# Spherical harmonics: rotation about the x axis

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.5.14.

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Here’s another example of using spherical harmonics to study the behaviour of wave functions in 3-d. Under a rotation by ${\theta_{x}}$ about the ${x}$ axis, the coordinates transform using the rotation matrix

$\displaystyle R\left(\theta_{x}\hat{\mathbf{x}}\right)=\left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & \cos\theta_{x} & -\sin\theta_{x}\\ 0 & \sin\theta_{x} & \cos\theta_{x} \end{array}\right] \ \ \ \ \ (1)$

This results in the coordinate transformations

 $\displaystyle x$ $\displaystyle \rightarrow$ $\displaystyle x\ \ \ \ \ (2)$ $\displaystyle y$ $\displaystyle \rightarrow$ $\displaystyle y\cos\theta_{x}-z\sin\theta_{x}\ \ \ \ \ (3)$ $\displaystyle z$ $\displaystyle \rightarrow$ $\displaystyle z\cos\theta_{x}+y\sin\theta_{x} \ \ \ \ \ (4)$

Using similar techniques to those for translations, it is found that the wave function ${\psi\left(x,y,z\right)}$ transforms into the wave function at the position obtained by rotating by ${-\theta_{x}}$ (that is, by rotating by ${\theta_{x}}$ in the opposite direction):

$\displaystyle \psi\left(x,y,z\right)\rightarrow\psi_{R}=\psi\left(x,y\cos\theta_{x}+z\sin\theta_{x},z\cos\theta_{x}-y\sin\theta_{x}\right) \ \ \ \ \ (5)$

Suppose we have a wave function given by

$\displaystyle \psi=Aze^{-r^{2}/a^{2}} \ \ \ \ \ (6)$

for some constants ${a}$ and ${A}$. Under this rotation, using 5 it transforms to

$\displaystyle \psi_{R}=A\left(z\cos\theta_{x}-y\sin\theta_{x}\right)e^{-r^{2}/a^{2}} \ \ \ \ \ (7)$

[Note that ${r^{2}}$ remains invariant under rotations about the origin, since the distance of a point from the origin is not affected by a rotation. You can verify this directly if you like by working out ${r^{2}=x^{2}+y^{2}+z^{2}}$ after the rotation.]

Equation 7 differs from the equation given in Shankar, which is

$\displaystyle \psi_{R}=A\left(z\cos\theta_{x}+y\sin\theta_{x}\right)e^{-r^{2}/a^{2}} \ \ \ \ \ (8)$

Curiously, in the errata for Shankar’s book (2006 edition) 7 is listed as the incorrect version, which is ‘corrected’ to 8. In my copy of the book (which doesn’t have a date on the title page), 8 is printed, but I don’t think this is right. In any case, we’ll proceed with the problem.

First, we write 6 in terms of spherical harmonics, using

 $\displaystyle x$ $\displaystyle =$ $\displaystyle \frac{1}{2}\sqrt{\frac{8\pi}{3}}r\left(Y_{1}^{-1}-Y_{1}^{1}\right)\ \ \ \ \ (9)$ $\displaystyle y$ $\displaystyle =$ $\displaystyle -\frac{1}{2i}\sqrt{\frac{8\pi}{3}}r\left(Y_{1}^{-1}+Y_{1}^{1}\right)\ \ \ \ \ (10)$ $\displaystyle z$ $\displaystyle =$ $\displaystyle \sqrt{\frac{4\pi}{3}}rY_{1}^{0} \ \ \ \ \ (11)$

We have

$\displaystyle \psi=A\sqrt{\frac{4\pi}{3}}rY_{1}^{0}e^{-r^{2}/a^{2}} \ \ \ \ \ (12)$

With the three spherical harmonics ${Y_{1}^{1}}$, ${Y_{1}^{0}}$ and ${Y_{1}^{-1}}$ as the basis, we can write this in vector notation as

$\displaystyle \psi=A\sqrt{\frac{4\pi}{3}}re^{-r^{2}/a^{2}}\left[\begin{array}{c} 0\\ 1\\ 0 \end{array}\right] \ \ \ \ \ (13)$

A rotation in 3-d for ${\ell=1}$ is given by

$\displaystyle D^{\left(1\right)}\left[R\right]=I^{\left(1\right)}+\frac{\left(\hat{\boldsymbol{\theta}}\cdot\mathbf{J}^{\left(1\right)}\right)^{2}}{\hbar^{2}}\left(\cos\theta-1\right)-\frac{i\hat{\boldsymbol{\theta}}\cdot\mathbf{J}^{\left(1\right)}}{\hbar}\sin\theta \ \ \ \ \ (14)$

For ${\hat{\boldsymbol{\theta}}=\theta_{x}\hat{\mathbf{x}}}$, this works out to

$\displaystyle D^{\left(1\right)}\left[R\left(\theta_{x}\hat{\mathbf{x}}\right)\right]=\frac{1}{2}\left[\begin{array}{ccc} 1+\cos\theta_{x} & -\sqrt{2}i\sin\theta_{x} & \cos\theta_{x}-1\\ -\sqrt{2}i\sin\theta_{x} & 2\cos\theta_{x} & -\sqrt{2}i\sin\theta_{x}\\ \cos\theta_{x}-1 & -\sqrt{2}i\sin\theta_{x} & 1+\cos\theta_{x} \end{array}\right] \ \ \ \ \ (15)$

We can use this to transform 13 to get

 $\displaystyle \psi_{R}$ $\displaystyle =$ $\displaystyle D^{\left(1\right)}\left[R\left(\theta_{x}\hat{\mathbf{x}}\right)\right]\psi\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle A\sqrt{\frac{4\pi}{3}}re^{-r^{2}/a^{2}}\frac{1}{2}\left[\begin{array}{ccc} 1+\cos\theta_{x} & -\sqrt{2}i\sin\theta_{x} & \cos\theta_{x}-1\\ -\sqrt{2}i\sin\theta_{x} & 2\cos\theta_{x} & -\sqrt{2}i\sin\theta_{x}\\ \cos\theta_{x}-1 & -\sqrt{2}i\sin\theta_{x} & 1+\cos\theta_{x} \end{array}\right]\left[\begin{array}{c} 0\\ 1\\ 0 \end{array}\right]\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle A\sqrt{\frac{\pi}{3}}re^{-r^{2}/a^{2}}\left[\begin{array}{c} -\sqrt{2}i\sin\theta_{x}\\ 2\cos\theta_{x}\\ -\sqrt{2}i\sin\theta_{x} \end{array}\right]\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle Ae^{-r^{2}/a^{2}}\left\{ \sqrt{\frac{4\pi}{3}}r\left[\begin{array}{c} 0\\ 1\\ 0 \end{array}\right]\cos\theta_{x}-\sqrt{\frac{2\pi}{3}}ri\sin\theta_{x}\left(\left[\begin{array}{c} 1\\ 0\\ 0 \end{array}\right]+\left[\begin{array}{c} 0\\ 0\\ 1 \end{array}\right]\right)\right\} \ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle Ae^{-r^{2}/a^{2}}\left\{ \sqrt{\frac{4\pi}{3}}rY_{1}^{0}\cos\theta_{x}+\frac{1}{2i}\sqrt{\frac{8\pi}{3}}r\left(Y_{1}^{1}+Y_{1}^{-1}\right)\sin\theta_{x}\right\} \ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle Ae^{-r^{2}/a^{2}}\left(z\cos\theta_{x}-y\sin\theta_{x}\right) \ \ \ \ \ (21)$

where we used 10 to get the last line. This result agrees with 7 and not with the equation 8 given in Shankar, so (provided I got the signs right) it looks like Shankar’s equation is wrong.