Angular momentum operators on a wave function

[gasar8]

Homework Statement

Particle is in a state with wave function \psi (r) = A z (x+y)e^{-\lambda r}.
a) What is the probability that the result of the L_z measurement is 0?
b) What are possilble results and what are their probabilities of a L^2 measurement?
c) What are possilble results and what are their probabilities of a L_x measurement?

The Attempt at a Solution

Firstly, I tried to write wave function in spherical harmonics form:
<br /> \begin{align*}<br /> \psi (r) &amp;= \alpha z (x+y); \ \ \alpha= A e^{-\lambda r}\\ &amp;= \alpha r \cos\theta (r \sin\theta\cos\phi+r \sin\theta \sin \phi)\\ &amp;= \alpha r^2 \cos\theta \sin \theta (\cos\phi + \sin \phi) \\ &amp;=\alpha r^2\cos\theta \sin\theta((\frac{1}{2}+\frac{i}{2})e^{-i \phi}+(\frac{1}{2}-\frac{i}{2})e^{i \phi}) \\ &amp;= -{\alpha \over 2} r^2 \sqrt{\frac{8 \pi}{15}} ((1+i)Y_{2,-1}+(1-i)Y_{2,1})<br /> \end{align*}<br />

Then, I normalized this function:
|\psi,r\rangle = {\alpha \over {2c}} r^2 \sqrt{\frac{8 \pi}{15}} ((1+i) c Y_{2,-1}+(1-i) c Y_{2,1}) \\ \Longrightarrow c^2 |(1+i)|^2+c^2 |(1-i)|^2=1 \\ c={1 \over 2}
So finally, I get my wave function:
\psi (r) = \alpha r^2 \sqrt{\frac{8 \pi}{15}} ({1 \over 2}(1+i) Y_{2,-1}+{1 \over 2}(1-i)Y_{2,1})
Can someone check this normalization, I am not sure if I did it correctly?

a) Form definition L_z|l,m\rangle=\hbar m |l,m\rangle I can't get 0 as a result, because there isn't any element of a wave function with m=0. So the answer is 0 probability.

b) Form definition L^2|l,m\rangle=\hbar^2 l(l+1) |l,m\rangle I can only get the result 6 \hbar^2 with 100% probability, since l is in both elements of a wave function 2.

c) Form definition L_x|l,m\rangle=\frac{L_+ + L_-}{2} |l,m\rangle, I get:
<br /> \begin{align*}<br /> L_+ |2,1\rangle &amp;= 2 \hbar |2,2\rangle \\<br /> L_+ |2,-1\rangle &amp;= \sqrt{6} \hbar |2,0\rangle \\<br /> L_- |2,1\rangle &amp;= \sqrt{6} \hbar |2,0\rangle \\<br /> L_- |2,-1\rangle &amp;= 2 \hbar |2,-2\rangle\\<br /> \end{align*}<br />
<br /> \begin{align*}<br /> \Longrightarrow L_x |2,1\rangle &amp;= \hbar (|2,2\rangle + \frac{\sqrt{6}}{2} |2,0\rangle)\\<br /> \Longrightarrow L_x |2,-1\rangle &amp;= \hbar (|2,-2\rangle + \frac{\sqrt{6}}{2} |2,0\rangle)<br /> \end{align*}<br />

So the possible results are \hbar and \frac{\sqrt{6}}{2} \hbar? But how can I now find their probabilities? I tried to write:
L_x |\psi\rangle = \sqrt{\frac{8 \pi}{15}} \alpha \hbar r^2(\frac{1+i}{2}(|2,2\rangle+{\sqrt{6} \over2} |2,0\rangle)+\frac{1-i}{2}(|2,-2\rangle +{\sqrt{6} \over2} |2,0\rangle)),
but the squares of absolute values of coefficients are above 1.

 

[DrClaude]

<br /> \begin{align*}<br /> \psi (r) &amp;= \alpha z (x+y); \ \ \alpha= A e^{-\lambda r}\\ &amp;= \alpha r \cos\theta (r \sin\theta\cos\phi+r \sin\theta \sin \phi)\\ &amp;= \alpha r^2 \cos\theta \sin \theta (\cos\phi + \sin \phi) \\ &amp;=\alpha r^2\cos\theta \sin\theta((\frac{1}{2}+\frac{i}{2})e^{-i \phi}+(\frac{1}{2}-\frac{i}{2})e^{i \phi}) \\ &amp;= -{\alpha \over 2} r^2 \sqrt{\frac{8 \pi}{15}} ((1+i)Y_{2,-1}+(1-i)Y_{2,1})<br /> \end{align*}<br />

Be careful with the sign in the last line.

Then, I normalized this function:
|\psi,r\rangle = {\alpha \over {2c}} r^2 \sqrt{\frac{8 \pi}{15}} ((1+i) c Y_{2,-1}+(1-i) c Y_{2,1}) \\ \Longrightarrow c^2 |(1+i)|^2+c^2 |(1-i)|^2=1 \\ c={1 \over 2}
So finally, I get my wave function:
\psi (r) = \alpha r^2 \sqrt{\frac{8 \pi}{15}} ({1 \over 2}(1+i) Y_{2,-1}+{1 \over 2}(1-i)Y_{2,1})
Can someone check this normalization, I am not sure if I did it correctly?

I don't understand what you did here. Actually, compare that last line with the last line above; do you see any difference? Also, what about $\alpha$ (and the $A$ it contains)?

 

[gasar8]

Aha, I forgot that Y_l^{-m}=(-1)^m Y_l^m, so I get:
<br /> |\psi\rangle = {\alpha \over 2} r^2 \sqrt{\frac{8 \pi}{15}} \big((1+i)Y_2^{-1}-(1-i)Y_2^{1}\big)<br />

Ok, here I tried something we did at lectures, but apparently I didn't understand it well. So if I go the long way:
<br /> \langle \psi,r | \psi,r \rangle = \frac{A A^\ast e^{-2\lambda r}}{4} {8\pi \over 15}\bigg[ \big((1-i) \langle 2, -1| - (1+i)\langle 2,1|)((1+i)|2,1\rangle-(1-i)|2,1\rangle \big) \bigg] =1\\<br /> \frac{2 A A^\ast \pi}{15}e^{-2\lambda r} (2+2) = 1\\<br /> |A|=\sqrt{\frac{15}{8 \pi}}e^{-\lambda r}<br />
So I get my wave function:
|\psi\rangle={r^2 \over 2} \bigg[(1+i) Y_2^{-1}-(1-i)Y_2^1\bigg].

EDIT: Forgot square root over fraction in |A|.

 

[gasar8]

I've got one similar exercise, but instead of angular momentum I've got spins. And I've got the same problems at S_x measurements.
So the exercise goes:
We have got two particles with S_1=1 and S_2=1. We know that S_{1z}|\psi_1\rangle=\hbar |\psi_1\rangle and S_{2x}|\psi_2\rangle = \hbar |\psi_2\rangle.

a) Find wave function |\psi_1\rangle in S_{1z} basis and |\psi_2\rangle in S_{2z} basis.
b) We measure S^2 of total spin. What are possible outcomes and what are their probabilities?
c) Find expectation value and uncertainty of S^2.
d) We measure x component of total spin. What are possible outcomes and what are their probabilities?a) |\psi_1\rangle = |11\rangle \\ |\psi_2\rangle = {1 \over 2} |1-1\rangle + {1 \over \sqrt{2}} |10\rangle+ {1 \over 2} |11\rangle. Can someone just check this?
b)<br /> \begin{align*}<br /> |\psi_{12}\rangle&amp;={1 \over 2}|1\rangle|-1\rangle+{1 \over \sqrt{2}} |1\rangle|0\rangle+{1 \over 2}|1\rangle|1\rangle=\\<br /> &amp;={1 \over \sqrt{24}}|20\rangle+{1 \over \sqrt{12}}|00\rangle+{1 \over 2}|21\rangle+{1 \over 2}|11\rangle+{1 \over 2}|22\rangle<br /> \end{align*}<br />
For S^2|\psi_{12}\rangle=\hbar^2 s(s+1)|\psi_{12}\rangle, we get:
<br /> \begin{align*}<br /> &amp;Results \ \ \ \ &amp;Probability\\<br /> &amp;6\hbar^2 &amp;{13\over24}\\<br /> &amp;2\hbar^2 &amp;{3 \over 8}\\<br /> &amp;0 &amp;{1 \over 12}<br /> \end{align*}<br />

c) Expectation value is \langle S^2 \rangle = \langle \psi|S^2|\psi\rangle=4\hbar^2, but I can't find uncertainty? I am thinking in this way:
\delta_{S^2}=\sqrt{\langle S^2\rangle- \langle S \rangle ^2} or \\<br /> \delta_{S^2}=\sqrt{\langle S^4\rangle- \langle S^2 \rangle ^2}?

d) Now, same problem as at the angular momentum. How do I find outcomes and probabilities? I tried with S_x=\frac{S_++S_-}{2}, but got some weird result, from which I can't find anything (As mentioned - like at angular momentum in previous post). Then I was thinking about Pauli matrices, so that possible outcomes would be \pm {\hbar \over 2}, but how can I apply this matrix to my wavefunction of 1x1 spins. I found something on wiki for such spins and tried but got nothing... (This two attempts are in my handwriting downthere):