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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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