We start with the exact energy eigenvalue problem, given by \(\hat{H}|n\rangle=E_{n}|n\rangle\). But, perhaps this problem is too difficult to solve. Then, if \(\hat{H}=\hat{H}_{0}+\hat{V}\), where \(\hat{H}_{0}\) can be solved exactly, but \(\hat{H}\) cannot and \(\hat{V}\) is in some sense “small,” then I can try to find how \(\hat{V}\) perturbs \(\hat{H}_0\) to arrive at \(\hat{H}\).
The unperturbed problem is \(\hat{H}_{0}|n\rangle_{0}=E_{n}^{0}|n\rangle_{0}\), whose exact solution we know by assumption. Furthermore, we assume the system is nondegenerate, so we have \(E_{n}^{0} \neq E_{m}^{0}\) unless \(m=n\). This is known to be the case for all one-dimensional problems on the infinite one-dimensional spatial domain via the so-called node theorem.
We want to find \(E_{n}\) and \(|n\rangle\) as a power series in \(\hat{V}\). We start with what we know, which is \[\left(\hat{H}_{0}+\hat{V}\right)|n\rangle=\left(E_{n}^{0}+\Delta E_{n}\right)|n\rangle, \quad\text{with}\quad E_{n}=E_{n}^{0}+\Delta E_{n}.\] We then re-arrange this expression to \[\left(E_{n}^{0}-\hat{H}_{0}\right)|n\rangle=\left(\hat{V}-\Delta E_{n}\right)|n\rangle.\]
We want to somehow "invert" this, in the sense that we wish to multiply by the inverse of the operator on the far left to obtain the perturbed state \(|n\rangle\). Unfortunately, that operator cannot be inverted, because it can involve a divide by zero.
Let us examine how to work with nontraditional operators such as \(\frac{1}{E_{n}^{0}-\hat{H}_{0}}\). This operator can be best expressed in terms of the eigenbasis of \(\hat{H}_0\) as \[\frac{1}{E_{n}^{0}-\hat{H}_{0}} \underbrace{\sum_{m}|m\rangle_{00} \langle m|}_{\text{complete set of states} = \mathbb{I}}\] by a multiply by one. But, \(\hat{H}_{0}|m\rangle_{0}=E_{m}^{0}|m\rangle_{0}\), so \[\frac{1}{E_{n}^{0}-\hat{H}_{0}} \sum_{m}|m\rangle_{00}\langle m|=\sum_{m} \frac{1}{E_{n}^{0}-E_{m}^{0}}|m\rangle_{00}\langle m|\] becomes singular for the one term in the sum, where \(m=n\).
We deal with this by introducing projection operators.
Define \(\hat{P}_{n}=|n\rangle_{00}\langle n| \quad \text{and}\quad\hat{Q}_{n}=\mathbb{I}-\hat{P}_{n}=\sum_{m \neq n}|m\rangle_{00}\langle m|\).
Let us first note some properties of projection operators: \[\begin{aligned} & \hat{P}_{n}+\hat{Q}_{n}=\mathbb{I} \\ & \hat{P}_{n}^{2}=\hat{P}_{n} \quad \text { check } \quad \hat{P}_{n}^{2}=|n\rangle_{00}\underbrace{\langle n \mid n\rangle_{00}}_{1} \langle n|=| n\rangle_{00}\langle n|=\hat{P}_{n}\\ & \hat{Q}_{n}^{2}=\hat{Q}_{n}\quad \text{ check } \quad \hat{Q}_{n}^{2}=\left(1-\hat{P}_{n}\right)^{2}=1-\hat{P}_{n}-\hat{P}_{n}+\hat{P}_{n}^{2}=1-\hat{P}_{n}-\hat{P}_{n}+\hat{P}_{n}=1-\hat{P}_{n}=\hat{Q}_{n}\\ & \hat{P}_{n} \hat{Q}_{n}=0\quad\text{check}\quad \hat{P}_n\hat{Q}_n=|n\rangle_{00}\langle n| \sum_{m+n}|m\rangle_{00}\langle m|, \quad \text { but } \text{}_{0}\langle n | m\rangle_{0}=0 \text { if } m \neq n, \quad\text{so}\quad \hat{Q}_{n}\hat{P}_n=0\\& \text{ and then } \left[\hat{P}_{n}, \hat{Q}_{n}\right]=0. \end{aligned}\]
How do projection operators act on an arbitrary operator? We determine this by representing the operator in the energy eigenbasis of \(\hat{H}_0\), where we can immediately apply the projection operators: \[\begin{aligned} & \hat{O}=\sum_{m m^{\prime}}{O_{m m^{\prime}}|m\rangle_{00}\left\langle m^{\prime}\right|} \\ & \hat{P}_{n} \hat{O}=\sum_{m^{\prime}} O_{n m^{\prime}} | n \rangle_{00}\left\langle m^{\prime}\right|, \quad \hat{P}_{n} \hat{O} \hat{P}_{n}=O_{n n}|n\rangle_{00}\langle n| \\ & \hat{Q}_{n} \hat{O}=\sum_{m \neq n} \sum_{m^{\prime}} O_{m m^{\prime}}|m\rangle_{00} \langle m^{\prime}| \\ & \hat{Q}_{n} \hat{O} \hat{Q}_{n}=\sum_{m \neq n} \sum_{m^{\prime} \neq n} O_{m m^{\prime}}|m\rangle_{0}{ }_{0}\langle m^{\prime}|, \text { and } \text { so on } \\ & \hat{Q}_{n} \hat{O} \hat{P_{n}}=\sum_{m\neq} O_{m n}|m\rangle_{00}\langle n|, \text { etc. } \end{aligned}\] In words, we say \(\hat{P}_{n}\) projects parallel to \(| n \rangle_{0}\) and \(\hat{Q}_n\) projects perpendicular to \(|n\rangle_0\).
Claim: \(\left[\hat{P}_{n},
\hat{H}_{0}\right]=0.\)
Proof: \[\begin{aligned}
&\hat{P}_{n} \hat{H}_{0}=|n\rangle_{00} \langle n|
\hat{H}_{0}=E_{n}^{0}|n\rangle_{00} \langle n|\\
&\hat{H}_{0} \hat{P}_{n}=\hat{H}_{0}|n\rangle_{00}\langle n| =
E_{n}^{0}|n\rangle_{00}\langle n |\\
\text{ so } & \hat{P}_{n} \hat{H}_{0}-\hat{H}_{0} \hat{P}_{n}=0.\\
\end{aligned}\] Since \(\hat{Q}_{n}=1-\hat{P}_{n}\) we have \(\left[\hat{Q}_{n} , \hat{H}_{0}\right]
=\left[1-\hat{P}_{n}, \hat{H}_{0}\right]=0\) as well.
Now examine the original equation \[(E_{n}^{0}-\hat{H}_{0}) |n \rangle=(\hat{V}-\Delta E_{n})|n\rangle.\] Then \[\hat{Q}_{n}|n\rangle=\frac{\hat{Q}_{n}}{E_{n}^{0}-\hat{H}_{0}}\left(\hat{V}-\Delta E_{n}\right)|n\rangle.\]
Check: \[\hat{Q}_{n}|n\rangle=\sum_{m \neq
n}|m\rangle_{00} \langle m | n\rangle,\] \[\frac{\hat{Q}_{n}}{E_{n}^{0}-\hat{H}_{0}}(\hat{V}-\Delta
E_{n}) |n\rangle=\sum_{m \neq n}|m\rangle_{00}\langle m|
\frac{\hat{V}-\Delta E_{n}}{E_{n}^{0}-E_{m}^{0}}|n\rangle.\] But,
\[\sum_{m \neq n}|m\rangle_{00}\langle m|
(E_{n}^{0}-\hat{H}_{0}) | n\rangle=\sum_{m \neq n}|m\rangle_{00}\langle
m| (\hat{V}-\Delta E_{n})|n\rangle,\] from our first equations
above. Hence, \[\text{LHS}=\sum_{m\neq
n}\left(E_{n}^{0}-E_{m}^{0}\right)|m\rangle_{00} \langle m |
n\rangle=\sum_{m \neq n}|m\rangle_{00}\langle m| (\hat{V}-\Delta
E_{n})|n\rangle\]
since each coefficient of \(|m\rangle_{0}\) must be equal and \(E_{n}^{0}-E_{m}^{0} \neq 0\), we then have
\[\sum_{m \neq n}|m\rangle_{00}\langle m |
n\rangle=\sum_{m \neq n}|m\rangle_{00}\langle m| \frac{\hat{V}-\Delta
E_{n}}{E_{n}^{0}-E_{m}^{0}}|n\rangle\] as claimed.
But \(|n\rangle=\left(\hat{P}_{n}+\hat{Q}_{n}\right)|n\rangle
\quad \text{and}\quad \hat{P}_{n}|n\rangle=|n\rangle_{00}\langle n
|n\rangle,\) So \[|n\rangle=|n\rangle_{00}\langle n |
n\rangle+\frac{\hat{Q}_{n}}{E_{n}^{0}-\hat{H}_{0}}(\hat{V}- \Delta
E_{n}\rangle|n\rangle\] We move the rightmost term to the left
hand side to get \[\left [\mathbb{I}
-\frac{\hat{Q}_{n}}{E_{n}^{0}-\hat{H}}(\hat{V}-\Delta
E_{n})\right]|n\rangle=|n\rangle_{00} \langle n | n\rangle.\]
Now, we can multiply by the inverse operator, because it never vanishes
(which is where we use the fact that \(\hat{V}\) is small), so we have \[|n\rangle=\text{}_{0}\langle n|
n\rangle\left[\mathbb{I}-\frac{\hat{Q}_{n}}{E_{n}^{0}-\hat{H}_{0}}\left(\hat{V}-\Delta
E_{n}\right)\right]^{-1}|n\rangle_{0}.\] As a convention we
choose \(\text{}_{0}\langle n
|n\rangle=1\) and normalize the true wavefunction \(|n\rangle\) only at the end. This
simplifies many places in the calculation, but one needs to remember
that \(\langle n | n\rangle \neq 1\)
now.
\[\text { so }\boxed{|n\rangle=[\mathbb{I}-\frac{\hat{Q}_{n}}{E_{n}^{0}-\hat{H}_{0}}(\hat{V}-\Delta E_{n})]^{-1}|n\rangle_{0}.}\]
By expanding the inverse operator as a geometric series, we generate the perturbation theory expansion. This gives us \[\begin{aligned} |n\rangle= & |n\rangle_{0}+\frac{\hat{Q}_{n}}{E_{n}^{0}-\hat{H}_{0}}\left(\hat{V}-\underbrace{\Delta E_{n}}_{\text{drop}}\right)|n\rangle_ 0+\frac{\hat{Q}_{n}}{E_{n}^{0}-\hat{H}_{0}}\left(\hat{V}-\Delta E_{n}\right) \frac{\hat{Q}}{E_{n}^{0}-\hat{H}_{0}}\left(\hat{V}-\underbrace{\Delta E_{n}}_{\text{drop}}\right)|n\rangle_{0} \\ & +\frac{\hat{Q}_{n}}{E_{n}^{0}-\hat{H}_{0}}\left(\hat{V}-\Delta E_{n}\right) \frac{\hat{Q}_{n}}{E_{n}^{0}-\hat{H}_{0}}\left(\hat{V}-\Delta E_{n}\right) \frac{\hat{Q}_{n}}{E_{n}^{0}-\hat{H}_{0}}\left(\hat{V}-\underbrace{\Delta E_{n}}_{\text{drop}}\right)|n\rangle_{0}+\cdots \end{aligned}\]
We can always drop the last \(\Delta E_{n}\) term since \[\Delta E_{n}=\text { number } \quad \text{ and } \quad \hat{Q}_{n} \Delta E_{n}|n\rangle_{0}=0 \text { always. }\]
To perform the perturbation theory expansion, we write \[\begin{aligned} & |n\rangle=\sum_{m=0}^{\infty}|n\rangle^{(m)},\quad\text{with} \quad|n\rangle^{(0)}=|n\rangle_{0} \quad \text { and the index denotes the powers of } \hat{V} \quad\text{in the expression}\\ & E_{n}=\sum_{m=0}^{\infty} E_{n}^{(m)} \quad E_{n}^{(0)}=E_{n}^{0} \\ & \Delta E_{n}=\sum_{m=1}^{\infty} E_{n}^{(m)} \end{aligned}\] We need to know the results up to \(E_{n}^{(m)}\) in order to find \(|n\rangle^{(m+1)}\).
The first few terms in the expansion are then \[\begin{aligned} &|n\rangle^{(0)}=|n\rangle_{0} \\ &|n\rangle^{(1)}= \frac{\hat{Q_{n}}}{\hat{E}_{n}^{0}-\hat{H}_{0}} \hat{V}|n\rangle_{0}, \quad|n\rangle^{(2)}=\frac{\hat{Q}_{n}}{E_{n}^{0}-\hat{H}_{0}}\left(\hat{V}-E_{n}^{(1)}\right) \frac{\hat{Q}_{n}}{E_{n}^{0}-H_{0}} \hat{V}|n\rangle_{0}, \\ &(n\rangle^{(3)}=\frac{\hat{Q}_{n}}{E_{n}^{0}-\hat{H}_{0}}\left(-E_{n}^{(2)}\right) \frac{\hat{Q}_{n}}{E_{n}^{0}-\hat{H}_{0}} \hat{V}|n\rangle_{0}+\frac{\hat{Q}_{n}}{E_{n}^{0}-H_{0}}\left(\hat{V}-E_{n}^{(1)}\right) \frac{\hat{Q}_{n}}{E_{n}^{0}- \hat{H}_{0}}\left(\hat{V}- E_{n}^{(0)}\right) \frac{\hat{Q}_{n}}{E_{n}^{0}-\hat{H}_{0}} \hat{V}|n\rangle_{0}, \end{aligned}\] and so on.
How do we find \(E_{n}^{(m)}\) ? Note that we have \[\begin{aligned} &\left(E_{n}^{0}-\hat{H} _0\right)|n\rangle=\left(\hat{V}-\Delta E_{n}\right)|n\rangle \quad \text { multiply by } {}_0\langle n| \\ &{ }_{0}\langle n|( E_{n}^{0}- \hat{H}_{0})|n\rangle={}_0\langle n| (\hat{V}-\Delta E_{n})|n\rangle,\quad\text{which gives us} \\ &\Delta E_n = \frac{{}_0\langle n|\hat{V}|n\rangle}{{}_0 \langle n|n \rangle} = {}_0 \langle n|\hat{V}|n, \rangle, \quad \text{since } {}_0\langle n|n\rangle =1. \end{aligned}\] So \(\quad \sum_{m=1}^{\infty} E_{n}^{(m)}=\sum_{m=0}^{\infty}\langle n| \hat{V}|n\rangle^{(m)}\). By matching powers of \(\hat{V}\), we get \[\boxed{ E_{n}^{(m)}={}_0\langle n| \hat{V}|n\rangle^{(m-1)}}\]
So, we work out the first few orders: \[\begin{aligned} & E_{n}^{(1)}={}_0\langle n| \hat{V}|n\rangle_{0}= \boxed{V_{n n}} \\ & |n\rangle^{(1)}=\frac{\hat{Q}_{n}}{E_{n}^{0}-\hat{H}_{0}} \hat{V}|n\rangle_{0}=\sum_{m \neq n} \frac{V_{m n}}{E_{n}^{0}-E_{m}^{0}}|m\rangle_{0}. \end{aligned}\] For \(m=2\) and 3, we have \[\begin{aligned} E_{n}^{(2)} & =\sum_{m \neq n} \frac{V_{n m} V_{m n}}{E_{n}^{0}-E_{m}^{0}}=\sum_{m \neq n} \frac{\left|V_{n m}\right|^{2}}{E_{n}^{0}-E_{m}^{0}}= \boxed{\sum_{m}{}^{\prime} \frac{\left|V_{n m}\right|^{2}}{E_{n}^{0}-E_{m}^{0}}}\text { prime } \Rightarrow m \neq n \\ |n\rangle^{(2)} & =\frac{\hat{Q}_{n}\left(\hat{V}-E_{n}^{(1)}\right)}{E_{n}^{0}-\hat{H}_{0}} \frac{\hat{Q}_{n}}{E_{n}^{0}-\hat{H}_{0}} \hat{V}|n\rangle_{0} \\ & =\sum_{m \neq n} \sum_{m^{\prime} \neq n}\left|m^{\prime}\right\rangle_{0} \frac{(V_{m^{\prime} m}-E_{n}^{(1)} \delta_{m m \prime})}{E_{n}^{0}-E_{m^{\prime} }^{0}} \frac{V_{m n}}{E_{n}^{0}-E_{m}^{0}}, \quad E_{n}^{(1)}=V_{n n}\\ E_{n}^{(3)} & =\sum_{m \neq n} \sum_{m^{\prime} \neq n} \frac{V_{n m'} V_{m^{\prime} m} V_{m n}}{\left(E_{n}^{0}-E_{m^{\prime}}^{0}\right)\left(E_{n}^{0}-E_{m}^{0}\right)}-V_{n n} \sum_{m \neq n} \frac{\left|V_{n m}\right|^{2}}{\left(E_{n}^{0}-E_{m}^{0}\right)^{2}}\\ &= \boxed{\sum_{m}{}^{\prime} \sum_{m^{\prime}}{}^{\prime \prime} \frac{V_{n m^\prime} V_{m^\prime m} V_{m n}}{\left(E_{n}^{0}-E_{m '}^{0}\right)\left(E_{n}^{0}-E_{m}^{0}\right)}-V_{n n} \sum_{m}{}^{\prime} \frac{\left|V_{n m}\right|^{2}}{\left(E_{n}^{0}-E_{m}^{0}\right)^{2}}} \end{aligned}\]
This process can be continued to arbitrary order (on the HW you will examine through 4th order.)