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.

Linear combinations of spherical harmonics; probabilities

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.5.13.

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If we can express a 3-d quantum state in terms of the spherical harmonics, we can calculate directly the probabilities of ${L_{z}}$ having one of its eigenvalues. That is, if we can write a state ${\psi}$ as

$\displaystyle \psi\left(r,\theta,\phi\right)=f\left(r\right)\sum_{m}C_{l}^{m}Y_{l}^{m} \ \ \ \ \ (1)$

for some constant coefficients ${C_{l}^{m}}$ and ${f}$ is some function of ${r}$ alone, then

$\displaystyle P\left(l_{z}=m\hbar\right)=\frac{\left|C_{l}^{m}\right|^{2}}{\sum_{n}\left|C_{l}^{n}\right|^{2}} \ \ \ \ \ (2)$

As an example, suppose we have

$\displaystyle \psi=N\left(x+y+2z\right)e^{-\alpha r} \ \ \ \ \ (3)$

where ${N}$ is a normalization constant. We start by expressing ${x}$, ${y}$ and ${z}$ in terms of ${Y_{1}^{m}}$. We have

 $\displaystyle Y_{1}^{\pm1}$ $\displaystyle =$ $\displaystyle \mp\sqrt{\frac{3}{8\pi}}\sin\theta e^{\pm i\phi}\ \ \ \ \ (4)$ $\displaystyle Y_{1}^{0}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{3}{4\pi}}\cos\theta \ \ \ \ \ (5)$

Using standard spherical-to-rectangular conversions

 $\displaystyle x$ $\displaystyle =$ $\displaystyle r\sin\theta\cos\phi\ \ \ \ \ (6)$ $\displaystyle y$ $\displaystyle =$ $\displaystyle r\sin\theta\sin\phi\ \ \ \ \ (7)$ $\displaystyle z$ $\displaystyle =$ $\displaystyle r\cos\theta \ \ \ \ \ (8)$

Therefore

 $\displaystyle \cos\phi$ $\displaystyle =$ $\displaystyle \frac{x}{r\sin\theta}\ \ \ \ \ (9)$ $\displaystyle \sin\phi$ $\displaystyle =$ $\displaystyle \frac{y}{r\sin\theta}\ \ \ \ \ (10)$ $\displaystyle \cos\theta$ $\displaystyle =$ $\displaystyle \frac{z}{r} \ \ \ \ \ (11)$

Plugging these into 4 we have

 $\displaystyle Y_{1}^{\pm1}$ $\displaystyle =$ $\displaystyle \mp\sqrt{\frac{3}{8\pi}}\sin\theta\left(\cos\phi\pm i\sin\phi\right)\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \mp\sqrt{\frac{3}{8\pi}}\frac{x\pm iy}{r}\ \ \ \ \ (13)$ $\displaystyle Y_{1}^{0}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{3}{4\pi}}\frac{z}{r}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{2}\sqrt{\frac{3}{8\pi}}\frac{z}{r} \ \ \ \ \ (15)$

Inverting these, we have

 $\displaystyle x$ $\displaystyle =$ $\displaystyle \frac{1}{2}\sqrt{\frac{8\pi}{3}}r\left(Y_{1}^{-1}-Y_{1}^{1}\right)\ \ \ \ \ (16)$ $\displaystyle y$ $\displaystyle =$ $\displaystyle -\frac{1}{2i}\sqrt{\frac{8\pi}{3}}r\left(Y_{1}^{-1}+Y_{1}^{1}\right)\ \ \ \ \ (17)$ $\displaystyle z$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\sqrt{\frac{8\pi}{3}}rY_{1}^{0} \ \ \ \ \ (18)$

Thus 3 becomes

$\displaystyle \psi=\sqrt{\frac{8\pi}{3}}Nre^{-\alpha r}\left[Y_{1}^{1}\left(-\frac{1}{2}-\frac{1}{2i}\right)+Y_{1}^{-1}\left(\frac{1}{2}-\frac{1}{2i}\right)+Y_{1}^{0}\sqrt{2}\right] \ \ \ \ \ (19)$

Comparing with 1 we find

 $\displaystyle C_{1}^{1}$ $\displaystyle =$ $\displaystyle -\frac{1}{2}-\frac{1}{2i}\ \ \ \ \ (20)$ $\displaystyle C_{1}^{-1}$ $\displaystyle =$ $\displaystyle \frac{1}{2}-\frac{1}{2i}\ \ \ \ \ (21)$ $\displaystyle C_{1}^{0}$ $\displaystyle =$ $\displaystyle \sqrt{2} \ \ \ \ \ (22)$

We have

 $\displaystyle \sum_{n}\left|C_{1}^{n}\right|^{2}$ $\displaystyle =$ $\displaystyle \frac{1}{2}+\frac{1}{2}+2=3\ \ \ \ \ (23)$ $\displaystyle P\left(l_{z}=0\right)$ $\displaystyle =$ $\displaystyle \frac{\left|C_{l}^{0}\right|^{2}}{\sum_{n}\left|C_{l}^{n}\right|^{2}}=\frac{2}{3}\ \ \ \ \ (24)$ $\displaystyle P\left(l_{z}=\hbar\right)$ $\displaystyle =$ $\displaystyle \frac{\left|C_{l}^{1}\right|^{2}}{\sum_{n}\left|C_{l}^{n}\right|^{2}}=\frac{1}{6}\ \ \ \ \ (25)$ $\displaystyle P\left(l_{z}=-\hbar\right)$ $\displaystyle =$ $\displaystyle \frac{\left|C_{l}^{-1}\right|^{2}}{\sum_{n}\left|C_{l}^{n}\right|^{2}}=\frac{1}{6} \ \ \ \ \ (26)$

Angular momentum and parity

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.5.12.

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The parity operator in 3-d reflects every point directly through the origin, so that a position vector ${\mathbf{r}\rightarrow-\mathbf{r}}$. In rectangular coordinates this means replacing each coordinate by its negative. In spherical coordinates, the angular coordinates change according to

 $\displaystyle \theta$ $\displaystyle \rightarrow$ $\displaystyle \pi-\theta\ \ \ \ \ (1)$ $\displaystyle \phi$ $\displaystyle \rightarrow$ $\displaystyle \pi+\phi \ \ \ \ \ (2)$

If this isn’t obvious, picture reflecting a vector ${\mathbf{r}}$ through the origin. If the original vector makes an angle ${\theta}$ with the ${z}$ (vertical) axis, then the reflected vector makes an angle ${\theta}$ with the ${-z}$ axis, which is equivalent to an angle of ${\pi-\theta}$ with the ${+z}$ axis. The azimuthal angle ${\phi}$ just gets rotated by ${\pi}$ to lie on the other side of the ${z}$ axis.

Using this, we can see that the parity operator ${\Pi}$ commutes with both ${L^{2}}$ and ${L_{z}}$, as follows. Since neither of these operators involves the radial coordinate, we can consider their effect on a function ${f\left(\theta,\phi\right)}$. Under parity, we have

$\displaystyle \Pi f\left(\theta,\phi\right)\rightarrow f\left(\pi-\theta,\pi+\phi\right) \ \ \ \ \ (3)$

Thus the derivatives transform under parity according to

 $\displaystyle \frac{\partial f\left(\theta,\phi\right)}{\partial\theta}$ $\displaystyle \rightarrow$ $\displaystyle -\frac{\partial f\left(\pi-\theta,\pi+\phi\right)}{\partial\theta}\ \ \ \ \ (4)$ $\displaystyle \frac{\partial f\left(\theta,\phi\right)}{\partial\phi}$ $\displaystyle \rightarrow$ $\displaystyle \frac{\partial f\left(\pi-\theta,\pi+\phi\right)}{\partial\phi} \ \ \ \ \ (5)$
 $\displaystyle L^{2}$ $\displaystyle =$ $\displaystyle -\hbar^{2}\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{\sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}}\right]\ \ \ \ \ (6)$ $\displaystyle L_{z}$ $\displaystyle =$ $\displaystyle -i\hbar\frac{\partial}{\partial\phi} \ \ \ \ \ (7)$

Thus the combined operation gives

 $\displaystyle L^{2}\Pi f\left(\theta,\phi\right)$ $\displaystyle \rightarrow$ $\displaystyle -\hbar^{2}\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{\sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}}\right]f\left(\pi-\theta,\pi+\phi\right)\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\hbar^{2}\left[\frac{1}{\sin\theta}\left(-\frac{\partial}{\partial\theta}\right)\left(\sin\theta\left(-\frac{\partial}{\partial\theta}\right)\right)+\frac{1}{\sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}}\right]f\left(\pi-\theta,\pi+\phi\right)\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\hbar^{2}\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{\sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}}\right]f\left(\pi-\theta,\pi+\phi\right)\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle L^{2}f\left(\pi-\theta,\pi+\phi\right) \ \ \ \ \ (11)$

If we apply ${\Pi}$ to ${L^{2}}$, we have

 $\displaystyle \Pi\left[L^{2}f\left(\theta,\phi\right)\right]$ $\displaystyle =$ $\displaystyle -\hbar^{2}\left[\frac{1}{\sin\left(\pi-\theta\right)}\left(-\frac{\partial}{\partial\theta}\right)\left(\sin\left(\pi-\theta\right)\left(-\frac{\partial}{\partial\theta}\right)\right)+\frac{1}{\sin^{2}\left(\pi-\theta\right)}\frac{\partial^{2}}{\partial\phi^{2}}\right]f\left(\pi-\theta,\pi+\phi\right)\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\hbar^{2}\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{\sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}}\right]f\left(\pi-\theta,\pi+\phi\right)\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle L^{2}f\left(\pi-\theta,\pi+\phi\right) \ \ \ \ \ (14)$

Thus

$\displaystyle \left[\Pi,L^{2}\right]=0 \ \ \ \ \ (15)$

where in the first line we used ${\sin\left(\pi-\theta\right)=\sin\theta}$.

Since ${L_{z}}$ involves only a derivative with respect to ${\phi}$ which doesn’t change under parity, we have

$\displaystyle \left[\Pi,L_{z}\right]=0 \ \ \ \ \ (16)$

Since ${\Pi}$ commutes with both ${L^{2}}$ and ${L_{z}}$ it is possible to find a set of functions that are simultaneous eigenfunctions of all three operators. These functions turn out to be the same spherical harmonics that we’ve been using all along. We can show this by starting with the top spherical harmonic

$\displaystyle Y_{l}^{l}=\left(-1\right)^{l}\sqrt{\frac{(2l+1)!}{4\pi}}\frac{1}{2^{l}l!}e^{il\phi}\sin^{l}\theta \ \ \ \ \ (17)$

where we’ve included the ${\left(-1\right)^{l}}$ to be consistent with Shankar’s equation 12.5.32. Under parity, this transforms as

 $\displaystyle \Pi Y_{l}^{l}$ $\displaystyle =$ $\displaystyle \left(-1\right)^{l}\sqrt{\frac{(2l+1)!}{4\pi}}\frac{1}{2^{l}l!}e^{il\left(\pi+\phi\right)}\sin^{l}\left(\pi-\theta\right)\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(-1\right)^{l}e^{il\pi}\sqrt{\frac{(2l+1)!}{4\pi}}\frac{1}{2^{l}l!}e^{il\phi}\sin^{l}\theta\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(-1\right)^{l}Y_{l}^{l} \ \ \ \ \ (20)$

where we used ${e^{il\pi}=\left(-1\right)^{l}}$ in the second line. Thus ${Y_{l}^{l}}$ is an eigenfunction of ${\Pi}$ with eigenvalue ${\left(-1\right)^{l}}$.

To show that the other spherical harmonics are also eigenfunctions, we can use the lowering operator ${L_{-}}$. In spherical coordinates, we have

$\displaystyle L_{-}Y_{l}^{m}=\hbar\sqrt{(\ell+m)(\ell-m+1)}Y_{l}^{m-1} \ \ \ \ \ (21)$

The operator can be expressed as

$\displaystyle L_{-}=-\hbar e^{-i\phi}\left[\frac{\partial}{\partial\theta}-i\cot\theta\frac{\partial}{\partial\phi}\right] \ \ \ \ \ (22)$

Under parity, we can transform 22 using ${\sin\left(\pi-\theta\right)=\sin\theta}$ and ${\cos\left(\pi-\theta\right)=-\cos\theta}$, so that ${\cot\left(\pi-\theta\right)=-\cot\theta}$. We therefore have

 $\displaystyle \Pi L_{-}$ $\displaystyle =$ $\displaystyle -\hbar e^{-i\left(\pi+\phi\right)}\left[-\frac{\partial}{\partial\theta}+i\cot\theta\frac{\partial}{\partial\phi}\right]\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\hbar e^{-i\phi}\left[\frac{\partial}{\partial\theta}-i\cot\theta\frac{\partial}{\partial\phi}\right]\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle L_{-} \ \ \ \ \ (25)$

Thus ${L_{-}}$ is unchanged by parity, which means that from 21, ${Y_{l}^{m-1}}$ has the same parity as ${Y_{l}^{m}}$. Starting with ${Y_{l}^{l}}$ and using the lowering operator successively to reduce the superscript index, we have therefore

$\displaystyle \Pi Y_{l}^{m}=\left(-1\right)^{l}Y_{l}^{m} \ \ \ \ \ (26)$

Thus all spherical harmonics are also eigenfunctions of parity.

Spherical harmonics using the lowering operator

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.5.11.

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The raising and lowering operators for angular momentum are

$\displaystyle L_{\pm}\equiv L_{x}\pm iL_{y} \ \ \ \ \ (1)$

On a state ${\left|\ell m\right\rangle }$ in the basis of eigenstates of ${L^{2}}$ and ${L_{z}}$, they have the effect:

$\displaystyle L_{\pm}\left|\ell m\right\rangle =\hbar\sqrt{(\ell\mp m)(\ell\pm m+1)}\left|\ell,m\pm1\right\rangle \ \ \ \ \ (2)$

This means that, if we can find the top state ${\left|\ell\ell\right\rangle }$, we can find the state for all lower values of ${m}$ by applying ${L_{-}}$ successively. To illustrate the process we’ll derive the 3 states for ${\ell=1}$. The top state ${\left|11\right\rangle }$ can be obtained by following the derivation given in Shankar from his equation 12.5.28 onwards. In spherical coordinates, the raising and lowering operators have the form

$\displaystyle L_{\pm}=\pm\hbar e^{\pm i\phi}\left[\frac{\partial}{\partial\theta}\pm i\cot\theta\frac{\partial}{\partial\phi}\right] \ \ \ \ \ (3)$

Applying ${L_{+}}$ to the top state ${\left|11\right\rangle }$ must give zero, so if ${\psi_{1}^{1}}$ is the representation of this state in spherical coordinates, we must solve the differential equation

$\displaystyle \left[\frac{\partial}{\partial\theta}+i\cot\theta\frac{\partial}{\partial\phi}\right]\psi_{1}^{1}=0 \ \ \ \ \ (4)$

Since ${\psi_{1}^{1}}$ is also an eigenfunction of ${L_{z}}$ with eigenvalue ${\hbar}$, we know that

$\displaystyle \psi_{1}^{1}=U_{1}^{1}\left(r,\theta\right)e^{i\phi} \ \ \ \ \ (5)$

Thus 4 becomes

$\displaystyle \left(\frac{\partial}{\partial\theta}-\cot\theta\right)U_{1}^{1}=0 \ \ \ \ \ (6)$

This can be solved by writing it in the form

 $\displaystyle \frac{dU_{1}^{1}}{U_{1}^{1}}$ $\displaystyle =$ $\displaystyle \frac{d\left(\sin\theta\right)}{\sin\theta}\ \ \ \ \ (7)$ $\displaystyle \ln U_{1}^{1}$ $\displaystyle =$ $\displaystyle \ln\left(\sin\theta\right)+\ln R\left(r\right)+\ln A \ \ \ \ \ (8)$

where ${R}$ is some unspecified function of ${r}$, and ${A}$ is a constant. We therefore have

 $\displaystyle U_{1}^{1}\left(r,\theta\right)$ $\displaystyle =$ $\displaystyle R\left(r\right)\left(A\sin\theta\right) \ \ \ \ \ (9)$

If we ignore ${R}$ for now, we can normalize over the angular coordinates by requiring

$\displaystyle \int\left|A\sin\theta\right|^{2}d\Omega=1 \ \ \ \ \ (10)$

The element ${d\Omega}$ of solid angle is

$\displaystyle d\Omega=\sin\theta\;d\phi\;d\theta \ \ \ \ \ (11)$

so we have

 $\displaystyle \left|A\right|^{2}\int_{0}^{\pi}\int_{0}^{2\pi}\sin^{3}\theta\;d\phi\;d\theta$ $\displaystyle =$ $\displaystyle 2\pi\left|A\right|^{2}\int_{0}^{\pi}\sin\theta\left(1-\cos^{2}\theta\right)d\theta\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{8\pi}{3}\left|A\right|^{2}\ \ \ \ \ (13)$ $\displaystyle A$ $\displaystyle =$ $\displaystyle \sqrt{\frac{3}{8\pi}} \ \ \ \ \ (14)$

Thus the spherical harmonic ${Y_{1}^{1}}$ is (using Shankar’s normalization convention of multiplying by ${\left(-1\right)^{\ell}}$):

$\displaystyle Y_{1}^{1}=-\sqrt{\frac{3}{8\pi}}\sin\theta e^{i\phi} \ \ \ \ \ (15)$

We can now get ${Y_{1}^{0}}$ by applying ${L_{-}}$ to ${Y_{1}^{1}}$. From 2 we have

 $\displaystyle L_{-}Y_{1}^{1}$ $\displaystyle =$ $\displaystyle \hbar\sqrt{\left(1+1\right)\left(1-1+1\right)}Y_{1}^{0}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{2}\hbar Y_{1}^{0} \ \ \ \ \ (17)$

From 3 we have

 $\displaystyle L_{-}Y_{1}^{1}$ $\displaystyle =$ $\displaystyle -\hbar e^{-i\phi}\left[\frac{\partial}{\partial\theta}-i\cot\theta\frac{\partial}{\partial\phi}\right]Y_{1}^{1}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\hbar e^{-i\phi}\left(-\sqrt{\frac{3}{8\pi}}\right)\left[\cos\theta-i\cot\theta\left(i\sin\theta\right)\right]e^{i\phi}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\hbar\sqrt{\frac{3}{8\pi}}\cos\theta \ \ \ \ \ (20)$

Comparing the last two results gives

 $\displaystyle \sqrt{2}\hbar Y_{1}^{0}$ $\displaystyle =$ $\displaystyle 2\sqrt{\frac{3}{8\pi}}\cos\theta\ \ \ \ \ (21)$ $\displaystyle Y_{1}^{0}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{3}{4\pi}}\cos\theta \ \ \ \ \ (22)$

Repeating the process, we get

 $\displaystyle L_{-}Y_{1}^{0}$ $\displaystyle =$ $\displaystyle \hbar\sqrt{\left(1+0\right)\left(1-0+1\right)}Y_{1}^{-1}\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{2}\hbar Y_{1}^{-1} \ \ \ \ \ (24)$

Also

 $\displaystyle L_{-}Y_{1}^{0}$ $\displaystyle =$ $\displaystyle -\hbar e^{-i\phi}\left[\frac{\partial}{\partial\theta}-i\cot\theta\frac{\partial}{\partial\phi}\right]Y_{1}^{0}\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\hbar e^{-i\phi}\sqrt{\frac{3}{4\pi}}\left(-\sin\theta-0\right)\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \hbar\sqrt{\frac{3}{4\pi}}\sin\theta e^{-i\phi} \ \ \ \ \ (27)$

Thus

 $\displaystyle \sqrt{2}\hbar Y_{1}^{-1}$ $\displaystyle =$ $\displaystyle \hbar\sqrt{\frac{3}{4\pi}}\sin\theta e^{-i\phi}\ \ \ \ \ (28)$ $\displaystyle Y_{1}^{-1}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{3}{8\pi}}\sin\theta e^{-i\phi} \ \ \ \ \ (29)$

Comparing these results with Shankar’s equation 12.5.39 we see that they match. [This exercise is similar to one we did earlier, where we used the raising operator to generate spherical harmonics with higher values of ${m}$.]

Spherical harmonics from power series – examples for m=0

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.5.10.

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The total angular momentum operator ${L^{2}}$ can be written in spherical coordinates as

$\displaystyle L^{2}=-\hbar^{2}\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{\sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}}\right] \ \ \ \ \ (1)$

Since ${\left[L^{2},L_{z}\right]=0}$, we can find a basis consisting of simultaneous eigenfunctions of ${L^{2}}$ and ${L_{z}}$. Suppose we call these states ${\left|\alpha\beta\right\rangle }$, where ${\alpha}$ is the eigenvalue of ${L^{2}}$ and ${\beta}$ is the eigenvalue of ${L_{z}}$. In spherical coordinates, we know that

$\displaystyle L_{z}=-i\hbar\frac{\partial}{\partial\phi} \ \ \ \ \ (2)$

and that its eigenvalues are ${m\hbar}$ for integer values of ${m}$. Thus we can separate the ${\theta}$ and ${\phi}$ dependence in the eigenstates and write

$\displaystyle \psi_{\alpha m}\left(\theta,\phi\right)=P_{\alpha}^{m}\left(\theta\right)e^{im\phi} \ \ \ \ \ (3)$

We therefore have the eigenvalue equation

 $\displaystyle L^{2}\left|\alpha m\right\rangle$ $\displaystyle =$ $\displaystyle \alpha\left|\alpha m\right\rangle \ \ \ \ \ (4)$ $\displaystyle L^{2}\psi_{\alpha m}\left(\theta,\phi\right)$ $\displaystyle =$ $\displaystyle \alpha\psi_{\alpha m}\left(\theta,\phi\right) \ \ \ \ \ (5)$

Combining 3 with 1, we have

 $\displaystyle \frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial\psi_{\alpha m}}{\partial\theta}\right)+\frac{1}{\sin^{2}\theta}\frac{\partial^{2}\psi_{\alpha m}}{\partial\phi^{2}}$ $\displaystyle =$ $\displaystyle \alpha\psi_{\alpha m}\ \ \ \ \ (6)$ $\displaystyle \frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial P_{\alpha}^{m}}{\partial\theta}\right)-\frac{m^{2}}{\sin^{2}\theta}P_{\alpha}^{m}+\frac{\alpha}{\hbar^{2}}P_{\alpha}^{m}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (7)$

We’d like to show that solutions of this equation require that (1)

 $\displaystyle \alpha$ $\displaystyle =$ $\displaystyle \hbar^{2}\ell\left(\ell+1\right)\ \ \ \ \ (8)$ $\displaystyle \left|m\right|$ $\displaystyle \le$ $\displaystyle \ell \ \ \ \ \ (9)$

for ${\ell=0,1,2,\ldots}$. In the problem given in Shankar, we tackle the less demanding case of ${m=0}$ and demonstrate only the result for ${\alpha}$. We begin by transforming 7 using the variable substitution:

$\displaystyle u\equiv\cos\theta \ \ \ \ \ (10)$

This gives us

$\displaystyle du=-\sin\theta\;d\theta \ \ \ \ \ (11)$

so that 7 becomes

 $\displaystyle \frac{-\sin\theta}{\sin\theta}\frac{\partial}{\partial u}\left(-\sin^{2}\theta\frac{\partial P_{\alpha}^{0}}{\partial u}\right)+\frac{\alpha}{\hbar^{2}}P_{\alpha}^{0}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (12)$ $\displaystyle \frac{\partial}{\partial u}\left(\left(1-u^{2}\right)\frac{\partial P_{\alpha}^{0}}{\partial u}\right)+\frac{\alpha}{\hbar^{2}}P_{\alpha}^{0}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (13)$ $\displaystyle \left(1-u^{2}\right)\frac{\partial^{2}P_{\alpha}^{0}}{\partial u^{2}}-2u\frac{\partial P_{\alpha}^{0}}{\partial u}+\frac{\alpha}{\hbar^{2}}P_{\alpha}^{0}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (14)$

We can use a power series to solve this by defining

 $\displaystyle P_{\alpha}^{0}\left(u\right)$ $\displaystyle =$ $\displaystyle \sum_{n=0}^{\infty}C_{n}u^{n}\ \ \ \ \ (15)$ $\displaystyle \frac{\partial P_{\alpha}^{0}}{\partial u}$ $\displaystyle =$ $\displaystyle \sum_{n=0}^{\infty}C_{n}nu^{n-1}\ \ \ \ \ (16)$ $\displaystyle \frac{\partial^{2}P_{\alpha}^{0}}{\partial u^{2}}$ $\displaystyle =$ $\displaystyle \sum_{n=0}^{\infty}C_{n}n\left(n-1\right)u^{n-2}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{n=0}^{\infty}C_{n+2}\left(n+2\right)\left(n+1\right)u^{n} \ \ \ \ \ (18)$

Plugging these into 14 and collecting terms, we get

$\displaystyle P_{\alpha}^{0}\left(u\right)=\sum_{n=0}^{\infty}\left[C_{n+2}\left(n+2\right)\left(n+1\right)+C_{n}\left(-n\left(n-1\right)-2n+\frac{\alpha}{\hbar^{2}}\right)\right]u^{n}=0 \ \ \ \ \ (19)$

If a power series equals zero, the coefficient of each power of ${u}$ must be zero (power series theorem from math), so we get the recurrence relation

 $\displaystyle C_{n+2}$ $\displaystyle =$ $\displaystyle \frac{n\left(n-1\right)+2n-\frac{\alpha}{\hbar^{2}}}{\left(n+2\right)\left(n+1\right)}C_{n}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{n^{2}+n-\frac{\alpha}{\hbar^{2}}}{n^{2}+3n+2}C_{n} \ \ \ \ \ (21)$

For large ${n}$ we have

$\displaystyle C_{n+2}\rightarrow\frac{n^{2}}{n^{2}}C_{n}=C_{n} \ \ \ \ \ (22)$

Since ${u=\cos\theta}$, ${u\in\left[-1,1\right]}$ and the series must converge for all these values. Although the power series ${\sum_{n=0}^{\infty}u^{n}}$ converges if ${\left|u\right|<1}$ (that’s the standard geometric series), it clearly diverges if ${u=1}$. Thus we require the series to terminate, which imposes a condition on ${\alpha}$:

$\displaystyle \alpha=\ell\left(\ell+1\right)\hbar^{2} \ \ \ \ \ (23)$

for some integer value ${\ell=0,1,2,\ldots}$. Since choosing a value for ${\ell}$ can be done only once in any given series, and the recursion relation relates every second coefficient, this implies that either all even coefficients or all odd coefficients must be zero. Thus ${P_{\alpha}^{0}\left(u\right)}$ is either a sum of even powers (making it an even function) or of odd powers (making it an odd function) only.

The first few values of ${P_{\alpha}^{0}\left(u\right)}$ are found by choosing values for ${C_{0}}$ and ${C_{1}}$ and then generating all higher coefficients using 21. If we take

 $\displaystyle C_{0}$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (24)$ $\displaystyle C_{1}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (25)$

then if we choose ${\ell=0}$ we get

$\displaystyle P_{0}^{0}=1 \ \ \ \ \ (26)$

Taking

 $\displaystyle C_{0}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (27)$ $\displaystyle C_{1}$ $\displaystyle =$ $\displaystyle 1 \ \ \ \ \ (28)$

and ${\ell=1}$ gives

$\displaystyle P_{1}^{0}=u=\cos\theta \ \ \ \ \ (29)$

Reverting to an even series and taking ${\ell=2}$ we have from 21

 $\displaystyle C_{2}$ $\displaystyle =$ $\displaystyle -\frac{\alpha}{2\hbar^{2}}C_{0}=-\frac{\ell\left(\ell+1\right)}{2}\left(1\right)=-3\ \ \ \ \ (30)$ $\displaystyle P_{2}^{0}$ $\displaystyle =$ $\displaystyle 1-3u^{2}=1-3\cos^{2}\theta \ \ \ \ \ (31)$

These values for ${P_{\ell}^{0}}$ agree with the spherical harmonics ${Y_{\ell}^{0}}$ apart from the constant scaling factors in each case. See Shankar’s equation 12.5.39 for comparison.

Total angular momentum is Hermitian

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.5.9.

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The total angular momentum operator ${L^{2}}$ can be written in spherical coordinates as

$\displaystyle L^{2}=-\hbar^{2}\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{\sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}}\right] \ \ \ \ \ (1)$

As ${L^{2}}$ is an observable, it should be Hermitian. We can verify this by showing that

$\displaystyle \left\langle \psi_{2}\left|L^{2}\right|\psi_{1}\right\rangle =\left\langle \psi_{1}\left|L^{2}\right|\psi_{2}\right\rangle ^* \ \ \ \ \ (2)$

In spherical coordinates, this becomes

$\displaystyle \int\psi_{2}^*\left(L^{2}\psi_{1}\right)d\Omega=\left[\int\psi_{1}^*\left(L^{2}\psi_{2}\right)d\Omega\right]^* \ \ \ \ \ (3)$

The element of solid angle ${d\Omega=\sin\theta\;d\theta\;d\phi}$, so the full integral is

$\displaystyle \int\psi_{2}^*\left(L^{2}\psi_{1}\right)d\Omega=\int_{0}^{2\pi}\int_{0}^{\pi}\psi_{2}^*\left(L^{2}\psi_{1}\right)\sin\theta\;d\theta\;d\phi \ \ \ \ \ (4)$

We can verify 3 by showing that it is true for each of the two terms in 1 separately. As usual for these sorts of integrals, we need to use integration by parts. To simplify things, we’ll consider ${-L^{2}/\hbar^{2}}$ so we can deal only with the terms in the brackets in 1. We’ll also use the shorthand notation

 $\displaystyle s$ $\displaystyle \equiv$ $\displaystyle \sin\theta\ \ \ \ \ (5)$ $\displaystyle c$ $\displaystyle \equiv$ $\displaystyle \cos\theta \ \ \ \ \ (6)$

Also, a prime indicates a derivative with respect to ${\theta}$: ${\psi_{1}^{\prime}\equiv\frac{\partial\psi_{1}}{\partial\theta}}$, etc.

For the first term, we have, considering only the integration over ${\theta}$:

 $\displaystyle \int_{0}^{\pi}\psi_{2}^*\frac{1}{s}\frac{\partial}{\partial\theta}\left(s\frac{\partial\psi_{1}}{\partial\theta}\right)s\;d\theta$ $\displaystyle =$ $\displaystyle \int_{0}^{\pi}\left[\psi_{2}^*c\psi_{1}^{\prime}+\psi_{2}^*s\psi_{1}^{\prime\prime}\right]d\theta\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left.\psi_{2}^*c\psi_{1}\right|_{0}^{\pi}+\left.\psi_{2}^*s\psi_{1}^{\prime}\right|_{0}^{\pi}-\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle$ $\displaystyle \int_{0}^{\pi}\left[\left(\psi_{2}^*\right)^{\prime}c\psi_{1}-\psi_{2}^*s\psi_{1}\right]d\theta-\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle$ $\displaystyle \int_{0}^{\pi}\left[\left(\psi_{2}^*\right)^{\prime}s\psi_{1}^{\prime}+\psi_{2}^*c\psi_{1}^{\prime}\right]d\theta \ \ \ \ \ (10)$

The second term in 8 is zero since ${\sin0=\sin\pi=0}$, but we can’t ignore the first term, which is not, in general, zero. Thus we are left with

 $\displaystyle \int_{0}^{\pi}\psi_{2}^*\frac{\partial}{\partial\theta}\left(s\frac{\partial\psi_{1}}{\partial\theta}\right)\;d\theta$ $\displaystyle =$ $\displaystyle \left.\psi_{2}^*c\psi_{1}\right|_{0}^{\pi}-\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle$ $\displaystyle \int_{0}^{\pi}\left[\left(\psi_{2}^*\right)^{\prime}c\psi_{1}-\psi_{2}^*s\psi_{1}\right]d\theta-\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle$ $\displaystyle \int_{0}^{\pi}\left[\left(\psi_{2}^*\right)^{\prime}s\psi_{1}^{\prime}+\psi_{2}^*c\psi_{1}^{\prime}\right]d\theta \ \ \ \ \ (13)$

We can now integrate the last line by parts again to get rid of the derivatives of ${\psi_{1}}$:

 $\displaystyle -\int_{0}^{\pi}\left[\left(\psi_{2}^*\right)^{\prime}s\psi_{1}^{\prime}+\psi_{2}^*c\psi_{1}^{\prime}\right]d\theta$ $\displaystyle =$ $\displaystyle -\left.\left(\psi_{2}^*\right)^{\prime}s\psi_{1}\right|_{0}^{\pi}-\left.\psi_{2}^*c\psi_{1}\right|_{0}^{\pi}+\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle$ $\displaystyle \int_{0}^{\pi}\left[\psi_{1}\left(\psi_{2}^*\right)^{\prime\prime}s+\left(\psi_{2}^*\right)^{\prime}c\psi_{1}\right]d\theta+\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle$ $\displaystyle \int_{0}^{\pi}\left[\psi_{1}\left(\psi_{2}^*\right)^{\prime}c-\psi_{2}^*s\psi_{1}\right]d\theta\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\left.\psi_{2}^*c\psi_{1}\right|_{0}^{\pi}+\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle$ $\displaystyle \int_{0}^{\pi}\left[\psi_{1}\left(\psi_{2}^*\right)^{\prime\prime}s+\left(\psi_{2}^*\right)^{\prime}c\psi_{1}\right]d\theta+\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle$ $\displaystyle \int_{0}^{\pi}\left[\psi_{1}\left(\psi_{2}^*\right)^{\prime}c-\psi_{2}^*s\psi_{1}\right]d\theta \ \ \ \ \ (19)$

Inserting this back into 11 and cancelling terms, we have

$\displaystyle \int_{0}^{\pi}\psi_{2}^*\frac{\partial}{\partial\theta}\left(s\frac{\partial\psi_{1}}{\partial\theta}\right)\;d\theta=\int_{0}^{\pi}\left[\psi_{1}\left(\psi_{2}^*\right)^{\prime\prime}s+\left(\psi_{2}^*\right)^{\prime}c\psi_{1}\right]d\theta \ \ \ \ \ (20)$

Comparing this with 7, we see that

$\displaystyle \int_{0}^{\pi}\psi_{2}^*\frac{\partial}{\partial\theta}\left(s\frac{\partial\psi_{1}}{\partial\theta}\right)\;d\theta=\left[\int_{0}^{\pi}\psi_{1}^*\frac{\partial}{\partial\theta}\left(s\frac{\partial\psi_{2}}{\partial\theta}\right)\;d\theta\right]^* \ \ \ \ \ (21)$

Thus the first term in 1 is Hermitian. (As this first term involves no derivatives with respect to ${\phi}$, the integration over ${\phi}$ is automatically Hermitian.)

For the second term in 1, we need to consider only the integral over ${\phi}$, so we have

$\displaystyle \int_{0}^{2\pi}\psi_{2}^*\frac{1}{\sin^{2}\theta}\frac{\partial^{2}\psi_{1}}{\partial\phi^{2}}\sin\theta\;d\phi=\frac{1}{s}\int_{0}^{2\pi}\psi_{2}^*\frac{\partial^{2}\psi_{1}}{\partial\phi^{2}}\;d\phi \ \ \ \ \ (22)$

(As we’re integrating over ${\phi}$, terms in ${\theta}$ act as constants and can be taken outside the integral.) The first integration by parts gives (where a prime now indicates a derivative with respect to ${\phi}$):

 $\displaystyle \int_{0}^{2\pi}\psi_{2}^*\psi_{1}^{\prime\prime}\;d\phi$ $\displaystyle =$ $\displaystyle \left.\psi_{2}^*\psi_{1}^{\prime}\right|_{0}^{2\pi}-\int_{0}^{2\pi}\left(\psi_{2}^*\right)^{\prime}\psi_{1}^{\prime}\;d\phi\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\int_{0}^{2\pi}\left(\psi_{2}^*\right)^{\prime}\psi_{1}^{\prime}\;d\phi \ \ \ \ \ (24)$

This time, we’re able to set the integrated term to zero, since ${\phi=0}$ and ${\phi=2\pi}$ refer to the same angle. A second integration by parts gives

 $\displaystyle -\int_{0}^{2\pi}\left(\psi_{2}^*\right)^{\prime}\psi_{1}^{\prime}\;d\phi$ $\displaystyle =$ $\displaystyle -\left.\left(\psi_{2}^*\right)^{\prime}\psi_{1}\right|_{0}^{2\pi}+\int_{0}^{2\pi}\left(\psi_{2}^*\right)^{\prime\prime}\psi_{1}\;d\phi\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{0}^{2\pi}\left(\psi_{2}^*\right)^{\prime\prime}\psi_{1}\;d\phi\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\int_{0}^{2\pi}\psi_{1}^*\psi_{2}^{\prime\prime}\;d\phi\right]^* \ \ \ \ \ (27)$

Thus both terms in 1 are Hermitian, so the complete operator ${L^{2}}$ is also Hermitian.

Angular momentum – raising and lowering operators from rectangular coordinates

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.5.8.

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To calculate the eigenfunctions of angular momentum, we will need expressions for the raising and lowering operators ${L_{\pm}}$ in spherical coordinates. We’ve seen one way of getting these by working with the gradient in spherical coordinates from the start, but it is also possible to convert the rectangular forms of ${L_{\pm}}$ to spherical coordinates by using the chain rule from calculus. This method is similar to one we used earlier in 2-d. To set the scene, we need the conversion formulas between rectangular and spherical coordinates:

 $\displaystyle x$ $\displaystyle =$ $\displaystyle r\sin\theta\cos\phi\ \ \ \ \ (1)$ $\displaystyle y$ $\displaystyle =$ $\displaystyle r\sin\theta\sin\phi\ \ \ \ \ (2)$ $\displaystyle z$ $\displaystyle =$ $\displaystyle r\cos\theta\ \ \ \ \ (3)$ $\displaystyle r$ $\displaystyle =$ $\displaystyle \sqrt{x^{2}+y^{2}+z^{2}}\ \ \ \ \ (4)$ $\displaystyle \theta$ $\displaystyle =$ $\displaystyle \arctan\frac{\sqrt{x^{2}+y^{2}}}{z}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \arctan\frac{q}{z}\ \ \ \ \ (6)$ $\displaystyle \phi$ $\displaystyle =$ $\displaystyle \arctan\frac{y}{x} \ \ \ \ \ (7)$

To simplify the notation, we’ve defined

$\displaystyle q\equiv\sqrt{x^{2}+y^{2}}=r\sin\theta \ \ \ \ \ (8)$

We’ll also use shorthand notation for sines and cosines so that

 $\displaystyle s_{\theta}$ $\displaystyle \equiv$ $\displaystyle \sin\theta\ \ \ \ \ (9)$ $\displaystyle c_{\theta}$ $\displaystyle \equiv$ $\displaystyle \cos\theta \ \ \ \ \ (10)$

and similarly for ${\phi}$. We’ll also use the notation ${\partial_{r}}$ to mean the partial derivative with respect to ${r}$, with a similar notation for other derivatives.

The required derivatives are

 $\displaystyle \partial_{x}$ $\displaystyle =$ $\displaystyle \partial_{x}r\cdot\partial_{r}+\partial_{x}\theta\cdot\partial_{\theta}+\partial_{x}\phi\cdot\partial_{\phi}\ \ \ \ \ (11)$ $\displaystyle \partial_{y}$ $\displaystyle =$ $\displaystyle \partial_{y}r\cdot\partial_{r}+\partial_{y}\theta\cdot\partial_{\theta}+\partial_{y}\phi\cdot\partial_{\phi}\ \ \ \ \ (12)$ $\displaystyle \partial_{z}$ $\displaystyle =$ $\displaystyle \partial_{z}r\cdot\partial_{r}+\partial_{z}\theta\cdot\partial_{\theta} \ \ \ \ \ (13)$

The required derivatives are

 $\displaystyle \partial_{x}r$ $\displaystyle =$ $\displaystyle \frac{x}{r}\ \ \ \ \ (14)$ $\displaystyle \partial_{y}r$ $\displaystyle =$ $\displaystyle \frac{y}{r}\ \ \ \ \ (15)$ $\displaystyle \partial_{z}r$ $\displaystyle =$ $\displaystyle \frac{z}{r}\ \ \ \ \ (16)$ $\displaystyle \partial_{x}\theta$ $\displaystyle =$ $\displaystyle \frac{x/q}{z\left(1+\frac{q^{2}}{z^{2}}\right)}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{xz}{qr^{2}}\ \ \ \ \ (18)$ $\displaystyle \partial_{y}\theta$ $\displaystyle =$ $\displaystyle \frac{yz}{qr^{2}}\ \ \ \ \ (19)$ $\displaystyle \partial_{z}\theta$ $\displaystyle =$ $\displaystyle -\frac{q}{r^{2}}\ \ \ \ \ (20)$ $\displaystyle \partial_{x}\phi$ $\displaystyle =$ $\displaystyle \frac{-y/x^{2}}{1+y^{2}/x^{2}}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{y}{q^{2}}\ \ \ \ \ (22)$ $\displaystyle \partial_{y}\phi$ $\displaystyle =$ $\displaystyle \frac{x}{q^{2}}\ \ \ \ \ (23)$ $\displaystyle \partial_{z}\phi$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (24)$

Plugging all these into 11 to 13 we have

 $\displaystyle \partial_{x}$ $\displaystyle =$ $\displaystyle \frac{x}{r}\partial_{r}+\frac{xz}{qr^{2}}\partial_{\theta}-\frac{y}{q^{2}}\partial_{\phi}\ \ \ \ \ (25)$ $\displaystyle \partial_{y}$ $\displaystyle =$ $\displaystyle \frac{y}{r}\partial_{r}+\frac{yz}{qr^{2}}\partial_{\theta}+\frac{x}{q^{2}}\partial_{\phi}\ \ \ \ \ (26)$ $\displaystyle \partial_{z}$ $\displaystyle =$ $\displaystyle \frac{z}{r}\partial_{r}-\frac{q}{r^{2}}\partial_{\theta} \ \ \ \ \ (27)$

We can now calculate the components ${L_{x}}$ and ${L_{y}}$:

 $\displaystyle L_{x}$ $\displaystyle =$ $\displaystyle -i\hbar\left(y\partial_{z}-z\partial_{y}\right)\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\hbar\left[\frac{yz}{r}\partial_{r}-\frac{yq}{r^{2}}\partial_{\theta}-\frac{yz}{r}\partial_{r}-\frac{yz^{2}}{qr^{2}}\partial_{\theta}-\frac{xz}{q^{2}}\partial_{\phi}\right]\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar\left[\left(\frac{yq}{r^{2}}+\frac{yz^{2}}{qr^{2}}\right)\partial_{\theta}+\frac{xz}{q^{2}}\partial_{\phi}\right]\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar\left[\left(s_{\theta}^{2}s_{\phi}+\frac{s_{\theta}s_{\phi}c_{\theta}^{2}}{s_{\theta}}\right)\partial_{\theta}+\frac{s_{\theta}c_{\theta}c_{\phi}}{s_{\theta}^{2}}\partial_{\phi}\right]\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar\left(\sin\phi\frac{\partial}{\partial\theta}+\cos\phi\cot\theta\frac{\partial}{\partial\phi}\right)\ \ \ \ \ (32)$ $\displaystyle L_{y}$ $\displaystyle =$ $\displaystyle -i\hbar\left(z\partial_{x}-x\partial_{z}\right)\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\hbar\left[\frac{xz}{r}\partial_{r}+\frac{xz^{2}}{qr^{2}}\partial_{\theta}-\frac{xz}{r}\partial_{r}+\frac{xq}{r^{2}}\partial_{\theta}-\frac{yz}{q^{2}}\partial_{\phi}\right]\ \ \ \ \ (34)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar\left[\left(-\frac{xz^{2}}{qr^{2}}-\frac{xq}{r^{2}}\right)\partial_{\theta}+\frac{yz}{q^{2}}\partial_{\phi}\right]\ \ \ \ \ (35)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar\left[\left(-\frac{s_{\theta}c_{\phi}c_{\theta}^{2}}{s_{\theta}}-s_{\theta}^{2}c_{\phi}\right)\partial_{\theta}+\frac{s_{\theta}c_{\theta}s_{\phi}}{s_{\theta}^{2}}\partial_{\phi}\right]\ \ \ \ \ (36)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar\left(-\cos\phi\frac{\partial}{\partial\theta}+\sin\phi\cot\theta\frac{\partial}{\partial\phi}\right) \ \ \ \ \ (37)$

From this we get the raising and lowering operators

 $\displaystyle L_{\pm}$ $\displaystyle =$ $\displaystyle L_{x}\pm iL_{y}\ \ \ \ \ (38)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar\left(\sin\phi\frac{\partial}{\partial\theta}+\cos\phi\cot\theta\frac{\partial}{\partial\phi}\right)\mp\ \ \ \ \ (39)$ $\displaystyle$ $\displaystyle$ $\displaystyle \hbar\left(-\cos\phi\frac{\partial}{\partial\theta}+\sin\phi\cot\theta\frac{\partial}{\partial\phi}\right)\ \ \ \ \ (40)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \hbar e^{\pm i\phi}\frac{\partial}{\partial\theta}\pm i\hbar e^{\pm i\phi}\cot\theta\frac{\partial}{\partial\phi}\ \ \ \ \ (41)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \hbar e^{\pm i\phi}\left(\frac{\partial}{\partial\theta}\pm i\cot\theta\frac{\partial}{\partial\phi}\right) \ \ \ \ \ (42)$

[Admittedly, it’s probably easier and more elegant to use spherical coordinates from the start, but it’s instructive to see how it’s done starting with rectangular coordinates.]

Rotations in 3-d: Euler angles

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.5.7.

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Any 3-d rotation can be expressed in terms of the Euler angles. These angles specify a sequence of three successive rotations about the rectangular axes. There are various definitions of Euler angles involving different sets of rotations, but the set used by Shankar in this problem consists of (1) a rotation by ${\gamma}$ about the ${z}$ axis, followed by (2) a rotation by ${\beta}$ about the ${y}$ axis and concluding with (3) a rotation by ${\alpha}$ about the ${z}$ axis. The proof that any rotation can be expressed this way would take us too far afield at this point, so we’ll just accept this for now.

We can see how these work in quantum mechanics by considering the special case of ${j=1}$, for which we derived the formula for a finite rotation here. A state ${\left|\psi\right\rangle }$ is transformed by a rotation ${\boldsymbol{\theta}}$ according to

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

For our purposes below, we’ll need the three components of ${\mathbf{J}^{\left(1\right)}}$, which can be copied from Shankar’s equations 12.5.23 and 12.5.24:

 $\displaystyle J_{x}^{\left(1\right)}$ $\displaystyle =$ $\displaystyle \frac{\hbar}{\sqrt{2}}\left[\begin{array}{ccc} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end{array}\right]\ \ \ \ \ (3)$ $\displaystyle J_{y}^{\left(1\right)}$ $\displaystyle =$ $\displaystyle \frac{i\hbar}{\sqrt{2}}\left[\begin{array}{ccc} 0 & -1 & 0\\ 1 & 0 & -1\\ 0 & 1 & 0 \end{array}\right]\ \ \ \ \ (4)$ $\displaystyle \left(J_{y}^{\left(1\right)}\right)^{2}$ $\displaystyle =$ $\displaystyle \frac{\hbar^{2}}{2}\left[\begin{array}{ccc} 1 & 0 & -1\\ 0 & 2 & 0\\ -1 & 0 & 1 \end{array}\right]\ \ \ \ \ (5)$ $\displaystyle J_{z}^{\left(1\right)}$ $\displaystyle =$ $\displaystyle \hbar\left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end{array}\right]\ \ \ \ \ (6)$ $\displaystyle \left(J_{z}^{\left(1\right)}\right)^{2}$ $\displaystyle =$ $\displaystyle \hbar^{2}\left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 1 \end{array}\right] \ \ \ \ \ (7)$

Evaluating 2 for the three Euler rotations, we have

 $\displaystyle D_{\gamma}$ $\displaystyle =$ $\displaystyle I^{\left(1\right)}+\frac{\left(J_{z}^{\left(1\right)}\right)^{2}}{\hbar^{2}}\left(\cos\gamma-1\right)-\frac{iJ_{z}^{\left(1\right)}}{\hbar}\sin\gamma\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{ccc} \cos\gamma-i\sin\gamma & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & \cos\gamma+i\sin\gamma \end{array}\right]\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{ccc} e^{-i\gamma} & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & e^{i\gamma} \end{array}\right]\ \ \ \ \ (10)$ $\displaystyle D_{\beta}$ $\displaystyle =$ $\displaystyle I^{\left(1\right)}+\frac{\left(J_{y}^{\left(1\right)}\right)^{2}}{\hbar^{2}}\left(\cos\beta-1\right)-\frac{iJ_{y}^{\left(1\right)}}{\hbar}\sin\beta\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{ccc} 1+\cos\beta & -\sqrt{2}\sin\beta & 1-\cos\beta\\ \sqrt{2}\sin\beta & 2\cos\beta & -\sqrt{2}\sin\beta\\ 1-\cos\beta & \sqrt{2}\sin\beta & 1+\cos\beta \end{array}\right]\ \ \ \ \ (12)$ $\displaystyle D_{\alpha}$ $\displaystyle =$ $\displaystyle I^{\left(1\right)}+\frac{\left(J_{z}^{\left(1\right)}\right)^{2}}{\hbar^{2}}\left(\cos\alpha-1\right)-\frac{iJ_{z}^{\left(1\right)}}{\hbar}\sin\alpha\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{ccc} \cos\alpha-i\sin\alpha & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & \cos\alpha+i\sin\alpha \end{array}\right]\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{ccc} e^{-i\alpha} & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & e^{i\alpha} \end{array}\right] \ \ \ \ \ (15)$

The complete rotation is the product of the three matrices:

 $\displaystyle D_{total}$ $\displaystyle =$ $\displaystyle D_{\alpha}D_{\beta}D_{\gamma}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle D_{\alpha}\frac{1}{2}\left[\begin{array}{ccc} \left(1+\cos\beta\right)e^{-i\gamma} & -\sqrt{2}\sin\beta & \left(1-\cos\beta\right)e^{i\gamma}\\ \sqrt{2}\sin\beta e^{-i\gamma} & 2\cos\beta & -\sqrt{2}\sin\beta e^{i\gamma}\\ \left(1-\cos\beta\right)e^{-i\gamma} & \sqrt{2}\sin\beta & \left(1+\cos\beta\right)e^{i\gamma} \end{array}\right]\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{ccc} \left(1+\cos\beta\right)e^{-i\gamma}e^{-i\alpha} & -\sqrt{2}\sin\beta e^{-i\alpha} & \left(1-\cos\beta\right)e^{i\gamma}e^{-i\alpha}\\ \sqrt{2}\sin\beta e^{-i\gamma} & 2\cos\beta & -\sqrt{2}\sin\beta e^{i\gamma}\\ \left(1-\cos\beta\right)e^{-i\gamma}e^{i\alpha} & \sqrt{2}\sin\beta e^{i\alpha} & \left(1+\cos\beta\right)e^{i\gamma}e^{i\alpha} \end{array}\right] \ \ \ \ \ (18)$

In the ${\left|jm\right\rangle }$ basis, the state ${\left|11\right\rangle }$ is represented by

$\displaystyle \left|11\right\rangle =\left[\begin{array}{c} 1\\ 0\\ 0 \end{array}\right] \ \ \ \ \ (19)$

Applying the rotation 18 to this state, we get

 $\displaystyle \left|11^{\prime}\right\rangle$ $\displaystyle =$ $\displaystyle D_{total}\left|11\right\rangle \ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{ccc} \left(1+\cos\beta\right)e^{-i\gamma}e^{-i\alpha} & -\sqrt{2}\sin\beta e^{-i\alpha} & \left(1-\cos\beta\right)e^{i\gamma}e^{-i\alpha}\\ \sqrt{2}\sin\beta e^{-i\gamma} & 2\cos\beta & -\sqrt{2}\sin\beta e^{i\gamma}\\ \left(1-\cos\beta\right)e^{-i\gamma}e^{i\alpha} & \sqrt{2}\sin\beta e^{i\alpha} & \left(1+\cos\beta\right)e^{i\gamma}e^{i\alpha} \end{array}\right]\left[\begin{array}{c} 1\\ 0\\ 0 \end{array}\right]\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{c} \left(1+\cos\beta\right)e^{-i\gamma}e^{-i\alpha}\\ \sqrt{2}\sin\beta e^{-i\gamma}\\ \left(1-\cos\beta\right)e^{-i\gamma}e^{i\alpha} \end{array}\right]\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{e^{-i\gamma}}{2}\left[\begin{array}{c} \left(1+\cos\beta\right)e^{-i\alpha}\\ \sqrt{2}\sin\beta\\ \left(1-\cos\beta\right)e^{i\alpha} \end{array}\right] \ \ \ \ \ (23)$

We can work out the average values of the components of ${\mathbf{J}}$ in the rotated state in the same way as in the previous problem, by using 3, 4 and 6. Note that ${\gamma}$ disappears from the matrix elements as it enters only in an overall phase factor. We get (I used Maple to do the tedious matrix multiplications):

 $\displaystyle \left\langle J_{x}\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle 11^{\prime}\left|J_{x}\right|11^{\prime}\right\rangle \ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{4\sqrt{2}}\left[\begin{array}{ccc} \left(1+\cos\beta\right)e^{i\alpha} & \sqrt{2}\sin\beta & \left(1-\cos\beta\right)e^{-i\alpha}\end{array}\right]\left[\begin{array}{ccc} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end{array}\right]\left[\begin{array}{c} \left(1+\cos\beta\right)e^{-i\alpha}\\ \sqrt{2}\sin\beta\\ \left(1-\cos\beta\right)e^{i\alpha} \end{array}\right]\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \hbar\sin\beta\cos\alpha\ \ \ \ \ (26)$ $\displaystyle \left\langle J_{y}\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle 11^{\prime}\left|J_{y}\right|11^{\prime}\right\rangle \ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{i\hbar}{4\sqrt{2}}\left[\begin{array}{ccc} \left(1+\cos\beta\right)e^{i\alpha} & \sqrt{2}\sin\beta & \left(1-\cos\beta\right)e^{-i\alpha}\end{array}\right]\left[\begin{array}{ccc} 0 & -1 & 0\\ 1 & 0 & -1\\ 0 & 1 & 0 \end{array}\right]\left[\begin{array}{c} \left(1+\cos\beta\right)e^{-i\alpha}\\ \sqrt{2}\sin\beta\\ \left(1-\cos\beta\right)e^{i\alpha} \end{array}\right]\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \hbar\sin\beta\sin\alpha\ \ \ \ \ (29)$ $\displaystyle \left\langle J_{z}\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle 11^{\prime}\left|J_{z}\right|11^{\prime}\right\rangle \ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{4}\left[\begin{array}{ccc} \left(1+\cos\beta\right)e^{i\alpha} & \sqrt{2}\sin\beta & \left(1-\cos\beta\right)e^{-i\alpha}\end{array}\right]\left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end{array}\right]\left[\begin{array}{c} \left(1+\cos\beta\right)e^{-i\alpha}\\ \sqrt{2}\sin\beta\\ \left(1-\cos\beta\right)e^{i\alpha} \end{array}\right]\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \hbar\cos\beta \ \ \ \ \ (32)$

Going back to 23, we can confirm our earlier result that it is impossible to rotate the state ${\left|11\right\rangle }$ into just ${\left|10\right\rangle }$. To do so, the state in 23 would have to be a multiple of ${\left[\begin{array}{c} 0\\ 1\\ 0 \end{array}\right]}$ which could happen only if both ${1+\cos\beta}$ and ${1-\cos\beta}$ were zero, which is impossible.

However, if we take ${\beta=\pi}$, then the rotated state 23 becomes

$\displaystyle \left|11^{\prime}\right\rangle =e^{i\left(\alpha-\gamma\right)}\left[\begin{array}{c} 0\\ 0\\ 1 \end{array}\right]=e^{i\left(\alpha-\gamma\right)}\left|1,-1\right\rangle \ \ \ \ \ (33)$

so apart from a phase factor, it is possible to rotate one eigenstate of ${J_{z}}$ into another.

The only values of ${\beta}$ that produce zero elements in 23 are ${\beta=0}$ and ${\beta=\pi}$, (the values of ${\alpha}$ and ${\gamma}$ produce only phase factors), so for any other value of ${\beta}$, all three elements of 23 are non-zero. Thus a general rotation from any starting state can always be made to produce a rotated state containing all three eigenstates of ${J_{z}}$: ${\left|11\right\rangle }$, ${\left|10\right\rangle }$ and ${\left|1,-1\right\rangle }$.

Rotations in 3-d: classical and quantum rotations compared

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.5.6.

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

This is an example of how risky it can be to attempt to derive quantum behaviour by using logic based on classical mechanics. In classical mechanics, if we have a system with some total angular momentum with magnitude ${\left|\mathbf{J}\right|}$ and rotate this sytem through any angle, the magnitude of the angular momentum remains the same (although, of course, its direction changes). Based on this fact, we might think that if we start with a quantum state such as ${\left|jm\right\rangle }$ (where ${j}$ is the total angular momentum number and ${m}$ is the number for ${J_{z}}$), we should be able to obtain the other states with the same total angular momentum number ${j}$ by rotating this state through various angles about the appropriate rotation axis. To see that this won’t work, suppose we consider a state with ${j=1}$ and ${m=1}$, that is ${\left|jm\right\rangle =\left|11\right\rangle }$. Classically, such a system has its angular momentum aligned along the ${z}$ axis, so we might think that if we rotate this system by ${\frac{\pi}{2}}$ about, say, the ${x}$ axis, we should get a state with ${m=0}$, since the angular momentum is now aligned along the ${-y}$ axis.

To see if this works, we can use the formula for a finite rotation for a ${j=1}$ state. Since ${j}$ remains constant, a rotation of a state ${\left|jm\right\rangle }$ is given by

$\displaystyle D^{\left(1\right)}\left[R\right]\left|jm\right\rangle \ \ \ \ \ (1)$

where

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

For a rotation by an angle ${\theta}$ about the ${x}$ axis, this formula reduces to

$\displaystyle D^{\left(1\right)}\left[R\left(\theta\hat{\mathbf{x}}\right)\right]=I^{\left(1\right)}+\frac{\left(J_{x}^{\left(1\right)}\right)^{2}}{\hbar^{2}}\left(\cos\theta-1\right)-\frac{iJ_{x}^{\left(1\right)}}{\hbar}\sin\theta \ \ \ \ \ (3)$

We can copy the matrix ${J_{x}^{\left(1\right)}}$ from Shankar’s equation 12.5.23:

$\displaystyle J_{x}^{\left(1\right)}=\frac{\hbar}{\sqrt{2}}\left[\begin{array}{ccc} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end{array}\right] \ \ \ \ \ (4)$

We therefore have

$\displaystyle \left(J_{x}^{\left(1\right)}\right)^{2}=\frac{\hbar^{2}}{2}\left[\begin{array}{ccc} 1 & 0 & 1\\ 0 & 2 & 0\\ 1 & 0 & 1 \end{array}\right] \ \ \ \ \ (5)$

Plugging these into 3 we have

$\displaystyle D^{\left(1\right)}\left[R\left(\theta\hat{\mathbf{x}}\right)\right]=\frac{1}{2}\left[\begin{array}{ccc} 1+\cos\theta & -\sqrt{2}i\sin\theta & \cos\theta-1\\ -\sqrt{2}i\sin\theta & 2\cos\theta & -\sqrt{2}i\sin\theta\\ \cos\theta-1 & -\sqrt{2}i\sin\theta & 1+\cos\theta \end{array}\right] \ \ \ \ \ (6)$

In the ${\left|jm\right\rangle }$ basis, the state ${\left|11\right\rangle }$ is represented by

$\displaystyle \left|11\right\rangle =\left[\begin{array}{c} 1\\ 0\\ 0 \end{array}\right] \ \ \ \ \ (7)$

Thus a rotation about the ${x}$ axis rotates this state into:

 $\displaystyle \left|\psi\right\rangle$ $\displaystyle =$ $\displaystyle D^{\left(1\right)}\left[R\left(\theta\hat{\mathbf{x}}\right)\right]\left|11\right\rangle \ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{ccc} 1+\cos\theta & -\sqrt{2}i\sin\theta & \cos\theta-1\\ -\sqrt{2}i\sin\theta & 2\cos\theta & -\sqrt{2}i\sin\theta\\ \cos\theta-1 & -\sqrt{2}i\sin\theta & 1+\cos\theta \end{array}\right]\left[\begin{array}{c} 1\\ 0\\ 0 \end{array}\right]\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{c} 1+\cos\theta\\ -\sqrt{2}i\sin\theta\\ \cos\theta-1 \end{array}\right] \ \ \ \ \ (10)$

If rotation by some angle ${\theta}$ could change ${\left|11\right\rangle }$ into the state ${\left|10\right\rangle }$, this result would need to be a multiple of ${\left[\begin{array}{c} 0\\ 1\\ 0 \end{array}\right]}$, so we’d need to find ${\theta}$ such that both of the following equations are true:

 $\displaystyle 1+\cos\theta$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (11)$ $\displaystyle \cos\theta-1$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (12)$

This is impossible, so no rotation about the ${x}$ axis can change ${\left|11\right\rangle }$ into the state ${\left|10\right\rangle }$.

However, there is still a correspondence between classical and quantum mechanics if we compare the average values of the components of ${\mathbf{J}}$. That is, we want to find ${\left\langle \mathbf{J}\right\rangle }$ for the state 10. To do this, we need the other two matrix components ${J_{y}^{\left(1\right)}}$ and ${J_{z}^{\left(1\right)}}$. We can get ${J_{y}^{\left(1\right)}}$ from Shankar’s equation 12.5.24:

$\displaystyle J_{y}^{\left(1\right)}=\frac{i\hbar}{\sqrt{2}}\left[\begin{array}{ccc} 0 & -1 & 0\\ 1 & 0 & -1\\ 0 & 1 & 0 \end{array}\right] \ \ \ \ \ (13)$

${J_{z}^{\left(1\right)}}$ is just the diagonal matrix:

$\displaystyle J_{z}^{\left(1\right)}=\hbar\left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end{array}\right] \ \ \ \ \ (14)$

We can now calculate the averages for the state 10:

 $\displaystyle \left\langle J_{x}^{\left(1\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle \psi\left|J_{x}^{\left(1\right)}\right|\psi\right\rangle \ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{4\sqrt{2}}\left[\begin{array}{ccc} 1+\cos\theta & \sqrt{2}i\sin\theta & \cos\theta-1\end{array}\right]\left[\begin{array}{ccc} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end{array}\right]\left[\begin{array}{c} 1+\cos\theta\\ -\sqrt{2}i\sin\theta\\ \cos\theta-1 \end{array}\right]\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{4}\left[\begin{array}{ccc} i\sin\theta & \sqrt{2}\cos\theta & i\sin\theta\end{array}\right]\left[\begin{array}{c} 1+\cos\theta\\ -\sqrt{2}i\sin\theta\\ \cos\theta-1 \end{array}\right]\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (18)$ $\displaystyle \left\langle J_{y}^{\left(1\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle \psi\left|J_{y}^{\left(1\right)}\right|\psi\right\rangle \ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{i\hbar}{4\sqrt{2}}\left[\begin{array}{ccc} 1+\cos\theta & \sqrt{2}i\sin\theta & \cos\theta-1\end{array}\right]\left[\begin{array}{ccc} 0 & -1 & 0\\ 1 & 0 & -1\\ 0 & 1 & 0 \end{array}\right]\left[\begin{array}{c} 1+\cos\theta\\ -\sqrt{2}i\sin\theta\\ \cos\theta-1 \end{array}\right]\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{4}\left[\begin{array}{ccc} -\sin\theta & -i\sqrt{2} & \sin\theta\end{array}\right]\left[\begin{array}{c} 1+\cos\theta\\ -\sqrt{2}i\sin\theta\\ \cos\theta-1 \end{array}\right]\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\hbar\sin\theta\ \ \ \ \ (22)$ $\displaystyle \left\langle J_{z}^{\left(1\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle \psi\left|J_{z}^{\left(1\right)}\right|\psi\right\rangle \ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{4}\left[\begin{array}{ccc} 1+\cos\theta & \sqrt{2}i\sin\theta & \cos\theta-1\end{array}\right]\left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end{array}\right]\left[\begin{array}{c} 1+\cos\theta\\ -\sqrt{2}i\sin\theta\\ \cos\theta-1 \end{array}\right]\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{4}\left[\begin{array}{ccc} 1+\cos\theta & 0 & 1-\cos\theta\end{array}\right]\left[\begin{array}{c} 1+\cos\theta\\ -\sqrt{2}i\sin\theta\\ \cos\theta-1 \end{array}\right]\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \hbar\cos\theta \ \ \ \ \ (26)$

Thus for the average, we have

$\displaystyle \left\langle \mathbf{J}\right\rangle =-\hbar\sin\theta\hat{\mathbf{y}}+\hbar\cos\theta\hat{\mathbf{z}} \ \ \ \ \ (27)$

In this case, a rotation by ${\theta=\frac{\pi}{2}}$ does indeed rotate ${\left\langle \mathbf{J}\right\rangle }$ so that it points along the ${-y}$ axis, just as it would in classical mechanics.

Total angular momentum – finite rotations

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercises 12.5.4 – 12.5.5.

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

For infinitesimal 3-d rotations, we’ve seen that the generator is ${\hat{\boldsymbol{\theta}}\cdot\mathbf{L}}$ where ${\hat{\boldsymbol{\theta}}}$ is a unit vector along the axis of rotation. Generalizing this to the total angular momentum ${\mathbf{J}}$ we have the operator for a general 3-d rotation through an infinitesimal angle:

$\displaystyle U\left[R\left(\delta\boldsymbol{\theta}\right)\right]=I-\frac{i\delta\boldsymbol{\theta}\cdot\mathbf{J}}{\hbar} \ \ \ \ \ (1)$

In principle ‘all’ we need to do to get the operator for a finite 3-d rotation is take the exponential, in the form

$\displaystyle e^{-i\boldsymbol{\theta}\cdot\mathbf{J}/\hbar} \ \ \ \ \ (2)$

The problem is that in this case, ${\mathbf{J}}$ is infinite dimensional, so the exponential of such a matrix cannot be calculated. However, because the components of ${\mathbf{J}}$ are block diagonal (see Shankar’s equations 12.5.23 and 12.5.24), all powers of these components are also block diagonal, and thus so is the exponential. For a given value of the total momentum quantum number ${j}$, the corresponding block is a ${\left(2j+1\right)\times\left(2j+1\right)}$ sub-matrix ${J_{i}^{\left(j\right)}}$ (where the suffix ${i}$ refers to ${x}$, ${y}$ or ${z}$), so the block in the exponential, defined as ${D^{\left(j\right)}\left[R\left(\boldsymbol{\theta}\right)\right]}$ is calculated as

$\displaystyle D^{\left(j\right)}\left[R\left(\boldsymbol{\theta}\right)\right]=\sum_{n=0}^{\infty}\frac{1}{n!}\left(\frac{-i\theta}{\hbar}\right)^{n}\left(\hat{\boldsymbol{\theta}}\cdot\mathbf{J}\right)^{n} \ \ \ \ \ (3)$

This may still look pretty hopeless in terms of actual calculation, but for small values of ${j}$, we can actually get closed-form solutions.

First, we look at the eigenvalues of ${\hat{\boldsymbol{\theta}}\cdot\mathbf{J}}$. If we review the calculations by which we found that the eigenvalues of ${L_{z}}$ (and thus also ${J_{z}}$) were ${-j,-j+1,\ldots,j-1,j}$ (multiplied by ${\hbar}$), we see that there’s nothing special about the fact that we chose the ${z}$ direction over any other direction as the component of ${\mathbf{J}}$ for which we calculated the eigenvalues. We could, for example, go through exactly the same calculations taking ${L_{x}}$ to be the chosen component. We would then define raising and lowering operators as ${J_{\pm}=J_{y}\pm iJ_{z}}$ and come out with the conclusion that the eigenvalues of ${L_{x}}$ are also ${-j,-j+1,\ldots,j-1,j}$ (multiplied by ${\hbar}$). We can generalize even further and choose the ‘special’ direction to be the axis of rotation, however that axis may be oriented in space. This would lead us to the conclusion that the eigenvalues of ${\hat{\boldsymbol{\theta}}\cdot\mathbf{J}}$ are the same as those of ${J_{z}}$.

Now consider the operator (where ${J\equiv\hat{\boldsymbol{\theta}}\cdot\mathbf{J}}$):

$\displaystyle \left(J-j\hbar\right)\left(J-\left(j-1\right)\hbar\right)\left(J-\left(j-2\right)\hbar\right)\ldots\left(J+\left(j-1\right)\hbar\right)\left(J+j\hbar\right) \ \ \ \ \ (4)$

First, suppose that ${J=J_{z}}$ (so that ${\hat{\boldsymbol{\theta}}}$ is along the ${z}$ axis). Then if we’re in an eigenstate ${\left|jm\right\rangle }$ of ${J_{z}}$, the term ${\left(J-m\hbar\right)}$ in this operator will give zero when operating on this state. Thus the operator 4 will always give zero when operating on an eigenstate of ${J_{z}}$. However, since the set of eigenstates of ${J_{z}}$ span the space in which the total angular momentum number is ${j}$, any state in this space can be expressed as a linear combination of eigenstates of ${J_{z}}$, so when 4 operates on this state, there is always one factor in the operator that gives zero for each term in the linear combination. Thus this operator always gives zero when operating on any state with angular momentum ${j}$. [Note that the order in which we write the factors in 4 doesn’t matter; the only operator in the expression is ${J}$, so all the factors commute with each other.] That is, we have

$\displaystyle \left(J-j\hbar\right)\left(J-\left(j-1\right)\hbar\right)\left(J-\left(j-2\right)\hbar\right)\ldots\left(J+\left(j-1\right)\hbar\right)\left(J+j\hbar\right)=0 \ \ \ \ \ (5)$

If we multiply out this operator, we get a polynomial of degree ${2j+1}$ in ${J}$. The highest power can thus be written as a linear combination of lower powers:

$\displaystyle J^{2j+1}=\sum_{n=0}^{2j}a_{n}J^{n} \ \ \ \ \ (6)$

where the coefficients ${a_{n}}$ can be found by expanding the formula (which we won’t need to do here). But this implies that all higher powers of ${J}$ can also be written as linear combinations of powers up to ${J^{2j}}$. To see this, consider

 $\displaystyle J^{2j+2}$ $\displaystyle =$ $\displaystyle J\times J^{2j+1}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{n=0}^{2j}a_{n}J^{n+1}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{n=1}^{2j}a_{n-1}J^{n}+a_{2j}J^{2j+1}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{n=1}^{2j}a_{n-1}J^{n}+a_{2j}\sum_{n=0}^{2j}a_{n}J^{n} \ \ \ \ \ (10)$

Thus ${J^{2j+2}}$ can be written as a linear combination of powers of ${J}$ up to ${J^{2j}}$. By iterating this process, we can express all higher powers of ${J}$ as a linear combination of powers of ${J}$ up to ${J^{2j}}$. Here are a couple of examples. [Shankar marks these as ‘hard’, though I can’t see that they are any more difficult than most of his other problems, so hopefully I’m not missing anything.]

Consider ${D^{\left(1/2\right)}\left[R\right]}$, starting from 3. We first use 5 with ${j=\frac{1}{2}}$:

 $\displaystyle \left(J-\frac{\hbar}{2}\right)\left(J+\frac{\hbar}{2}\right)$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (11)$ $\displaystyle J^{2}$ $\displaystyle =$ $\displaystyle \frac{\hbar^{2}}{4} \ \ \ \ \ (12)$

We can now iterate this formula as described above to get (to be accurate, all the ${I}$ and ${J}$ terms should have a superscript ${\left(1/2\right)}$ to indicate that they refer to the subspace with ${j=\frac{1}{2}}$, but this would clutter the notation).

 $\displaystyle J^{0}$ $\displaystyle =$ $\displaystyle I\ \ \ \ \ (13)$ $\displaystyle J^{1}$ $\displaystyle =$ $\displaystyle J\ \ \ \ \ (14)$ $\displaystyle J^{2}$ $\displaystyle =$ $\displaystyle \left(\frac{\hbar}{2}\right)^{2}I\ \ \ \ \ (15)$ $\displaystyle J^{3}$ $\displaystyle =$ $\displaystyle \left(\frac{\hbar}{2}\right)^{2}J\ \ \ \ \ (16)$ $\displaystyle J^{4}$ $\displaystyle =$ $\displaystyle \left(\frac{\hbar}{2}\right)^{4}I\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle \vdots$

From 3 we have

 $\displaystyle D^{\left(1/2\right)}\left[R\right]$ $\displaystyle =$ $\displaystyle \sum_{n=0}^{\infty}\frac{1}{n!}\left(\frac{-i\theta}{\hbar}\right)^{n}J^{n} \ \ \ \ \ (18)$

We can consider the even and odd terms in this sum separately. For the evens:

 $\displaystyle \left(D^{\left(1/2\right)}\left[R\right]\right)_{even}$ $\displaystyle =$ $\displaystyle \sum_{n\;even}\frac{1}{n!}\left(\frac{-i\theta}{\hbar}\right)^{n}\left(\frac{\hbar}{2}\right)^{n}I\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{n\;even}\frac{1}{n!}\left(\frac{-i\theta}{2}\right)^{n}I\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[1-\left(\frac{\theta}{2}\right)^{2}\frac{1}{2!}+\left(\frac{\theta}{2}\right)^{4}\frac{1}{4!}-\ldots\right]I\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle I\cos\frac{\theta}{2} \ \ \ \ \ (22)$

For the odds:

 $\displaystyle \left(D^{\left(1/2\right)}\left[R\right]\right)_{odd}$ $\displaystyle =$ $\displaystyle \sum_{n\;odd}\frac{1}{n!}\left(\frac{-i\theta}{\hbar}\right)^{n}\left(\frac{\hbar}{2}\right)^{n-1}J\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2J}{\hbar}\sum_{n\;odd}\frac{\left(-i\right)^{n}}{n!}\left(\frac{\theta}{2}\right)^{n}\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2J}{\hbar}\left[-\frac{\theta}{2}i+\left(\frac{\theta}{2}\right)^{3}\frac{i}{3!}-\left(\frac{\theta}{2}\right)^{5}\frac{i}{5!}+\ldots\right]\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{2iJ}{\hbar}\sin\frac{\theta}{2} \ \ \ \ \ (26)$

Thus we get

 $\displaystyle D^{\left(1/2\right)}\left[R\right]$ $\displaystyle =$ $\displaystyle I\cos\frac{\theta}{2}-\frac{2iJ}{\hbar}\sin\frac{\theta}{2}\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle I^{\left(1/2\right)}\cos\frac{\theta}{2}-\frac{2i\hat{\boldsymbol{\theta}}\cdot\mathbf{J}^{\left(1/2\right)}}{\hbar}\sin\frac{\theta}{2} \ \ \ \ \ (28)$

(I’ve restored the superscript ${\left(1/2\right)}$.)

Going through the same process for ${j=1}$, we first look at 5 to get

 $\displaystyle \left(J-\hbar\right)J\left(J+\hbar\right)$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (29)$ $\displaystyle J^{3}$ $\displaystyle =$ $\displaystyle \hbar^{2}J \ \ \ \ \ (30)$

Again, by iterating we find the pattern:

 $\displaystyle J^{0}$ $\displaystyle =$ $\displaystyle I\ \ \ \ \ (31)$ $\displaystyle J^{1}$ $\displaystyle =$ $\displaystyle J\ \ \ \ \ (32)$ $\displaystyle J^{2}$ $\displaystyle =$ $\displaystyle J^{2}\ \ \ \ \ (33)$ $\displaystyle J^{3}$ $\displaystyle =$ $\displaystyle \hbar^{2}J\ \ \ \ \ (34)$ $\displaystyle J^{4}$ $\displaystyle =$ $\displaystyle \hbar^{2}J^{2}\ \ \ \ \ (35)$ $\displaystyle$ $\displaystyle \vdots$

We then have

$\displaystyle D^{\left(1\right)}\left[R\right]=\sum_{n=0}^{\infty}\frac{1}{n!}\left(\frac{-i\theta}{\hbar}\right)^{n}J^{n} \ \ \ \ \ (36)$

Again, we can consider evens and odds separately:

 $\displaystyle \left(D^{\left(1\right)}\left[R\right]\right)_{even}$ $\displaystyle =$ $\displaystyle \sum_{n\;even}\frac{1}{n!}\left(\frac{-i\theta}{\hbar}\right)^{n}J^{n}\ \ \ \ \ (37)$ $\displaystyle$ $\displaystyle =$ $\displaystyle I+\sum_{n=2,4,\ldots}\frac{1}{n!}\left(\frac{-i\theta}{\hbar}\right)^{n}\hbar^{n-2}J^{2}\ \ \ \ \ (38)$ $\displaystyle$ $\displaystyle =$ $\displaystyle I+\frac{J^{2}}{\hbar^{2}}\sum_{n=2,4,\ldots}\frac{\left(-i\theta\right)^{n}}{n!}\ \ \ \ \ (39)$ $\displaystyle$ $\displaystyle =$ $\displaystyle I+\frac{J^{2}}{\hbar^{2}}\left(\cos\theta-1\right) \ \ \ \ \ (40)$

For the odds:

 $\displaystyle \left(D^{\left(1\right)}\left[R\right]\right)_{even}$ $\displaystyle =$ $\displaystyle \sum_{n\;odd}\frac{1}{n!}\left(\frac{-i\theta}{\hbar}\right)^{n}\hbar^{n-1}J\ \ \ \ \ (41)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{J}{\hbar}\sum_{n\;odd}\frac{\left(-i\theta\right)^{n}}{n!}\ \ \ \ \ (42)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{iJ}{\hbar}\sin\theta \ \ \ \ \ (43)$

We have

 $\displaystyle D^{\left(1\right)}\left[R\right]$ $\displaystyle =$ $\displaystyle I+\frac{J^{2}}{\hbar^{2}}\left(\cos\theta-1\right)-\frac{iJ}{\hbar}\sin\theta\ \ \ \ \ (44)$ $\displaystyle$ $\displaystyle =$ $\displaystyle I^{\left(1\right)}+\frac{\left(\hat{\boldsymbol{\theta}}\cdot\mathbf{J}^{\left(1\right)}\right)^{2}}{\hbar^{2}}\left(\cos\theta-1\right)-\frac{i\hat{\boldsymbol{\theta}}\cdot\mathbf{J}^{\left(1\right)}}{\hbar}\sin\theta \ \ \ \ \ (45)$

[I’m not sure why Shankar restricts this problem to the ${x}$ axis, or, for that matter, why he expects us to use the matrix for ${J_{x}}$.]