diff options
Diffstat (limited to 'latex')
| -rw-r--r-- | latex/know/concept/time-independent-perturbation-theory/source.md | 319 |
1 files changed, 319 insertions, 0 deletions
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 $\hat{L}$. |