\[\begin{aligned} \hat{H} = \hat{H}_0 + \lambda \hat{H}_1 \end{aligned}\]
-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 \(\lambda\):
+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 \(\lambda\):
\[\begin{aligned} \ket{\psi_n} - &= \ket{\psi_n^{(0)}} + \lambda \ket{\psi_n^{(1)}} + \lambda^2 \ket{\psi_n^{(2)}} + ... + &= \ket*{\psi_n^{(0)}} + \lambda \ket*{\psi_n^{(1)}} + \lambda^2 \ket*{\psi_n^{(2)}} + ... \\ E_n &= E_n^{(0)} + \lambda E_n^{(1)} + \lambda^2 E_n^{(2)} + ... \end{aligned}\]
-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:
+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:
\[\begin{aligned} \hat{H} \ket{\psi_n} - &= \hat{H}_0 \ket{\psi_n^{(0)}} - + \lambda \big( \hat{H}_1 \ket{\psi_n^{(0)}} + \hat{H}_0 \ket{\psi_n^{(1)}} \big) \\ - &\qquad + \lambda^2 \big( \hat{H}_1 \ket{\psi_n^{(1)}} + \hat{H}_0 \ket{\psi_n^{(2)}} \big) + ... + &= \hat{H}_0 \ket*{\psi_n^{(0)}} + + \lambda \big( \hat{H}_1 \ket*{\psi_n^{(0)}} + \hat{H}_0 \ket*{\psi_n^{(1)}} \big) \\ + &\qquad + \lambda^2 \big( \hat{H}_1 \ket*{\psi_n^{(1)}} + \hat{H}_0 \ket*{\psi_n^{(2)}} \big) + ... \\ E_n \ket{\psi_n} - &= E_n^{(0)} \ket{\psi_n^{(0)}} - + \lambda \big( E_n^{(1)} \ket{\psi_n^{(0)}} + E_n^{(0)} \ket{\psi_n^{(1)}} \big) \\ - &\qquad + \lambda^2 \big( E_n^{(2)} \ket{\psi_n^{(0)}} + E_n^{(1)} \ket{\psi_n^{(1)}} + E_n^{(0)} \ket{\psi_n^{(2)}} \big) + ... + &= E_n^{(0)} \ket*{\psi_n^{(0)}} + + \lambda \big( E_n^{(1)} \ket*{\psi_n^{(0)}} + E_n^{(0)} \ket*{\psi_n^{(1)}} \big) \\ + &\qquad + \lambda^2 \big( E_n^{(2)} \ket*{\psi_n^{(0)}} + E_n^{(1)} \ket*{\psi_n^{(1)}} + E_n^{(0)} \ket*{\psi_n^{(2)}} \big) + ... \end{aligned}\]
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:
\[\begin{aligned} - \hat{H}_0 \ket{\psi_n^{(0)}} - &= E_n^{(0)} \ket{\psi_n^{(0)}} + \hat{H}_0 \ket*{\psi_n^{(0)}} + &= E_n^{(0)} \ket*{\psi_n^{(0)}} \\ - \hat{H}_1 \ket{\psi_n^{(0)}} + \hat{H}_0 \ket{\psi_n^{(1)}} - &= E_n^{(1)} \ket{\psi_n^{(0)}} + E_n^{(0)} \ket{\psi_n^{(1)}} + \hat{H}_1 \ket*{\psi_n^{(0)}} + \hat{H}_0 \ket*{\psi_n^{(1)}} + &= E_n^{(1)} \ket*{\psi_n^{(0)}} + E_n^{(0)} \ket*{\psi_n^{(1)}} \\ - \hat{H}_1 \ket{\psi_n^{(1)}} + \hat{H}_0 \ket{\psi_n^{(2)}} - &= E_n^{(2)} \ket{\psi_n^{(0)}} + E_n^{(1)} \ket{\psi_n^{(1)}} + E_n^{(0)} \ket{\psi_n^{(2)}} + \hat{H}_1 \ket*{\psi_n^{(1)}} + \hat{H}_0 \ket*{\psi_n^{(2)}} + &= E_n^{(2)} \ket*{\psi_n^{(0)}} + E_n^{(1)} \ket*{\psi_n^{(1)}} + E_n^{(0)} \ket*{\psi_n^{(2)}} \\ ... &= ... \end{aligned}\]
-The first equation is the unperturbed problem, which we assume has already been solved, with eigenvalues \(E_n^{(0)} = \varepsilon_n\) and eigenvectors \(\ket{\psi_n^{(0)}} = \ket{n}\):
+The first equation is the unperturbed problem, which we assume has already been solved, with eigenvalues \(E_n^{(0)} = \varepsilon_n\) and eigenvectors \(\ket*{\psi_n^{(0)}} = \ket{n}\):
\[\begin{aligned} \hat{H}_0 \ket{n} = \varepsilon_n \ket{n} \end{aligned}\]
@@ -96,17 +96,17 @@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:
\[\begin{aligned} - (\hat{H}_1 - E_n^{(1)}) \ket{n} + (\hat{H}_0 - \varepsilon_n) \ket{\psi_n^{(1)}} = 0 + (\hat{H}_1 - E_n^{(1)}) \ket{n} + (\hat{H}_0 - \varepsilon_n) \ket*{\psi_n^{(1)}} = 0 \end{aligned}\]
-Since \(\ket{n}\) form a complete basis, we can express \(\ket{\psi_n^{(1)}}\) in terms of them:
+Since \(\ket{n}\) form a complete basis, we can express \(\ket*{\psi_n^{(1)}}\) in terms of them:
\[\begin{aligned} - \ket{\psi_n^{(1)}} = \sum_{m \neq n} c_m \ket{m} + \ket*{\psi_n^{(1)}} = \sum_{m \neq n} c_m \ket{m} \end{aligned}\]
-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:
+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:
\[\begin{aligned} - (\hat{H}_1 - E_n^{(1)}) \ket{n} + (\hat{H}_0 - \varepsilon_n) \ket{\psi_n^{(1)}} - (\varepsilon_n - \varepsilon_n) c_n \ket{n} = 0 + (\hat{H}_1 - E_n^{(1)}) \ket{n} + (\hat{H}_0 - \varepsilon_n) \ket*{\psi_n^{(1)}} - (\varepsilon_n - \varepsilon_n) c_n \ket{n} = 0 \end{aligned}\]
-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:
+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:
\[\begin{aligned} (\hat{H}_1 - E_n^{(1)}) \ket{n} + \sum_{m \neq n} c_m (\varepsilon_m - \varepsilon_n) \ket{m} = 0 \end{aligned}\]
@@ -125,10 +125,10 @@\[\begin{aligned} \matrixel{k}{\hat{H}_1}{n} + c_k (\varepsilon_k - \varepsilon_n) = 0 \end{aligned}\]
-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:
+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:
\[\begin{aligned} \boxed{ - \ket{\psi_n^{(1)}} + \ket*{\psi_n^{(1)}} = \sum_{m \neq n} \frac{\matrixel{m}{\hat{H}_1}{n}}{\varepsilon_n - \varepsilon_m} \ket{m} } \end{aligned}\]
@@ -143,7 +143,7 @@ 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:
+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:
\[\begin{aligned} \boxed{ E_n^{(2)} @@ -196,7 +196,7 @@ c_1 \\ \vdots \\ c_D \end{bmatrix} \end{aligned}\]
-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 \(M\):
+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}
@@ -205,18 +205,18 @@
Which is the same as in the non-degenerate case! Even better, the first-order wave function correction is also unchanged: \[\begin{aligned}
\boxed{
- \ket{\psi_{n,g}^{(1)}}
+ \ket*{\psi_{n,g}^{(1)}}
= \sum_{m \neq (n, g)} \frac{\matrixel{m}{\hat{H}_1}{n, g}}{\varepsilon_n - \varepsilon_m} \ket{m}
}
\end{aligned}\] 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 \(\hat{H}_1\): 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 \(\hat{H}_1\): \[\begin{aligned}
-= [\hat{L}, \hat{H}_1] = 0
+ [\hat{L}, \hat{H}_0] = [\hat{L}, \hat{H}_1] = 0
\end{aligned}\] 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 \(\hat{L}\). 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, b_1} = \ell_1 \ket{n, b_1}
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