0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ but if the the Eigenvalue is discrete, isnt the probability just the coefficient squared? (in this case, its the absolute of C(r) squared) $\endgroup$ – I. Farhan Dec 14 '18 at 4:01
  • $\begingroup$ @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. $\endgroup$ – ZeroTheHero Dec 14 '18 at 4:19
  • $\begingroup$ 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?? $\endgroup$ – I. Farhan Dec 14 '18 at 4:38
  • $\begingroup$ i mean |C(r)|^2.r^2.dr $\endgroup$ – I. Farhan Dec 14 '18 at 5:45
  • $\begingroup$ @I.Farhan yes that’s right. Don’t forget the $r^2$ factor from the spherical volume element. $\endgroup$ – ZeroTheHero Dec 14 '18 at 6:14
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – I. Farhan Dec 14 '18 at 3:58
  • $\begingroup$ 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.) $\endgroup$ – Hanting Zhang Dec 14 '18 at 4:07
  • $\begingroup$ 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 $\endgroup$ – I. Farhan Dec 14 '18 at 4:11
  • $\begingroup$ 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 $C_l^{m'}, \, -l \leq m' \leq l$ $\endgroup$ – Hanting Zhang Dec 14 '18 at 4:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.