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" --- content/know/category/quantum-mechanics.md | 3 + content/know/concept/index.md | 3 + .../time-independent-perturbation-theory/source.md | 319 +++++++++++++++++++++ .../index.html | 230 +++++++++++++++ 4 files changed, 555 insertions(+) create mode 100644 latex/know/concept/time-independent-perturbation-theory/source.md create mode 100644 static/know/concept/time-independent-perturbation-theory/index.html diff --git a/content/know/category/quantum-mechanics.md b/content/know/category/quantum-mechanics.md index d302dfe..944e491 100644 --- a/content/know/category/quantum-mechanics.md +++ b/content/know/category/quantum-mechanics.md @@ -14,3 +14,6 @@ Alphabetical list of concepts in this category. ## P * [Probability current](/know/concept/probability-current/) + +## T +* [Time-independent perturbation theory](/know/concept/time-independent-perturbation-theory/) diff --git a/content/know/concept/index.md b/content/know/concept/index.md index 19f2027..db1c81c 100644 --- a/content/know/concept/index.md +++ b/content/know/concept/index.md @@ -14,3 +14,6 @@ Alphabetical list of concepts in this knowledge base. ## P * [Probability current](/know/concept/probability-current/) + +## T +* [Time-independent perturbation theory](/know/concept/time-independent-perturbation-theory/) diff --git a/latex/know/concept/time-independent-perturbation-theory/source.md b/latex/know/concept/time-independent-perturbation-theory/source.md new file mode 100644 index 0000000..d076457 --- /dev/null +++ b/latex/know/concept/time-independent-perturbation-theory/source.md @@ -0,0 +1,319 @@ +% 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 $\hat{L}$. diff --git a/static/know/concept/time-independent-perturbation-theory/index.html b/static/know/concept/time-independent-perturbation-theory/index.html new file mode 100644 index 0000000..eeb53a3 --- /dev/null +++ b/static/know/concept/time-independent-perturbation-theory/index.html @@ -0,0 +1,230 @@ + + + + + + + Prefetch | Time-independent perturbation theory + + + + + + +
+
PREFETCH
+
+blog  +code  +know +
+
+
+

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 \(\hat{L}\).

+
+© "Prefetch". Licensed under CC BY-SA 4.0. + + -- cgit v1.2.3