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| -rw-r--r-- | latex/know/concept/time-independent-perturbation-theory/source.md | 86 | ||||
| -rw-r--r-- | static/know/concept/time-independent-perturbation-theory/index.html | 66 |
2 files changed, 75 insertions, 77 deletions
diff --git a/latex/know/concept/time-independent-perturbation-theory/source.md b/latex/know/concept/time-independent-perturbation-theory/source.md index d076457..a3167cd 100644 --- a/latex/know/concept/time-independent-perturbation-theory/source.md +++ b/latex/know/concept/time-independent-perturbation-theory/source.md @@ -16,31 +16,31 @@ $$\begin{aligned} Where $\hat{H}_0$ is a Hamiltonian for which the time-independent Schrödinger equation has a known solution, and $\hat{H}_1$ is a small perturbing Hamiltonian. The eigenenergies $E_n$ and eigenstates -$\ket{\psi_n}$ of the composite problem are expanded accordingly in the +$\ket{\psi_n}$ of the composite problem are expanded in the perturbation "bookkeeping" parameter $\lambda$: $$\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}$$ -Where $E_n^{(1)}$ and $\ket{\psi_n^{(1)}}$ are called the *first-order +Where $E_n^{(1)}$ and $\ket*{\psi_n^{(1)}}$ are called the *first-order corrections*, and so on for higher orders. We insert this into the Schrödinger equation: $$\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}$$ If we collect the terms according to the order of $\lambda$, we arrive @@ -48,14 +48,14 @@ at the following endless series of equations, of which in practice only the first three are typically used: $$\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)}} \\ ... &= ... @@ -63,7 +63,7 @@ $$\begin{aligned} The first equation is the unperturbed problem, which we assume has already been solved, with eigenvalues $E_n^{(0)} = \varepsilon_n$ and -eigenvectors $\ket{\psi_n^{(0)}} = \ket{n}$: +eigenvectors $\ket*{\psi_n^{(0)}} = \ket{n}$: $$\begin{aligned} \hat{H}_0 \ket{n} = \varepsilon_n \ket{n} @@ -79,28 +79,27 @@ $\varepsilon_n$ 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: +$\ket*{\psi_n^{(1)}}$ 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 +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: $$\begin{aligned} (\hat{H}_1 - E_n^{(1)}) \ket{n} + \sum_{m \neq n} c_m (\varepsilon_m - \varepsilon_n) \ket{m} = 0 @@ -135,12 +134,12 @@ $$\begin{aligned} \end{aligned}$$ We isolate this result for $c_k$ and insert it into the series form of -$\ket{\psi_n^{(1)}}$ to get the full first-order correction to the wave +$\ket*{\psi_n^{(1)}}$ to get the full first-order correction to the wave function: $$\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}$$ @@ -166,9 +165,9 @@ $$\begin{aligned} = \matrixel{n}{\hat{H}_1}{\psi_n^{(1)}} - E_n^{(1)} \braket{n}{\psi_n^{(1)}} \end{aligned}$$ -We explicitly removed the $\ket{n}$-dependence of $\ket{\psi_n^{(1)}}$, +We explicitly removed the $\ket{n}$-dependence of $\ket*{\psi_n^{(1)}}$, so the last term is zero. By simply inserting our result for -$\ket{\psi_n^{(1)}}$, we thus arrive at: +$\ket*{\psi_n^{(1)}}$, we thus arrive at: $$\begin{aligned} \boxed{ @@ -257,9 +256,8 @@ $$\begin{aligned} This is an eigenvalue problem for $E_n^{(1)}$, where $c_d$ are the components of the eigenvectors which represent the "good" states. -Suppose that this eigenvalue problem has been solved, and that -$\ket{n, g}$ are the resulting "good" states. Then, as long as -$E_n^{(1)}$ is a non-degenerate eigenvalue of $M$: +After solving this, let $\ket{n, g}$ be the resulting "good" states. +Then, as long as $E_n^{(1)}$ is a non-degenerate eigenvalue of $M$: $$\begin{aligned} \boxed{ @@ -272,36 +270,36 @@ 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 +$\ket{n}$-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 some information missing about the components $c_d$ of the -"good" states, in which case we must find these states some other way. +is still information missing about the components $c_d$ of the +"good" states, in which case we must find them 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 +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 method is to find a Hermitian operator $\hat{L}$ (usually using -symmetry) which commutes with both $\hat{H}_0$ and $\hat{H}_1$: +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 +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 diff --git a/static/know/concept/time-independent-perturbation-theory/index.html b/static/know/concept/time-independent-perturbation-theory/index.html index eeb53a3..aeaa570 100644 --- a/static/know/concept/time-independent-perturbation-theory/index.html +++ b/static/know/concept/time-independent-perturbation-theory/index.html @@ -54,41 +54,41 @@ <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>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 in the perturbation “bookkeeping” parameter <span class="math inline">\(\lambda\)</span>:</p> <p><span class="math display">\[\begin{aligned} \ket{\psi_n} - &= \ket{\psi_n^{(0)}} + \lambda \ket{\psi_n^{(1)}} + \lambda^2 \ket{\psi_n^{(2)}} + ... + &= \ket*{\psi_n^{(0)}} + \lambda \ket*{\psi_n^{(1)}} + \lambda^2 \ket*{\psi_n^{(2)}} + ... \\ E_n &= E_n^{(0)} + \lambda E_n^{(1)} + \lambda^2 E_n^{(2)} + ... \end{aligned}\]</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>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">\[\begin{aligned} \hat{H} \ket{\psi_n} - &= \hat{H}_0 \ket{\psi_n^{(0)}} - + \lambda \big( \hat{H}_1 \ket{\psi_n^{(0)}} + \hat{H}_0 \ket{\psi_n^{(1)}} \big) \\ - &\qquad + \lambda^2 \big( \hat{H}_1 \ket{\psi_n^{(1)}} + \hat{H}_0 \ket{\psi_n^{(2)}} \big) + ... + &= \hat{H}_0 \ket*{\psi_n^{(0)}} + + \lambda \big( \hat{H}_1 \ket*{\psi_n^{(0)}} + \hat{H}_0 \ket*{\psi_n^{(1)}} \big) \\ + &\qquad + \lambda^2 \big( \hat{H}_1 \ket*{\psi_n^{(1)}} + \hat{H}_0 \ket*{\psi_n^{(2)}} \big) + ... \\ E_n \ket{\psi_n} - &= E_n^{(0)} \ket{\psi_n^{(0)}} - + \lambda \big( E_n^{(1)} \ket{\psi_n^{(0)}} + E_n^{(0)} \ket{\psi_n^{(1)}} \big) \\ - &\qquad + \lambda^2 \big( E_n^{(2)} \ket{\psi_n^{(0)}} + E_n^{(1)} \ket{\psi_n^{(1)}} + E_n^{(0)} \ket{\psi_n^{(2)}} \big) + ... + &= E_n^{(0)} \ket*{\psi_n^{(0)}} + + \lambda \big( E_n^{(1)} \ket*{\psi_n^{(0)}} + E_n^{(0)} \ket*{\psi_n^{(1)}} \big) \\ + &\qquad + \lambda^2 \big( E_n^{(2)} \ket*{\psi_n^{(0)}} + E_n^{(1)} \ket*{\psi_n^{(1)}} + E_n^{(0)} \ket*{\psi_n^{(2)}} \big) + ... \end{aligned}\]</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">\[\begin{aligned} - \hat{H}_0 \ket{\psi_n^{(0)}} - &= E_n^{(0)} \ket{\psi_n^{(0)}} + \hat{H}_0 \ket*{\psi_n^{(0)}} + &= E_n^{(0)} \ket*{\psi_n^{(0)}} \\ - \hat{H}_1 \ket{\psi_n^{(0)}} + \hat{H}_0 \ket{\psi_n^{(1)}} - &= E_n^{(1)} \ket{\psi_n^{(0)}} + E_n^{(0)} \ket{\psi_n^{(1)}} + \hat{H}_1 \ket*{\psi_n^{(0)}} + \hat{H}_0 \ket*{\psi_n^{(1)}} + &= E_n^{(1)} \ket*{\psi_n^{(0)}} + E_n^{(0)} \ket*{\psi_n^{(1)}} \\ - \hat{H}_1 \ket{\psi_n^{(1)}} + \hat{H}_0 \ket{\psi_n^{(2)}} - &= E_n^{(2)} \ket{\psi_n^{(0)}} + E_n^{(1)} \ket{\psi_n^{(1)}} + E_n^{(0)} \ket{\psi_n^{(2)}} + \hat{H}_1 \ket*{\psi_n^{(1)}} + \hat{H}_0 \ket*{\psi_n^{(2)}} + &= E_n^{(2)} \ket*{\psi_n^{(0)}} + E_n^{(1)} \ket*{\psi_n^{(1)}} + E_n^{(0)} \ket*{\psi_n^{(2)}} \\ ... &= ... \end{aligned}\]</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>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> @@ -96,17 +96,17 @@ <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">\[\begin{aligned} - (\hat{H}_1 - E_n^{(1)}) \ket{n} + (\hat{H}_0 - \varepsilon_n) \ket{\psi_n^{(1)}} = 0 + (\hat{H}_1 - E_n^{(1)}) \ket{n} + (\hat{H}_0 - \varepsilon_n) \ket*{\psi_n^{(1)}} = 0 \end{aligned}\]</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>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">\[\begin{aligned} - \ket{\psi_n^{(1)}} = \sum_{m \neq n} c_m \ket{m} + \ket*{\psi_n^{(1)}} = \sum_{m \neq n} c_m \ket{m} \end{aligned}\]</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>Importantly, <span class="math inline">\(n\)</span> has been removed from the summation to prevent dividing by zero later. We are allowed to do this, because <span class="math inline">\(\ket*{\psi_n^{(1)}} - c_n \ket{n}\)</span> also satisfies the order-<span class="math inline">\(\lambda^1\)</span> equation for any value of <span class="math inline">\(c_n\)</span>, as demonstrated here:</p> <p><span class="math display">\[\begin{aligned} - (\hat{H}_1 - E_n^{(1)}) \ket{n} + (\hat{H}_0 - \varepsilon_n) \ket{\psi_n^{(1)}} - (\varepsilon_n - \varepsilon_n) c_n \ket{n} = 0 + (\hat{H}_1 - E_n^{(1)}) \ket{n} + (\hat{H}_0 - \varepsilon_n) \ket*{\psi_n^{(1)}} - (\varepsilon_n - \varepsilon_n) c_n \ket{n} = 0 \end{aligned}\]</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>Where we used <span class="math inline">\(\hat{H}_0 \ket{n} = \varepsilon_n \ket{n}\)</span>. We insert the series form of <span class="math inline">\(\ket*{\psi_n^{(1)}}\)</span> into the <span class="math inline">\(\lambda^1\)</span>-equation:</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> @@ -125,10 +125,10 @@ <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>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">\[\begin{aligned} \boxed{ - \ket{\psi_n^{(1)}} + \ket*{\psi_n^{(1)}} = \sum_{m \neq n} \frac{\matrixel{m}{\hat{H}_1}{n}}{\varepsilon_n - \varepsilon_m} \ket{m} } \end{aligned}\]</span></p> @@ -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}\]</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>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">\[\begin{aligned} \boxed{ E_n^{(2)} @@ -196,7 +196,7 @@ c_1 \\ \vdots \\ c_D \end{bmatrix} \end{aligned}\]</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>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. After solving this, let <span class="math inline">\(\ket{n, g}\)</span> be 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} @@ -205,18 +205,18 @@ <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">\[\begin{aligned} \boxed{ - \ket{\psi_{n,g}^{(1)}} + \ket*{\psi_{n,g}^{(1)}} = \sum_{m \neq (n, g)} \frac{\matrixel{m}{\hat{H}_1}{n, g}}{\varepsilon_n - \varepsilon_m} \ket{m} } \end{aligned}\]</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>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>, which is explicitly excluded), then 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 information missing about the components <span class="math inline">\(c_d\)</span> of the “good” states, in which case we must find them some other way.</p> +<p>Such an alternative way of determining these “good” states is also of interest even if there is no degeneracy in <span class="math inline">\(M\)</span>, since such a shortcut would allow us to use the formulae from non-degenerate perturbation theory straight away.</p> +<p>The trick is to find a Hermitian operator <span class="math inline">\(\hat{L}\)</span> (usually using symmetries of the system) 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">\[\begin{aligned} -= [\hat{L}, \hat{H}_1] = 0 + [\hat{L}, \hat{H}_0] = [\hat{L}, \hat{H}_1] = 0 \end{aligned}\]</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>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 all 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} |