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Let's take a direction eigenket $|{\bf\hat{n}}\rangle$ in 3-dimensional space oriented with angles $\theta\in\left[0,\pi\right]$ and $\phi\in\left[0,2\pi\right]$ in spherical coordinates. Next take the $|{\bf\hat{z}}\rangle$ direction eigenket.

There are infinite rotations which take $|{\bf\hat{z}}\rangle$ to $|{\bf\hat{n}}\rangle$, but in particular we can consider a specific rotation given by the following operator (Wigner D-Matrix) written with Euler angles $(\alpha,\beta,\gamma)$:

\begin{align} &\mathcal{D}(R)=\mathcal{D}(\alpha=\phi,\beta=\theta,\gamma=0)\\ &\implies|{\bf\hat{n}}\rangle=\mathcal{D}(R)|{\bf\hat{z}}\rangle \end{align}

Notice that if we change the value of $\gamma$ the equation remains valid (because $\gamma$ represents a rotation about ${\bf \hat{n}}$ axis). This is the same notation that appears on Sakurai and Wikipedia. Next it is shown a relation with spherical harmonics (from Sakurai & Napolitano, sec. 3.6 pp 205-206):

\begin{align} &|{\bf\hat{n}}\rangle=\mathcal{D}(R)|{\bf\hat{z}}\rangle\\ \implies &|{\bf\hat{n}}\rangle=\sum_{l',m'}\mathcal{D}(R)|l',m'\rangle\langle l',m'|{\bf\hat{z}}\rangle\\ \implies \langle l,m &|{\bf\hat{n}}\rangle=\langle l,m|\sum_{l',m'}\mathcal{D}(R)|l',m'\rangle\langle l',m'|{\bf\hat{z}}\rangle\\ \text{Using that}&\ \text{$\mathcal{D}(R)$ doesn't change the $l'$ number:}\\ \implies Y_{l}^{m*}(\theta,\phi)&=\sum_{m'}\langle l,m|\mathcal{D}(R)|l,m'\rangle\langle l,m'|{\bf\hat{z}}\rangle\\ \implies Y_{l}^{m*}(\theta,\phi)&=\sum_{m'}\mathcal{D}^{(l)}_{m,m'}(R) Y_l^{m'}(\theta=0,\phi)\\ \text{And using that}&\ \text{$Y_l^{m'}$ vanishes at $\theta=0$ for $m'\neq 0$:}\\ \implies Y_{l}^{m*}(\theta,\phi)&=\sqrt{\frac{2l+1}{4\pi}}\mathcal{D}^{(l)}_{m,0}(R)\\ \end{align}

This gives a relation between the rotation operator $\mathcal{D}(R)$ and spherical harmonics. It does not seem in any step of the derivation that $\gamma=0$ is necessary. I think that if we choose another rotation $R$ with $\gamma\neq 0$ such that $|{\bf\hat{n}}\rangle=\mathcal{D}(R)|{\bf\hat{z}}\rangle$ the relation remains valid, but this is weird because if $R$ changes then the operator $\mathcal{D}(R)$ changes (and $Y_l^{m*}$ doesn't change).

My question is: Why the spherical harmonics represent rotations with $\gamma=0$ and not another value? and, if we have a rotation with $\gamma\neq 0$, can we write it with spherical harmonics?

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  • $\begingroup$ I do not understand your final equation: $\mathcal{D}$ is an operator acting on spherical harmonics, how can it be equal to a spherical harmonic itself? That's like saying a matrix is equal to a vector, which is nonsensical. Why does the l.h.s. of your equation depend on $\theta$ and $\phi$ and the r.h.s. does not (another mathematical impossiblity, all sides of an equation must have the same number of free variables)? $\endgroup$ – ACuriousMind Oct 4 '17 at 10:40
  • $\begingroup$ $\mathcal{D}$ is not equal to a spherical harmonic, but its elements of the first column $\mathcal{D}_{m,0}^{(l)}$ are spherical harmonics. And $(\theta,\phi)$ are fixed because ${\bf \hat{n}}$ is fixed $\endgroup$ – adiselann Oct 4 '17 at 15:54
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The simplest explanation starts by writing explicitly the first Euler rotation about the $\boldsymbol{\hat z}$ as $$ R_z(\gamma)= \left(\begin{array}{ccc} \cos\gamma&\sin\gamma & 0 \\ -\sin\gamma &\cos\gamma &0 \\ 0&0&1 \end{array} \right) $$ Clearly $$ R_z(\gamma)\hat z = R_z(\gamma) \left(\begin{array}{c} 0 \\ 0 \\ 1\end{array}\right)= \left(\begin{array}{c}0 \\ 0 \\ 1\end{array}\right) $$ independent of the angle $\gamma$. Thus, your function must also be independent of $\gamma$; this is achieved by setting $\gamma=0$.

[By the way this rotation by $\gamma$ is about $\boldsymbol{\hat z}$, not ${\boldsymbol{\hat n}}$.]

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