Calculating Orbital Angular Momentum: What Is the Correct Answer?

[Martin89]

Homework Statement

Homework Equations

The Attempt at a Solution

Hi All,

My problem is that when I calculate this integral or use software to do it for me I get (3*i*pi)/16, when I've been told that the answer is 1/2i giving a probability of 1/4. Would someone be able to point out where my mistake is.

Thanks

 

[kuruman]

You are missing a $\sin\theta$ in the integrand, $d\Omega =\sin \theta d\theta d\phi$.

 

[Martin89]

You are missing a $\sin\theta$ in the integrand, $d\Omega =\sin \theta d\theta d\phi$.

So I have, thanks for the spot!

 

[kuruman]

At some point you will learn to do such integrals without actually integrating. You can easily show with a bit of algebra that$$ -\sqrt{\frac{3}{8\pi}}\sin\theta \sin \phi=\frac{1}{2i} \left(Y_{1,1}+Y_{1,-1}\right)~;~~~-\sqrt{\frac{3}{8\pi}}i\cos\theta=-\frac{i}{\sqrt{2}}Y_{1,0}$$so that $$X_{1,1}=\frac{1}{2i} \left(Y_{1,1}+Y_{1,-1}\right)-\frac{i}{\sqrt{2}}Y_{1,0}$$ and then use the orthonormality property of the spherical harmonics, $\int Y^*_{l,m}Y_{l',m'}d\Omega=\delta_{ll'}\delta_{mm'}$ to get the answer. Clearly, only the integral involving $Y_{1,1}$ is non-zero.

 

[Martin89]

At some point you will learn to do such integrals without actually integrating. You can easily show with a bit of algebra that$$ -\sqrt{\frac{3}{8\pi}}\sin\theta \sin \phi=\frac{1}{2i} \left(Y_{1,1}+Y_{1,-1}\right)~;~~~-\sqrt{\frac{3}{8\pi}}i\cos\theta=-\frac{i}{\sqrt{2}}Y_{1,0}$$so that $$X_{1,1}=\frac{1}{2i} \left(Y_{1,1}+Y_{1,-1}\right)-\frac{i}{\sqrt{2}}Y_{1,0}$$ and then use the orthonormality property of the spherical harmonics, $\int Y^*_{l,m}Y_{l',m'}d\Omega=\delta_{ll'}\delta_{mm'}$ to get the answer. Clearly, only the integral involving $Y_{1,1}$ is non-zero.

Or I can calculate the probability directly from the the sum of the squares of the coefficients as they must equal unity. Thanks again for the help!