- #1
gasar8
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Homework Statement
Particle is in a state with wave function [itex]\psi (r) = A z (x+y)e^{-\lambda r}.[/itex]
a) What is the probability that the result of the [itex]L_z[/itex] measurement is 0?
b) What are possilble results and what are their probabilities of a [itex]L^2[/itex] measurement?
c) What are possilble results and what are their probabilities of a [itex]L_x[/itex] measurement?
The Attempt at a Solution
Firstly, I tried to write wave function in spherical harmonics form:
[tex]
\begin{align*}
\psi (r) &= \alpha z (x+y); \ \ \alpha= A e^{-\lambda r}\\ &= \alpha r \cos\theta (r \sin\theta\cos\phi+r \sin\theta \sin \phi)\\ &= \alpha r^2 \cos\theta \sin \theta (\cos\phi + \sin \phi) \\ &=\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}) \\ &= -{\alpha \over 2} r^2 \sqrt{\frac{8 \pi}{15}} ((1+i)Y_{2,-1}+(1-i)Y_{2,1})
\end{align*}
[/tex]
Then, I normalized this function:
[tex] |\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}[/tex]
So finally, I get my wave function:
[tex]\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})[/tex]
Can someone check this normalization, I am not sure if I did it correctly?
a) Form definition [itex]L_z|l,m\rangle=\hbar m |l,m\rangle[/itex] 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 [itex]L^2|l,m\rangle=\hbar^2 l(l+1) |l,m\rangle[/itex] I can only get the result [itex]6 \hbar^2[/itex] with 100% probability, since [itex]l[/itex] is in both elements of a wave function 2.
c) Form definition [itex]L_x|l,m\rangle=\frac{L_+ + L_-}{2} |l,m\rangle[/itex], I get:
[tex]
\begin{align*}
L_+ |2,1\rangle &= 2 \hbar |2,2\rangle \\
L_+ |2,-1\rangle &= \sqrt{6} \hbar |2,0\rangle \\
L_- |2,1\rangle &= \sqrt{6} \hbar |2,0\rangle \\
L_- |2,-1\rangle &= 2 \hbar |2,-2\rangle\\
\end{align*}
[/tex]
[tex]
\begin{align*}
\Longrightarrow L_x |2,1\rangle &= \hbar (|2,2\rangle + \frac{\sqrt{6}}{2} |2,0\rangle)\\
\Longrightarrow L_x |2,-1\rangle &= \hbar (|2,-2\rangle + \frac{\sqrt{6}}{2} |2,0\rangle)
\end{align*}
[/tex]
So the possible results are [itex]\hbar[/itex] and [itex]\frac{\sqrt{6}}{2} \hbar[/itex]? But how can I now find their probabilities? I tried to write:
[tex]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)), [/tex]
but the squares of absolute values of coefficients are above 1.