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.
-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 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.
-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 \(\hat{H}_1\):
-\[\begin{aligned} - [\hat{L}, \hat{H}_0] = [\hat{L}, \hat{H}_1] = 0 -\end{aligned}\]
-So that it shares its eigenstates with \(\hat{H}_0\) (and \(\hat{H}_1\)), meaning 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 \(d_1 \neq d_2\):
-\[\begin{aligned} - \hat{L} \ket{n, b_1} = \ell_1 \ket{n, b_1} - \qquad - \hat{L} \ket{n, b_2} = \ell_2 \ket{n, b_2} -\end{aligned}\]
-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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