Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

Posts Tagged ‘degeneracy’

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.)]


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})
\caption{Arbitrary potential in an infinite well.}

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)


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


\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}


\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)


\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)


\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)


\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)


\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.


\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})

\caption{Energy level splitting}

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)


\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)


\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)


\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)


\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}


\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)


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

Posted in Math and Physics Learning. | Tagged: , , , , | Leave a Comment »

Time independent perturbation theory with degeneracy

Posted by peeterjoot on September 30, 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.)]

Time independent perturbation with degeneracy.

In class it was claimed that if we repeated the derivation of the first order pertubation with degenerate states, then we’d get into (divide by zero) trouble if the state we were perturbing had degeneracy. Here I alter the previous derivation to show this explicitly.

The setup

Like the non-degenerate case, we are covering the time independent perturbation methods from section 16.1 of the text [1].

We start with a known Hamiltonian H_0, and alter it with the addition of a “small” perturbation

\begin{aligned}H = H_0 + \lambda H', \qquad \lambda \in [0,1]\end{aligned} \hspace{\stretch{1}}(1.1)

For the original operator, we assume that a complete set of eigenvectors and eigenkets is known

\begin{aligned}H_0 {\lvert {{\psi_{s \alpha}}^{(0)}} \rangle} = {E_s}^{(0)} {\lvert {{\psi_{s \alpha}}^{(0)}} \rangle}\end{aligned} \hspace{\stretch{1}}(1.2)

We seek the perturbed eigensolution

\begin{aligned}H {\lvert {\psi_{s \alpha}} \rangle} = E_{s \alpha} {\lvert {\psi_{s \alpha}} \rangle}\end{aligned} \hspace{\stretch{1}}(1.3)

and assumed a perturbative series representation for the energy eigenvalues in the new system

\begin{aligned}E_{s \alpha} = {E_s}^{(0)} + \lambda {E_{s \alpha}}^{(1)} + \lambda^2 {E_{s \alpha}}^{(2)} + \cdots\end{aligned} \hspace{\stretch{1}}(1.4)

Note that we do not assume that the perturbed energy states, if degenerate in the original system, are still degenerate after pertubation.

Given an assumed representation for the new eigenkets in terms of the known basis

\begin{aligned}{\lvert {\psi_{s \alpha}} \rangle} = \sum_{n, \beta} c_{ns;\beta \alpha} {\lvert {{\psi_{n \beta}}^{(0)}} \rangle} \end{aligned} \hspace{\stretch{1}}(1.5)

and a pertubative series representation for the probability coefficients

\begin{aligned}c_{ns;\beta \alpha} = {c_{ns;\beta \alpha}}^{(0)} + \lambda {c_{ns;\beta \alpha}}^{(1)} + \lambda^2 {c_{ns;\beta \alpha}}^{(2)},\end{aligned} \hspace{\stretch{1}}(1.6)

so that

\begin{aligned}{\lvert {\psi_{s \alpha}} \rangle} = \sum_{n, \beta} {c_{ns;\beta \alpha}}^{(0)} {\lvert {{\psi_{n \beta}}^{(0)}} \rangle} +\lambda\sum_{n, \beta} {c_{ns;\beta \alpha}}^{(1)} {\lvert {{\psi_{n \beta}}^{(0)}} \rangle} + \lambda^2\sum_{n, \beta} {c_{ns;\beta \alpha}}^{(2)} {\lvert {{\psi_{n \beta}}^{(0)}} \rangle} + \cdots\end{aligned} \hspace{\stretch{1}}(1.7)

Setting \lambda = 0 requires

\begin{aligned}{c_{ns;\beta \alpha}}^{(0)} = \delta_{ns;\beta \alpha},\end{aligned} \hspace{\stretch{1}}(1.8)


\begin{aligned}\begin{aligned}{\lvert {\psi_{s \alpha}} \rangle} &= {\lvert {{\psi_{s \alpha}}^{(0)}} \rangle} +\lambda\sum_{n, \beta} {c_{ns;\beta \alpha}}^{(1)} {\lvert {{\psi_{n \beta}}^{(0)}} \rangle} + \lambda^2\sum_{n, \beta} {c_{ns;\beta \alpha}}^{(2)} {\lvert {{\psi_{n \beta}}^{(0)}} \rangle} + \cdots \\ &=\left(1 + \lambda {c_{ss ; \alpha \alpha}}^{(1)} + \lambda^2 {c_{ss ; \alpha \alpha}}^{(2)} + \cdots\right){\lvert {{\psi_{s \alpha}}^{(0)}} \rangle} + \lambda\sum_{n\beta \ne s\alpha} {c_{ns;\beta \alpha}}^{(1)} {\lvert {{\psi_{n \beta}}^{(0)}} \rangle} +\lambda^2\sum_{n\beta \ne s\alpha} {c_{ns;\beta \alpha}}^{(2)} {\lvert {{\psi_{n \beta}}^{(0)}} \rangle} + \cdots\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.9)

We rescale our kets

\begin{aligned}{\lvert {\bar{\psi}_{s \alpha}} \rangle} ={\lvert {{\psi_{s \alpha}}^{(0)}} \rangle} + \lambda\sum_{n\beta \ne s\alpha} {\bar{c}_{ns;\beta \alpha}}^{(1)} {\lvert {{\psi_{n \beta}}^{(0)}} \rangle} +\lambda^2\sum_{n\beta \ne s\alpha} {\bar{c}_{ns;\beta \alpha}}^{(2)} {\lvert {{\psi_{n \beta}}^{(0)}} \rangle} + \cdots\end{aligned} \hspace{\stretch{1}}(1.10)


\begin{aligned}{\bar{c}_{ns;\beta \alpha}}^{(j)} = \frac{{c_{ns;\beta \alpha}}^{(j)}}{1 + \lambda {c_{ss ; \alpha \alpha}}^{(1)} + \lambda^2 {c_{ss ; \alpha \alpha}}^{(2)} + \cdots}\end{aligned} \hspace{\stretch{1}}(1.11)

The normalization of the rescaled kets is then

\begin{aligned}\left\langle{{\bar{\psi}_{s \alpha}}} \vert {{\bar{\psi}_{s \alpha}}}\right\rangle =1+ \lambda^2\sum_{n\beta \ne s\alpha} {\left\lvert{{\bar{c}_{ss}}^{(1)}}\right\rvert}^2+\cdots\equiv \frac{1}{{Z_{s \alpha}}},\end{aligned} \hspace{\stretch{1}}(1.12)

One can then construct a renormalized ket if desired

\begin{aligned}{\lvert {\bar{\psi}_{s \alpha}} \rangle}_R = Z_{s \alpha}^{1/2} {\lvert {\bar{\psi}_{s \alpha}} \rangle},\end{aligned} \hspace{\stretch{1}}(1.13)

so that

\begin{aligned}({\lvert {\bar{\psi}_{s \alpha}} \rangle}_R)^\dagger {\lvert {\bar{\psi}_{s \alpha}} \rangle}_R = Z_{s \alpha} \left\langle{{\bar{\psi}_{s \alpha}}} \vert {{\bar{\psi}_{s \alpha}}}\right\rangle = 1.\end{aligned} \hspace{\stretch{1}}(1.14)

The meat.

We continue by renaming terms in 1.10

\begin{aligned}{\lvert {\bar{\psi}_{s \alpha}} \rangle} ={\lvert {{\psi_{s \alpha}}^{(0)}} \rangle} + \lambda {\lvert {{\psi_{s \alpha}}^{(1)}} \rangle} + \lambda^2 {\lvert {{\psi_{s \alpha}}^{(2)}} \rangle} + \cdots\end{aligned} \hspace{\stretch{1}}(1.15)


\begin{aligned}{\lvert {{\psi_{s \alpha}}^{(j)}} \rangle} = \sum_{n\beta \ne s\alpha} {\bar{c}_{ns;\beta \alpha}}^{(j)} {\lvert {{\psi_{n \beta}}^{(0)}} \rangle}.\end{aligned} \hspace{\stretch{1}}(1.16)

Now we act on this with the Hamiltonian

\begin{aligned}H {\lvert {\bar{\psi}_{s \alpha}} \rangle} = E_{s \alpha} {\lvert {\bar{\psi}_{s \alpha}} \rangle},\end{aligned} \hspace{\stretch{1}}(1.17)


\begin{aligned}H {\lvert {\bar{\psi}_{s \alpha}} \rangle} - E_{s \alpha} {\lvert {\bar{\psi}_{s \alpha}} \rangle} = 0.\end{aligned} \hspace{\stretch{1}}(1.18)

Expanding this, we have

\begin{aligned}\begin{aligned}&(H_0 + \lambda H') \left({\lvert {{\psi_{s \alpha}}^{(0)}} \rangle} + \lambda {\lvert {{\psi_{s \alpha}}^{(1)}} \rangle} + \lambda^2 {\lvert {{\psi_{s \alpha}}^{(2)}} \rangle} + \cdots\right) \\ &\quad - \left( {E_s}^{(0)} + \lambda {E_{s \alpha}}^{(1)} + \lambda^2 {E_{s \alpha}}^{(2)} + \cdots \right)\left({\lvert {{\psi_{s \alpha}}^{(0)}} \rangle} + \lambda {\lvert {{\psi_{s \alpha}}^{(1)}} \rangle} + \lambda^2 {\lvert {{\psi_{s \alpha}}^{(2)}} \rangle} + \cdots\right)= 0.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.19)

We want to write this as

\begin{aligned}{\lvert {A} \rangle} + \lambda {\lvert {B} \rangle} + \lambda^2 {\lvert {C} \rangle} + \cdots = 0.\end{aligned} \hspace{\stretch{1}}(1.20)

This is

\begin{aligned}\begin{aligned}0 &=\lambda^0(H_0 - E_s^{(0)}) {\lvert {{\psi_{s \alpha}}^{(0)}} \rangle}  \\ &+ \lambda\left((H_0 - E_s^{(0)}) {\lvert {{\psi_{s \alpha}}^{(1)}} \rangle} +(H' - E_{s \alpha}^{(1)}) {\lvert {{\psi_{s \alpha}}^{(0)}} \rangle} \right) \\ &+ \lambda^2\left((H_0 - E_s^{(0)}) {\lvert {{\psi_{s \alpha}}^{(2)}} \rangle} +(H' - E_{s \alpha}^{(1)}) {\lvert {{\psi_{s \alpha}}^{(1)}} \rangle} -E_{s \alpha}^{(2)} {\lvert {{\psi_{s \alpha}}^{(0)}} \rangle} \right) \\ &\cdots\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.21)

So we form

\begin{aligned}{\lvert {A} \rangle} &=(H_0 - E_s^{(0)}) {\lvert {{\psi_{s \alpha}}^{(0)}} \rangle} \\ {\lvert {B} \rangle} &=(H_0 - E_s^{(0)}) {\lvert {{\psi_{s \alpha}}^{(1)}} \rangle} +(H' - E_{s \alpha}^{(1)}) {\lvert {{\psi_{s \alpha}}^{(0)}} \rangle} \\ {\lvert {C} \rangle} &=(H_0 - E_s^{(0)}) {\lvert {{\psi_{s \alpha}}^{(2)}} \rangle} +(H' - E_{s \alpha}^{(1)}) {\lvert {{\psi_{s \alpha}}^{(1)}} \rangle} -E_{s \alpha}^{(2)} {\lvert {{\psi_{s \alpha}}^{(0)}} \rangle},\end{aligned} \hspace{\stretch{1}}(1.22)

and so forth.

Zeroth order in \lambda

Since H_0 {\lvert {{\psi_{s \alpha}}^{(0)}} \rangle} = E_s^{(0)} {\lvert {{\psi_{s \alpha}}^{(0)}} \rangle}, this first condition on {\lvert {A} \rangle} is not much more than a statement that 0 - 0 = 0.

First order in \lambda

How about {\lvert {B} \rangle} = 0? For this to be zero we require that both of the following are simultaneously zero

\begin{aligned}\left\langle{{{\psi_{s \alpha}}^{(0)}}} \vert {{B}}\right\rangle &= 0 \\ \left\langle{{{\psi_{m \beta}}^{(0)}}} \vert {{B}}\right\rangle &= 0, \qquad m \beta \ne s \alpha\end{aligned} \hspace{\stretch{1}}(1.25)

This first condition is

\begin{aligned}{\langle {{\psi_{s \alpha}}^{(0)}} \rvert} (H' - E_{s \alpha}^{(1)}) {\lvert {{\psi_{s \alpha}}^{(0)}} \rangle} = 0.\end{aligned} \hspace{\stretch{1}}(1.27)


\begin{aligned}{\langle {{\psi_{m \beta}}^{(0)}} \rvert} H' {\lvert {{\psi_{s \alpha}}^{(0)}} \rangle} \equiv {H_{ms ; \beta \alpha}}',\end{aligned} \hspace{\stretch{1}}(1.28)


\begin{aligned}{H_{ss ; \alpha \alpha}}' = E_{s \alpha}^{(1)}.\end{aligned} \hspace{\stretch{1}}(1.29)

From the second condition we have

\begin{aligned}0 = {\langle {{\psi_{m \beta}}^{(0)}} \rvert} (H_0 - E_s^{(0)}) {\lvert {{\psi_{s \alpha}}^{(1)}} \rangle} +{\langle {{\psi_{m \beta}}^{(0)}} \rvert} (H' - E_{s \alpha}^{(1)}) {\lvert {{\psi_{s \alpha}}^{(0)}} \rangle} \end{aligned} \hspace{\stretch{1}}(1.30)

Utilizing the Hermitian nature of H_0 we can act backwards on {\langle {{\psi_m}^{(0)}} \rvert}

\begin{aligned}{\langle {{\psi_{m \beta}}^{(0)}} \rvert} H_0=E_m^{(0)} {\langle {{\psi_{m \beta}}^{(0)}} \rvert}.\end{aligned} \hspace{\stretch{1}}(1.31)

We note that \left\langle{{{\psi_{m \beta}}^{(0)}}} \vert {{{\psi_{s \alpha}}^{(0)}}}\right\rangle = 0, m \beta \ne s \alpha. We can also expand the \left\langle{{{\psi_{m \beta}}^{(0)}}} \vert {{{\psi_{s \alpha}}^{(1)}}}\right\rangle, which is

\begin{aligned}\left\langle{{{\psi_{m \beta}}^{(0)}}} \vert {{{\psi_{s \alpha}}^{(1)}}}\right\rangle &={\langle {{\psi_{m \beta}}^{(0)}} \rvert}\left(\sum_{n\delta \ne s\alpha} {\bar{c}_{ns;\delta \alpha}}^{(1)} {\lvert {{\psi_{n \delta}}^{(0)}} \rangle}\right) \\ \end{aligned}

I found that reducing this sum wasn’t obvious until some actual integers were plugged in. Suppose that s = 3\,1, and m \beta = 2\,2, then this is

\begin{aligned}\left\langle{{{\psi_{2\,2}}^{(0)}}} \vert {{{\psi_{3\,1}}^{(1)}}}\right\rangle &={\langle {{\psi_{2\,2}}^{(0)}} \rvert}\left(\sum_{n \delta \in \{1\,1, 1\,2, \cdots, 2\,1, 2\,2, 2\,3, \cdots, 3\,2, 3\,3, \cdots \} } {\bar{c}_{n 3; \delta 1}}^{(1)} {\lvert {{\psi_{n \delta}}^{(0)}} \rangle}\right) \\ &={\bar{c}_{2\,3 ; 2\,1}}^{(1)} \left\langle{{{\psi_{2\,2}}^{(0)}}} \vert {{{\psi_{2\,2}}^{(0)}}}\right\rangle \\ &={\bar{c}_{2\,3; 2\,1}}^{(1)}.\end{aligned}

Observe that we can also replace the superscript (1) with (j) in the above manipulation without impacting anything else. That and putting back in the abstract indexes, we have the general result

\begin{aligned}\left\langle{{{\psi_{m \beta}}^{(0)}}} \vert {{{\psi_{s \alpha}}^{(j)}}}\right\rangle ={\bar{c}_{ms ; \beta \alpha}}^{(j)}.\end{aligned} \hspace{\stretch{1}}(1.32)

Utilizing this gives us, for {red}

\begin{aligned}0 = ( E_m^{(0)} - E_s^{(0)}) {\bar{c}_{ms ; \beta \alpha}}^{(1)}+{H_{ms ; \beta \alpha}}' \end{aligned} \hspace{\stretch{1}}(1.33)

Here we see our first sign of the trouble hinted at in lecture 5. Just because m \beta \ne s \alpha does not mean that m \ne s. For example, with m \beta = 1\,1 and s\alpha = 1\,2 we would have

\begin{aligned}E_{1 2}^{(1)} &= {H_{1\,1 ; 2 2}}' \\ {\bar{c}_{1\,1 ; 1 2}}^{(1)}&=\frac{{H_{1\,1 ; 1 2}}' }{ E_1^{(0)} - E_1^{(0)} }\end{aligned} \hspace{\stretch{1}}(1.34)

We’ve got a {red} unless additional restrictions are imposed!

If we return to 1.33, we see that, for the result to be valid, when m = s, and there exists degeneracy for the s state, we require

\begin{aligned}{H_{ms ; \beta \alpha}}' = 0\end{aligned} \hspace{\stretch{1}}(1.36)

(then 1.33 becomes a 0 = 0 equality, and all is still okay)

And summarizing what we learn from our {\lvert {B} \rangle} = 0 conditions we have

\begin{aligned}E_{s \alpha}^{(1)} &= {H_{ss ; \alpha \alpha}}' \\ {\bar{c}_{ms ; \beta \alpha}}^{(1)}&=\frac{{H_{ms ; \beta \alpha}}' }{ E_s^{(0)} - E_m^{(0)} }, \qquad {m \ne s} \\ {H_{ss ; \beta \alpha}}' &= 0, \qquad \beta \alpha \ne 1\,1\end{aligned} \hspace{\stretch{1}}(1.37)

Second order in \lambda

Doing the same thing for {\lvert {C} \rangle} = 0 we form (or assume)

\begin{aligned}\left\langle{{{\psi_{s \alpha}}^{(0)}}} \vert {{C}}\right\rangle = 0 \end{aligned} \hspace{\stretch{1}}(1.40)

\begin{aligned}0 &= \left\langle{{{\psi_{s \alpha}}^{(0)}}} \vert {{C}}\right\rangle  \\ &={\langle {{\psi_{s \alpha}}^{(0)}} \rvert}\left((H_0 - E_s^{(0)}) {\lvert {{\psi_{s \alpha}}^{(2)}} \rangle} +(H' - E_{s \alpha}^{(1)}) {\lvert {{\psi_{s \alpha}}^{(1)}} \rangle} -E_{s \alpha}^{(2)} {\lvert {{\psi_{s \alpha}}^{(0)}} \rangle}  \right) \\ &=(E_s^{(0)} - E_s^{(0)}) \left\langle{{{\psi_{s \alpha}}^{(0)}}} \vert {{{\psi_{s \alpha}}^{(2)}}}\right\rangle +{\langle {{\psi_{s \alpha}}^{(0)}} \rvert}(H' - E_{s \alpha}^{(1)}) {\lvert {{\psi_{s \alpha}}^{(1)}} \rangle} -E_{s \alpha}^{(2)} \left\langle{{{\psi_{s \alpha}}^{(0)}}} \vert {{{\psi_{s \alpha}}^{(0)}}}\right\rangle \end{aligned}

We need to know what the \left\langle{{{\psi_{s \alpha}}^{(0)}}} \vert {{{\psi_{s \alpha}}^{(1)}}}\right\rangle is, and find that it is zero

\begin{aligned}\left\langle{{{\psi_{s \alpha}}^{(0)}}} \vert {{{\psi_{s \alpha}}^{(1)}}}\right\rangle={\langle {{\psi_{s \alpha}}^{(0)}} \rvert}\sum_{n\beta \ne s\alpha} {\bar{c}_{ns;\beta \alpha}}^{(1)} {\lvert {{\psi_{n \beta}}^{(0)}} \rangle} = 0\end{aligned} \hspace{\stretch{1}}(1.41)

Utilizing that we have

\begin{aligned}E_{s \alpha}^{(2)} &={\langle {{\psi_{s \alpha}}^{(0)}} \rvert} H' {\lvert {{\psi_{s \alpha}}^{(1)}} \rangle}  \\ &={\langle {{\psi_{s \alpha}}^{(0)}} \rvert} H' \sum_{m \beta \ne s \alpha} {\bar{c}_{ms}}^{(1)} {\lvert {{\psi_{m \beta}}^{(0)}} \rangle} \\ &=\sum_{m \beta \ne s \alpha} {\bar{c}_{ms ; \beta \alpha}}^{(1)} {H_{sm ; \alpha \beta}}'\end{aligned}

From 1.37, treating the {red} case carefully, we have

\begin{aligned}E_{s \alpha}^{(2)} =\sum_{\beta \ne \alpha} {\bar{c}_{ss ; \beta \alpha}}^{(1)} {H_{ss ; \alpha \beta}}'+\sum_{m \beta \ne s \alpha, m \ne s} \frac{{H_{ms ; \beta \alpha}}' }{ E_s^{(0)} - E_m^{(0)} }{H_{sm ; \alpha \beta}}'\end{aligned} \hspace{\stretch{1}}(1.42)

Again, only if H_{ss ; \alpha \beta} = 0 for \beta \ne \alpha do we have a result we can use. If that is the case, the first sum is killed without a divide by zero, leaving

\begin{aligned}E_{s \alpha}^{(2)} =\sum_{m \beta \ne s \alpha, m \ne s} \frac{{\left\lvert{{H_{ms ; \beta \alpha}}'}\right\rvert}^2 }{ E_s^{(0)} - E_m^{(0)} }.\end{aligned} \hspace{\stretch{1}}(1.43)

We can now summarize by forming the first order terms of the perturbed energy and the corresponding kets

\begin{aligned}E_{s \alpha} &= E_s^{(0)} + \lambda {H_{ss ; \alpha \alpha}}' + \lambda^2 \sum_{m \ne s, m \beta \ne s \alpha} \frac{{\left\lvert{{H_{ms ; \beta \alpha}}'}\right\rvert}^2 }{ E_s^{(0)} - E_m^{(0)} } + \cdots\\ {\lvert {\bar{\psi}_{s \alpha}} \rangle} &= {\lvert {{\psi_{s \alpha}}^{(0)}} \rangle} + \lambda\sum_{m \ne s, m \beta \ne s \alpha} \frac{{H_{ms ; \beta \alpha}}'}{ E_s^{(0)} - E_m^{(0)} } {\lvert {{\psi_{m \beta}}^{(0)}} \rangle}+ \cdots \\ {H_{ss ; \beta \alpha}}' &= 0, \qquad \beta \alpha \ne 1\,1\end{aligned} \hspace{\stretch{1}}(1.44)

Notational discrepency: OOPS. It looks like I used different notation than in class for our matrix elements for the placement of the indexes.


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

Posted in Math and Physics Learning. | Tagged: , , , | Leave a Comment »