From f648dab1359c72731711b61b8ad3e0b62615c0ce Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Sat, 20 Feb 2021 17:53:24 +0100
Subject: Add "Time-independent perturbation theory"

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+% Time-independent perturbation theory
+
+
+# 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 accordingly 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. This is allowed, because
+$\ket{\psi_n^{(1)}} - c_n \ket{n}$ also satisfies the $\lambda^1$-order
+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 $\hat{H}_0 \ket{n} = \varepsilon_n \ket{n}$. Inserting the
+series form of $\ket{\psi_n^{(1)}}$ into the order-$\lambda^1$ equation
+gives us:
+
+$$\begin{aligned}
+    (\hat{H}_1 - E_n^{(1)}) \ket{n} + \sum_{m \neq n} c_m (\varepsilon_m - \varepsilon_n) \ket{m} = 0
+\end{aligned}$$
+
+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 correction to the energy $E_n^{(2)}$, 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.
+Suppose that this eigenvalue problem has been solved, and that
+$\ket{n, g}$ are 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 $\ket{n,g}$ of course, which is
+explicitly excluded), then conveniently 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 some information missing about the components $c_d$ of the
+"good" states, in which case we must find these states some other way.
+
+An alternative way of determining these "good" states is also of
+interest if there is no degeneracy in $M$, since such a shortcut would
+allow us use the formulae from non-degenerate perturbation theory
+straight away.
+
+The method is to find a Hermitian operator $\hat{L}$ (usually using
+symmetry) which commutes with both $\hat{H}_0$ and $\hat{H}_1$:
+
+$$\begin{aligned}
+= [\hat{L}, \hat{H}_1] = 0
+\end{aligned}$$
+
+So that it shares its eigenstates with $\hat{H}_0$ (and $\hat{H}_1$),
+meaning at least $D$ of 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 $d_1 \neq d_2$:
+
+$$\begin{aligned}
+    \hat{L} \ket{n, b_1} = \ell_1 \ket{n, b_1}
+    \qquad
+    \hat{L} \ket{n, b_2} = \ell_2 \ket{n, b_2}
+\end{aligned}$$
+
+When this 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 $\hat{L}$.
-- 
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