Z-Component of Angular Momentum.

[FLms]

Homework Statement

A particle is in a state described by the wave function:

\Psi = \frac{1}{\sqrt{4}}(e^{i\phi} sin \theta + cos \theta) g(r),

where

\int\limits_0^\infty dr r^{2} |g(r)|^{2} = 1

and \phi and \theta are the azimuth and polar angle, respectively.

OBS: The first spherical harmonics are:

Y_{0,0} = \frac{1}{sqrt{4 \pi}, Y_{1,0} = \frac{sqrt{3}}{sqrt{4 \pi} cos \theta and Y_{1,\pm1} = \mp \frac{sqrt{3}}{sqrt{8 \pi}} e^{\pm i \phi} sin \theta

a - What are the possibles results of a measurement of the z-componet L_{z} 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 L_{z}?

Homework Equations

L_{z} = -i\hbar \frac{\partial}{\partial \phi}

L_{z} \Psi = m_{l} \hbar \Psi

The Attempt at a Solution

I start by writing the wavefunction in terms of the spherical harmonics.

\Psi = (- \frac{\sqrt{2}{\sqrt{3}}} Y_{1,1} + \frac{1}{sqrt{3} Y_{1,0}) g(r)

So, the possibles values for m_{l} are 0 and 1; and l_{z} = 0\hbar and l_{z} = 1 \hbar

Thus, the probabilities should be P = \frac{2}{3} for l_{z} = 1 \hbar and P = 1/3 for l_{z} = 0 \hbar.

For the expectation value, I should evaluate the integral:

<L_{z}> = \int \Psi L_{z} \Psi \,d^{3}r = \int \Psi \frac{\partial \Psi}{\partial \phi} \,d^{3}r

<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

The Y_{1,0} does not depend on the azimuth angle, so:

<L_{z}> = \frac{2}{3} \int Y_{1,1} (-i \hbar \frac{\partial}{\partial \phi}) Y_{1,1} \,d^{3}r

<L_{z}> = \frac{2}{3} \hbar \int_0^\pi sin^{2} \theta \,d\theta \int_0^{2 \pi} e^{2 i \phi} \,d\phi


However, the result of this integral is zero.

Is that right or am I doing something wrong?

 

[TSny]

For the expectation value, I should evaluate the integral:

<L_{z}> = \int \Psi L_{z} \Psi \,d^{3}r = \int \Psi \frac{\partial \Psi}{\partial \phi} \,d^{3}r

Did you forget to take the complex conjugate somewhere?

You can get the answer for $<L_{z}>$ without doing these integrations by just using the probabilities that you have found.

 

[FLms]

Did you forget to take the complex conjugate somewhere?

Yes. The integral is actually:\int \Psi^{*} L_{z} \Psi \,d^{3}r = -i \hbar \int \Psi^{*} \frac{\partial \Psi}{\partial \phi} \,d^{3}r

And,

Y_{l,m}^{*} = (-1)^{m}Y_{l,-m}

You can get the answer for $<L_{z}>$ without doing these integrations by just using the probabilities that you have found.

&lt;L_{z}&gt; = P_{m_{l} = 1} (m_{l} \hbar) + P_{m_{l} = -1} (m_{l} \hbar)
&lt;L_{z}&gt; = \frac{2}{3}(1 \hbar) + \frac{1}{3} (0 \hbar)
&lt;L_{z}&gt; = \frac{2}{3} \hbar

Is that it?

By solvind the integral, I get &lt;L_{z}&gt; = \frac{2}{3} \pi^{2} \hbar.

 

[TSny]

&lt;L_{z}&gt; = P_{m_{l} = 1} (m_{l} \hbar) + P_{m_{l} = -1} (m_{l} \hbar)
&lt;L_{z}&gt; = \frac{2}{3}(1 \hbar) + \frac{1}{3} (0 \hbar)
&lt;L_{z}&gt; = \frac{2}{3} \hbar

Is that it?

Yes. (Maybe a typographical error in your first line.)

By solvind the integral, I get &lt;L_{z}&gt; = \frac{2}{3} \pi^{2} \hbar.

It seems to come out ok for me. I don't think your original wave function $\Psi$ is normalized.
(Your integrals over the spherical harmonics should not have $d^3r$ since you have presumably already integrated over $r$.)