_0}} \vert {{\psi_0}}\right\rangle = 1\end{aligned} \hspace{\stretch{1}}(2.30)

We are going to concentrate on the unentagled state of 2.26.

\begin{itemize}
\item
How about with

\begin{aligned}{\left\lvert{\alpha}\right\rvert}^2 = 1, \beta = 0\end{aligned} \hspace{\stretch{1}}(2.31)

{\left\lvert {\chi} \right\rangle} is an eigenket of S_z with eigenvalue \hbar/2.
\item

\begin{aligned}{\left\lvert{\beta}\right\rvert}^2 = 1, \alpha = 0\end{aligned} \hspace{\stretch{1}}(2.32)

{\left\lvert {\chi} \right\rangle} is an eigenket of S_z with eigenvalue -\hbar/2.

\item
What is {\left\lvert {\chi} \right\rangle} if it is an eigenket of \hat{\mathbf{n}} \cdot \mathbf{S}?
\end{itemize}

FIXME: F1: standard spherical projection picture, with \hat{\mathbf{n}} projected down onto the x,y plane at angle \phi and at an angle \theta from the z axis.

The eigenvalues will still be \pm \hbar/2 since there is nothing special about the z direction.

\begin{aligned}\begin{aligned}\hat{\mathbf{n}} \cdot \mathbf{S} &= n_x S_x+n_y S_y+n_z S_z \\ &\rightarrow\frac{\hbar}{2} \begin{bmatrix}n_z & n_x - i n_y \\ n_x + i n_y & -n_z\end{bmatrix} \\ &=\frac{\hbar}{2} \begin{bmatrix}\cos\theta & \sin\theta e^{-i\phi}\sin\theta e^{i\phi} & -\cos\theta\end{bmatrix}\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.33)

To find the eigenkets we diagonalize this, and we find representations of the eigenkets are

\begin{aligned}{\left\lvert {\hat{\mathbf{n}}+} \right\rangle} &\rightarrow \begin{bmatrix}\cos\left(\frac{\theta}{2}\right) e^{-i\phi/2} \\ \sin\left(\frac{\theta}{2}\right) e^{i\phi/2} \end{bmatrix} \\ {\left\lvert {\hat{\mathbf{n}}-} \right\rangle} &\rightarrow \begin{bmatrix}-\sin\left(\frac{\theta}{2}\right) e^{-i\phi/2} \\ \cos\left(\frac{\theta}{2}\right) e^{i\phi/2} \end{bmatrix},\end{aligned} \hspace{\stretch{1}}(2.34)

with eigenvalues \hbar/2 and -\hbar/2 respectively.

So in the abstract notation, tossing the specific representation, we have

\begin{aligned}{\left\lvert {\hat{\mathbf{n}}+} \right\rangle} &\rightarrow \cos\left(\frac{\theta}{2}\right) e^{-i\phi/2} {\left\lvert {+} \right\rangle}\sin\left(\frac{\theta}{2}\right) e^{i\phi/2} {\left\lvert {-} \right\rangle} \\ {\left\lvert {\hat{\mathbf{n}}-} \right\rangle} &\rightarrow -\sin\left(\frac{\theta}{2}\right) e^{-i\phi/2} {\left\lvert {+} \right\rangle}\cos\left(\frac{\theta}{2}\right) e^{i\phi/2} {\left\lvert {-} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.36)

Representation of two state kets

Every ket

\begin{aligned}{\left\lvert {\chi} \right\rangle} \rightarrow \begin{bmatrix}\alpha \\ \beta\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.38)

for which

\begin{aligned}{\left\lvert{\alpha}\right\rvert}^2 + {\left\lvert{\beta}\right\rvert}^2 = 1\end{aligned} \hspace{\stretch{1}}(3.39)

can be written in the form 2.34 for some \theta and \phi, neglecting an overall phase factor.

For any ket in H_s, that ket is “spin up” in some direction.

FIXME: show this.

Pauli spin matrices.

It is useful to write

\begin{aligned}S_x &= \frac{\hbar}{2} \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} \equiv \frac{\hbar}{2} \sigma_x \\ S_y &= \frac{\hbar}{2} \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix} \equiv \frac{\hbar}{2} \sigma_y \\ &= \frac{\hbar}{2} \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} \equiv \frac{\hbar}{2} \sigma_z \end{aligned} \hspace{\stretch{1}}(4.40)

where

\begin{aligned}\sigma_x &= \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} \\ \sigma_y &= \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix} \\ \sigma_z &= \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(4.43)

These are the Pauli spin matrices.

Interesting properties.

\begin{itemize}
\item

\begin{aligned}\left[{\sigma_i},{\sigma_j}\right] = \sigma_i \sigma_j + \sigma_j \sigma_i = 0, \qquad \mbox{ if latex i < j$}\end{aligned} \hspace{\stretch{1}}(4.46)$

\item

\begin{aligned}\sigma_x \sigma_y = i \sigma_z\end{aligned} \hspace{\stretch{1}}(4.47)

(and cyclic permutations)

\item

\begin{aligned}\text{Tr}(\sigma_i) = 0\end{aligned} \hspace{\stretch{1}}(4.48)

\item

\begin{aligned}(\hat{\mathbf{n}} \cdot \boldsymbol{\sigma})^2 = \sigma_0\end{aligned} \hspace{\stretch{1}}(4.49)

where

\begin{aligned}\hat{\mathbf{n}} \cdot \boldsymbol{\sigma} \equiv n_x \sigma_x + n_y \sigma_y + n_z \sigma_z,\end{aligned} \hspace{\stretch{1}}(4.50)

and

\begin{aligned}\sigma_0 = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(4.51)

(note \text{Tr}(\sigma_0) \ne 0)

\item

\begin{aligned}\left[{\sigma_i},{\sigma_j}\right] &= 2 \delta_{ij} \sigma_0 \\ \left[{\sigma_x},{\sigma_y}\right] &= 2 i \sigma_z\end{aligned} \hspace{\stretch{1}}(4.52)

(and cyclic permutations of the latter).

Can combine these to show that

\begin{aligned}(\mathbf{A} \cdot \boldsymbol{\sigma})(\mathbf{B} \cdot \boldsymbol{\sigma})=(\mathbf{A} \cdot \mathbf{B}) \sigma_0 + i (\mathbf{A} \times \mathbf{B}) \cdot \boldsymbol{\sigma}\end{aligned} \hspace{\stretch{1}}(4.54)

where \mathbf{A} and \mathbf{B} are vectors (or more generally operators that commute with the \boldsymbol{\sigma} matrices).

\item

\begin{aligned}\text{Tr}(\sigma_i \sigma_j) = 2 \delta_{ij}\end{aligned} \hspace{\stretch{1}}(4.55)

\item

\begin{aligned}\text{Tr}(\sigma_\alpha \sigma_\beta) = 2 \delta_{\alpha \beta},\end{aligned} \hspace{\stretch{1}}(4.56)

where \alpha, \beta = 0, x, y, z
\end{itemize}

Note that any complext matrix M can be written as

\begin{aligned}\begin{aligned}M &= \sum_\alpha m_a \sigma_\alpha \\ &=\begin{bmatrix}m_0 + m_z & m_x - i m_y \\ m_x + i m_y & m_0 - m_z\end{bmatrix}\end{aligned}\end{aligned} \hspace{\stretch{1}}(4.57)

for any four complex numbers m_0, m_x, m_y, m_z

where

\begin{aligned}m_\beta = \frac{1}{{2}} \text{Tr}(M \sigma_\beta).\end{aligned} \hspace{\stretch{1}}(4.58)

References

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

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new version of geometric algebra notes compilation

Posted by peeterjoot on October 25, 2011

Newly added to the compilation (version 10) since last update are:

April 30, 2011 A cylindrical Lienard-Wiechert potential calculation using multivector matrix products.

Dec 27, 2010 Vector form of Julia fractal.

Some typos are probably fixed too.

Changelog for version 9 was here.

Posted in geometric algebra | Tagged: , | 1 Comment »

PHY456H1F: Quantum Mechanics II. Lecture 13 (Taught by Prof J.E. Sipe). Spin and spinors (cont.)

Posted by peeterjoot on October 24, 2011

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

Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

Multiple wavefunction spaces.

Reading: See section 26.5 in the text [1].

We identified

\begin{aligned}\psi(\mathbf{r}) = \left\langle{{ \mathbf{r}}} \vert {{\psi}}\right\rangle\end{aligned} \hspace{\stretch{1}}(2.1)

with improper basis kets

\begin{aligned}{\left\lvert {\mathbf{r}} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.2)

Now introduce many function spaces

\begin{aligned}\begin{bmatrix}\psi_1(\mathbf{r}) \\ \psi_2(\mathbf{r}) \\ \dot{v}s \\ \psi_\gamma(\mathbf{r})\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.3)

with improper (unnormalizable) basis kets

\begin{aligned}{\left\lvert {\mathbf{r} \alpha} \right\rangle}, \qquad \alpha \in 1, 2, ... \gamma\end{aligned} \hspace{\stretch{1}}(2.4)

\begin{aligned}\psi_\alpha(\mathbf{r}) = \left\langle{{ \mathbf{r}\alpha}} \vert {{\psi}}\right\rangle\end{aligned} \hspace{\stretch{1}}(2.5)

for an abstract ket {\left\lvert {\psi} \right\rangle}

We will try taking this Hilbert space

\begin{aligned}H = H_o \otimes H_s\end{aligned} \hspace{\stretch{1}}(2.6)

Where H_o is the Hilbert space of “scalar” QM, “o” orbital and translational motion, associated with kets {\left\lvert {\mathbf{r}} \right\rangle} and H_s is the Hilbert space associated with the \gamma components {\left\lvert {\alpha} \right\rangle}. This latter space we will label the “spin” or “internal physics” (class suggestion: or perhaps intrinsic). This is “unconnected” with translational motion.

We build up the basis kets for H by direct products

\begin{aligned}{\left\lvert {\mathbf{r} \alpha} \right\rangle} = {\left\lvert {\mathbf{r}} \right\rangle} \otimes {\left\lvert {\alpha} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.7)

Now, for a rotated ket we seek a general angular momentum operator \mathbf{J} such that

\begin{aligned}{\left\lvert {\psi'} \right\rangle} = e^{-i \theta \hat{\mathbf{n}} \cdot \mathbf{J}/\hbar} {\left\lvert {\psi} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.8)

where

\begin{aligned}\mathbf{J} = \mathbf{L} + \mathbf{S},\end{aligned} \hspace{\stretch{1}}(2.9)

where \mathbf{L} acts over kets in H_o, “orbital angular momentum”, and \mathbf{S} is the “spin angular momentum”, acting on kets in H_s.

Strictly speaking this would be written as direct products involving the respective identities

\begin{aligned}\mathbf{J} = \mathbf{L} \otimes I_s + I_o \otimes \mathbf{S}.\end{aligned} \hspace{\stretch{1}}(2.10)

We require

\begin{aligned}\left[{J_i},{J_j}\right] = i \hbar \sum \epsilon_{i j k} J_k\end{aligned} \hspace{\stretch{1}}(2.11)

Since \mathbf{L} and \mathbf{S} “act over separate Hilbert spaces”. Since these come from legacy operators

\begin{aligned}\left[{L_i},{S_j}\right] = 0\end{aligned} \hspace{\stretch{1}}(2.12)

We also know that

\begin{aligned}\left[{L_i},{L_j}\right] = i \hbar \sum \epsilon_{i j k} L_k\end{aligned} \hspace{\stretch{1}}(2.13)

so

\begin{aligned}\left[{S_i},{S_j}\right] = i \hbar \sum \epsilon_{i j k} S_k, \end{aligned} \hspace{\stretch{1}}(2.14)

as expected. We could, in principle, have more complicated operators, where this would not be true. This is a proposal of sorts. Given such a definition of operators, let’s see where we can go with it.

For matrix elements of \mathbf{L} we have

\begin{aligned}{\left\langle {\mathbf{r}} \right\rvert} L_x {\left\lvert {\mathbf{r}'} \right\rangle} = -i \hbar \left( y \frac{\partial {}}{\partial {z}}-z \frac{\partial {}}{\partial {y}} \right) \delta(\mathbf{r}- \mathbf{r}')\end{aligned} \hspace{\stretch{1}}(2.15)

What are the matrix elements of {\left\langle {\alpha} \right\rvert} S_i {\left\lvert {\alpha'} \right\rangle}? From the commutation relationships we know

\begin{aligned}\sum_{\alpha'' = 1}^\gamma {\left\langle {\alpha} \right\rvert} S_i {\left\lvert {\alpha''} \right\rangle}{\left\langle {\alpha''} \right\rvert} S_j {\left\lvert {\alpha'} \right\rangle}-\sum_{\alpha'' = 1}^\gamma {\left\langle {\alpha} \right\rvert} S_j {\left\lvert {\alpha''} \right\rangle}{\left\langle {\alpha''} \right\rvert} S_i {\left\lvert {\alpha'} \right\rangle}=i \hbar \sum_k \epsilon_{ijk} {\left\langle {\alpha} \right\rvert} S_k {\left\lvert {\alpha''} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.16)

We see that our matrix element is tightly constrained by our choice of commutator relationships. We have \gamma^2 such matrix elements, and it turns out that it is possible to choose (or find) matrix elements that satisfy these constraints?

The {\left\langle {\alpha} \right\rvert} S_i {\left\lvert {\alpha'} \right\rangle} matrix elements that satisfy these constraints are found by imposing the commutation relations

\begin{aligned}\left[{S_i},{S_j}\right] = i \hbar \sum \epsilon_{i j k} S_k, \end{aligned} \hspace{\stretch{1}}(2.17)

and with

\begin{aligned}S^2 = \sum_j S_j^2,\end{aligned} \hspace{\stretch{1}}(2.18)

(this is just a definition). We find

\begin{aligned}\left[{S^2},{S_i}\right] = 0\end{aligned} \hspace{\stretch{1}}(2.19)

and seeking eigenkets

\begin{aligned}S^2 {\left\lvert {s m_s} \right\rangle} &= s(s+1) \hbar^2 {\left\lvert {s m_s} \right\rangle} \\ S_z {\left\lvert {s m_s} \right\rangle} &= \hbar m_s {\left\lvert {s m_s} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.20)

Find solutions for s = 1/2, 1, 3/2, 2, \cdots, where m_s \in \{-s, \cdots, s\}. ie. 2 s + 1 possible vectors {\left\lvert {s m_s} \right\rangle} for a given s.

\begin{aligned}s = \frac{1}{{2}} &\implies \gamma = 2 \\ s = 1 &\implies \gamma = 3 \\ s = \frac{3}{2} &\implies \gamma = 4 \end{aligned}

We start with the algebra (mathematically the Lie algebra), and one can compute the Hilbert spaces that are consistent with these algebraic constraints.

We assume that for any type of given particle S is fixed, where this has to do with the nature of the particle.

\begin{aligned}s = \frac{1}{{2}} &\qquad \text{A spin latex 1/2$ particle} \\ s = 1 &\qquad \text{A spin 1 particle} \\ s = \frac{3}{2} &\qquad \text{A spin 3/2 particle}\end{aligned} $

S is fixed once we decide that we are talking about a specific type of particle.

A non-relativistic particle in this framework has two nondynamical quantities. One is the mass m and we now introduce a new invariant, the spin s of the particle.

This has been introduced as a kind of strategy. It is something that we are going to try, and it turns out that it does. This agrees well with experiment.

In 1939 Wigner asked, “what constraints do I get if I constrain the constraints of quantum mechanics with special relativity.” It turns out that in the non-relativistic limit, we get just this.

There’s a subtlety here, because we get into some logical trouble with the photon with a rest mass of zero (m = 0 is certainly allowed as a value of our invariant m above). We can’t stop or slow down a photon, so orbital angular momentum is only a conceptual idea. Really, the orbital angular momentum and the spin angular momentum cannot be separated out for a photon, so talking of a spin 1 particle really means spin as in \mathbf{J}, and not spin as in \mathbf{L}.

Spin 1/2 particles

Reading: See section 26.6 in the text [1].

Let’s start talking about the simplest case. This includes electrons, all leptons (integer spin particles like photons and the weakly interacting W and Z bosons), and quarks.

\begin{aligned}s &= \frac{1}{{2}} \\ m_s &= \pm \frac{1}{{2}}\end{aligned} \hspace{\stretch{1}}(2.22)

states

\begin{aligned}{\left\lvert {s m_s} \right\rangle} = {\left\lvert { \frac{1}{{2}}, \frac{1}{{2}} } \right\rangle},{\left\lvert { \frac{1}{{2}}, -\frac{1}{{2}} } \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.24)

Note there is a convention

\begin{aligned}{\left\lvert { \frac{1}{{2}} \bar{\frac{1}{{2}}} } \right\rangle} &= {\left\lvert { \frac{1}{{2}}, -\frac{1}{{2}} } \right\rangle} \\ {\left\lvert { \frac{1}{{2}} \frac{1}{{2}} } \right\rangle} &= {\left\lvert { \frac{1}{{2}} \frac{1}{{2}} } \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.25)

\begin{aligned}\begin{aligned}S^2 {\left\lvert {\frac{1}{{2}} m_s} \right\rangle} &= \frac{1}{{2}} \left( \frac{1}{{2}} + 1 \right) \hbar^2 {\left\lvert {\frac{1}{{2}} m_s} \right\rangle} \\ &=\frac{3}{4} \hbar^2 {\left\lvert {\frac{1}{{2}} m_s} \right\rangle} \\ \end{aligned}\end{aligned} \hspace{\stretch{1}}(2.27)

\begin{aligned}S_z {\left\lvert {\frac{1}{{2}} m_s} \right\rangle} = m_s \hbar {\left\lvert {\frac{1}{{2}} m_s} \right\rangle} \end{aligned} \hspace{\stretch{1}}(2.28)

For shorthand

\begin{aligned}{\left\lvert { \frac{1}{{2}} \frac{1}{{2}} } \right\rangle} &= {\left\lvert { + } \right\rangle} \\ {\left\lvert { \frac{1}{{2}} \bar{\frac{1}{{2}}} } \right\rangle} &= {\left\lvert { - } \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.29)

\begin{aligned}S^2 \rightarrow \frac{3}{4} \hbar^2 \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.31)

\begin{aligned}S_z \rightarrow \frac{\hbar}{2}\begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.32)

One can easily work out from the commutation relationships that

\begin{aligned}S_x \rightarrow \frac{\hbar}{2}\begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.33)

\begin{aligned}S_y \rightarrow \frac{\hbar}{2}\begin{bmatrix}0 & -i \\ i & 0\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.34)

We’ll start with adding \mathbf{L} into the mix on Wednesday.

References

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

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PHY456H1F: Quantum Mechanics II. Lecture 12 (Taught by Mr. Federico Duque Gomez). WKB Method

Posted by peeterjoot on October 21, 2011

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

Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

WKB (Wentzel-Kramers-Brillouin) Method.

This is covered in section 24 in the text [1]. Also section 8 of [2].

We start with the 1D time independent Schr\”{o}dinger equation

\begin{aligned}-\frac{\hbar^2}{2m} \frac{d^2 U}{dx^2} + V(x) U(x) = E U(x)\end{aligned} \hspace{\stretch{1}}(2.1)

which we can write as

\begin{aligned}\frac{d^2 U}{dx^2} + \frac{2m}{\hbar^2} (E - V(x)) U(x) = 0\end{aligned} \hspace{\stretch{1}}(2.2)

Consider a finite well potential as in figure (\ref{fig:qmTwoL13:qmTwoL12fig1})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL12fig1}
\caption{Finite well potential}
\end{figure}

With

\begin{aligned}k &= \frac{2m (E - V)}{\hbar},\qquad E > V \\ \kappa &= \frac{2m (V - E)}{\hbar}, \qquad V > E,\end{aligned} \hspace{\stretch{1}}(2.3)

we have for a bound state within the well

\begin{aligned}U \propto e^{\pm i k x}\end{aligned} \hspace{\stretch{1}}(2.5)

and for that state outside the well

\begin{aligned}U \propto e^{\pm \kappa x}\end{aligned} \hspace{\stretch{1}}(2.6)

In general we can hope for something similar. Let’s look for that something, but allow the constants k and \kappa to be functions of position

\begin{aligned}k^2(x) &= \frac{2m (E - V(x))}{\hbar},\qquad E > V \\ \kappa^2(x) &= \frac{2m (V(x) - E)}{\hbar}, \qquad V > E.\end{aligned} \hspace{\stretch{1}}(2.7)

In terms of k Schr\”{o}dinger’s equation is just

\begin{aligned}\frac{d^2 U(x)}{dx^2} + k^2(x) U(x) = 0.\end{aligned} \hspace{\stretch{1}}(2.9)

We use the trial solution

\begin{aligned}U(x) = A e^{i \phi(x)},\end{aligned} \hspace{\stretch{1}}(2.10)

allowing \phi(x) to be complex

\begin{aligned}\phi(x) = \phi_R(x) + i \phi_I(x).\end{aligned} \hspace{\stretch{1}}(2.11)

We need second derivatives

\begin{aligned}(e^{i \phi})'' &=(i \phi' e^{i \phi})' \\ &=(i \phi')^2 e^{i \phi} + i \phi'' e^{i \phi},\end{aligned}

and plug back into our Schr\”{o}dinger equation to obtain

\begin{aligned}- (\phi'(x))^2 + i \phi''(x) + k^2(x) = 0.\end{aligned} \hspace{\stretch{1}}(2.12)

For the first round of approximation we assume

\begin{aligned}\phi''(x) \approx 0,\end{aligned} \hspace{\stretch{1}}(2.13)

and obtain

\begin{aligned}(\phi'(x))^2 = k^2(x),\end{aligned} \hspace{\stretch{1}}(2.14)

or

\begin{aligned}\phi'(x) = \pm k(x).\end{aligned} \hspace{\stretch{1}}(2.15)

A second round of approximation we use 2.15 and obtain

\begin{aligned}\phi''(x) = \pm k'(x)\end{aligned} \hspace{\stretch{1}}(2.16)

Plugging back into 2.12 we have

\begin{aligned}-(\phi'(x))^2 \pm i k'(x) + k^2(x) = 0,\end{aligned} \hspace{\stretch{1}}(2.17)

or

\begin{aligned}\begin{aligned}\phi'(x) &= \pm \sqrt{ \pm i k'(x) + k^2(x) } \\ &= \pm k(x) \sqrt{ 1 \pm i \frac{k'(x)}{k^2(x)} } .\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.18)

If k' is small compared to k^2

\begin{aligned}\frac{k'(x)}{k^2(x)} \ll 1, \end{aligned} \hspace{\stretch{1}}(2.19)

we have

\begin{aligned}\phi'(x) = \pm k(x) \pm i \frac{k'(x)}{2 k(x)} \end{aligned} \hspace{\stretch{1}}(2.20)

Integrating

\begin{aligned}\phi(x) &= \pm \int dx k(x) \pm i \int dx \frac{k'(x)}{2 k(x)} + \text{const} \\ &= \pm \int dx k(x) \pm i \frac{1}{{2}} \ln k(x) + \text{const} \end{aligned}

Going back to our wavefunction, if E > V(x) we have

\begin{aligned}U(x) &\sim A e^{i \phi(x)} \\ &= \exp \left(i\left( \pm \int dx k(x) \pm i \frac{1}{{2}} \ln k(x) + \text{const} \right)\right) \\ &\sim \exp \left(i\left( \pm \int dx k(x) \pm i \frac{1}{{2}} \ln k(x) \right)\right) \\ &= e^{\pm i \int dx k(x)} e^{\mp \frac{1}{{2}} \ln k(x)} \\ \end{aligned}

or

\begin{aligned}U(x) \propto \frac{1}{{\sqrt{k(x)}}} e^{\pm i \int dx k(x)} \end{aligned} \hspace{\stretch{1}}(2.21)

FIXME: Question: the \pm on the real exponential got absorbed here, but would not U(x) \propto \sqrt{k(x)} e^{\pm i \int dx k(x)} also be a solution? If so, why is that one excluded?

Similarly for the E < V(x) case we can find

\begin{aligned}U(x) \propto \frac{1}{{\sqrt{\kappa(x)}}} e^{\pm i \int dx \kappa(x)}.\end{aligned} \hspace{\stretch{1}}(2.22)

Validity
\begin{enumerate}
\item V(x) changes very slowly \implies k'(x) small, and k(x) = \sqrt{2 m (E - V(x))}/\hbar.
\item E very far away from the potential {\left\lvert{(E - V(x))/V(x)}\right\rvert} \gg 1.
\end{enumerate}

Examples

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL12fig2}
\caption{Example of a general potential}
\end{figure}

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL12fig3}
\caption{Turning points where WKB won’t work}
\end{figure}

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL12fig4}
\caption{Diagram for patching method discussion}
\end{figure}

WKB won’t work at the turning points in this figure since our main assumption was that

\begin{aligned}{\left\lvert{\frac{k'(x)}{k^2(x)}}\right\rvert} \ll 1\end{aligned} \hspace{\stretch{1}}(3.23)

so we get into trouble where k(x) \sim 0. There are some methods for dealing with this. Our text as well as Griffiths give some examples, but they require Bessel functions and more complex mathematics.

The idea is that one finds the WKB solution in the regions of validity, and then looks for a polynomial solution in the patching region where we are closer to the turning point, probably requiring lookup of various special functions.

This power series method is also outlined in [3], where solutions to connect the regions are expressed in terms of Airy functions.

References

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

[2] D.J. Griffiths. Introduction to quantum mechanics, volume 1. Pearson Prentice Hall, 2005.

[3] Wikipedia. Wkb approximation — wikipedia, the free encyclopedia, 2011. [Online; accessed 19-October-2011]. http://en.wikipedia.org/w/index.php?title=WKB_approximation&oldid=453833635.

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PHY456H1F: Quantum Mechanics II. Lecture 11 (Taught by Prof J.E. Sipe). Spin and Spinors

Posted by peeterjoot on October 17, 2011

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

Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

Generators.

Covered in section 26 of the text [1].

Example: Time translation

\begin{aligned}{\lvert {\psi(t)} \rangle} = e^{-i H t/\hbar} {\lvert {\psi(0)} \rangle} .\end{aligned} \hspace{\stretch{1}}(2.1)

The Hamiltonian “generates” evolution (or translation) in time.

Example: Spatial translation

\begin{aligned}{\lvert {\mathbf{r} + \mathbf{a}} \rangle} = e^{-i \mathbf{a} \cdot \mathbf{P}/\hbar} {\lvert {\mathbf{r}} \rangle}.\end{aligned} \hspace{\stretch{1}}(2.2)

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL11fig1}
\caption{Vector translation.}

\end{figure}

\mathbf{P} is the operator that generates translations. Written out, we have

\begin{aligned}\begin{aligned}e^{-i \mathbf{a} \cdot \mathbf{P}/\hbar} &= e^{- i (a_x P_x + a_y P_y + a_z P_z)/\hbar} \\ &= e^{- i a_x P_x/\hbar}e^{- i a_y P_y/\hbar}e^{- i a_z P_z/\hbar},\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.3)

where the factorization was possible because P_x, P_y, and P_z commute

\begin{aligned}\left[{P_i},{P_j}\right] = 0,\end{aligned} \hspace{\stretch{1}}(2.4)

for any i, j (including i = i as I dumbly questioned in class … this is a commutator, so \left[{P_i},{P_j}\right] = P_i P_i - P_i P_i = 0).

The fact that the P_i commute means that successive translations can be done in any orderr and have the same result.

In class we were rewarded with a graphic demo of translation component commutation as Professor Sipe pulled a giant wood carving of a cat (or tiger?) out from beside the desk and proceeded to translate it around on the desk in two different orders, with the cat ending up in the same place each time.

Exponential commutation.

Note that in general

\begin{aligned}e^{A + B} \ne e^A e^B,\end{aligned} \hspace{\stretch{1}}(2.5)

unless \left[{A},{B}\right] = 0. To show this one can compare

\begin{aligned}\begin{aligned}e^{A + B} &= 1 + A + B + \frac{1}{{2}}(A + B)^2 + \cdots \\ &= 1 + A + B + \frac{1}{{2}}(A^2 + A B + BA + B^2) + \cdots \\ \end{aligned}\end{aligned} \hspace{\stretch{1}}(2.6)

and

\begin{aligned}\begin{aligned}e^A e^B &= \left(1 + A + \frac{1}{{2}}A^2 + \cdots\right)\left(1 + B + \frac{1}{{2}}B^2 + \cdots\right) \\ &= 1 + A + B + \frac{1}{{2}}( A^2 + 2 A B + B^2 ) + \cdots\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.7)

Comparing the second order (for example) we see that we must have for equality

\begin{aligned}A B + B A = 2 A B,\end{aligned} \hspace{\stretch{1}}(2.8)

or

\begin{aligned}B A = A B,\end{aligned} \hspace{\stretch{1}}(2.9)

or

\begin{aligned}\left[{A},{B}\right] = 0\end{aligned} \hspace{\stretch{1}}(2.10)

Translating a ket

If we consider the quantity

\begin{aligned}e^{-i \mathbf{a} \cdot \mathbf{P}/\hbar} {\lvert {\psi} \rangle} = {\lvert {\psi'} \rangle} ,\end{aligned} \hspace{\stretch{1}}(2.11)

does this ket “translated” by \mathbf{a} make any sense? The vector \mathbf{a} lives in a 3D space and our ket {\lvert {\psi} \rangle} lives in Hilbert space. A quantity like this deserves some careful thought and is the subject of some such thought in the Interpretations of Quantum mechanics course. For now, we can think of the operator and ket as a “gadget” that prepares a state.

A student in class pointed out that {\lvert {\psi} \rangle} can be dependent on many degress of freedom, for example, the positions of eight different particles. This translation gadget in such a case acts on the whole kit and kaboodle.

Now consider the matrix element

\begin{aligned}\left\langle{\mathbf{r}} \vert {{\psi'}}\right\rangle = {\langle {\mathbf{r}} \rvert} e^{-i \mathbf{a} \cdot \mathbf{P}/\hbar} {\lvert {\psi} \rangle}.\end{aligned} \hspace{\stretch{1}}(2.12)

Note that

\begin{aligned}{\langle {\mathbf{r}} \rvert} e^{-i \mathbf{a} \cdot \mathbf{P}/\hbar} &= \left( e^{i \mathbf{a} \cdot \mathbf{P}/\hbar} {\lvert {\mathbf{r}} \rangle} \right)^\dagger \\ &= \left( {\lvert {\mathbf{r} - \mathbf{a}} \rangle} \right)^\dagger,\end{aligned}

so

\begin{aligned}\left\langle{\mathbf{r}} \vert {{\psi'}}\right\rangle = \left\langle{{\mathbf{r} -\mathbf{a}}} \vert {{\psi}}\right\rangle,\end{aligned} \hspace{\stretch{1}}(2.13)

or

\begin{aligned}\psi'(\mathbf{r}) = \psi(\mathbf{r} - \mathbf{a})\end{aligned} \hspace{\stretch{1}}(2.14)

This is what we expect of a translated function, as illustrated in figure (\ref{fig:qmTwoL11:qmTwoL11fig2})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL11fig2}
\caption{Active spatial translation.}
\end{figure}

Example: Spatial rotation

We’ve been introduced to the angular momentum operator

\begin{aligned}\mathbf{L} = \mathbf{R} \times \mathbf{P},\end{aligned} \hspace{\stretch{1}}(2.15)

where

\begin{aligned}L_x &= Y P_z - Z P_y \\ L_y &= Z P_x - X P_z \\ L_z &= X P_y - Y P_x.\end{aligned} \hspace{\stretch{1}}(2.16)

We also found that

\begin{aligned}\left[{L_i},{L_j}\right] = i \hbar \sum_k \epsilon_{ijk} L_k.\end{aligned} \hspace{\stretch{1}}(2.19)

These non-zero commutators show that the components of angular momentum do not commute.

Define

\begin{aligned}{\lvert {\mathcal{R}(\mathbf{r})} \rangle} = e^{-i \theta \hat{\mathbf{n}} \cdot \mathbf{L}/\hbar}{\lvert {\mathbf{r}} \rangle} .\end{aligned} \hspace{\stretch{1}}(2.20)

This is the vecvtor that we get by actively rotating the vector \mathbf{r} by an angule \theta counterclockwise about \hat{\mathbf{n}}, as in figure (\ref{fig:qmTwoL11:qmTwoL11fig3})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL11fig3}
\caption{Active vector rotations}
\end{figure}

An active rotation rotates the vector, leaving the coordinate system fixed, whereas a passive rotation is one for which the coordinate system is rotated, and the vector is left fixed.

Note that rotations do not commute. Suppose that we have a pair of rotations as in figure (\ref{fig:qmTwoL11:qmTwoL11fig4})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL11fig4}
\caption{A example pair of non-commuting rotations.}
\end{figure}

Again, we get the graphic demo, with Professor Sipe rotating the big wooden cat sculpture. Did he bring that in to class just to make this point (too bad I missed the first couple minutes of the lecture).

Rather amusingly, he points out that most things in life do not commute. We get much different results if we apply the operations of putting water into the teapot and turning on the stove in different orders.

Rotating a ket

With a rotation gadget

\begin{aligned}{\lvert {\psi'} \rangle} = e^{-i \theta \hat{\mathbf{n}} \cdot \mathbf{L}/\hbar }{\lvert {\psi} \rangle},\end{aligned} \hspace{\stretch{1}}(2.21)

we can form the matrix element

\begin{aligned}\left\langle{\mathbf{r}} \vert {{\psi'}}\right\rangle = {\langle {\mathbf{r}} \rvert} e^{-i \theta \hat{\mathbf{n}} \cdot \mathbf{L}/\hbar }{\lvert {\psi} \rangle}.\end{aligned} \hspace{\stretch{1}}(2.22)

In this we have

\begin{aligned}{\langle {\mathbf{r}} \rvert} e^{-i \theta \hat{\mathbf{n}} \cdot \mathbf{L}/\hbar }&=\left( e^{i \theta \hat{\mathbf{n}} \cdot \mathbf{L}/\hbar } {\lvert {\mathbf{r}} \rangle} \right)^\dagger \\ &=\left( {\lvert {\mathcal{R}^{-1}(\mathbf{r}) } \rangle} \right)^\dagger,\end{aligned}

so

\begin{aligned}\left\langle{\mathbf{r}} \vert {{\psi'}}\right\rangle = \left\langle{{\mathcal{R}^{-1}(\mathbf{r}) }} \vert {{\psi'}}\right\rangle,\end{aligned} \hspace{\stretch{1}}(2.23)

or

\begin{aligned}\psi'(\mathbf{r}) = \psi( \mathcal{R}^{-1}(\mathbf{r}) )\end{aligned} \hspace{\stretch{1}}(2.24)

Generalizations.

Recall what you did last year, where H, \mathbf{P}, and \mathbf{L} were defined mechaniccally. We found

\begin{itemize}
\item H generates time evolution (or translation in time).
\item \mathbf{P} generates spatial translation.
\item \mathbf{L} generates spatial rotation.
\end{itemize}

For our mechanical definitions we have

\begin{aligned}\left[{P_i},{P_j}\right] = 0,\end{aligned} \hspace{\stretch{1}}(3.25)

and

\begin{aligned}\left[{L_i},{L_j}\right] = i \hbar \sum_k \epsilon_{ijk} L_k.\end{aligned} \hspace{\stretch{1}}(3.26)

These are the relations that show us the way translations and rotations combine. We want to move up to a higher plane, a new level of abstraction. To do so we define H as the operator that generates time evolution. If we have a theory that covers the behaviour of how anything evolves in time, H encodes the rules for this time evolution.

Define \mathbf{P} as the operator that generates translations in space.

Define \mathbf{J} as the operator that generates rotations in space.

In order that these match expectations, we require

\begin{aligned}\left[{P_i},{P_j}\right] = 0,\end{aligned} \hspace{\stretch{1}}(3.27)

and

\begin{aligned}\left[{J_i},{J_j}\right] = i \hbar \sum_k \epsilon_{ijk} J_k.\end{aligned} \hspace{\stretch{1}}(3.28)

In the simple theory of a spinless particle we have

\begin{aligned}\mathbf{J} \equiv \mathbf{L} = \mathbf{R} \times \mathbf{P}.\end{aligned} \hspace{\stretch{1}}(3.29)

We actually need a generalization of this since this is, in fact, not good enought, even for low energy physics.

Many component wave functions.

We are free to construct tuples of spatial vector functions like

\begin{aligned}\begin{bmatrix}\Psi_I(\mathbf{r}, t) \\ \Psi_{II}(\mathbf{r}, t)\end{bmatrix},\end{aligned} \hspace{\stretch{1}}(3.30)

or

\begin{aligned}\begin{bmatrix}\Psi_I(\mathbf{r}, t) \\ \Psi_{II}(\mathbf{r}, t) \\ \Psi_{III}(\mathbf{r}, t)\end{bmatrix},\end{aligned} \hspace{\stretch{1}}(3.31)

etc.

We will see that these behave qualitatively different than one component wave functions. We also don’t have to be considering multiple particle wave functions, but just one particle that requires three functions in \mathbb{R}^{3} to describe it (ie: we are moving in on spin).

Question: Do these live in the same vector space?
Answer: We will get to this.

A classical analogy.

“There’s only bad analogies, since if the are good they’d be describing the same thing. We can however, produce some useful bad analogies”

\begin{enumerate}
\item A temperature field

\begin{aligned}T(\mathbf{r})\end{aligned} \hspace{\stretch{1}}(3.32)

\item Electric field

\begin{aligned}\begin{bmatrix}E_x(\mathbf{r}) \\ E_y(\mathbf{r}) \\ E_z(\mathbf{r}) \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.33)

\end{enumerate}

These behave in a much different way. If we rotate a scalar field like T(\mathbf{r}) as in figure (\ref{fig:qmTwoL11:qmTwoL11fig5})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL11fig5}
\caption{Rotated temperature (scalar) field}
\end{figure}

Suppose we have a temperature field generated by, say, a match. Rotating the match above, we have

\begin{aligned}T'(\mathbf{r}) = T(\mathcal{R}^{-1}(\mathbf{r})).\end{aligned} \hspace{\stretch{1}}(3.34)

Compare this to the rotation of an electric field, perhaps one produced by a capacitor, as in figure (\ref{fig:qmTwoL11:qmTwoL11fig6})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL11fig6}
\caption{Rotating a capacitance electric field}
\end{figure}

Is it true that we have

\begin{aligned}\begin{bmatrix}E_x(\mathbf{r}) \\ E_y(\mathbf{r}) \\ E_z(\mathbf{r}) \end{bmatrix}\stackrel{?}{=}\begin{bmatrix}E_x(\mathcal{R}^{-1}(\mathbf{r})) \\ E_y(\mathcal{R}^{-1}(\mathbf{r})) \\ E_z(\mathcal{R}^{-1}(\mathbf{r})) \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.35)

No. Because the components get mixed as well as the positions at which those components are evaluated.

We will work with many component wave functions, some of which will behave like vectors, and will have to develope the methods and language to tackle this.

References

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

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PHY456H1F: Quantum Mechanics II. Lecture 10 (Taught by Prof J.E. Sipe). Fermi’s golden rule (cont.)

Posted by peeterjoot on October 14, 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.)]

Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

Recap. Where we got to on Fermi’s golden rule.

We are continuing on the topic of Fermi golden rule, as also covered in section 17.2 of the text [1]. Utilizing a wave train with peaks separation \Delta t = 2\pi/\omega_0, zero before some initial time (\ref{fig:qmTwoL10:unitStepSine}).

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{unitStepSine}
\caption{Sine only after an initial time.}
\end{figure}

Pertubing a state in the ith energy level, and looking at the states for the mth energy level as illustrated in figure (\ref{fig:qmTwoL10:2})

\begin{figure}[htp]
\centering
\def\svgwidth{0.3\columnwidth}
\caption{Perturbation from i to mth energy levels}
\end{figure}

Our matrix element was

\begin{aligned}\begin{aligned}H_{mi}'(t) &= 2 A_mi \sin(\omega_0 t) \theta(t) \\ &= i A_{mi} ( e^{-i \omega_0 t} - e^{i \omega_0 t} ) \theta(t),\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.1)

and we found

\begin{aligned}c_m^{(1)}(t) = \frac{A_{mi}}{\hbar} \int_0^t dt' \left( e^{i (\omega_{mi} - \omega_0) t'} -e^{i (\omega_{mi} + \omega_0) t'} \right),\end{aligned} \hspace{\stretch{1}}(2.2)

and argued that

\begin{aligned}{\left\lvert{ c_m^{(1)}(t) }\right\rvert}^2 \sim \left( \frac{A_{mi}}{\hbar} \right)^2 t^2 + \cdots\end{aligned} \hspace{\stretch{1}}(2.3)

where \omega_0 t \gg 1 for \omega_{mi} \sim \pm \omega_0.

We can also just integrate 2.2 directly

\begin{aligned}\begin{aligned}c_m^{(1)}(t) &= \frac{A_{mi}}{\hbar} \left( \frac{e^{i (\omega_{mi} - \omega_0) t} - 1}{ i (\omega_{mi} - \omega_0) }-\frac{e^{i (\omega_{mi} + \omega_0) t} - 1}{ i (\omega_{mi} + \omega_0) }\right) \\ &\equiv A_{mi}(\omega_0, t) - A_{mi}(-\omega_0, t),\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.4)

where

\begin{aligned}A_{mi}(\omega_0, t) =\frac{A_{mi}}{\hbar} \frac{e^{i (\omega_{mi} - \omega_0) t} - 1}{ i (\omega_{mi} - \omega_0) }\end{aligned} \hspace{\stretch{1}}(2.5)

Factoring out the phase term, we have

\begin{aligned}A_{mi}(\omega_0, t) =\frac{2 A_{mi}}{\hbar} e^{i (\omega_{mi} - \omega_0) t/2}\frac{\sin((\omega_{mi} - \omega_0) t/2)}{ (\omega_{mi} - \omega_0) }\end{aligned} \hspace{\stretch{1}}(2.6)

We we will have two lobes, centered on \pm \omega_0, as illustrated in figure (\ref{fig:qmTwoL10:qmTwoL10fig3})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL10fig3}
\caption{Two sinc lobes}
\end{figure}

Fermi’s Golden rule.

Fermi’s Golden rule applies to a continuum of states (there are other forms of Fermi’s golden rule, but this is the one we will talk about, and is the one in the book). One example is the ionized states of an atom, where the energy level separation becomes so small that we can consider it continuous

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumEnergyLevels}
\caption{Continuum of energy levels for ionized states of an atom.}
\end{figure}

Another example are the unbound states in a semiconductor well as illustrated in figure (\ref{fig:qmTwoL10:semiConductorWell})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{semiConductorWell}
\caption{Semi-conductor well}
\end{figure}

Note that we can have reflection from the well even in the continuum states where we would have no such reflection classically. However, with enough energy, states are approximately plane waves. In one dimension

\begin{aligned}\begin{aligned}\left\langle{{x}} \vert {{\psi_p}}\right\rangle &\approx \frac{e^{i p x/\hbar}}{\sqrt{2 \pi \hbar}} \\ \left\langle{{\psi_p}} \vert {{\psi_p'}}\right\rangle &= \delta(p - p')\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.7)

or in 3d

\begin{aligned}\begin{aligned}\left\langle{\mathbf{r}} \vert {{\psi_{\mathbf{p}}}}\right\rangle &\approx \frac{e^{i \mathbf{p} \cdot \mathbf{r}/\hbar}}{(2 \pi \hbar)^{3/2} } \\ \left\langle{{\psi_{\mathbf{p}}}} \vert {{\psi_{\mathbf{p}'}}}\right\rangle &= \delta^3(\mathbf{p} - \mathbf{p}')\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.8)

Let’s consider the 1d model for the quantum well in more detail. Including both discrete and continuous states we have

\begin{aligned}{\lvert {\psi(t)} \rangle} = \sum_n c_n(t) e^{-i \omega_n t} {\lvert {\psi_n} \rangle} + \int dp c_p(t) e^{-i \omega_p t} {\lvert {\psi_p} \rangle} \end{aligned} \hspace{\stretch{1}}(3.9)

Imagine at t=0 that the wave function started in some discrete state, and look at the probability that we “kick the electron out of the well”. Calculate

\begin{aligned}\mathcal{P} = \int dp {\left\lvert{c_p^{(1)}(t)}\right\rvert}^2\end{aligned} \hspace{\stretch{1}}(3.10)

Now, we assume that our matrix element has the following form

\begin{aligned}H_{pi}'(t) = \left( \bar{A}_{pi} e^{-i \omega_0 t}+\bar{B}_{pi} e^{i \omega_0 t} \right) \theta(t)\end{aligned} \hspace{\stretch{1}}(3.11)

generalizing the wave train matrix element that we had previously

\begin{aligned}H_{mi}'(t) = i A_{mi} \left( e^{-i \omega_0 t}- e^{i \omega_0 t} \right) \theta(t)\end{aligned} \hspace{\stretch{1}}(3.12)

Doing the perturbation we have

\begin{aligned}\mathcal{P} = \int dp {\left\lvert{A_{pi}(\omega_0, t)+ B_{pi}(-\omega_0, t)}\right\rvert}^2\end{aligned} \hspace{\stretch{1}}(3.13)

where

\begin{aligned}A_{pi}(\omega_0, t) = \frac{2 \bar{A}_{pi}}{i \hbar } e^{i (\omega_{pi} - \omega_0) t/2}\frac{\sin((\omega_{pi} - \omega_0) t/2)}{\omega_{pi} - \omega_0}\end{aligned} \hspace{\stretch{1}}(3.14)

which is peaked at \omega_{pi} = \omega_0, and

\begin{aligned}B_{pi}(\omega_0, t) = \frac{2 \bar{B}_{pi}}{i \hbar } e^{i (\omega_{pi} + \omega_0) t/2}\frac{\sin((\omega_{pi} + \omega_0) t/2)}{\omega_{pi} + \omega_0}\end{aligned} \hspace{\stretch{1}}(3.15)

which is peaked at \omega_{pi} = -\omega_0.

FIXME: show that this is the perturbation result.

In 3.13 at t \gg 0 the only significant contribution is from the A portion as illustrated in figure (\ref{fig:qmTwoL10:qmTwoL10fig6}) where we are down in the wiggles of A_{pi}.

\begin{figure}[htp]
\centering

\includegraphics[totalheight=0.4\textheight]{qmTwoL10fig6}

\end{figure}

Our probability to find the particle in the continuum range is now approximately

\begin{aligned}\mathcal{P} = \int dp {\left\lvert{A_{pi}(\omega_0, t)}\right\rvert}^2\end{aligned} \hspace{\stretch{1}}(3.16)

With

\begin{aligned}\omega_{pi} - \omega_0 = \frac{1}{{\hbar}}\left( \frac{p^2}{2m} - E_i \right) - \omega_0,\end{aligned} \hspace{\stretch{1}}(3.17)

define \bar{p} so that

\begin{aligned}0 = \frac{1}{{\hbar}}\left( \frac{\bar{p}^2}{2m} - E_i \right) - \omega_0.\end{aligned} \hspace{\stretch{1}}(3.18)

In momentum space, we know have the sinc functions peaked at \pm \bar{p} as in figure (\ref{fig:qmTwoL10:qmTwoL10fig7})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL10fig7}
\caption{Momentum space view}
\end{figure}

The probability that the electron goes to the right is then

\begin{aligned}\begin{aligned}\mathcal{P}_{+} &= \int_0^\infty dp {\left\lvert{c_p^{(1)}(t)}\right\rvert}^2 \\ &=\int_0^\infty dp {\left\lvert{\bar{A}_{pi}}\right\rvert}^2 \frac{\sin^2\left( (\omega_{pi} - \omega_0) t/2 \right)}{\left( \omega_{pi} - \omega_0 \right)^2},\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.19)

with

\begin{aligned}\omega_{pi} = \frac{1}{{\hbar}}\left( \frac{p^2}{2m} - E_i\right)\end{aligned} \hspace{\stretch{1}}(3.20)

we have with a change of variables

\begin{aligned}\mathcal{P}_{+} =\frac{4}{\hbar^2}\int_{-E_i/\hbar}^\infty d\omega_{pi}{\left\lvert{\bar{A}_{pi}}\right\rvert}^2 \frac{dp}{d\omega_{pi}}\frac{\sin^2\left( (\omega_{pi} - \omega_0) t/2 \right)}{\left( \omega_{pi} - \omega_0 \right)^2}.\end{aligned} \hspace{\stretch{1}}(3.21)

Now suppose we have t small enough so that \mathcal{P}_{+} \ll 1 and t large enough so

\begin{aligned}{\left\lvert{\bar{A}_{pi}}\right\rvert}^2 \frac{dp}{d\omega_{pi}}\end{aligned} \hspace{\stretch{1}}(3.22)

is roughly constant over \Delta \omega. This is a sort of “Goldilocks condition”, a time that can’t be too small, and can’t be too large, but instead has to be “just right”. Given such a condition

\begin{aligned}\mathcal{P}_{+} =\frac{4}{\hbar^2}{\left\lvert{\bar{A}_{pi}}\right\rvert}^2 \frac{dp}{d\omega_{pi}}\int_{-E_i/\hbar}^\infty d\omega_{pi}\frac{\sin^2\left( (\omega_{pi} - \omega_0) t/2 \right)}{\left( \omega_{pi} - \omega_0 \right)^2},\end{aligned} \hspace{\stretch{1}}(3.23)

where we can pull stuff out of the integral since the main contribution is at the peak. Provided \bar{p} is large enough, then

\begin{aligned}\begin{aligned}\int_{-E_i/\hbar}^\infty d\omega_{pi}\frac{\sin^2\left( (\omega_{pi} - \omega_0) t/2 \right)}{\left( \omega_{pi} - \omega_0 \right)^2}&\approx \int_{-\infty}^\infty d\omega_{pi}\frac{\sin^2\left( (\omega_{pi} - \omega_0) t/2 \right)}{\left( \omega_{pi} - \omega_0 \right)^2} \\ &=\frac{t}{2} \pi,\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.24)

leaving the probability of the electron with a going right continuum state as

\begin{aligned}\mathcal{P}_{+} =\frac{4}{\hbar^2}\underbrace{{\left\lvert{\bar{A}_{pi}}\right\rvert}^2 }{\text{matrix element}}\underbrace{{\left.{{\frac{dp}{d\omega_{pi}}}}\right\vert}_{{\bar{p}}}}{\text{density of states}}\frac{t}{2} \pi.\end{aligned} \hspace{\stretch{1}}(3.25)

The dp/d\omega_{pi} is something like “how many continuous states are associated with a transition from a discrete frequency interval.”

We can also get this formally from 3.23 with

\begin{aligned}\frac{\sin^2\left( (\omega_{pi} - \omega_0) t/2 \right)}{\left( \omega_{pi} - \omega_0 \right)^2}\rightarrow \frac{t}{2} \pi \delta(\omega_{pi} - \omega_0),\end{aligned} \hspace{\stretch{1}}(3.26)

so

\begin{aligned}\begin{aligned}c_p^{(1)}(t) &\rightarrow \frac{2 \pi t}{\hbar^2} {\left\lvert{\bar{A}_{pi}}\right\rvert}^2 \delta(\omega_{pi} - \omega_0) \\ &=\frac{2 \pi t}{\hbar} {\left\lvert{\bar{A}_{pi}}\right\rvert}^2 \delta(E_{pi} - \hbar \omega_0)\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.27)

where \delta(ax) = \delta(x)/{\left\lvert{a}\right\rvert} has been used to pull in a factor of \hbar into the delta.

The ratio of the coefficient to time is then

\begin{aligned}\frac{c_p^{(1)}(t) }{t}=\frac{2 \pi}{\hbar} {\left\lvert{\bar{A}_{pi}}\right\rvert}^2 \delta(E_{pi} - \hbar \omega_0).\end{aligned} \hspace{\stretch{1}}(3.28)

or “between friends”

\begin{aligned}''\frac{dc_p^{(1)}(t) }{dt}''=\frac{2 \pi}{\hbar} {\left\lvert{\bar{A}_{pi}}\right\rvert}^2 \delta(E_{pi} - \hbar \omega_0),\end{aligned} \hspace{\stretch{1}}(3.29)

roughly speaking we have a “rate” or transitions from the discrete into the continuous. Here “rate” is in quotes since it doesn’t hold for small t.

This has been worked out for \mathcal{P}_{+}. This can also be done for \mathcal{P}_{-}, the probability that the electron will end up in a left trending continuum state.

While the above is not a formal derivation, but illustrates the form of what is called Fermi’s golden rule. Namely that such a rate has the structure

\begin{aligned}\frac{2 \pi}{\hbar} \times (\text{matrix element})^2 \times \text{energy conservation}\end{aligned} \hspace{\stretch{1}}(3.30)

References

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

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