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## 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.