2
$\begingroup$

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?

Improve this question
$\endgroup$
  • $\begingroup$ 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 $g(r)$? $\endgroup$ – Deschele Schilder Jan 11 '18 at 5:40
3
$\begingroup$

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 \begin{equation} O = J_+ J_- = (L_+ + S_+)(L_- + S_-) \end{equation} 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 \begin{equation} \langle\psi|O|\psi\rangle = \langle\psi|J_+ J_-|\psi\rangle =\langle\phi|\phi\rangle . \end{equation} So it's all about computing $J_- |\psi\rangle = (L_- + S_-) |\psi\rangle$ and finding the inner product with itself.

Improve this answer
$\endgroup$
  • $\begingroup$ And what if I have to find the medium value of $J^2$ in a situation like this? $\endgroup$ – pter26 Jan 14 '18 at 18:02
  • $\begingroup$ 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 $J_z = L_z + S_z$. $\endgroup$ – secavara Jan 14 '18 at 18:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.