# Probability to get an Eigenvalue of Angular Momentum Operator on an Arbitrary Ket

Hello physics SE community, I am currently working on Principles of Quantum Mechanics by Shankar and i get stuck in page 336 (its not even an exercise).

It basically said that "we may expand any $$\psi(r,\theta,\phi)$$ as $$\psi(r,\theta,\phi) = \sum_{l=0}^{\infty} \sum_{m=-l}^{l} C_l^m(r) Y_l^m(\theta, \phi)$$ (where $$C_l^m(r) = \int Y_l^{m*}(\theta, \phi) \psi(r,\theta,\phi) d\Omega$$)

and if we compute $$\langle \psi |L^2| \psi \rangle$$ (assuming $$\psi$$ is normalizable to unity) and interpret the result as a weighted average, we can readily see that $$P(L^2=l(l+1)\hbar^2, L_z=m\hbar) = \int _0^{\infty} |C_l^m(r)|^2 r^2 dr$$" How can you get the probability?? At first i thought the probablity is just $$|C_l^m(r)|^2$$ since the eigen value is discrete, but how does it involve integral over the radius?

Any hints or answer is appreciated.

Let's look first at a problem in 1d. The probability of finding a particle described by $$\psi(x)$$ anywhere between $$x_1$$ and $$x_2$$ is $$P(x_1\le x\le x_2)=\int_{x_1}^{x_2} dx \vert\psi(x)\vert^2\, .$$ You're just doing a slightly more sophisticated version of this: $$\int_0^{\infty} \vert C^\ell_m(r)\vert^2 r^2 dr$$ is the probability of finding the particle in an angular momentum state with $$\ell$$ and $$m$$ at any radius (since you're integrating over all $$r$$'s).

• but if the the Eigenvalue is discrete, isnt the probability just the coefficient squared? (in this case, its the absolute of C(r) squared) – I. Farhan Dec 14 '18 at 4:01
• @I.Farhan let me turn this around: how can the probability depend on $r$? which value of $r$ would you choose to evaluate $C(r)$? See en.m.wikipedia.org/wiki/Marginal_distribution. – ZeroTheHero Dec 14 '18 at 4:19
• oh yea haha thanks! so is it correct to assume that |C(r)|^2.dr is the probability to get the respective angular momentum in the region r + dr?? – I. Farhan Dec 14 '18 at 4:38
• i mean |C(r)|^2.r^2.dr – I. Farhan Dec 14 '18 at 5:45
• @I.Farhan yes that’s right. Don’t forget the $r^2$ factor from the spherical volume element. – ZeroTheHero Dec 14 '18 at 6:14

If you're computing $$\langle\psi | L^2 | \psi \rangle$$, then we also have to include the integral over the radius, since the distribution of $$|\psi\rangle$$ includes the radial component.

• Well i try to do that, and all i got is $\sum_{l=0}^{\infty}\sum_{m=-l}^l l(l+1)\hbar\int_0^{\infty}|C_l^m(r)|^2 r^2 dr$. I don't know how to relate it to the probability – I. Farhan Dec 14 '18 at 3:58
• Well, from your equation above, each $l(l+1)$ term has $\int_0^{\infty}|C_l^m(r)|^2 r^2 dr$ attached to it. Hence the probability that measuring $L^2$ will return a value $l(l+1)$ is $\int_0^{\infty}|C_l^m(r)|^2 r^2 dr$. (Sharkar says $P(L^2=l(l+1)\hbar^2, L_z=m\hbar) = \int _0^{\infty} |C_l^m(r)|^2 r^2 dr$ because we also need to specify the component of $L_z$ in order to avoid degeneracy.) – Hanting Zhang Dec 14 '18 at 4:07
• Ok i see! but how about the Lz operator? the probability supposed to be to get L^2 = l(l+1)hbar and Lz = m.hbar – I. Farhan Dec 14 '18 at 4:11
• As I said, it's to avoid degeneracy. $C_l^m(r)$ depends on both $l$ and $m$, which in turn depend on $L^2$ and $L_z$, respectively. If I didn't specify $L_z$, (and hence the $z$-component of angular momentum), then we could be talking about any of the coefficients $C_l^{m'}, \, -l \leq m' \leq l$ – Hanting Zhang Dec 14 '18 at 4:15