Possible outcomes of measuring ##L^2## and ##L_z##?

  • Thread starter Thread starter Kara386
  • Start date Start date
  • Tags Tags
    Measuring
Kara386
Messages
204
Reaction score
2

Homework Statement


A particle has the wavefunction

##\psi(\theta, \phi) = \sqrt{\frac{5}{2\pi}}\sin(\theta)cos(\frac{\theta}{2})^2\cos(\phi)##

What are the possible outcomes of measuring ##L^2## and ##L_z##? And the relative probabilities of each outcome?

Homework Equations

The Attempt at a Solution


I think I should try and write the wavefunction as a combination of spherical harmonics so I can find the eigenvalues more easily. Substituting in ##cos(\frac{\theta}{2})^2 = \frac{1}{2} +\frac{1}{2}\cos(\theta)##,

##\psi(\theta, \phi) = \sqrt{\frac{5}{8\pi}}\sin(\theta)\cos(\phi)+\sqrt{\frac{5}{8\pi}}\cos(\theta)cos(\phi)##

I found that ##\psi(\theta, \phi) = \sqrt{\frac{5}{12\pi}}(Y_{1,-1}-Y_{1,1}) + \sqrt{\frac{1}{12}}(Y_{2,-1}-Y_{2,1})##. Now I don't really know what to do with that! I know that the coefficient of each spherical harmonic squared will be the probability of that particular outcome. And if you square every coefficient and add them then they do add to 1.

The possible outcomes are the eigenvalues of the operator in question, which for ##L^2## are ##\hbar^2 l(l+1)## and for ##L_z## are ##\hbar m##. So what I don't understand is, every spherical harmonic will have an eigenvalue of ##L^2## and one for ##L_z##, but when I square the coefficients am I finding the probability of the ##L^2## outcome, the ##L_z## outcome or both?
 
Physics news on Phys.org
Kara386 said:
The possible outcomes are the eigenvalues of the operator in question, which for ##L^2## are ##\hbar^2 l(l+1)## and for ##L_z## are ##\hbar m##. So what I don't understand is, every spherical harmonic will have an eigenvalue of ##L^2## and one for ##L_z##, but when I square the coefficients am I finding the probability of the ##L^2## outcome, the ##L_z## outcome or both?
What you found is that, for instance, the probability of being in state ##Y_{1,-1}## is 5/12. That state will lead to defined values of ##L^2## and ##L_z##. You need to go through all states the same way, and figure out the probability of individual outcomes, and you must then combine them to find, say, the total probability of measuring ##L = 2 \hbar^2##.
 
  • Like
Likes Kara386
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top