Successive Measurements of angular momentum

[Sekonda]

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):

J_{y}=\frac{\hbar}{\sqrt{2}i}\begin{pmatrix} 0 & -1 & 0\\ 1 & 0 & -1\\ 0 & -1 & 0 \end{pmatrix}

Is this correct?

I got the eigenvalues of this matrix to be i√2, -i√2 and 0 which I believe correspond, respectively to,
\hbar, -\hbar, 0

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 non-zero eigenvector, we are then asked to find the probability of finding each value :
\hbar, -\hbar, 0
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

 

[Sekonda]

I found the eigenvectors of J(y), shown above, to be:

\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}

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 h-bar, negative h-bar and zero.

 

[Bill_K]

we are told the system is in a state of the J(y) corresponding to the positive non-zero eigenvector, we are then asked to find the probability of finding each value 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?

The probability amplitude of finding each value of J_z in an eigenstate of J_y is the overlap, <J_z= m_z|J_y= m_y>, and the probability itself is that thing squared, |<|>|².

 

[soothsayer]

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 J_y = \hbar measurement, which corresponds to your state, \psi. Then, you know the eigenvalues of the J_z matrix,
J_z = \hbar \begin{pmatrix}<br /> 1 &amp; 0 &amp; 0\\ <br /> 0 &amp; 0 &amp; 0\\ <br /> 0 &amp; 0 &amp; -1<br /> \end{pmatrix}

has eigenvalues -\hbar, 0, \hbar

find the eigenvectors corresponding to each of those values, then to find the probabilities, you do

P_\lambda = |\langle \psi_z \mid \psi_y \rangle|^2

where P_\lambda is the probability of a measurement of eigenvalue λ for J_z, \psi_z is the eigenvector for that eigenvalue of J_z, and \psi_y is the eigenvector corresponding to the \hbar 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!

 

[Sekonda]

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!

 

[Sekonda]

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:

J_{y}=\frac{\hbar}{\sqrt{2}i}\begin{pmatrix} 0 &amp;1 &amp; 0\\ -1 &amp; 0 &amp; 1\\ 0 &amp; -1 &amp; 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}

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:

\hbar: \begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix}\; ,\; -\hbar:\begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix}\; ,\; 0:\begin{pmatrix} 0\\ 1\\ 0 \end{pmatrix}

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

 

[soothsayer]

Yeah, you got it!

 

[Sekonda]

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

 

[soothsayer]

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.

 

[Sekonda]

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