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Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough.

These are notes for the UofT course PHY1520, Graduate Quantum Mechanics, taught by Prof. Paramekanti, covering [1] ch. 5 content.

Perturbation theory

Given a $2 \times 2$ Hamiltonian $H = H_0 + V$, where

$$\begin{equation}\label{eqn:qmLecture20:20} H = \begin{bmatrix} a & c \\ c^\conj & b \end{bmatrix} \end{equation}$$

which has eigenvalues

$$\begin{equation}\label{eqn:qmLecture20:40} \lambda_\pm = \frac{a + b}{2} \pm \sqrt{ \lr{ \frac{a – b}{2}}^2 + \Abs{c}^2 }. \end{equation}$$

If $c = 0$,

$$\begin{equation}\label{eqn:qmLecture20:60} H_0 = \begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix}, \end{equation}$$

so

$$\begin{equation}\label{eqn:qmLecture20:80} V = \begin{bmatrix} 0 & c \\ c^\conj & 0 \end{bmatrix}. \end{equation}$$

Suppose that $\Abs{c} \ll \Abs{a – b}$, then

$$\begin{equation}\label{eqn:qmLecture20:100} \lambda_\pm \approx \frac{a + b}{2} \pm \Abs{ \frac{a – b}{2} } \lr{ 1 + 2 \frac{\Abs{c}^2}{\Abs{a – b}^2} }. \end{equation}$$

If $a > b$, then

$$\begin{equation}\label{eqn:qmLecture20:120} \lambda_\pm \approx \frac{a + b}{2} \pm \frac{a – b}{2} \lr{ 1 + 2 \frac{\Abs{c}^2}{\lr{a – b}^2} }. \end{equation}$$

$$\begin{equation}\label{eqn:qmLecture20:140} \begin{aligned} \lambda_{+} &= \frac{a + b}{2} + \frac{a – b}{2} \lr{ 1 + 2 \frac{\Abs{c}^2}{\lr{a – b}^2} } \\ &= a + \lr{a – b} \frac{\Abs{c}^2}{\lr{a – b}^2} \\ &= a + \frac{\Abs{c}^2}{a – b}, \end{aligned} \end{equation}$$

and
$$\begin{equation}\label{eqn:qmLecture20:680} \begin{aligned} \lambda_{-} &= \frac{a + b}{2} – \frac{a – b}{2} \lr{ 1 + 2 \frac{\Abs{c}^2}{\lr{a – b}^2} } \\ &= b + \lr{a – b} \frac{\Abs{c}^2}{\lr{a – b}^2} \\ &= b + \frac{\Abs{c}^2}{a – b}. \end{aligned} \end{equation}$$

This adiabatic evolution displays a “level repulsion”, quadradic in $\Abs{c}$ as sketched in fig. 1, and is described as a non-degenerate perbutation.

fig. 1. Adiabatic (non-degenerate) perturbation

If $\Abs{c} \gg \Abs{a -b}$, then

$$\begin{equation}\label{eqn:qmLecture20:160} \begin{aligned} \lambda_\pm &= \frac{a + b}{2} \pm \Abs{c} \sqrt{ 1 + \inv{\Abs{c}^2} \lr{ \frac{a – b}{2}}^2 } \\ &\approx \frac{a + b}{2} \pm \Abs{c} \lr{ 1 + \inv{2 \Abs{c}^2} \lr{ \frac{a – b}{2}}^2 } \\ &= \frac{a + b}{2} \pm \Abs{c} \pm \frac{\lr{a – b}^2}{8 \Abs{c}}. \end{aligned} \end{equation}$$

Here we loose the adiabaticity, and have “level repulsion” that is linear in $\Abs{c}$, as sketched in fig. 2. We no longer have the sign of $a – b$ in the expansion. This is described as a degenerate perbutation.

fig. 2. Degenerate perbutation

General non-degenerate perturbation

Given an unperturbed system with solutions of the form

$$\begin{equation}\label{eqn:qmLecture20:180} H_0 \ket{n^{(0)}} = E_n^{(0)} \ket{n^{(0)}}, \end{equation}$$

we want to solve the perturbed Hamiltonian equation

$$\begin{equation}\label{eqn:qmLecture20:200} \lr{ H_0 + \lambda V } \ket{ n } = \lr{ E_n^{(0)} + \Delta n } \ket{n}. \end{equation}$$

Here $\Delta n$ is an energy shift as that goes to zero as $\lambda \rightarrow 0$. We can write this as

$$\begin{equation}\label{eqn:qmLecture20:220} \lr{ E_n^{(0)} – H_0 } \ket{ n } = \lr{ \lambda V – \Delta_n } \ket{n}. \end{equation}$$

We are hoping to iterate with application of the inverse to an initial estimate of $\ket{n}$

$$\begin{equation}\label{eqn:qmLecture20:240} \ket{n} = \lr{ E_n^{(0)} – H_0 }^{-1} \lr{ \lambda V – \Delta_n } \ket{n}. \end{equation}$$

This gets us into trouble if $\lambda \rightarrow 0$, which can be fixed by using

$$\begin{equation}\label{eqn:qmLecture20:260} \ket{n} = \lr{ E_n^{(0)} – H_0 }^{-1} \lr{ \lambda V – \Delta_n } \ket{n} + \ket{ n^{(0)} }, \end{equation}$$

which can be seen to be a solution to $\ref{eqn:qmLecture20:220}$. We want to ask if

$$\begin{equation}\label{eqn:qmLecture20:280} \lr{ \lambda V – \Delta_n } \ket{n} , \end{equation}$$

contains a bit of $\ket{ n^{(0)} }$? To determine this act with $\bra{n^{(0)}}$ on the left

$$\begin{equation}\label{eqn:qmLecture20:300} \begin{aligned} \bra{ n^{(0)} } \lr{ \lambda V – \Delta_n } \ket{n} &= \bra{ n^{(0)} } \lr{ E_n^{(0)} – H_0 } \ket{n} \\ &= \lr{ E_n^{(0)} – E_n^{(0)} } \braket{n^{(0)}}{n} \\ &= 0. \end{aligned} \end{equation}$$

This shows that $\ket{n}$ is entirely orthogonal to $\ket{n^{(0)}}$.

Define a projection operator

$$\begin{equation}\label{eqn:qmLecture20:320} P_n = \ket{n^{(0)}}\bra{n^{(0)}}, \end{equation}$$

which has the idempotent property $P_n^2 = P_n$ that we expect of a projection operator.

Define a rejection operator
$$\begin{equation}\label{eqn:qmLecture20:340} \overline{{P}}_n = 1 – \ket{n^{(0)}}\bra{n^{(0)}} = \sum_{m \ne n} \ket{m^{(0)}}\bra{m^{(0)}}. \end{equation}$$

Because $\ket{n}$ has no component in the direction $\ket{n^{(0)}}$, the rejection operator can be inserted much like we normally do with the identity operator, yielding

$$\begin{equation}\label{eqn:qmLecture20:360} \ket{n}’ = \lr{ E_n^{(0)} – H_0 }^{-1} \overline{{P}}_n \lr{ \lambda V – \Delta_n } \ket{n} + \ket{ n^{(0)} }, \end{equation}$$

valid for any initial $\ket{n}$.

Power series perturbation expansion

Instead of iterating, suppose that the unknown state and unknown energy difference operator can be expanded in a $\lambda$ power series, say

$$\begin{equation}\label{eqn:qmLecture20:380} \ket{n} = \ket{n_0} + \lambda \ket{n_1} + \lambda^2 \ket{n_2} + \lambda^3 \ket{n_3} + \cdots \end{equation}$$

and

$$\begin{equation}\label{eqn:qmLecture20:400} \Delta_{n} = \Delta_{n_0} + \lambda \Delta_{n_1} + \lambda^2 \Delta_{n_2} + \lambda^3 \Delta_{n_3} + \cdots \end{equation}$$

We usually interpret functions of operators in terms of power series expansions. In the case of $\lr{ E_n^{(0)} – H_0 }^{-1}$, we have a concrete interpretation when acting on one of the unpertubed eigenstates

$$\begin{equation}\label{eqn:qmLecture20:420} \inv{ E_n^{(0)} – H_0 } \ket{m^{(0)}} = \inv{ E_n^{(0)} – E_m^0 } \ket{m^{(0)}}. \end{equation}$$

This gives

$$\begin{equation}\label{eqn:qmLecture20:440} \ket{n} = \inv{ E_n^{(0)} – H_0 } \sum_{m \ne n} \ket{m^{(0)}}\bra{m^{(0)}} \lr{ \lambda V – \Delta_n } \ket{n} + \ket{ n^{(0)} }, \end{equation}$$

or

$$\begin{equation}\label{eqn:qmLecture20:460} \boxed{ \ket{n} = \ket{ n^{(0)} } + \sum_{m \ne n} \frac{\ket{m^{(0)}}\bra{m^{(0)}}} { E_n^{(0)} – E_m^{(0)} } \lr{ \lambda V – \Delta_n } \ket{n}. } \end{equation}$$

From $\ref{eqn:qmLecture20:220}$, note that

$$\begin{equation}\label{eqn:qmLecture20:500} \Delta_n = \frac{\bra{n^{(0)}} \lambda V \ket{n}}{\braket{n^0}{n}}, \end{equation}$$

however, we will normalize by setting $\braket{n^0}{n} = 1$, so

$$\begin{equation}\label{eqn:qmLecture20:521} \boxed{ \Delta_n = \bra{n^{(0)}} \lambda V \ket{n}. } \end{equation}$$

to $O(\lambda^0)$

If all $\lambda^n, n > 0$ are zero, then we have

\label{eqn:qmLecture20:780}
$$\begin{equation}\label{eqn:qmLecture20:740} \ket{n_0} = \ket{ n^{(0)} } + \sum_{m \ne n} \frac{\ket{m^{(0)}}\bra{m^{(0)}}} { E_n^{(0)} – E_m^{(0)} } \lr{ – \Delta_{n_0} } \ket{n_0} \end{equation}$$
$$\begin{equation}\label{eqn:qmLecture20:800} \Delta_{n_0} \braket{n^{(0)}}{n_0} = 0 \end{equation}$$

so

$$\begin{equation}\label{eqn:qmLecture20:540} \begin{aligned} \ket{n_0} &= \ket{n^{(0)}} \\ \Delta_{n_0} &= 0. \end{aligned} \end{equation}$$

to $O(\lambda^1)$

Requiring identity for all $\lambda^1$ terms means

$$\begin{equation}\label{eqn:qmLecture20:760} \ket{n_1} \lambda = \sum_{m \ne n} \frac{\ket{m^{(0)}}\bra{m^{(0)}}} { E_n^{(0)} – E_m^{(0)} } \lr{ \lambda V – \Delta_{n_1} \lambda } \ket{n_0}, \end{equation}$$

so

$$\begin{equation}\label{eqn:qmLecture20:560} \ket{n_1} = \sum_{m \ne n} \frac{ \ket{m^{(0)}} \bra{ m^{(0)}} } { E_n^{(0)} – E_m^{(0)} } \lr{ V – \Delta_{n_1} } \ket{n_0}. \end{equation}$$

With the assumption that $\ket{n^{(0)}}$ is normalized, and with the shorthand

$$\begin{equation}\label{eqn:qmLecture20:600} V_{m n} = \bra{ m^{(0)}} V \ket{n^{(0)}}, \end{equation}$$

that is

$$\begin{equation}\label{eqn:qmLecture20:580} \begin{aligned} \ket{n_1} &= \sum_{m \ne n} \frac{ \ket{m^{(0)}} } { E_n^{(0)} – E_m^{(0)} } V_{m n} \\ \Delta_{n_1} &= \bra{ n^{(0)} } V \ket{ n^0} = V_{nn}. \end{aligned} \end{equation}$$

to $O(\lambda^2)$

The second order perturbation states are found by selecting only the $\lambda^2$ contributions to

$$\begin{equation}\label{eqn:qmLecture20:820} \lambda^2 \ket{n_2} = \sum_{m \ne n} \frac{\ket{m^{(0)}}\bra{m^{(0)}}} { E_n^{(0)} – E_m^{(0)} } \lr{ \lambda V – (\lambda \Delta_{n_1} + \lambda^2 \Delta_{n_2}) } \lr{ \ket{n_0} + \lambda \ket{n_1} }. \end{equation}$$

Because $\ket{n_0} = \ket{n^{(0)}}$, the $\lambda^2 \Delta_{n_2}$ is killed, leaving

$$\begin{equation}\label{eqn:qmLecture20:840} \begin{aligned} \ket{n_2} &= \sum_{m \ne n} \frac{\ket{m^{(0)}}\bra{m^{(0)}}} { E_n^{(0)} – E_m^{(0)} } \lr{ V – \Delta_{n_1} } \ket{n_1} \\ &= \sum_{m \ne n} \frac{\ket{m^{(0)}}\bra{m^{(0)}}} { E_n^{(0)} – E_m^{(0)} } \lr{ V – \Delta_{n_1} } \sum_{l \ne n} \frac{ \ket{l^{(0)}} } { E_n^{(0)} – E_l^{(0)} } V_{l n}, \end{aligned} \end{equation}$$

which can be written as

$$\begin{equation}\label{eqn:qmLecture20:620} \ket{n_2} = \sum_{l,m \ne n} \ket{m^{(0)}} \frac{V_{m l} V_{l n}} { \lr{ E_n^{(0)} – E_m^{(0)} } \lr{ E_n^{(0)} – E_l^{(0)} } } – \sum_{m \ne n} \ket{m^{(0)}} \frac{V_{n n} V_{m n}} { \lr{ E_n^{(0)} – E_m^{(0)} }^2 }. \end{equation}$$

For the second energy perturbation we have

$$\begin{equation}\label{eqn:qmLecture20:860} \lambda^2 \Delta_{n_2} = \bra{n^{(0)}} \lambda V \lr{ \lambda \ket{n_1} }, \end{equation}$$

or

$$\begin{equation}\label{eqn:qmLecture20:880} \begin{aligned} \Delta_{n_2} &= \bra{n^{(0)}} V \ket{n_1} \\ &= \bra{n^{(0)}} V \sum_{m \ne n} \frac{ \ket{m^{(0)}} } { E_n^{(0)} – E_m^{(0)} } V_{m n}. \end{aligned} \end{equation}$$

That is

$$\begin{equation}\label{eqn:qmLecture20:900} \Delta_{n_2} = \sum_{m \ne n} \frac{V_{n m} V_{m n} }{E_n^{(0)} – E_m^{(0)}}. \end{equation}$$

to $O(\lambda^3)$

Similarily, it can be shown that

$$\begin{equation}\label{eqn:qmLecture20:640} \Delta_{n_3} = \sum_{l, m \ne n} \frac{V_{n m} V_{m l} V_{l n} }{ \lr{ E_n^{(0)} – E_m^{(0)} } \lr{ E_n^{(0)} – E_l^{(0)} } } – \sum_{ m \ne n} \frac{V_{n m} V_{n n} V_{m n} }{ \lr{ E_n^{(0)} – E_m^{(0)} }^2 }. \end{equation}$$

In general, the energy perturbation is given by

$$\begin{equation}\label{eqn:qmLecture20:660} \Delta_n^{(l)} = \bra{n^{(0)}} V \ket{n^{(l-1)}}. \end{equation}$$

References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.