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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 @@ +<!DOCTYPE html> +<html xmlns="http://www.w3.org/1999/xhtml" lang="" xml:lang=""> +<head> + <meta charset="utf-8" /> + <meta name="generator" content="pandoc" /> + <meta name="viewport" content="width=device-width, initial-scale=1.0, user-scalable=yes" /> + <title>Prefetch | Time-independent perturbation theory</title> + <link rel="icon" href="data:,"> + <style> + body { + background:#ddd; + color:#222; + max-width:80ch; + text-align:justify; + margin:auto; + padding:1em 0; + font-family:sans-serif; + line-height:1.3; + } + a {text-decoration:none;color:#00f;} + h1,h2,h3 {text-align:center} + h1 {font-size:200%;} + h2 {font-size:160%;} + h3 {font-size:120%;} + .nav {height:3rem;font-size:250%;} + .nav a:link,a:visited {color:#222;} + .nav a:hover,a:focus,a:active {color:#00f;} + .navl {width:30%;float:left;text-align:left;} + .navr {width:70%;float:left;text-align:right;} + pre {filter:invert(100%);} + @media (prefers-color-scheme: dark) { + body {background:#222;filter:invert(100%);} + } </style> + <script> + MathJax = { + loader: {load: ["[tex]/physics"]}, + tex: {packages: {"[+]": ["physics"]}} + }; + </script> + <script src="/mathjax/tex-svg.js" type="text/javascript"></script> + </head> +<body> +<div class="nav"> +<div class="navl"><a href="/">PREFETCH</a></div> +<div class="navr"> +<a href="/blog/">blog</a>  +<a href="/code/">code</a>  +<a href="/know/">know</a> +</div> +</div> +<hr> +<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> +<hr> +© "Prefetch". 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