Notes on separation of variables

Sat, 08 Feb 2025 00:00:00 +0000

Arbitrary boundary conditions for spherical polar coordinates

In the lectures you learned that the Laplacian in 3D has the eigenfunctions $$ \Delta u_{nlm}(r,\theta, \phi) = \lambda_{nlm} u(r,\theta, \phi), $$ where $$ u_{nlm}(r,\theta,\phi) = R_{nl}(r) \tilde P_{lm}(\theta) \Phi_m(\phi), $$ where $\tilde P_{lm}(\theta) = C_{lm} P_{lm}(\cos \theta)$. As it is separable, it can be thought of a tensor product of three vectors living in separate spaces $1,2,3$, as $$ \ket{u_{nlm}} = \ket{R_{nl}}_1 \ket{\tilde P_{lm}}_2 \ket{\Phi_m}_3. $$ with $u_{nlm}(r, \theta, \phi) = \langle r \vert R_{nl} \rangle_1 \langle \theta \vert \tilde P_{lm} \rangle_2 \langle \phi \vert \Phi_m \rangle_3$. Since inner product on the joint space is defined as $$ \langle u \vert u’ \rangle := \int_0^\infty r^2 dr \int_0^\pi \sin\theta d \theta \int_0^{2\pi} d \phi ,u^*(r, \theta, \phi) u(r, \theta, \phi), $$ The individual factors can be normalised under the inner products $$ \langle R \vert R’ \rangle_1 := \int_0^\infty r^2 dr R^*(r) R’(r), $$ $$ \langle \tilde P \vert \tilde P’ \rangle_2 := \int_0^\pi \sin \theta d \theta \tilde{P}^*(\theta) \tilde{P}’(\theta), $$ and $$ \langle \Phi \vert \Phi’ \rangle_3 := \int_0^{2\pi} d\phi \Phi^*(r) \Phi’(r). $$ ${u_{nlm}}_{nlm}$ form a CONS (complete set of orthonomal basis) and therefore any solution can be written as $$ \psi(r, \theta, \phi) = \sum_{nlm} A_{nlm} R_{nl}(r) \tilde P_{lm}(\theta) \Phi_m(\phi), $$ or abstractly as $$ \ket{\psi} = \sum_{nlm} A_{nlm} \ket{R_{nl}}_1 \ket{\tilde P_{lm}}_2 \ket{\Phi_m}_3. $$ We can evaluate this abstract vector on a subset of variables, for example only for $r$, as: $$ \ket{\psi(r)} := \bra{r}_1\ket{\psi} = \sum_{nlm} A_{nlm} R_{nl}(r) \ket{\tilde P_{lm}}_2 \ket{\Phi_m}_3. $$ In the lectures, you had a simple boundary condition $u(R, \theta, \phi) = \cos(\theta)$. What if it is more complicated? Consider we have a boundary condition like $u(R, \theta, \phi) = f(\theta, \phi)$.

This can be written as an abstract vector in spaces $2, 3$ as $$ \ket{u(R)} = \sum_{nlm} B_{nlm} R_{nl}(R) R_{nl}(r) \ket{\tilde P_{lm}}_2 \ket{\Phi_m}_3. $$ The coefficient $B_{nlm}$ can be found by taking the inner product of $\ket{u(R)}$ with basis vectors for spaces $2, 3$, as $$ B_{nlm} = \frac{1}{R_{nl}(R)} \int_0^\pi \sin\theta d \theta \int_0^{2\pi} d \phi , \tilde P^*_{lm}(\theta) \Phi^*_m(\phi) f(\theta, \phi). $$ Then to obey the boundary conditions, we must have that $B_{nlm} = A_{nlm}$, such that $\psi(R, \theta, \phi) = u(R,\theta,\phi) = f(\theta, \phi)$.

For example, our boundary condition $u(R, \theta, \phi) = \cos(\theta)$ is proportional to $\tilde P_{10}(\theta) \Phi_m(\phi)$. Therefore $B_{nlm} = \delta_{l,1} \delta_{m,0}$.

Notes on group and representation theory

Mon, 12 Feb 2024 00:00:00 +0000

Here are some questions and solutions on group theory. I tried to be as detailed as possible. Click on the question to see the solution. Thanks to Gabriele Pinna for working on these problems with me.

1. Prove that the alternating group (even permutations) $A_n$ has order $\frac{n!}{2}$.

We know that the order of the symmetric group (aka all permutations) $S_n$, is $n!$. We also learned that any permutation $\pi \in S_n$ can be written as a product of transpositions, which, if there are even number of them, then $\pi$ is an even permutation.

Apart from the fact that we call $A_n$ the alternating group, we can see that it is one, since identity $e=(12)(21)$ and a product of two even permutations are even, since the number of transpositions will add. We know that there are odd permutations too, so we know that $A_n$ is a subgroup of $S_n$, i.e. $A_n \subset S_n$.

But, we have not shown how many elements in $A_n$ there are. How to show this?

Our strategy will be to do a coset decomposition of $S_n$, then use Lagrange’s theorem. We will find that there are only two coset representatives (these are elements that label the disjoint cosets), so that there are exactly half the elements of $S_n$ in $A_n$.

First, we know that $A_n e = A_n$ is a valid right coset, and will contain all the even permutations in $S_n$. Let’s choose some odd permutation $d \in S_n$ as the next coset representative. We want to check that $A_n d$ contains the rest of $S_n$. Consider some other odd element $d’ \in S_n$. Then $A_n d’ = A_n d’ d^{-1} d$. But $d’ d^{-1}$ is an even permutation, since the number of transpositions will add and an odd number plus another odd number is an even number. Therefore $A_n d’ = A_n d$. So we see that $A_n d$ contains all the odd elements in $S_n$. All elements in $A_n d$ must be odd. So $A_n e$ and $A_n d$ are disjoint since the former contains only even elements and $A_n d$ contain only the odd elements, and $S_n = A_n e \cup A_n d$. There are two coset representatives, so $\lvert S_n \rvert = 2 \lvert A_n \rvert \implies \lvert A_n \rvert = n!/2$.

2. Express the permutation $(abc \cdots k)(al)$, operating right to left, as a product of disjoint cycles.

Consider the first operation, $(al)$. This means that $a \rightarrow k$ and $k \rightarrow a$. Let us write this in the two-line notation. Even though this is operating on elements $a$ and $l$ only, we can still write in elements $b c \cdots k$, as long as they stay in place. So we have: $$ (al) \leftrightarrow \begin{pmatrix} a & b & c & \cdots & k & l \ l & b & c & \cdots & k & a \end{pmatrix}. $$ Then, let’s apply the second operation, $(abc \cdots k)$. Here we just have $a \rightarrow b$, $b \rightarrow c, \cdots, k \rightarrow a$. Applying this on the previous cycle/permutation, we have $$ (abc \cdots k) (al) \leftrightarrow \begin{pmatrix} a & b & c & \cdots & k & l \ l & c & d & \cdots & a & b \end{pmatrix}. $$ Written in this form, it is not clear that it is a product of disjoint cycles. Let’s chase how each element is mapped. We have $a \rightarrow l$, $l \rightarrow b$. From $b$ to $k$ in fact we have the mapping we’d expect, $b \rightarrow c$, $c \rightarrow d$, and so on, until $j \rightarrow k$. Then we have $k \rightarrow a$. But we’re now back to $a$, so we have completed a cycle! We’ve also gone through all the elements, so it must be that there is only one cycle in this permutation. Summarising the above, we can write it as a single (and therefore disjoint) cycle: $$ (abc \cdots k) (al) = (a l b c \cdots k). $$

3. For a group $G$ that has an action on a finite set $S$, such that $\forall g \in G$, $g(i) \in S$, let the stabiliser subgroup of $x$ given group $G$ be defined as $$\mathrm{Stab}_G(i) := \left\{ \phi \in G \vert \phi(i) = i\right\}.$$ Let the orbit of of an element $i \in S$ given group $G$ be $$\mathrm{Orb}_G(i) := \left\{ g(i) \vert g \in G \right\}.$$ Prove that for any $i$, $$\lvert G \rvert = \lvert \mathrm{Orb}_G(i) \rvert \lvert \mathrm{Stab}_G(i) \rvert.$$

For fixed element $i$, let’s denote $H = \mathrm{Stab}_G(i)$. Since $H$ is a subgroup of $G$ (you should check this), let us consider the left coset decomposition of $G$: $$G = t_1 H \cup t_2 H \cup \cdots \cup t_n H,$$ where $t_1 = e$, the identity.

So we can write the orbit as $$\mathrm{Orb}_G(i) = \bigcup_{a=1}^n \left\{ g(i) | g \in t_a H \right\}.$$ We can also iterate over each coset like $$\mathrm{Orb}_G(i) = \bigcup_{a=1}^n \left\{ t_a(\phi(i)) | h \in H \right\}.$$ Since all $\phi(i) = i$, and sets only count unique elements, we only have one element contributing from each coset at most, so $$\mathrm{Orb}_G(i) = \bigcup_{a=1}^n \left\{ t_a(i) \right\} = \left\{ t_a(i) \vert a=1, \dots, n\right\}.$$ We’re almost there, since if $t_a(i)$ are unique with $a$, then we see that $\mathrm{Orb}_G(i)$ counts the number of cosets, so we can use Lagrange’s theorem to prove the result.

Let’s check this. Assume there exists $t_a(i) = t_b(i)$ where $a \neq b$. Then $t_b^{-1} (t_a(i)) = i$ so $t_b^{-1} t_a = h$. This means that $t_a = t_b h$ so $t_a H = t_b H$, so are the same cosets. But this is a contradiction because $t_a H$ and $t_b H$ were supposed to be disjoint. So $t_a(i)$ are unique.

4. Let $A$, $B$ be linear operators. For the case where both $A$ and $B$ commute with $[A,B]$ (for example $\hat x$ and $\hat p$!), prove the special case of the Baker-Campbell-Hausdorff formula, $$e^{A} e^{B} = e^{A+B+[A,B]/2}.$$ You might find it useful to prove that, for any linear operators $A$ and $B$, that $$ e^{tA} B e^{-tA} = e^{t[A, \, \cdot \,]} B, $$ where we define $[A, \, \cdot \,]$ as a 'superoperator' that acts on operators like $[A, \, \cdot \,] B = [A, B]$, and for example $[A, \, \cdot \,]^2 B = [A, [A, B]]$.

Let us prove the second statement first. Define $$ f(t) := e^{t A} B e^{-t A}, $$ where $t$ is just a number. If we take the derivative, we have $$ \begin{aligned} \frac{df}{d t} & = A e^{t A} B e^{-t A} - e^{t A} B e^{-t A} A \\ & = [A, \, \cdot \,] f. \end{aligned} $$ This is the same operator at every `time’ $t$. Since we have $f(0) = B$, we can use the usual knowledge of differential equations and find that $f(t) = e^{t[A, \, \cdot \,]} B$.

For our specific case where $[A,B]$ commutes with both $A$ and $B$, let’s write out $e^{t[A, \, \cdot \,]} B$: $$ \begin{aligned} e^{t[A, \, \cdot \,]} B & = \sum_{n=0}^{\infty} \frac{t^n}{n!} [A, \, \cdot \,]^n B \\ & = B + t [A, B] + \sum_{n=2}^{\infty} \frac{t^n}{n!} [A, \, \cdot \,]^{n-1} [A, B]. \end{aligned} $$ But from our assumptions we know that $[A,B]$ commutes with $A$, so all $n \geq 2$ terms vanish. Therefore we have that $$ e^{tA} B e^{-tA} = e^{t[A, \, \cdot \,]} B = B + t[A, B].$$ Now let’s tackle the main result. Define $$ g(t) := e^{tA} e^{tB}, $$ where $t$ is just a number. Again differentiating, you should be able to show that $$ \begin{aligned} \frac{dg}{dt} & = (A + e^{tA} B e^{-tA}) g = g (B + e^{-tB} A e^{tB}) \\ & = (A + e^{t[A, \, \cdot \,]} B) g = g (B + e^{-t[B, \, \cdot \,]} A). \end{aligned} $$ Therefore we have $$ \frac{dg}{dt} = (A + B + t[A, B]) g. $$ Let us denote $T(t) := A + B + t[A,B]$. It’s easy to show that $[T(t), T(t’)]=0$ for all $t, t’$. Therefore there exists some $U$ independently diagonalises $T(t)$; defining $\overline{M} := S M S^{-1}$, $\overline{T}(x)$ is diagonal. We also have that $[g(0)]_{ij} = \delta_{ij}$, which stays diagonal upon similarity transformation. Therefore in the rotated frame everything stays in the diagonal, and we have $$ \frac{d \overline{g}_{ii}}{dt} = \overline{T}_{ii}(t) \overline{g}_{ii}. $$ Using the initial condition $\overline{g}_{ii}(0) = 1$, we have $$ \begin{aligned} \overline{g}_{ii}(t) & = \exp\left( \int_{0}^t dt’ \overline{T}_{ii} (t’) \right) = \exp\left( \sum_{jk} S_{ij} S^{-1}_{ki}\int_{0}^t dt’ \left(A_{jk} + B_{jk} + x [A, B]_{jk} \right) \right) \\ & = \exp\left( \sum_{jk} S_{ij} S^{-1}_{ki}\left(tA_{jk} + tB_{jk} + \frac{t^2}{2} [A, B]_{jk} \right) \right) \\ & = \exp\left( t \overline{A} + t\overline{B} + \frac{t^2}{2} \overline{[A, B]} \right)_{ii}. \end{aligned} $$ Transforming back to the normal coordinates by applying $S^{-1} \left( \, \cdot \, \right) S$, then setting $t=1$, we have $$ g(1) = \exp\left( {A} + {B} + \frac{1}{2} {[A, B]} \right) = \exp(A) \exp(B). $$

Teaching | 김선우

Notes on separation of variables

Arbitrary boundary conditions for spherical polar coordinates

Notes on group and representation theory