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diff --git a/source/know/concept/time-independent-perturbation-theory/index.md b/source/know/concept/time-independent-perturbation-theory/index.md index ba1e89d..4d1ac2f 100644 --- a/source/know/concept/time-independent-perturbation-theory/index.md +++ b/source/know/concept/time-independent-perturbation-theory/index.md @@ -19,11 +19,11 @@ $$\begin{aligned} \hat{H} = \hat{H}_0 + \lambda \hat{H}_1 \end{aligned}$$ -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 $\lambda$: +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 $$\lambda$$: $$\begin{aligned} \Ket{\psi_n} @@ -33,7 +33,7 @@ $$\begin{aligned} &= 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)}}$$ are called the **first-order corrections**, and so on for higher orders. We insert this into the Schrödinger equation: @@ -49,7 +49,7 @@ $$\begin{aligned} &\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}$$ -If we collect the terms according to the order of $\lambda$, we arrive +If we collect the terms according to the order of $$\lambda$$, we arrive at the following endless series of equations, of which in practice only the first three are typically used: @@ -68,58 +68,58 @@ $$\begin{aligned} \end{aligned}$$ The first equation is the unperturbed problem, which we assume has -already been solved, with eigenvalues $E_n^{(0)} = \varepsilon_n$ and -eigenvectors $\ket{\psi_n^{(0)}} = \Ket{n}$: +already been solved, with eigenvalues $$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}$$ The approach to solving the other two equations varies depending on -whether this $\hat{H}_0$ has a degenerate spectrum or not. +whether this $$\hat{H}_0$$ has a degenerate spectrum or not. ## Without degeneracy We start by assuming that there is no degeneracy, in other words, each -$\varepsilon_n$ corresponds to one $\Ket{n}$. At order $\lambda^1$, we +$$\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}$$ -Since $\Ket{n}$ form a complete basis, we can express -$\ket{\psi_n^{(1)}}$ in terms of them: +Since $$\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}$$ -Importantly, $n$ has been removed from the summation to prevent dividing +Importantly, $$n$$ has been removed from the summation to prevent dividing by zero later. We are allowed to do this, because -$\ket{\psi_n^{(1)}} - c_n \Ket{n}$ also satisfies the order-$\lambda^1$ -equation for any value of $c_n$, as demonstrated here: +$$\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}$$ -Where we used $\hat{H}_0 \Ket{n} = \varepsilon_n \Ket{n}$. -We insert the series form of $\ket{\psi_n^{(1)}}$ into the $\lambda^1$-equation: +Where we used $$\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}$$ -We then put an arbitrary basis vector $\Bra{k}$ in front of this +We then put an arbitrary basis vector $$\Bra{k}$$ in front of this equation to get: $$\begin{aligned} \matrixel{k}{\hat{H}_1}{n} - E_n^{(1)} \Inprod{k}{n} + \sum_{m \neq n} c_m (\varepsilon_m - \varepsilon_n) \Inprod{k}{m} = 0 \end{aligned}$$ -Suppose that $k = n$. Since $\Ket{n}$ form an orthonormal basis, we end +Suppose that $$k = n$$. Since $$\Ket{n}$$ form an orthonormal basis, we end up with: $$\begin{aligned} @@ -128,20 +128,20 @@ $$\begin{aligned} } \end{aligned}$$ -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 $\Ket{n}$. +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 +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 -$c_l$: +$$c_l$$: $$\begin{aligned} \matrixel{k}{\hat{H}_1}{n} + c_k (\varepsilon_k - \varepsilon_n) = 0 \end{aligned}$$ -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 +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: $$\begin{aligned} @@ -154,27 +154,27 @@ $$\begin{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 energy correction $E_n^{(2)}$, -we take the corresponding equation and put $\Bra{n}$ in front of it: +Next, to find the second-order energy correction $$E_n^{(2)}$$, +we take the corresponding equation and put $$\Bra{n}$$ in front of it: $$\begin{aligned} \matrixel{n}{\hat{H}_1}{\psi_n^{(1)}} + \matrixel{n}{\hat{H}_0}{\psi_n^{(2)}} &= E_n^{(2)} \Inprod{n}{n} + E_n^{(1)} \inprod{n}{\psi_n^{(1)}} + \varepsilon_n \inprod{n}{\psi_n^{(2)}} \end{aligned}$$ -Because $\hat{H}_0$ is Hermitian, we know that -$\matrixel{n}{\hat{H}_0}{\psi_n^{(2)}} = \varepsilon_n \inprod{n}{\psi_n^{(2)}}$, +Because $$\hat{H}_0$$ is Hermitian, we know that +$$\matrixel{n}{\hat{H}_0}{\psi_n^{(2)}} = \varepsilon_n \inprod{n}{\psi_n^{(2)}}$$, i.e. we apply it to the bra, which lets us eliminate two terms. Also, -since $\Ket{n}$ is normalized, we find: +since $$\Ket{n}$$ is normalized, we find: $$\begin{aligned} E_n^{(2)} = \matrixel{n}{\hat{H}_1}{\psi_n^{(1)}} - E_n^{(1)} \inprod{n}{\psi_n^{(1)}} \end{aligned}$$ -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: $$\begin{aligned} \boxed{ @@ -188,8 +188,8 @@ In practice, it is not particulary useful to calculate more corrections. ## With degeneracy -If $\varepsilon_n$ is $D$-fold degenerate, then its eigenstate could be -any vector $\Ket{n, d}$ from the corresponding $D$-dimensional +If $$\varepsilon_n$$ is $$D$$-fold degenerate, then its eigenstate could be +any vector $$\Ket{n, d}$$ from the corresponding $$D$$-dimensional eigenspace: $$\begin{aligned} @@ -199,28 +199,28 @@ $$\begin{aligned} = \sum_{d = 1}^{D} c_{d} \Ket{n, d} \end{aligned}$$ -In general, adding the perturbation $\hat{H}_1$ will *lift* the +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 $\Ket{n}$. +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 $\hat{H}_1$, +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 $\Bra{n, \delta}$: +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 $$\Bra{n, \delta}$$: $$\begin{aligned} \matrixel{n, \delta}{\hat{H}_1}{n} + \matrixel{n, \delta}{\hat{H}_0}{\psi_n^{(1)}} &= E_n^{(1)} \Inprod{n, \delta}{n} + \varepsilon_n \inprod{n, \delta}{\psi_n^{(1)}} \end{aligned}$$ -Since $\hat{H}_0$ is Hermitian, we use the same trick as before to +Since $$\hat{H}_0$$ is Hermitian, we use the same trick as before to reduce the problem to: $$\begin{aligned} @@ -228,8 +228,8 @@ $$\begin{aligned} &= E_n^{(1)} \Inprod{n, \delta}{n} \end{aligned}$$ -We express $\Ket{n}$ as a linear combination of the eigenbasis vectors -$\Ket{n, d}$ to get: +We express $$\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} @@ -238,13 +238,13 @@ $$\begin{aligned} \end{aligned}$$ Let us now interpret the summation terms as matrix elements -$M_{\delta, d}$: +$$M_{\delta, d}$$: $$\begin{aligned} M_{\delta, d} = \matrixel{n, \delta}{\hat{H}_1}{n, d} \end{aligned}$$ -By varying the value of $\delta$ from $1$ to $D$, we end up with +By varying the value of $$\delta$$ from $$1$$ to $$D$$, we end up with equations of the form: $$\begin{aligned} @@ -262,10 +262,10 @@ $$\begin{aligned} \end{bmatrix} \end{aligned}$$ -This is an eigenvalue problem for $E_n^{(1)}$, where $c_d$ are the +This is an eigenvalue problem for $$E_n^{(1)}$$, where $$c_d$$ are the components of the eigenvectors which represent the "good" states. -After solving this, let $\Ket{n, g}$ be the resulting "good" states. -Then, as long as $E_n^{(1)}$ is a non-degenerate eigenvalue of $M$: +After solving this, let $$\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{ @@ -283,36 +283,36 @@ $$\begin{aligned} } \end{aligned}$$ -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 +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 +$$\matrixel{n, \gamma}{\hat{H}_1}{n, g} = M_{\gamma, g} = 0$$, 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 +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 +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 $\hat{H}_1$: +The trick is to find a Hermitian operator $$\hat{L}$$ (usually using +symmetries of the system) which commutes with both $$\hat{H}_0$$ and $$\hat{H}_1$$: $$\begin{aligned} \comm{\hat{L}}{\hat{H}_0} = \comm{\hat{L}}{\hat{H}_1} = 0 \end{aligned}$$ -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 $\hat{L}$. +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 $d_1 \neq d_2$: +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 $$d_1 \neq d_2$$: $$\begin{aligned} \hat{L} \Ket{n, d_1} = \ell_1 \Ket{n, d_1} @@ -320,9 +320,9 @@ $$\begin{aligned} \hat{L} \Ket{n, d_2} = \ell_2 \Ket{n, d_2} \end{aligned}$$ -When this condition holds for any orthogonal choice of $\Ket{n, d_1}$ and -$\Ket{n, d_2}$, then these specific eigenvectors of $\hat{L}$ are the -"good states", for any valid choice of $\hat{L}$. +When this condition holds for any orthogonal choice of $$\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}$$. |