---
title: "Time-independent perturbation theory"
firstLetter: "T"
publishDate: 2021-02-22
categories:
- Quantum mechanics
- Perturbation
- Physics

date: 2021-02-22T21:38:18+01:00
draft: false
markup: pandoc
---

# Time-independent perturbation theory

**Time-independent perturbation theory**, sometimes also called
**stationary state perturbation theory**, is a specific application of
perturbation theory to the time-independent Schrödinger
equation in quantum physics, for
Hamiltonians of the following form:

$$\begin{aligned}
    \hat{H} = \hat{H}_0 + \lambda \hat{H}_1
\end{aligned}$$

Where $\hat{H}_0$ is a Hamiltonian for which the time-independent
Schrödinger equation has a known solution, and $\hat{H}_1$ is a small
perturbing Hamiltonian. The eigenenergies $E_n$ and eigenstates
$\ket{\psi_n}$ of the composite problem are expanded in the
perturbation "bookkeeping" parameter $\lambda$:

$$\begin{aligned}
    \ket{\psi_n}
    &= \ket*{\psi_n^{(0)}} + \lambda \ket*{\psi_n^{(1)}} + \lambda^2 \ket*{\psi_n^{(2)}} + ...
    \\
    E_n
    &= E_n^{(0)} + \lambda E_n^{(1)} + \lambda^2 E_n^{(2)} + ...
\end{aligned}$$

Where $E_n^{(1)}$ and $\ket*{\psi_n^{(1)}}$ are called the **first-order
corrections**, and so on for higher orders. We insert this into the
Schrödinger equation:

$$\begin{aligned}
    \hat{H} \ket{\psi_n}
    &= \hat{H}_0 \ket*{\psi_n^{(0)}}
    + \lambda \big( \hat{H}_1 \ket*{\psi_n^{(0)}} + \hat{H}_0 \ket*{\psi_n^{(1)}} \big) \\
    &\qquad + \lambda^2 \big( \hat{H}_1 \ket*{\psi_n^{(1)}} + \hat{H}_0 \ket*{\psi_n^{(2)}} \big) + ...
    \\
    E_n \ket{\psi_n}
    &= E_n^{(0)} \ket*{\psi_n^{(0)}}
    + \lambda \big( E_n^{(1)} \ket*{\psi_n^{(0)}} + E_n^{(0)} \ket*{\psi_n^{(1)}} \big) \\
    &\qquad + \lambda^2 \big( E_n^{(2)} \ket*{\psi_n^{(0)}} + E_n^{(1)} \ket*{\psi_n^{(1)}} + E_n^{(0)} \ket*{\psi_n^{(2)}} \big) + ...
\end{aligned}$$

If we collect the terms according to the order of $\lambda$, we arrive
at the following endless series of equations, of which in practice only
the first three are typically used:

$$\begin{aligned}
    \hat{H}_0 \ket*{\psi_n^{(0)}}
    &= E_n^{(0)} \ket*{\psi_n^{(0)}}
    \\
    \hat{H}_1 \ket*{\psi_n^{(0)}} + \hat{H}_0 \ket*{\psi_n^{(1)}}
    &= E_n^{(1)} \ket*{\psi_n^{(0)}} + E_n^{(0)} \ket*{\psi_n^{(1)}}
    \\
    \hat{H}_1 \ket*{\psi_n^{(1)}} + \hat{H}_0 \ket*{\psi_n^{(2)}}
    &= E_n^{(2)} \ket*{\psi_n^{(0)}} + E_n^{(1)} \ket*{\psi_n^{(1)}} + E_n^{(0)} \ket*{\psi_n^{(2)}}
    \\
    ...
    &= ...
\end{aligned}$$

The first equation is the unperturbed problem, which we assume has
already been solved, with eigenvalues $E_n^{(0)} = \varepsilon_n$ and
eigenvectors $\ket*{\psi_n^{(0)}} = \ket{n}$:

$$\begin{aligned}
    \hat{H}_0 \ket{n} = \varepsilon_n \ket{n}
\end{aligned}$$

The approach to solving the other two equations varies depending on
whether this $\hat{H}_0$ has a degenerate spectrum or not.


## Without degeneracy

We start by assuming that there is no degeneracy, in other words, each
$\varepsilon_n$ corresponds to one $\ket{n}$. At order $\lambda^1$, we
rewrite the equation as follows:

$$\begin{aligned}
    (\hat{H}_1 - E_n^{(1)}) \ket{n} + (\hat{H}_0 - \varepsilon_n) \ket*{\psi_n^{(1)}} = 0
\end{aligned}$$

Since $\ket{n}$ form a complete basis, we can express
$\ket*{\psi_n^{(1)}}$ in terms of them:

$$\begin{aligned}
    \ket*{\psi_n^{(1)}} = \sum_{m \neq n} c_m \ket{m}
\end{aligned}$$

Importantly, $n$ has been removed from the summation to prevent dividing
by zero later. We are allowed to do this, because
$\ket*{\psi_n^{(1)}} - c_n \ket{n}$ also satisfies the order-$\lambda^1$
equation for any value of $c_n$, as demonstrated here:

$$\begin{aligned}
    (\hat{H}_1 - E_n^{(1)}) \ket{n} + (\hat{H}_0 - \varepsilon_n) \ket*{\psi_n^{(1)}} - (\varepsilon_n - \varepsilon_n) c_n \ket{n} = 0
\end{aligned}$$

Where we used $\hat{H}_0 \ket{n} = \varepsilon_n \ket{n}$.
We insert the series form of $\ket*{\psi_n^{(1)}}$ into the $\lambda^1$-equation:

$$\begin{aligned}
    (\hat{H}_1 - E_n^{(1)}) \ket{n} + \sum_{m \neq n} c_m (\varepsilon_m - \varepsilon_n) \ket{m} = 0
\end{aligned}$$

We then put an arbitrary basis vector $\bra{k}$ in front of this
equation to get:

$$\begin{aligned}
    \matrixel{k}{\hat{H}_1}{n} - E_n^{(1)} \braket{k}{n} + \sum_{m \neq n} c_m (\varepsilon_m - \varepsilon_n) \braket{k}{m} = 0
\end{aligned}$$

Suppose that $k = n$. Since $\ket{n}$ form an orthonormal basis, we end
up with:

$$\begin{aligned}
    \boxed{
        E_n^{(1)} = \matrixel{n}{\hat{H}_1}{n}
    }
\end{aligned}$$

In other words, the first-order energy correction $E_n^{(1)}$ is the
expectation value of the perturbation $\hat{H}_1$ for the unperturbed
state $\ket{n}$.

Suppose now that $k \neq n$, then only one term of the summation
survives, and we are left with the following equation, which tells us
$c_l$:

$$\begin{aligned}
    \matrixel{k}{\hat{H}_1}{n} + c_k (\varepsilon_k - \varepsilon_n) = 0
\end{aligned}$$

We isolate this result for $c_k$ and insert it into the series form of
$\ket*{\psi_n^{(1)}}$ to get the full first-order correction to the wave
function:

$$\begin{aligned}
    \boxed{
        \ket*{\psi_n^{(1)}}
        = \sum_{m \neq n} \frac{\matrixel{m}{\hat{H}_1}{n}}{\varepsilon_n - \varepsilon_m} \ket{m}
    }
\end{aligned}$$

Here it is clear why this is only valid in the non-degenerate case:
otherwise we would divide by zero in the denominator.

Next, to find the second-order energy correction $E_n^{(2)}$,
we take the corresponding equation and put $\bra{n}$ in front of it:

$$\begin{aligned}
    \matrixel*{n}{\hat{H}_1}{\psi_n^{(1)}} + \matrixel*{n}{\hat{H}_0}{\psi_n^{(2)}}
    &= E_n^{(2)} \braket{n}{n} + E_n^{(1)} \braket*{n}{\psi_n^{(1)}} + \varepsilon_n \braket*{n}{\psi_n^{(2)}}
\end{aligned}$$

Because $\hat{H}_0$ is Hermitian, we know that
$\matrixel*{n}{\hat{H}_0}{\psi_n^{(2)}} = \varepsilon_n \braket*{n}{\psi_n^{(2)}}$,
i.e. we apply it to the bra, which lets us eliminate two terms. Also,
since $\ket{n}$ is normalized, we find:

$$\begin{aligned}
    E_n^{(2)}
    = \matrixel*{n}{\hat{H}_1}{\psi_n^{(1)}} - E_n^{(1)} \braket*{n}{\psi_n^{(1)}}
\end{aligned}$$

We explicitly removed the $\ket{n}$-dependence of $\ket*{\psi_n^{(1)}}$,
so the last term is zero. By simply inserting our result for
$\ket*{\psi_n^{(1)}}$, we thus arrive at:

$$\begin{aligned}
    \boxed{
        E_n^{(2)}
        = \sum_{m \neq n} \frac{\big| \matrixel{m}{\hat{H}_1}{n} \big|^2}{\varepsilon_n - \varepsilon_m}
    }
\end{aligned}$$

In practice, it is not particulary useful to calculate more corrections.


## With degeneracy

If $\varepsilon_n$ is $D$-fold degenerate, then its eigenstate could be
any vector $\ket{n, d}$ from the corresponding $D$-dimensional
eigenspace:

$$\begin{aligned}
    \hat{H}_0 \ket{n} = \varepsilon_n \ket{n}
    \quad \mathrm{where} \quad
    \ket{n}
    = \sum_{d = 1}^{D} c_{d} \ket{n, d}
\end{aligned}$$

In general, adding the perturbation $\hat{H}_1$ will *lift* the
degeneracy, meaning the perturbed states will be non-degenerate. In the
limit $\lambda \to 0$, these $D$ perturbed states change into $D$
orthogonal states which are all valid $\ket{n}$.

However, the $\ket{n}$ that they converge to are not arbitrary: only
certain unperturbed eigenstates are "good" states. Without $\hat{H}_1$,
this distinction is irrelevant, but in the perturbed case it will turn
out to be important.

For now, we write $\ket{n, d}$ to refer to any orthonormal set of
vectors in the eigenspace of $\varepsilon_n$ (not necessarily the "good"
ones), and $\ket{n}$ to denote any linear combination of these. We then
take the equation at order $\lambda^1$ and prepend an arbitrary
eigenspace basis vector $\bra{n, \delta}$:

$$\begin{aligned}
    \matrixel{n, \delta}{\hat{H}_1}{n} + \matrixel*{n, \delta}{\hat{H}_0}{\psi_n^{(1)}}
    &= E_n^{(1)} \braket{n, \delta}{n} + \varepsilon_n \braket*{n, \delta}{\psi_n^{(1)}}
\end{aligned}$$

Since $\hat{H}_0$ is Hermitian, we use the same trick as before to
reduce the problem to:

$$\begin{aligned}
    \matrixel{n, \delta}{\hat{H}_1}{n}
    &= E_n^{(1)} \braket{n, \delta}{n}
\end{aligned}$$

We express $\ket{n}$ as a linear combination of the eigenbasis vectors
$\ket{n, d}$ to get:

$$\begin{aligned}
    \sum_{d = 1}^{D} c_d \matrixel{n, \delta}{\hat{H}_1}{n, d}
    = E_n^{(1)} \sum_{d = 1}^{D} c_d \braket{n, \delta}{n, d}
    = c_{\delta} E_n^{(1)}
\end{aligned}$$

Let us now interpret the summation terms as matrix elements
$M_{\delta, d}$:

$$\begin{aligned}
    M_{\delta, d} = \matrixel{n, \delta}{\hat{H}_1}{n, d}
\end{aligned}$$

By varying the value of $\delta$ from $1$ to $D$, we end up with
equations of the form:

$$\begin{aligned}
    \begin{bmatrix}
        M_{1, 1} & \cdots & M_{1, D} \\
        \vdots & \ddots & \vdots \\
        M_{D, 1} & \cdots & M_{D, D}
    \end{bmatrix}
    \begin{bmatrix}
        c_1 \\ \vdots \\ c_D
    \end{bmatrix}
    = E_n^{(1)}
    \begin{bmatrix}
        c_1 \\ \vdots \\ c_D
    \end{bmatrix}
\end{aligned}$$

This is an eigenvalue problem for $E_n^{(1)}$, where $c_d$ are the
components of the eigenvectors which represent the "good" states.
After solving this, let $\ket{n, g}$ be the resulting "good" states.
Then, as long as $E_n^{(1)}$ is a non-degenerate eigenvalue of $M$:

$$\begin{aligned}
    \boxed{
        E_{n, g}^{(1)} = \matrixel{n, g}{\hat{H}_1}{n, g}
    }
\end{aligned}$$

Which is the same as in the non-degenerate case! Even better, the
first-order wave function correction is also unchanged:

$$\begin{aligned}
    \boxed{
        \ket*{\psi_{n,g}^{(1)}}
        = \sum_{m \neq (n, g)} \frac{\matrixel{m}{\hat{H}_1}{n, g}}{\varepsilon_n - \varepsilon_m} \ket{m}
    }
\end{aligned}$$

This works because the matrix $M$ is diagonal in the $\ket{n, g}$-basis,
such that when $\ket{m}$ is any vector $\ket{n, \gamma}$ in the
$\ket{n}$-eigenspace (except for $\ket{n,g}$, which is
explicitly excluded), then the corresponding numerator
$\matrixel{n, \gamma}{\hat{H}_1}{n, g} = M_{\gamma, g} = 0$, so the term
does not contribute.

If any of the eigenvalues $E_n^{(1)}$ of $M$ are degenerate, then there
is still information missing about the components $c_d$ of the
"good" states, in which case we must find them some other way.

Such an alternative way of determining these "good" states is also of
interest even if there is no degeneracy in $M$, since such a shortcut would
allow us to use the formulae from non-degenerate perturbation theory
straight away.

The trick is to find a Hermitian operator $\hat{L}$ (usually using
symmetries of the system) which commutes with both $\hat{H}_0$ and $\hat{H}_1$:

$$\begin{aligned}
    \comm*{\hat{L}}{\hat{H}_0} = \comm*{\hat{L}}{\hat{H}_1} = 0
\end{aligned}$$

So that it shares its eigenstates with $\hat{H}_0$ (and $\hat{H}_1$),
meaning all the vectors of the $D$-dimensional
$\ket{n}$-eigenspace are also eigenvectors of $\hat{L}$.

The crucial part, however, is that $\hat{L}$ must be chosen such that
$\ket{n, d_1}$ and $\ket{n, d_2}$ have distinct eigenvalues
$\ell_1 \neq \ell_2$ for $d_1 \neq d_2$:

$$\begin{aligned}
    \hat{L} \ket{n, d_1} = \ell_1 \ket{n, d_1}
    \qquad
    \hat{L} \ket{n, d_2} = \ell_2 \ket{n, d_2}
\end{aligned}$$

When this condition holds for any orthogonal choice of $\ket{n, d_1}$ and
$\ket{n, d_2}$, then these specific eigenvectors of $\hat{L}$ are the
"good states", for any valid choice of $\hat{L}$.



## References
1.  D.J. Griffiths, D.F. Schroeter,
    *Introduction to quantum mechanics*, 3rd edition,
    Cambridge.