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PHY456H1F: Quantum Mechanics II. Lecture 4 (Taught by Prof J.E. Sipe). Time independent perturbation theory (continued)

Posted by peeterjoot on September 23, 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.

Time independent perturbation.

The setup

To recap, we were 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}}(2.1)

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

\begin{aligned}H_0 {\lvert {{\psi_0}^{(0)}} \rangle} = {E_s}^{(0)} {\lvert {{\psi_s}^{(0)}} \rangle}\end{aligned} \hspace{\stretch{1}}(2.2)

We seek the perturbed eigensolution

\begin{aligned}H {\lvert {\psi_s} \rangle} = E_s {\lvert {\psi_s} \rangle}\end{aligned} \hspace{\stretch{1}}(2.3)

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

\begin{aligned}E_s = {E_s}^{(0)} + \lambda {E_s}^{(1)} + \lambda^2 {E_s}^{(2)} + \cdots\end{aligned} \hspace{\stretch{1}}(2.4)

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

\begin{aligned}{\lvert {\psi_s} \rangle} = \sum_n c_{ns} {\lvert {{\psi_n}^{(0)}} \rangle} \end{aligned} \hspace{\stretch{1}}(2.5)

and a pertubative series representation for the probability coefficients

\begin{aligned}c_{ns} = {c_{ns}}^{(0)} + \lambda {c_{ns}}^{(1)} + \lambda^2 {c_{ns}}^{(2)},\end{aligned} \hspace{\stretch{1}}(2.6)

so that

\begin{aligned}{\lvert {\psi_s} \rangle} = \sum_n {c_{ns}}^{(0)} {\lvert {{\psi_n}^{(0)}} \rangle} +\lambda\sum_n {c_{ns}}^{(1)} {\lvert {{\psi_n}^{(0)}} \rangle} + \lambda^2\sum_n {c_{ns}}^{(2)} {\lvert {{\psi_n}^{(0)}} \rangle} + \cdots\end{aligned} \hspace{\stretch{1}}(2.7)

Setting \lambda = 0 requires

\begin{aligned}{c_{ns}}^{(0)} = \delta_{ns},\end{aligned} \hspace{\stretch{1}}(2.8)

for

\begin{aligned}\begin{aligned}{\lvert {\psi_s} \rangle} &= {\lvert {{\psi_s}^{(0)}} \rangle} +\lambda\sum_n {c_{ns}}^{(1)} {\lvert {{\psi_n}^{(0)}} \rangle} + \lambda^2\sum_n {c_{ns}}^{(2)} {\lvert {{\psi_n}^{(0)}} \rangle} + \cdots \\ &=\left(1 + \lambda {c_{ns}}^{(1)} + \lambda^2 {c_{ns}}^{(2)} + \cdots\right){\lvert {{\psi_s}^{(0)}} \rangle} + \lambda\sum_{n \ne s} {c_{ns}}^{(1)} {\lvert {{\psi_n}^{(0)}} \rangle} +\lambda^2\sum_{n \ne s} {c_{ns}}^{(2)} {\lvert {{\psi_n}^{(0)}} \rangle} + \cdots\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.9)

We rescaled our kets

\begin{aligned}{\lvert {\bar{\psi}_s} \rangle} ={\lvert {{\psi_s}^{(0)}} \rangle} + \lambda\sum_{n \ne s} {\bar{c}_{ns}}^{(1)} {\lvert {{\psi_n}^{(0)}} \rangle} +\lambda^2\sum_{n \ne s} {\bar{c}_{ns}}^{(2)} {\lvert {{\psi_n}^{(0)}} \rangle} + \cdots\end{aligned} \hspace{\stretch{1}}(2.10)

where

\begin{aligned}{\bar{c}_{ns}}^{(j)} = \frac{{c_{ns}}^{(j)}}{1 + \lambda {c_{ns}}^{(1)} + \lambda^2 {c_{ns}}^{(2)} + \cdots}\end{aligned} \hspace{\stretch{1}}(2.11)

The normalization of the rescaled kets is then

\begin{aligned}\left\langle{{\bar{\psi}_s}} \vert {{\bar{\psi}_s}}\right\rangle =1+ \lambda^2\sum_{n \ne s} {\left\lvert{{\bar{c}_{ns}}^{(1)}}\right\rvert}^2+\cdots\equiv \frac{1}{{Z_s}},\end{aligned} \hspace{\stretch{1}}(2.12)

One can then construct a renormalized ket if desired

\begin{aligned}{\lvert {\bar{\psi}_s} \rangle}_R = Z_s^{1/2} {\lvert {\bar{\psi}_s} \rangle},\end{aligned} \hspace{\stretch{1}}(2.13)

so that

\begin{aligned}({\lvert {\bar{\psi}_s} \rangle}_R)^\dagger {\lvert {\bar{\psi}_s} \rangle}_R = Z_s \left\langle{{\bar{\psi}_s}} \vert {{\bar{\psi}_s}}\right\rangle = 1.\end{aligned} \hspace{\stretch{1}}(2.14)

The meat.

That’s as far as we got last time. We continue by renaming terms in 2.10

\begin{aligned}{\lvert {\bar{\psi}_s} \rangle} ={\lvert {{\psi_s}^{(0)}} \rangle} + \lambda {\lvert {{\psi_s}^{(1)}} \rangle} + \lambda^2 {\lvert {{\psi_s}^{(2)}} \rangle} + \cdots\end{aligned} \hspace{\stretch{1}}(2.15)

where

\begin{aligned}{\lvert {{\psi_n}^{(j)}} \rangle} = \sum_{n \ne s} {\bar{c}_{ns}}^{(j)} {\lvert {{\psi_s}^{(0)}} \rangle}.\end{aligned} \hspace{\stretch{1}}(2.16)

Now we act on this with the Hamiltonian

\begin{aligned}H {\lvert {\bar{\psi}_s} \rangle} = E_s {\lvert {\bar{\psi}_s} \rangle},\end{aligned} \hspace{\stretch{1}}(2.17)

or

\begin{aligned}H {\lvert {\bar{\psi}_s} \rangle} - E_s {\lvert {\bar{\psi}_s} \rangle} = 0.\end{aligned} \hspace{\stretch{1}}(2.18)

Expanding this, we have

\begin{aligned}\begin{aligned}&(H_0 + \lambda H') \left({\lvert {{\psi_s}^{(0)}} \rangle} + \lambda {\lvert {{\psi_s}^{(1)}} \rangle} + \lambda^2 {\lvert {{\psi_s}^{(2)}} \rangle} + \cdots\right) \\ &\quad - \left( {E_s}^{(0)} + \lambda {E_s}^{(1)} + \lambda^2 {E_s}^{(2)} + \cdots \right)\left({\lvert {{\psi_s}^{(0)}} \rangle} + \lambda {\lvert {{\psi_s}^{(1)}} \rangle} + \lambda^2 {\lvert {{\psi_s}^{(2)}} \rangle} + \cdots\right)= 0.\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.19)

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}}(2.20)

This is

\begin{aligned}\begin{aligned}0 &=\lambda^0(H_0 - E_s^{(0)}) {\lvert {{\psi_s}^{(0)}} \rangle} \\ &+ \lambda\left((H_0 - E_s^{(0)}) {\lvert {{\psi_s}^{(1)}} \rangle} +(H' - E_s^{(1)}) {\lvert {{\psi_s}^{(0)}} \rangle} \right) \\ &+ \lambda^2\left((H_0 - E_s^{(0)}) {\lvert {{\psi_s}^{(2)}} \rangle} +(H' - E_s^{(1)}) {\lvert {{\psi_s}^{(1)}} \rangle} -E_s^{(2)} {\lvert {{\psi_s}^{(0)}} \rangle} \right) \\ &\cdots\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.21)

So we form

\begin{aligned}{\lvert {A} \rangle} &=(H_0 - E_s^{(0)}) {\lvert {{\psi_s}^{(0)}} \rangle} \\ {\lvert {B} \rangle} &=(H_0 - E_s^{(0)}) {\lvert {{\psi_s}^{(1)}} \rangle} +(H' - E_s^{(1)}) {\lvert {{\psi_s}^{(0)}} \rangle} \\ {\lvert {C} \rangle} &=(H_0 - E_s^{(0)}) {\lvert {{\psi_s}^{(2)}} \rangle} +(H' - E_s^{(1)}) {\lvert {{\psi_s}^{(1)}} \rangle} -E_s^{(2)} {\lvert {{\psi_s}^{(0)}} \rangle},\end{aligned} \hspace{\stretch{1}}(2.22)

and so forth.

Zeroth order in \lambda

Since H_0 {\lvert {{\psi_s}^{(0)}} \rangle} = E_s^{(0)} {\lvert {{\psi_s}^{(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}^{(0)}}} \vert {{B}}\right\rangle &= 0 \\ \left\langle{{{\psi_m}^{(0)}}} \vert {{B}}\right\rangle &= 0, \qquad m \ne s\end{aligned} \hspace{\stretch{1}}(2.25)

This first condition is

\begin{aligned}{\langle {{\psi_s}^{(0)}} \rvert} (H' - E_s^{(1)}) {\lvert {{\psi_s}^{(0)}} \rangle} = 0.\end{aligned} \hspace{\stretch{1}}(2.27)

With

\begin{aligned}{\langle {{\psi_m}^{(0)}} \rvert} H' {\lvert {{\psi_s}^{(0)}} \rangle} \equiv {H_{ms}}',\end{aligned} \hspace{\stretch{1}}(2.28)

or

\begin{aligned}{H_{ss}}' = E_s^{(1)}.\end{aligned} \hspace{\stretch{1}}(2.29)

From the second condition we have

\begin{aligned}0 = {\langle {{\psi_m}^{(0)}} \rvert} (H_0 - E_s^{(0)}) {\lvert {{\psi_s}^{(1)}} \rangle} +{\langle {{\psi_m}^{(0)}} \rvert} (H' - E_s^{(1)}) {\lvert {{\psi_s}^{(0)}} \rangle} \end{aligned} \hspace{\stretch{1}}(2.30)

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

\begin{aligned}{\langle {{\psi_m}^{(0)}} \rvert} H_0=E_m^{(0)} {\langle {{\psi_m}^{(0)}} \rvert}.\end{aligned} \hspace{\stretch{1}}(2.31)

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

\begin{aligned}\left\langle{{{\psi_m}^{(0)}}} \vert {{{\psi_s}^{(1)}}}\right\rangle &={\langle {{\psi_m}^{(0)}} \rvert}\left(\sum_{n \ne s} {\bar{c}_{ns}}^{(1)} {\lvert {{\psi_n}^{(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, and m = 5, then this is

\begin{aligned}\left\langle{{{\psi_5}^{(0)}}} \vert {{{\psi_3}^{(1)}}}\right\rangle &={\langle {{\psi_5}^{(0)}} \rvert}\left(\sum_{n = 0, 1, 2, 4, 5, \cdots} {\bar{c}_{n3}}^{(1)} {\lvert {{\psi_n}^{(0)}} \rangle}\right) \\ &={\bar{c}_{53}}^{(1)} \left\langle{{{\psi_5}^{(0)}}} \vert {{{\psi_5}^{(0)}}}\right\rangle \\ &={\bar{c}_{53}}^{(1)}.\end{aligned}

More generally that is

\begin{aligned}\left\langle{{{\psi_m}^{(0)}}} \vert {{{\psi_s}^{(1)}}}\right\rangle ={\bar{c}_{ms}}^{(1)}.\end{aligned} \hspace{\stretch{1}}(2.32)

Utilizing this gives us

\begin{aligned}0 = ( E_m^{(0)} - E_s^{(0)}) {\bar{c}_{ms}}^{(1)}+{H_{ms}}' \end{aligned} \hspace{\stretch{1}}(2.33)

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

\begin{aligned}E_s^{(1)} &= {H_{ss}}' \\ {\bar{c}_{ms}}^{(1)}&=\frac{{H_{ms}}' }{ E_s^{(0)} - E_m^{(0)} }\end{aligned} \hspace{\stretch{1}}(2.34)

Second order in \lambda

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

\begin{aligned}\left\langle{{{\psi_s}^{(0)}}} \vert {{C}}\right\rangle &= 0 \\ \left\langle{{{\psi_m}^{(0)}}} \vert {{C}}\right\rangle &= 0, \qquad m \ne s\end{aligned} \hspace{\stretch{1}}(2.36)

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

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

\begin{aligned}\left\langle{{{\psi_s}^{(0)}}} \vert {{{\psi_s}^{(1)}}}\right\rangle={\langle {{\psi_s}^{(0)}} \rvert}\sum_{n \ne s} {\bar{c}_{ns}}^{(1)} {\lvert {{\psi_n}^{(0)}} \rangle}\end{aligned} \hspace{\stretch{1}}(2.38)

Again, suppose that s = 3. Our sum ranges over all n \ne 3, so all the brakets are zero. Utilizing that we have

\begin{aligned}E_s^{(2)} &={\langle {{\psi_s}^{(0)}} \rvert} H' {\lvert {{\psi_s}^{(1)}} \rangle} \\ &={\langle {{\psi_s}^{(0)}} \rvert} H' \sum_{m \ne s} {\bar{c}_{ms}}^{(1)} {\lvert {{\psi_m}^{(0)}} \rangle} \\ &=\sum_{m \ne s} {\bar{c}_{ms}}^{(1)} {H_{sm}}'\end{aligned}

From 2.34 we have

\begin{aligned}E_s^{(2)} =\sum_{m \ne s} \frac{{H_{ms}}' }{ E_s^{(0)} - E_m^{(0)} }{H_{sm}}'=\sum_{m \ne s} \frac{{\left\lvert{{H_{ms}}'}\right\rvert}^2 }{ E_s^{(0)} - E_m^{(0)} }\end{aligned} \hspace{\stretch{1}}(2.39)

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

\begin{aligned}E_s &= E_s^{(0)} + \lambda {H_{ss}}' + \lambda^2 \sum_{m \ne s} \frac{{\left\lvert{{H_{ms}}'}\right\rvert}^2 }{ E_s^{(0)} - E_m^{(0)} } + \cdots\\ {\lvert {\bar{\psi}_s} \rangle} &= {\lvert {{\psi_s}^{(0)}} \rangle} + \lambda\sum_{m \ne s} \frac{{H_{ms}}'}{ E_s^{(0)} - E_m^{(0)} } {\lvert {{\psi_m}^{(0)}} \rangle}+ \cdots\end{aligned} \hspace{\stretch{1}}(2.40)

We can continue calculating, but are hopeful that we can stop the calculation without doing more work, even if \lambda = 1. If one supposes that the

\begin{aligned}\sum_{m \ne s} \frac{{H_{ms}}'}{ E_s^{(0)} - E_m^{(0)} } \end{aligned} \hspace{\stretch{1}}(2.42)

term is “small”, then we can hope that truncating the sum will be reasonable for \lambda = 1. This would be the case if

\begin{aligned}{H_{ms}}' \ll {\left\lvert{ E_s^{(0)} - E_m^{(0)} }\right\rvert},\end{aligned} \hspace{\stretch{1}}(2.43)

however, to put some mathematical rigor into making a statement of such smallness takes a lot of work. We are referred to [2]. Incidentally, these are loosely referred to as the first and second testaments, because of the author’s name, and the fact that they came as two volumes historically.

References

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

[2] A. Messiah, G.M. Temmer, and J. Potter. Quantum mechanics: two volumes bound as one. Dover Publications New York, 1999.

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This entry was posted on September 23, 2011 at 9:40 pm and is filed under Math and Physics Learning.. Tagged: , , , , , , , . You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

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