Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

Posts Tagged ‘WKB’

An updated compilation of notes, for ‘PHY452H1S Basic Statistical Mechanics’, Taught by Prof. Arun Paramekanti

Posted by peeterjoot on March 3, 2013

In A compilation of notes, so far, for ‘PHY452H1S Basic Statistical Mechanics’ I posted a link this compilation of statistical mechanics course notes.

That compilation now all of the following too (no further updates will be made to any of these) :

February 28, 2013 Rotation of diatomic molecules

February 28, 2013 Helmholtz free energy

February 26, 2013 Statistical and thermodynamic connection

February 24, 2013 Ideal gas

February 16, 2013 One dimensional well problem from Pathria chapter II

February 15, 2013 1D pendulum problem in phase space

February 14, 2013 Continuing review of thermodynamics

February 13, 2013 Lightning review of thermodynamics

February 11, 2013 Cartesian to spherical change of variables in 3d phase space

February 10, 2013 n SHO particle phase space volume

February 10, 2013 Change of variables in 2d phase space

February 10, 2013 Some problems from Kittel chapter 3

February 07, 2013 Midterm review, thermodynamics

February 06, 2013 Limit of unfair coin distribution, the hard way

February 05, 2013 Ideal gas and SHO phase space volume calculations

February 03, 2013 One dimensional random walk

February 02, 2013 1D SHO phase space

February 02, 2013 Application of the central limit theorem to a product of random vars

January 31, 2013 Liouville’s theorem questions on density and current

January 30, 2013 State counting

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One dimensional well problem from Pathria chapter II

Posted by peeterjoot on February 16, 2013

[Click here for a PDF of this post with nicer formatting]

Problem 2.5 [2] asks to show that

\begin{aligned}\oint p dq = \left( { n + \frac{1}{{2}} } \right) h,\end{aligned} \hspace{\stretch{1}}(1.0.1)

provided the particle’s potential is such that

\begin{aligned}m \hbar \left\lvert { \frac{dV}{dq} } \right\rvert \ll \left( { m ( E - V ) } \right)^{3/2}.\end{aligned} \hspace{\stretch{1}}(1.0.2)

I took a guess that this was actually the WKB condition

\begin{aligned}\frac{k'}{k^2} \ll 1,\end{aligned} \hspace{\stretch{1}}(1.0.3)

where the WKB solution was of the form

\begin{aligned}k^2(q) = 2 m (E - V(q))/\hbar^2\end{aligned} \hspace{\stretch{1}}(1.0.4a)

\begin{aligned}\psi(q) = \frac{1}{{\sqrt{k}}} e^{\pm i \int k(q) dq}.\end{aligned} \hspace{\stretch{1}}(1.0.4b)

The WKB validity condition is

\begin{aligned}1 \gg \frac{-2 m V'}{\hbar} \frac{1}{{2}} \frac{1}{{\sqrt{2 m (E - V)}}} \frac{\hbar^2}{2 m(E - V)}\end{aligned} \hspace{\stretch{1}}(1.0.5)

or

\begin{aligned}m \hbar \left\lvert {V'} \right\rvert \ll \left( {2 m (E - V)} \right)^{3/2}.\end{aligned} \hspace{\stretch{1}}(1.0.6)

This differs by a factor of 2 \sqrt{2} from the constraint specified in the problem, but I’m guessing that constant factors of that sort have just been dropped.

Even after figuring out that this question was referring to WKB, I didn’t know what to make of the oriented integral \int p dq. With p being an operator in the QM context, what did this even mean. I found the answer in [1] section 12.12. Here p just means

\begin{aligned}p(q) = \hbar k(q),\end{aligned} \hspace{\stretch{1}}(1.0.7)

where k(q) is given by eq. 1.0.4a. The rest of the problem can also be found there and relies on the WKB connection formulas, which aren’t derived in any text that I own. Quoting results based on other results that I don’t know the origin of it’s worthwhile, so that’s as far as I’ll attempt this question (but do plan to eventually look up and understand those WKB connection formulas, and then see how they can be applied in a problem like this).

References

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

[2] RK Pathria. Statistical mechanics. Butterworth Heinemann, Oxford, UK, 1996.

Posted in Math and Physics Learning. | Tagged: , , , , , | 2 Comments »

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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