Spherical harmonics: rotation about the x axis

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

[If some equations are too small to read easily, use your browser’s magnifying option (Ctrl + on Chrome, probably something similar on other browsers).]

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.

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