# Converting $\psi(x,y,z)$ to $f(r)Y_l^m$ and calculating expectation value of the angular momentum operators

I've been trying to solve a particular class of problems, where I'm given a wavefunction $$\psi (x,y,z)$$ and I'm asked to find the expectation value or eigenvalue related to the angular momentum operators $$L^2,L_z,L_+$$, etc. My intuition is that I need to break the wave function into a radial part and spherical harmonics.

For example, consider the following wavefunction :

$$\psi = (x+y+3z)f(r)$$

We need to find $$\langle L^2\rangle$$ for the above wavefunction. Converting into spherical coordinates, and ignoring the radial part, I get:

$$\psi(\theta,\phi)=\sin\theta \cos\phi +\sin\theta \sin\phi+ 3\cos\theta$$

From this, I obtain: $$\psi(\theta,\phi) = \frac{1}{2} \sin\theta e^{i\phi}+\frac{1}{2} \sin\theta e^{-i\phi}+\frac{1}{2i} \sin\theta e^{i\phi}-\frac{1}{2i} \sin\theta e^{-i\phi}+3\cos\theta$$

I can identify the spherical harmonics but they are not normalized. Hence I'm writing the above as:

$$\psi(\theta,\phi) = aY_1^1 + bY_1^{-1}+cY_1^1+dY_1^{-1}+eY_1^0$$ Here $$a,b,c,d,e$$ are constants.

My question is how can I figure out $$\langle L^2\rangle$$ from the above? \since I don't know the values of the constants, I'm not able to find out the expectation value.

However, intuitively, I noticed that all the terms have the same value for the angular quantum number $$l$$, and so in each of the terms, $$L^2$$ should have an expected value of $$l(l+1)\hbar^2 = 2\hbar^2$$, so overall the expectation value should come out to be $$2\hbar^2$$.

However, what about the cases, where all the terms in the superposition do not have the same value of $$l$$? Is there any way to solve the problem in that case, without memorizing the normalization constants for each of the spherical harmonics?

My intuition is that I need to break the wave function into a radial part and spherical harmonics.

Correct.

I can Identify the spherical harmonics but they are not normalized.

You know the spherical harmonics including their normalization factors. See for example at Table of spherical harmonics:

\begin{align} Y^{-1}_1(\theta,\phi) &=&\frac{1}{2}\sqrt{\frac{3}{2\pi}}e^{-i\phi}\sin\theta &=&\frac{1}{2}\sqrt{\frac{3}{2\pi}}\frac{x-iy}{r} \\ Y^{0}_1(\theta,\phi) &=&\frac{1}{2}\sqrt{\frac{3}{\pi}}\cos\theta &=&\frac{1}{2}\sqrt{\frac{3}{\pi}}\frac{z}{r} \\ Y^{+1}_1(\theta,\phi) &=&-\frac{1}{2}\sqrt{\frac{3}{2\pi}}e^{i\phi}\sin\theta &=&-\frac{1}{2}\sqrt{\frac{3}{2\pi}}\frac{x+iy}{r} \end{align}

You can resolve these to $$x,y,z$$: \begin{align} x&=\sqrt{\frac{2\pi}{3}}r\left(Y^{+1}_1(\theta,\phi)+Y^{-1}_1(\theta,\phi)\right) \\ y&=i\sqrt{\frac{2\pi}{3}}r\left(Y^{+1}_1(\theta,\phi)-Y^{-1}_1(\theta,\phi)\right) \\ z&=2\sqrt{\frac{\pi}{3}}r\ Y^{0}_1(\theta,\phi) \end{align}

since I don't know the values of the constants, I'm not able to find out the expectation value.

Using these expressions for $$x,y,z$$ you can rewrite your example wave function $$\psi(x,y,z)=(x+y+3z)f(r)$$ from cartesian coordinates $$(x,y,z)$$ to spherical coordinates $$(r,\theta,\phi)$$ with spherical harmonics and constants in front of them. $$\psi(r,\theta,\phi)=rf(r)\left( \sqrt{\frac{2\pi}{3}}(1+i)Y^{+1}_1(\theta,\phi) +6\sqrt{\frac{\pi}{3}}Y^{0}_1(\theta,\phi) +\sqrt{\frac{2\pi}{3}}(1-i)Y^{-1}_1(\theta,\phi) \right)$$

With the expression above for $$\psi$$ you are ready for calculating expectation values of angular momentum related operators, for example $$\langle\psi|L^2|\psi\rangle$$. Make use of $$L^2Y^m_\ell(\theta,\phi)=\hbar^2\ell(\ell+1)Y^m_\ell(\theta,\phi)$$ and the orthonormality of the $$Y^m_\ell$$ functions, so that you actually don't need to do the $$\theta,\phi$$ integrations by hand. You will end up with something like $$\langle\psi|L^2|\psi\rangle=\text{(some constant)}\int_0^\infty r^3|f(r)|^2dr$$

Is there any way to solve the problem in that case, without memorizing the normalization constants for each of the spherical harmonics?

Don't try to memorize the normalization constants. It is ok to use a table of spherical harmonics. Physicists do this all the time.