Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

«
»

PHY456H1F: Quantum Mechanics II. Lecture 3 (Taught by Prof J.E. Sipe). Perturbation methods

Posted by peeterjoot on September 19, 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.)]

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

States and wave functions

Suppose we have the following non-degenerate energy eigenstates

\begin{aligned}&\dot{v}s \\ E_1 &\sim {\lvert {\Psi_1} \rangle} \\ E_0 &\sim {\lvert {\Psi_0} \rangle}\end{aligned}

and consider a state that is “very close” to {\lvert {\Psi_n} \rangle}.

\begin{aligned}{\lvert {\Psi} \rangle} = {\lvert {\Psi_n} \rangle} + {\lvert {\delta \Psi_n} \rangle}\end{aligned} \hspace{\stretch{1}}(1.1)

We form projections onto {\lvert {\Psi_n} \rangle} “direction”. The difference from this projection will be written {\lvert {\Psi_{n \perp}} \rangle}, as depicted in figure (\ref{fig:qmTwoL3fig1}). This illustration cannot not be interpreted literally, but illustrates the idea nicely.

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.4\textheight]{qmTwoL3fig1}
\caption{Pictorial illustration of ket projections}
\end{figure}

For the amount along the projection onto {\lvert {\Psi_n} \rangle} we write

\begin{aligned}\left\langle{{\Psi_n}} \vert {{\delta \Psi_n}}\right\rangle = \delta \alpha\end{aligned} \hspace{\stretch{1}}(1.2)

so that the total deviation from the original state is

\begin{aligned}{\lvert {\delta \Psi_n} \rangle} = \delta \alpha {\lvert {\Psi_n} \rangle} + {\lvert {\delta \Psi_{n \perp}} \rangle} .\end{aligned} \hspace{\stretch{1}}(1.3)

The varied ket is then

\begin{aligned}{\lvert {\Psi} \rangle} = (1 + \delta \alpha ){\lvert {\Psi_n} \rangle} + {\lvert {\delta \Psi_{n \perp}} \rangle} \end{aligned} \hspace{\stretch{1}}(1.4)

where

\begin{aligned}(\delta \alpha)^2, \left\langle{{\delta \Psi_{n \perp}}} \vert {{\delta \Psi_{n \perp}}}\right\rangle \ll 1\end{aligned} \hspace{\stretch{1}}(1.5)

In terms of these projections our kets magnitude is

\begin{aligned}\left\langle{{\Psi}} \vert {{\Psi}}\right\rangle &= \Bigl((1 + {\delta \alpha}^{*} ){\langle {\Psi_n} \rvert} + {\langle {\delta \Psi_{n \perp}} \rvert} \Bigr)\Bigl((1 + \delta \alpha ){\lvert {\Psi_n} \rangle} + {\lvert {\delta \Psi_{n \perp}} \rangle} \Bigr) \\ &={\left\lvert{1 + \delta \alpha}\right\rvert}^2 \left\langle{{\Psi_n}} \vert {{\Psi_n}}\right\rangle+ \left\langle{{\delta \Psi_{n \perp}}} \vert {{\delta \Psi_{n \perp}}}\right\rangle \\ &\quad +(1 + {\delta \alpha}^{*} )\left\langle{{\Psi_n}} \vert {{\delta \Psi_{n \perp}}}\right\rangle +(1 + \delta \alpha )\left\langle{{\delta \Psi_{n \perp}}} \vert {{\delta \Psi_n}}\right\rangle \end{aligned}

Because \left\langle{{\delta \Psi_{n \perp}}} \vert {{\delta \Psi_n}}\right\rangle = 0 this is

\begin{aligned}\left\langle{{\Psi}} \vert {{\Psi}}\right\rangle= {\left\lvert{1 + \delta \alpha }\right\rvert}^2\left\langle{{\delta \Psi_{n \perp}}} \vert {{\delta \Psi_{n \perp}}}\right\rangle.\end{aligned} \hspace{\stretch{1}}(1.6)

Similarly for the energy expectation we have

\begin{aligned}\left\langle{{\Psi}} \vert {{\Psi}}\right\rangle &= \Bigl((1 + {\delta \alpha}^{*} ){\langle {\Psi_n} \rvert} + {\langle {\delta \Psi_{n \perp}} \rvert} \Bigr)H\Bigl((1 + \delta \alpha ){\lvert {\Psi_n} \rangle} + {\lvert {\delta \Psi_{n \perp}} \rangle} \Bigr) \\ &={\left\lvert{1 + \delta \alpha}\right\rvert}^2 E_n \left\langle{{\Psi_n}} \vert {{\Psi_n}}\right\rangle+ \braket{\delta \Psi_{n \perp}} H {\delta \Psi_{n \perp}} \\ &\quad + (1 + {\delta \alpha}^{*} ) E_n \left\langle{{\Psi_n}} \vert {{\delta \Psi_{n \perp}}}\right\rangle +(1 + \delta \alpha ) E_n \left\langle{{\delta \Psi_{n \perp}}} \vert {{\delta \Psi_n}}\right\rangle \end{aligned}

Or

\begin{aligned}{\langle {\Psi} \rvert} H {\lvert {\Psi} \rangle}= E_n {\left\lvert{1 + \delta \alpha }\right\rvert}^2+{\langle {\delta \Psi_{n \perp}} \rvert} H {\lvert {\delta \Psi_{n \perp}} \rangle}.\end{aligned} \hspace{\stretch{1}}(1.7)

This gives

\begin{aligned}E[\Psi] &= \frac{{\langle {\Psi} \rvert} H {\lvert {\Psi} \rangle}}{\left\langle{{\Psi}} \vert {{\Psi}}\right\rangle} \\ &=\frac{E_n {\left\lvert{1 + \delta \alpha }\right\rvert}^2 + {\langle {\delta \Psi_{n \perp}} \rvert} H {\lvert {\delta \Psi_{n \perp}} \rangle}}{{\left\lvert{1 + \delta \alpha }\right\rvert}^2\left\langle{{\delta \Psi_{n \perp}}} \vert {{\delta \Psi_{n \perp}}}\right\rangle } \\ &=\frac{E_n + \frac{{\langle {\delta \Psi_{n \perp}} \rvert} H {\lvert {\delta \Psi_{n \perp}} \rangle} }{{\left\lvert{1 + \delta \alpha }\right\rvert}^2}}{1+\frac{\left\langle{{\delta \Psi_{n \perp}}} \vert {{\delta \Psi_{n \perp}}}\right\rangle }{{\left\lvert{1 + \delta \alpha }\right\rvert}^2}} \\ &=E_n \left( 1 - \frac{\left\langle{{\delta \Psi_{n \perp}}} \vert {{\delta \Psi_{n \perp}}}\right\rangle }{{\left\lvert{1 + \delta \alpha }\right\rvert}^2}+ \cdots \right) + \cdots \\ &=E_n\left[1 + \mathcal{O}\left((\delta \Psi_{n \perp})^2\right)\right]\end{aligned}

where

\begin{aligned}(\delta \Psi_{n \perp})^2\sim\left\langle{{\delta \Psi_{n \perp}}} \vert {{\delta \Psi_{n \perp}}}\right\rangle\end{aligned} \hspace{\stretch{1}}(1.8)

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.4\textheight]{qmTwoL3fig2}
\caption{Illustration of variation of energy with variation of Hamiltonian}
\end{figure}

“small errors” in {\lvert {\Psi} \rangle} don’t lead to large errors in E[\Psi]

It is reasonably easy to get a good estimate and E_0, although it is reasonably hard to get a good estimate of {\lvert {\Psi_0} \rangle}. This is for the same reason, because E[] is not terribly sensitive.

Excited states.

\begin{aligned}&\dot{v}s \\ E_2 &\sim {\lvert {\Psi_2} \rangle} \\ E_1 &\sim {\lvert {\Psi_1} \rangle} \\ E_0 &\sim {\lvert {\Psi_0} \rangle}\end{aligned}

Suppose we wanted an estimate of E_1. If we knew the ground state {\lvert {\Psi_0} \rangle}. For any trial {\lvert {\Psi} \rangle} form

\begin{aligned}{\lvert {\Psi'} \rangle} = {\lvert {\Psi} \rangle} - {\lvert {\Psi_0} \rangle} \left\langle{{\Psi_0}} \vert {{\Psi}}\right\rangle\end{aligned} \hspace{\stretch{1}}(2.9)

We are taking out the projection of the ground state from an arbitrary trial function.

For a state written in terms of the basis states, allowing for an \alpha degeneracy

\begin{aligned}{\lvert {\Psi} \rangle} = c_0 {\lvert {\Psi_0} \rangle} +\sum_{n> 0, \alpha} c_{n \alpha} {\lvert {\Psi_{n \alpha}} \rangle}\end{aligned} \hspace{\stretch{1}}(2.10)

\begin{aligned}\left\langle{{\Psi_0}} \vert {{\Psi}}\right\rangle = c_0 \end{aligned} \hspace{\stretch{1}}(2.11)

and

\begin{aligned}{\lvert {\Psi'} \rangle} = \sum_{n> 0, \alpha} c_{n \alpha} {\lvert {\Psi_{n \alpha}} \rangle}\end{aligned} \hspace{\stretch{1}}(2.12)

(note that there are some theorems that tell us that the ground state is generally non-degenerate).

\begin{aligned}E[\Psi'] &= \frac{{\langle {\Psi'} \rvert} H {\lvert {\Psi'} \rangle}}{\left\langle{{\Psi'}} \vert {{\Psi'}}\right\rangle} \\ &=\frac{\sum_{n> 0, \alpha} {\left\lvert{c_{n \alpha}}\right\rvert}^2 E_n}{\sum_{m> 0, \beta} {\left\lvert{c_{m \beta}}\right\rvert}^2 }\ge E_1\end{aligned}

Often don’t know the exact ground state, although we might have a guess {\lvert {\tilde{\Psi}_0} \rangle}.

for

\begin{aligned}{\lvert {\Psi''} \rangle} = {\lvert {\Psi} \rangle} - {\lvert {\tilde{\Psi}_0} \rangle}\left\langle{{\tilde{\Psi}_0}} \vert {{\Psi}}\right\rangle\end{aligned} \hspace{\stretch{1}}(2.13)

but cannot prove that

\begin{aligned}\frac{{\langle {\Psi''} \rvert} H {\lvert {\Psi''} \rangle}}{\left\langle{{\Psi''}} \vert {{\Psi''}}\right\rangle} \ge E_1\end{aligned} \hspace{\stretch{1}}(2.14)

Then

FIXME: missed something here.

\begin{aligned}\frac{{\langle {\Psi'''} \rvert} H {\lvert {\Psi'''} \rangle}}{\left\langle{{\Psi'''}} \vert {{\Psi'''}}\right\rangle} \ge E_1\end{aligned} \hspace{\stretch{1}}(2.15)

Somewhat remarkably, this is often possible. We talked last time about the Hydrogen atom. In that case, you can guess that the excited state is in the 2s orbital and and therefore orthogonal to the 1s (?) orbital.

Time independent perturbation theory.

See section 16.1 of the text [1].

We can sometimes use this sort of physical insight to help construct a good approximation. This is provided that we have some of this physical insight, or that it is good insight in the first place.

This is the no-think (turn the crank) approach.

Here we split our Hamiltonian into two parts

\begin{aligned}H = H_0 + H'\end{aligned} \hspace{\stretch{1}}(3.16)

where H_0 is a Hamiltonian for which we know the energy eigenstates and the eigenkets. The H' is the “perturbation” that is supposed to be small “in some sense”.

Prof Sipe will provide some references later that provide a more specific meaning to this “smallness”. From some ad-hoc discussion in the class it sounds like one has to consider sequences of operators, and look at the convergence of those sequences (is this L2 measure theory?)

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.4\textheight]{qmTwoL3fig3}
\caption{Example of small perturbation from known Hamiltonian}
\end{figure}

We’d like to consider a range of problems of the form

\begin{aligned}H = H_0 + \lambda H'\end{aligned} \hspace{\stretch{1}}(3.17)

where

\begin{aligned}\lambda \in [0,1]\end{aligned} \hspace{\stretch{1}}(3.18)

So that when \lambda \rightarrow 0 we have

\begin{aligned}H \rightarrow H_0\end{aligned} \hspace{\stretch{1}}(3.19)

the problem that we already know, but for \lambda \rightarrow 1 we have

\begin{aligned}H = H_0 + H'\end{aligned} \hspace{\stretch{1}}(3.20)

the problem that we’d like to solve.

We are assuming that we know the eigenstates and eigenvalues for H_0. Assuming no degeneracy

\begin{aligned}H_0 {\lvert {\Psi_s^{(0)}} \rangle} = E_s^{(0)}{\lvert {\Psi_s^{(0)}} \rangle} \end{aligned} \hspace{\stretch{1}}(3.21)

We seek

\begin{aligned}(H_0 + H'){\lvert {\Psi_s} \rangle} = E_s{\lvert {\Psi_s} \rangle} \end{aligned} \hspace{\stretch{1}}(3.22)

(this is the \lambda = 1 case).

Once (if) found, when \lambda \rightarrow 0 we will have

\begin{aligned}E_s &\rightarrow E_s^{(0)} \\ {\lvert {\Psi_s} \rangle} &\rightarrow {\lvert {\Psi_s^{(0)}} \rangle}\end{aligned}

\begin{aligned}E_s = E_s^{(0)} + \lambda E_s^{(1)} + \frac{\lambda^2}{2} E_s^{(2)}\end{aligned} \hspace{\stretch{1}}(3.23)

\begin{aligned}\Psi_s = \sum_n c_{ns} {\lvert {\Psi_n^{(0)}} \rangle}\end{aligned} \hspace{\stretch{1}}(3.24)

This we know we can do because we are assumed to have a complete set of states.

with

\begin{aligned}c_{ns} = c_{ns}^{(0)} + \lambda c_{ns}^{(1)} + \frac{\lambda^2}{2} c_{ns}^{(2)}\end{aligned} \hspace{\stretch{1}}(3.25)

where

\begin{aligned}c_{ns}^{(0)} = \delta_{ns}\end{aligned} \hspace{\stretch{1}}(3.26)

There’s a subtlety here that will be treated differently from the text. We write

\begin{aligned}{\lvert {\Psi_s} \rangle}&={\lvert {\Psi_s^{(0)}} \rangle}+ \lambda \sum_nc_{ns}^{(1)} {\lvert {\Psi_n^{(0)}} \rangle}+ \frac{\lambda^2}{2} \sum_nc_{ns}^{(2)}{\lvert {\Psi_n^{(0)}} \rangle}+ \cdots \\ &=\left(1 + \lambda c_{ss}^{(1)} + \cdots\right){\lvert {\Psi_s^{(0)}} \rangle}+ \lambda \sum_{n \ne s} c_{ns}^{(1)} {\lvert {\Psi_n^{(0)}} \rangle}+ \cdots\end{aligned}

Take

\begin{aligned}{\lvert {\bar{\Psi}_s} \rangle}&={\lvert {\bar{\Psi}_s^{(0)}} \rangle}+ \lambda\frac{\sum_{n \ne s} c_{ns}^{(1)} {\lvert {\Psi_n^{(0)}} \rangle}}{1 + \lambda c_{ss}^{(1)}} + \cdots\\ &={\lvert {\bar{\Psi}_s^{(0)}} \rangle}+ \lambda\sum_{n \ne s} \bar{c}_{ns}^{(1)} {\lvert {\Psi_n^{(0)}} \rangle} + \cdots\end{aligned}

where

\begin{aligned}\bar{c}_{ns}^{(1)} =\frac{c_{ns}^{(1)} }{1 + \lambda c_{ss}^{(1)}} \end{aligned} \hspace{\stretch{1}}(3.27)

We have:

\begin{aligned}\bar{c}_{ns}^{(1)} &= c_{ns}^{(1)} \\ \bar{c}_{ns}^{(2)} &\ne c_{ns}^{(2)} \end{aligned}

FIXME: I missed something here.

Note that this is no longer normalized.

\begin{aligned}\left\langle{{\bar{\Psi}_s}} \vert {{\bar{\Psi}_s}}\right\rangle \ne 1\end{aligned} \hspace{\stretch{1}}(3.28)

References

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

Advertisements

This entry was posted on September 19, 2011 at 8:53 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.

Leave a Reply

Fill in your details below or click an icon to log in:

Gravatar
WordPress.com Logo

You are commenting using your WordPress.com account. ( Log Out / Change )

Twitter picture

You are commenting using your Twitter account. ( Log Out / Change )

Facebook photo

You are commenting using your Facebook account. ( Log Out / Change )

Google+ photo

You are commenting using your Google+ account. ( Log Out / Change )

Cancel

Connecting to %s

«
»
 
%d bloggers like this: