Successive Measurements of angular momentumby Sekonda Tags: angular, measurements, momentum, successive 

#1
Nov1912, 12:24 PM

P: 209

Hello,
I believe this to be a rather simple problem but I am not quite sure if my thinking is correct. We have a particle in a j=1 state of angular momentum J. I am first asked to find some eigenvectors of the matrix J(y): [tex]J_{y}=\frac{\hbar}{\sqrt{2}i}\begin{pmatrix} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end{pmatrix}[/tex] Is this correct? I got the eigenvalues of this matrix to be i√2, i√2 and 0 which I believe correspond, respectively to, [tex]\hbar, \hbar, 0[/tex] Though my main question (provided the above is correct) is that we are told the system is in a state of the J(y) corresponding to the positive nonzero eigenvector, we are then asked to find the probability of finding each value : [tex]\hbar, \hbar, 0[/tex] of the J(z) angular momentum. Surely this is just given by the normalized coefficients of the eigenvectors of J(z) and the fact it is in a state of J(y) is not relevant? I can post the eigenvectors I attained for the J(y) matrix if required. Thanks, SK 



#2
Nov1912, 02:13 PM

P: 209

I found the eigenvectors of J(y), shown above, to be:
[tex]\begin{pmatrix} 1\\ \sqrt{2}i\\ 1 \end{pmatrix} \begin{pmatrix} 1\\ \sqrt{2}i\\ 1 \end{pmatrix} \begin{pmatrix} \sqrt{2}\\ 0\\ \sqrt{2} \end{pmatrix}[/tex] Normalized by a factor 0.5 in front of each of them, these eigenvectors have the eigenvalues: i√2, i√2 and 0 respectively, I believe that they can be manipulated into hbar, negative hbar and zero. 



#3
Nov1912, 03:32 PM

Sci Advisor
Thanks
P: 3,864





#4
Nov1912, 03:38 PM

P: 398

Successive Measurements of angular momentum
First off, I think you meant for the bottom middle entry of the matrix to be a positive one, no? Otherwise, that J_{y} just has eigenvalues of zero.
I would think you take the eigenvector corresponding to the [itex]J_y = \hbar[/itex] measurement, which corresponds to your state, [itex]\psi[/itex]. Then, you know the eigenvalues of the J_{z} matrix, [itex]J_z = \hbar \begin{pmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 1 \end{pmatrix}[/itex] has eigenvalues [itex] \hbar, 0, \hbar[/itex] find the eigenvectors corresponding to each of those values, then to find the probabilities, you do [itex]P_\lambda = \langle \psi_z \mid \psi_y \rangle^2[/itex] where [itex]P_\lambda[/itex] is the probability of a measurement of eigenvalue λ for J_{z}, [itex]\psi_z[/itex] is the eigenvector for that eigenvalue of J_{z}, and [itex]\psi_y[/itex] is the eigenvector corresponding to the [itex]\hbar[/itex] eigenvalue for J_{y}. You may want a second opinion, because Angular Momentum was not my strongest subject in QM, but this seems right. Remember to normalize all eigenvectors! 



#5
Nov1912, 04:21 PM

P: 209

Thanks Bill_K and soothslayer, I was getting confused and I also thought that particular eigenvector component should of been a different sign.
This now makes sense, I'll try and apply it and come back soon if I'm having trouble. Thanks guys! 



#6
Nov2012, 01:02 PM

P: 209

Hey I'm back,
I think I have done this question now, does the following seem correct. Right so I determined the Jy angular momentum matrix with the following normalized eigenvectors and corresponding eigenvalues: [tex]J_{y}=\frac{\hbar}{\sqrt{2}i}\begin{pmatrix} 0 &1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end{pmatrix}\; ,\; \frac{\hbar}{\sqrt{2}i} : \frac{1}{2}\begin{pmatrix} 1\\ \sqrt2i\\ 1 \end{pmatrix}\; ,\; \frac{\hbar}{\sqrt{2}i}: \frac{1}{2}\begin{pmatrix} 1\\ \sqrt2i\\ 1 \end{pmatrix}\; ,\; 0: \frac{1}{2}\begin{pmatrix} \sqrt2\\ 0\\ \sqrt2 \end{pmatrix}[/tex] Though I'm not sure if this labelling of eigenvalues is correct! I don't know if I can just list them as +hbar, hbar and 0. Anyway, the eigenvectors and corresponding eigenvalues of the Jz matrix are: [tex]\hbar: \begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix}\; ,\; \hbar:\begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix}\; ,\; 0:\begin{pmatrix} 0\\ 1\\ 0 \end{pmatrix}[/tex] Using the +ve eigenstate of Jy I found the following probabilities of attaining states +hbar and hbar of Jz as 0.25 each and 0.5 for the 0 state of Jz, they sum to 1 so I'm guessing I've done it right? Is it? Thanks, SK 



#7
Nov2012, 01:41 PM

P: 398

Yeah, you got it!




#8
Nov2012, 01:59 PM

P: 209

Ahh good, thanks. I think the eigenvalues for the Jy matrix can be simplified to hbar, hbar and 0.




#9
Nov2012, 02:38 PM

P: 398

Yes, they can. You actually can't have imaginary numbers as eigenvalues in quantum mechanics, for the most part. Most operators in QM are Hermitian, because the eigenvalues must correspond to observables, which must be real. Not ALL the time, but certainly for something that should be observable, like angular momentum.




#10
Nov2012, 04:41 PM

P: 209

Ahh yes of course, I forgot about the whole operators are observables and the fact that the eigenvalue is the measured result etc. Thanks again soothslayer for your help, much appreciated!
SK 


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