# Problem about angular momentum in quantum mechanics

A particle with spin $\frac{1}{2}$ at $t=0$ is in a quantum state described by the wave function: $$\Psi=(|+\rangle +(1+\cos\theta) |-\rangle)f(r).$$ Temporal evolution is given by $$H=\frac{\omega} {\hbar} (L_x^2+L_y^2)$$

I have to calculate the expectation value of the operator $O=J_+J_-$

Because of the presence of the cosine I wrote the angular part of the quantum state with the spherical harmonics, I know that $$Y_1^0=\sqrt{\frac{3}{4\pi}}cos\theta$$ ($Y_\ell^m$) So $$cos\theta=\sqrt{\frac{4\pi}{3}}Y_1^0$$ I also know that $$Y_0^0=\sqrt{\frac{1}{4\pi}}$$ At the end I obtained (after a renormalization) $$\Psi=g(r) (|00\rangle|+\rangle+|00\rangle|-\rangle+\frac{1}{\sqrt{3}}|10\rangle|-\rangle)$$ Where $|L^2, L_z\rangle=|00\rangle=Y_0^0$ and $|10\rangle=Y_1^0$

With the temporal evolution obtained: $$\Psi_t=g(r) (|00\rangle|+\rangle+|00\rangle|-\rangle+\frac{1}{\sqrt{3}}e^{-\frac{i\omega t} {\sqrt{3}}}|10\rangle|-\rangle)$$ but now I should apply the composition of angular momenta, in order to calculate $\langle O\rangle$ and here is the problem, I have never applied the composition in a case of a wave function depending by two different values of $\ell$, I thought that I could treat separately the two parts with different $\ell$ but I don't know if it is possible! How can I solve this problem?

• How did you find, (after a renormalization), the factor $\frac{1}{\sqrt{3}}$? Shouldn't it be $\sqrt{1-\frac{1}{\pi}}$? And why did you replace $f(r)$ by $g(r)$? Jan 11 '18 at 5:40

I don't think it is necessary to go to the basis of total angular momentum to compute that expression. In principle you could just write $$O = J_+ J_- = (L_+ + S_+)(L_- + S_-)$$ and use the usual formulae for these ladder operators.
You can even simplify the computation by using the fact that the hermitian conjugate of $J_+$ is $J_-$ and viceversa. Which means that you can compute first the state $|\phi\rangle = J_- |\psi\rangle$. And then realize that $$\langle\psi|O|\psi\rangle = \langle\psi|J_+ J_-|\psi\rangle =\langle\phi|\phi\rangle .$$ So it's all about computing $J_- |\psi\rangle = (L_- + S_-) |\psi\rangle$ and finding the inner product with itself.
• And what if I have to find the medium value of $J^2$ in a situation like this? Jan 14 '18 at 18:02
• You can use similar tricks. In this case, since you have already computed the expectation value of $J_+ J_-$, you can use the identity $J^2 = J_+ J_- + J^2_z - \hbar J_z$. And then write $J_z = L_z + S_z$. Jan 14 '18 at 18:13