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<h1 id="time-independent-perturbation-theory">Time-independent perturbation theory</h1>
<p><em>Time-independent perturbation theory</em>, sometimes also called <em>stationary state perturbation theory</em>, is a specific application of perturbation theory to the time-independent Schrödinger equation in quantum physics, for Hamiltonians of the following form:</p>
<p><span class="math display">\[\begin{aligned}
\hat{H} = \hat{H}_0 + \lambda \hat{H}_1
\end{aligned}\]</span></p>
<p>Where <span class="math inline">\(\hat{H}_0\)</span> is a Hamiltonian for which the time-independent Schrödinger equation has a known solution, and <span class="math inline">\(\hat{H}_1\)</span> is a small perturbing Hamiltonian. The eigenenergies <span class="math inline">\(E_n\)</span> and eigenstates <span class="math inline">\(\ket{\psi_n}\)</span> of the composite problem are expanded accordingly in the perturbation “bookkeeping” parameter <span class="math inline">\(\lambda\)</span>:</p>
<p><span class="math display">\[\begin{aligned}
\ket{\psi_n}
&= \ket{\psi_n^{(0)}} + \lambda \ket{\psi_n^{(1)}} + \lambda^2 \ket{\psi_n^{(2)}} + ...
\\
E_n
&= E_n^{(0)} + \lambda E_n^{(1)} + \lambda^2 E_n^{(2)} + ...
\end{aligned}\]</span></p>
<p>Where <span class="math inline">\(E_n^{(1)}\)</span> and <span class="math inline">\(\ket{\psi_n^{(1)}}\)</span> are called the <em>first-order corrections</em>, and so on for higher orders. We insert this into the Schrödinger equation:</p>
<p><span class="math display">\[\begin{aligned}
\hat{H} \ket{\psi_n}
&= \hat{H}_0 \ket{\psi_n^{(0)}}
+ \lambda \big( \hat{H}_1 \ket{\psi_n^{(0)}} + \hat{H}_0 \ket{\psi_n^{(1)}} \big) \\
&\qquad + \lambda^2 \big( \hat{H}_1 \ket{\psi_n^{(1)}} + \hat{H}_0 \ket{\psi_n^{(2)}} \big) + ...
\\
E_n \ket{\psi_n}
&= E_n^{(0)} \ket{\psi_n^{(0)}}
+ \lambda \big( E_n^{(1)} \ket{\psi_n^{(0)}} + E_n^{(0)} \ket{\psi_n^{(1)}} \big) \\
&\qquad + \lambda^2 \big( E_n^{(2)} \ket{\psi_n^{(0)}} + E_n^{(1)} \ket{\psi_n^{(1)}} + E_n^{(0)} \ket{\psi_n^{(2)}} \big) + ...
\end{aligned}\]</span></p>
<p>If we collect the terms according to the order of <span class="math inline">\(\lambda\)</span>, we arrive at the following endless series of equations, of which in practice only the first three are typically used:</p>
<p><span class="math display">\[\begin{aligned}
\hat{H}_0 \ket{\psi_n^{(0)}}
&= E_n^{(0)} \ket{\psi_n^{(0)}}
\\
\hat{H}_1 \ket{\psi_n^{(0)}} + \hat{H}_0 \ket{\psi_n^{(1)}}
&= E_n^{(1)} \ket{\psi_n^{(0)}} + E_n^{(0)} \ket{\psi_n^{(1)}}
\\
\hat{H}_1 \ket{\psi_n^{(1)}} + \hat{H}_0 \ket{\psi_n^{(2)}}
&= E_n^{(2)} \ket{\psi_n^{(0)}} + E_n^{(1)} \ket{\psi_n^{(1)}} + E_n^{(0)} \ket{\psi_n^{(2)}}
\\
...
&= ...
\end{aligned}\]</span></p>
<p>The first equation is the unperturbed problem, which we assume has already been solved, with eigenvalues <span class="math inline">\(E_n^{(0)} = \varepsilon_n\)</span> and eigenvectors <span class="math inline">\(\ket{\psi_n^{(0)}} = \ket{n}\)</span>:</p>
<p><span class="math display">\[\begin{aligned}
\hat{H}_0 \ket{n} = \varepsilon_n \ket{n}
\end{aligned}\]</span></p>
<p>The approach to solving the other two equations varies depending on whether this <span class="math inline">\(\hat{H}_0\)</span> has a degenerate spectrum or not.</p>
<h2 id="without-degeneracy">Without degeneracy</h2>
<p>We start by assuming that there is no degeneracy, in other words, each <span class="math inline">\(\varepsilon_n\)</span> corresponds to one <span class="math inline">\(\ket{n}\)</span>. At order <span class="math inline">\(\lambda^1\)</span>, we rewrite the equation as follows:</p>
<p><span class="math display">\[\begin{aligned}
(\hat{H}_1 - E_n^{(1)}) \ket{n} + (\hat{H}_0 - \varepsilon_n) \ket{\psi_n^{(1)}} = 0
\end{aligned}\]</span></p>
<p>Since <span class="math inline">\(\ket{n}\)</span> form a complete basis, we can express <span class="math inline">\(\ket{\psi_n^{(1)}}\)</span> in terms of them:</p>
<p><span class="math display">\[\begin{aligned}
\ket{\psi_n^{(1)}} = \sum_{m \neq n} c_m \ket{m}
\end{aligned}\]</span></p>
<p>Importantly, <span class="math inline">\(n\)</span> has been removed from the summation to prevent dividing by zero later. This is allowed, because <span class="math inline">\(\ket{\psi_n^{(1)}} - c_n \ket{n}\)</span> also satisfies the <span class="math inline">\(\lambda^1\)</span>-order equation for any value of <span class="math inline">\(c_n\)</span>, as demonstrated here:</p>
<p><span class="math display">\[\begin{aligned}
(\hat{H}_1 - E_n^{(1)}) \ket{n} + (\hat{H}_0 - \varepsilon_n) \ket{\psi_n^{(1)}} - (\varepsilon_n - \varepsilon_n) c_n \ket{n} = 0
\end{aligned}\]</span></p>
<p>Where we used <span class="math inline">\(\hat{H}_0 \ket{n} = \varepsilon_n \ket{n}\)</span>. Inserting the series form of <span class="math inline">\(\ket{\psi_n^{(1)}}\)</span> into the order-<span class="math inline">\(\lambda^1\)</span> equation gives us:</p>
<p><span class="math display">\[\begin{aligned}
(\hat{H}_1 - E_n^{(1)}) \ket{n} + \sum_{m \neq n} c_m (\varepsilon_m - \varepsilon_n) \ket{m} = 0
\end{aligned}\]</span></p>
<p>We then put an arbitrary basis vector <span class="math inline">\(\bra{k}\)</span> in front of this equation to get:</p>
<p><span class="math display">\[\begin{aligned}
\matrixel{k}{\hat{H}_1}{n} - E_n^{(1)} \braket{k}{n} + \sum_{m \neq n} c_m (\varepsilon_m - \varepsilon_n) \braket{k}{m} = 0
\end{aligned}\]</span></p>
<p>Suppose that <span class="math inline">\(k = n\)</span>. Since <span class="math inline">\(\ket{n}\)</span> form an orthonormal basis, we end up with:</p>
<p><span class="math display">\[\begin{aligned}
\boxed{
E_n^{(1)} = \matrixel{n}{\hat{H}_1}{n}
}
\end{aligned}\]</span></p>
<p>In other words, the first-order energy correction <span class="math inline">\(E_n^{(1)}\)</span> is the expectation value of the perturbation <span class="math inline">\(\hat{H}_1\)</span> for the unperturbed state <span class="math inline">\(\ket{n}\)</span>.</p>
<p>Suppose now that <span class="math inline">\(k \neq n\)</span>, then only one term of the summation survives, and we are left with the following equation, which tells us <span class="math inline">\(c_l\)</span>:</p>
<p><span class="math display">\[\begin{aligned}
\matrixel{k}{\hat{H}_1}{n} + c_k (\varepsilon_k - \varepsilon_n) = 0
\end{aligned}\]</span></p>
<p>We isolate this result for <span class="math inline">\(c_k\)</span> and insert it into the series form of <span class="math inline">\(\ket{\psi_n^{(1)}}\)</span> to get the full first-order correction to the wave function:</p>
<p><span class="math display">\[\begin{aligned}
\boxed{
\ket{\psi_n^{(1)}}
= \sum_{m \neq n} \frac{\matrixel{m}{\hat{H}_1}{n}}{\varepsilon_n - \varepsilon_m} \ket{m}
}
\end{aligned}\]</span></p>
<p>Here it is clear why this is only valid in the non-degenerate case: otherwise we would divide by zero in the denominator.</p>
<p>Next, to find the second-order correction to the energy <span class="math inline">\(E_n^{(2)}\)</span>, we take the corresponding equation and put <span class="math inline">\(\bra{n}\)</span> in front of it:</p>
<p><span class="math display">\[\begin{aligned}
\matrixel{n}{\hat{H}_1}{\psi_n^{(1)}} + \matrixel{n}{\hat{H}_0}{\psi_n^{(2)}}
&= E_n^{(2)} \braket{n}{n} + E_n^{(1)} \braket{n}{\psi_n^{(1)}} + \varepsilon_n \braket{n}{\psi_n^{(2)}}
\end{aligned}\]</span></p>
<p>Because <span class="math inline">\(\hat{H}_0\)</span> is Hermitian, we know that <span class="math inline">\(\matrixel{n}{\hat{H}_0}{\psi_n^{(2)}} = \varepsilon_n \braket{n}{\psi_n^{(2)}}\)</span>, i.e. we apply it to the bra, which lets us eliminate two terms. Also, since <span class="math inline">\(\ket{n}\)</span> is normalized, we find:</p>
<p><span class="math display">\[\begin{aligned}
E_n^{(2)}
= \matrixel{n}{\hat{H}_1}{\psi_n^{(1)}} - E_n^{(1)} \braket{n}{\psi_n^{(1)}}
\end{aligned}\]</span></p>
<p>We explicitly removed the <span class="math inline">\(\ket{n}\)</span>-dependence of <span class="math inline">\(\ket{\psi_n^{(1)}}\)</span>, so the last term is zero. By simply inserting our result for <span class="math inline">\(\ket{\psi_n^{(1)}}\)</span>, we thus arrive at:</p>
<p><span class="math display">\[\begin{aligned}
\boxed{
E_n^{(2)}
= \sum_{m \neq n} \frac{\big| \matrixel{m}{\hat{H}_1}{n} \big|^2}{\varepsilon_n - \varepsilon_m}
}
\end{aligned}\]</span></p>
<p>In practice, it is not particulary useful to calculate more corrections.</p>
<h2 id="with-degeneracy">With degeneracy</h2>
<p>If <span class="math inline">\(\varepsilon_n\)</span> is <span class="math inline">\(D\)</span>-fold degenerate, then its eigenstate could be any vector <span class="math inline">\(\ket{n, d}\)</span> from the corresponding <span class="math inline">\(D\)</span>-dimensional eigenspace:</p>
<p><span class="math display">\[\begin{aligned}
\hat{H}_0 \ket{n} = \varepsilon_n \ket{n}
\quad \mathrm{where} \quad
\ket{n}
= \sum_{d = 1}^{D} c_{d} \ket{n, d}
\end{aligned}\]</span></p>
<p>In general, adding the perturbation <span class="math inline">\(\hat{H}_1\)</span> will <em>lift</em> the degeneracy, meaning the perturbed states will be non-degenerate. In the limit <span class="math inline">\(\lambda \to 0\)</span>, these <span class="math inline">\(D\)</span> perturbed states change into <span class="math inline">\(D\)</span> orthogonal states which are all valid <span class="math inline">\(\ket{n}\)</span>.</p>
<p>However, the <span class="math inline">\(\ket{n}\)</span> that they converge to are not arbitrary: only certain unperturbed eigenstates are “good” states. Without <span class="math inline">\(\hat{H}_1\)</span>, this distinction is irrelevant, but in the perturbed case it will turn out to be important.</p>
<p>For now, we write <span class="math inline">\(\ket{n, d}\)</span> to refer to any orthonormal set of vectors in the eigenspace of <span class="math inline">\(\varepsilon_n\)</span> (not necessarily the “good” ones), and <span class="math inline">\(\ket{n}\)</span> to denote any linear combination of these. We then take the equation at order <span class="math inline">\(\lambda^1\)</span> and prepend an arbitrary eigenspace basis vector <span class="math inline">\(\bra{n, \delta}\)</span>:</p>
<p><span class="math display">\[\begin{aligned}
\matrixel{n, \delta}{\hat{H}_1}{n} + \matrixel{n, \delta}{\hat{H}_0}{\psi_n^{(1)}}
&= E_n^{(1)} \braket{n, \delta}{n} + \varepsilon_n \braket{n, \delta}{\psi_n^{(1)}}
\end{aligned}\]</span></p>
<p>Since <span class="math inline">\(\hat{H}_0\)</span> is Hermitian, we use the same trick as before to reduce the problem to:</p>
<p><span class="math display">\[\begin{aligned}
\matrixel{n, \delta}{\hat{H}_1}{n}
&= E_n^{(1)} \braket{n, \delta}{n}
\end{aligned}\]</span></p>
<p>We express <span class="math inline">\(\ket{n}\)</span> as a linear combination of the eigenbasis vectors <span class="math inline">\(\ket{n, d}\)</span> to get:</p>
<p><span class="math display">\[\begin{aligned}
\sum_{d = 1}^{D} c_d \matrixel{n, \delta}{\hat{H}_1}{n, d}
= E_n^{(1)} \sum_{d = 1}^{D} c_d \braket{n, \delta}{n, d}
= c_{\delta} E_n^{(1)}
\end{aligned}\]</span></p>
<p>Let us now interpret the summation terms as matrix elements <span class="math inline">\(M_{\delta, d}\)</span>:</p>
<p><span class="math display">\[\begin{aligned}
M_{\delta, d} = \matrixel{n, \delta}{\hat{H}_1}{n, d}
\end{aligned}\]</span></p>
<p>By varying the value of <span class="math inline">\(\delta\)</span> from <span class="math inline">\(1\)</span> to <span class="math inline">\(D\)</span>, we end up with equations of the form:</p>
<p><span class="math display">\[\begin{aligned}
\begin{bmatrix}
M_{1, 1} & \cdots & M_{1, D} \\
\vdots & \ddots & \vdots \\
M_{D, 1} & \cdots & M_{D, D}
\end{bmatrix}
\begin{bmatrix}
c_1 \\ \vdots \\ c_D
\end{bmatrix}
= E_n^{(1)}
\begin{bmatrix}
c_1 \\ \vdots \\ c_D
\end{bmatrix}
\end{aligned}\]</span></p>
<p>This is an eigenvalue problem for <span class="math inline">\(E_n^{(1)}\)</span>, where <span class="math inline">\(c_d\)</span> are the components of the eigenvectors which represent the “good” states. Suppose that this eigenvalue problem has been solved, and that <span class="math inline">\(\ket{n, g}\)</span> are the resulting “good” states. Then, as long as <span class="math inline">\(E_n^{(1)}\)</span> is a non-degenerate eigenvalue of <span class="math inline">\(M\)</span>:</p>
<p><span class="math display">\[\begin{aligned}
\boxed{
E_{n, g}^{(1)} = \matrixel{n, g}{\hat{H}_1}{n, g}
}
\end{aligned}\]</span></p>
<p>Which is the same as in the non-degenerate case! Even better, the first-order wave function correction is also unchanged:</p>
<p><span class="math display">\[\begin{aligned}
\boxed{
\ket{\psi_{n,g}^{(1)}}
= \sum_{m \neq (n, g)} \frac{\matrixel{m}{\hat{H}_1}{n, g}}{\varepsilon_n - \varepsilon_m} \ket{m}
}
\end{aligned}\]</span></p>
<p>This works because the matrix <span class="math inline">\(M\)</span> is diagonal in the <span class="math inline">\(\ket{n, g}\)</span>-basis, such that when <span class="math inline">\(\ket{m}\)</span> is any vector <span class="math inline">\(\ket{n, \gamma}\)</span> in the <span class="math inline">\(\ket{n}\)</span>-eigenspace (except for <span class="math inline">\(\ket{n,g}\)</span> of course, which is explicitly excluded), then conveniently the corresponding numerator <span class="math inline">\(\matrixel{n, \gamma}{\hat{H}_1}{n, g} = M_{\gamma, g} = 0\)</span>, so the term does not contribute.</p>
<p>If any of the eigenvalues <span class="math inline">\(E_n^{(1)}\)</span> of <span class="math inline">\(M\)</span> are degenerate, then there is still some information missing about the components <span class="math inline">\(c_d\)</span> of the “good” states, in which case we must find these states some other way.</p>
<p>An alternative way of determining these “good” states is also of interest if there is no degeneracy in <span class="math inline">\(M\)</span>, since such a shortcut would allow us use the formulae from non-degenerate perturbation theory straight away.</p>
<p>The method is to find a Hermitian operator <span class="math inline">\(\hat{L}\)</span> (usually using symmetry) which commutes with both <span class="math inline">\(\hat{H}_0\)</span> and <span class="math inline">\(\hat{H}_1\)</span>:</p>
<p><span class="math display">\[\begin{aligned}
= [\hat{L}, \hat{H}_1] = 0
\end{aligned}\]</span></p>
<p>So that it shares its eigenstates with <span class="math inline">\(\hat{H}_0\)</span> (and <span class="math inline">\(\hat{H}_1\)</span>), meaning at least <span class="math inline">\(D\)</span> of the vectors of the <span class="math inline">\(D\)</span>-dimensional <span class="math inline">\(\ket{n}\)</span>-eigenspace are also eigenvectors of <span class="math inline">\(\hat{L}\)</span>.</p>
<p>The crucial part, however, is that <span class="math inline">\(\hat{L}\)</span> must be chosen such that <span class="math inline">\(\ket{n, d_1}\)</span> and <span class="math inline">\(\ket{n, d_2}\)</span> have distinct eigenvalues <span class="math inline">\(\ell_1 \neq \ell_2\)</span> for <span class="math inline">\(d_1 \neq d_2\)</span>:</p>
<p><span class="math display">\[\begin{aligned}
\hat{L} \ket{n, b_1} = \ell_1 \ket{n, b_1}
\qquad
\hat{L} \ket{n, b_2} = \ell_2 \ket{n, b_2}
\end{aligned}\]</span></p>
<p>When this holds for any orthogonal choice of <span class="math inline">\(\ket{n, d_1}\)</span> and <span class="math inline">\(\ket{n, d_2}\)</span>, then these specific eigenvectors of <span class="math inline">\(\hat{L}\)</span> are the “good states”, for any valid choice of <span class="math inline">\(\hat{L}\)</span>.</p>
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