Categories: Perturbation, Physics, Quantum mechanics.

# Time-independent perturbation theory

Time-independent perturbation theory, also known as 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.
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