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Homework Statement
A particle is in a state described by the wave function:
[tex]\Psi = \frac{1}{\sqrt{4}}(e^{i\phi} sin \theta + cos \theta) g(r)[/tex],
where
[tex]\int\limits_0^\infty dr r^{2} |g(r)|^{2} = 1[/tex]
and [itex]\phi[/itex] and [itex]\theta[/itex] are the azimuth and polar angle, respectively.
OBS: The first spherical harmonics are:
[itex]Y_{0,0} = \frac{1}{sqrt{4 \pi}[/itex], [itex]Y_{1,0} = \frac{sqrt{3}}{sqrt{4 \pi} cos \theta[/itex] and [itex]Y_{1,\pm1} = \mp \frac{sqrt{3}}{sqrt{8 \pi}} e^{\pm i \phi} sin \theta[/itex]
a - What are the possibles results of a measurement of the z-componet [itex]L_{z}[/itex] of the angular momentum of the particle in this state?
b - What is the probability of obtaining each of the possible results of part (a)?
c - What is the expectation value of [itex]L_{z}[/itex]?
Homework Equations
[tex]L_{z} = -i\hbar \frac{\partial}{\partial \phi}[/tex]
[tex]L_{z} \Psi = m_{l} \hbar \Psi[/tex]
The Attempt at a Solution
I start by writing the wavefunction in terms of the spherical harmonics.
[tex]\Psi = (- \frac{\sqrt{2}{\sqrt{3}}} Y_{1,1} + \frac{1}{sqrt{3} Y_{1,0}) g(r)[/tex]
So, the possibles values for [itex]m_{l}[/itex] are [itex]0[/itex] and [itex]1[/itex]; and [itex]l_{z} = 0\hbar[/itex] and [itex]l_{z} = 1 \hbar[/itex]
Thus, the probabilities should be [itex]P = \frac{2}{3}[/itex] for [itex]l_{z} = 1 \hbar[/itex] and [itex]P = 1/3[/itex] for [itex]l_{z} = 0 \hbar[/itex].
For the expectation value, I should evaluate the integral:
[tex]<L_{z}> = \int \Psi L_{z} \Psi \,d^{3}r = \int \Psi \frac{\partial \Psi}{\partial \phi} \,d^{3}r[/tex]
[tex]<L_{z}> = \frac{2}{3} \int Y_{1,1} (-i \hbar \frac{\partial}{\partial \phi}) Y_{1,1} \,d^{3}r + \frac{1}{3} \int Y_{1,0} (-i \hbar\frac{\partial}{\partial \phi}) Y_{1,0} \,d^{3}r[/tex]
The Y_{1,0} does not depend on the azimuth angle, so:
[tex]<L_{z}> = \frac{2}{3} \int Y_{1,1} (-i \hbar \frac{\partial}{\partial \phi}) Y_{1,1} \,d^{3}r[/tex]
[tex]<L_{z}> = \frac{2}{3} \hbar \int_0^\pi sin^{2} \theta \,d\theta \int_0^{2 \pi} e^{2 i \phi} \,d\phi[/tex]
However, the result of this integral is zero.
Is that right or am I doing something wrong?