Calculating the probability of a certain measurement

[Demon117]

Homework Statement

This really is not a homework problem but I am studying for the qualifying exam upcoming. I came across an objective that I am not familiar with. I'm given a wave function made of a linear combination of spherical harmonics with complex coefficients. I'm asked to calculate the possible values of measurement for $L^{2}$ and $L_{z}$ which is of course straight forward. What I am unsure of though is it asks me to calculate the probability of obtaining those values. Perhaps I missed this in my undergraduate coursework somehow, but I'll list the wavefunction.

Homework Equations

$\psi=A[(1+2i)Y_{3}^{-3}+(2-i)Y_{3}^{2}+\sqrt{10}Y_{2}^{2}]$

The Attempt at a Solution

I've already calculated the normalization constant A, and found it via integration and the orthogonality condition to be

$A=\frac{1}{4}\sqrt{\frac{7}{6\pi}}$

I know the values of angular momentum $L^{2}$ are $12\hbar^{2}$ corresponding to $\left|3,-3\right\rangle$, and $\left|3,2\right\rangle$. Also, $6\hbar^{2}$ corresponding to $\left|2,2\right\rangle$. I am utterly lost on how to calculate this probability. I have tried this:

$|\left\langle 3, m\right|A[(1+2i)\left|3,-3\right\rangle+(2-i)\left|3, 2\right\rangle+\sqrt{10}\left|2,2\right\rangle]|^{2}$ for m = -3, 2

&

$|\left\langle 2, 2\right|A[(1+2i)\left|3,-3\right\rangle+(2-i)\left|3, 2\right\rangle+\sqrt{10}\left|2,2\right\rangle]|^{2}$


This gives me values that do not add up to 1 as expected. Is this right, and if so perhaps I am doing it incorrectly? Any pointers would be appreciated.

 

[TSny]

Usually the functions $Y^m_l$ are normalized so they form an orthonormal set. If so, I don't see how you got the value for $A$.

 

[Demon117]

Usually the functions $Y^m_l$ are normalized so they form an orthonormal set. If so, I don't see how you got the value for $A$.

That is true but then how else would I calculate the value for A if not by the condition,

$\int\int Y_{l'}^{m'}\bar{Y_{l}^{m}}d\Omega = \frac{4\pi}{2l+1}\delta_{l,l'}\delta_{m,m'}$

where $d\Omega = sin(\theta)d\theta d\phi$ for $0≤\theta≤\pi$, and $0≤\phi≤2\pi$ ?

The condition, as always, is to calculate $\int |\psi|^{2} dV$=1 over all space, is it not? According to this, and by orthogonality of the spherical harmonics the cross terms will cancel and the resulting integration yields the number I obtained. Perhaps I miss-calculated something.

 

[TSny]

OK, you are using a normalization of the spherical harmonics such that they are orthogonal but not orthonormal. See here for some different choices of normalization. [EDIT: I agree with your result for A for your normalization.]

The answers for the individual probabilities will depend on how the functions $Y^m_l$ are assumed to be normalized in the statement of the problem. But, nevertheless, the sum of the probabilities should add to 1.

The probability for measuring $L^2 = 6\hbar^2$ would be $\frac{|<2,2|\psi>|^2}{|<\psi|\psi>|^2|<2,2|2,2>|^2}$ if $|2,2>$ denotes $Y^2_2$.

 

[Demon117]

OK, you are using a normalization of the spherical harmonics such that they are orthogonal but not orthonormal. See here for some different choices of normalization. [EDIT: I agree with your result for A for your normalization.]

The answers for the individual probabilities will depend on how the functions $Y^m_l$ are assumed to be normalized in the statement of the problem. But, nevertheless, the sum of the probabilities should add to 1.

The probability for measuring $L^2 = 6\hbar^2$ would be $\frac{|<2,2|\psi>|^2}{|<\psi|\psi>|^2|<2,2|2,2>|^2}$ if $|2,2>$ denotes $Y^2_2$.

So it turns out I used the wrong normalization. In that case my normalization constant becomes

$A=\frac{1}{2\sqrt{5}}$

Such that the probabilities become, 1/4 each for measuring the $12\hbar^{2}$ and 1/2 for measuring the $6\hbar^{2}$. This of course adds up to 1 as expected. Thanks for your help!

 

[TSny]

Looks good!