Changing the basis of Pauli spin matrices

  • #1
5
0

Homework Statement



Find the matrix representation of [itex]S_z[/itex] in the [itex]S_x[/itex] basis for spin [itex]1/2[/itex].

Homework Equations



I have the Pauli matrices, and I also have the respective kets derived in each basis. There aren't really any relevant equations, other than the eigenvalue equations for the operators.

The Attempt at a Solution



In the [itex]S_x[/itex] basis, the [itex]S_x[/itex] operator is just

[tex]
\frac{\hbar}{2}
\begin{pmatrix}
1 & 0\\
0 & -1
\end{pmatrix}
[/tex]

So isn't the [itex]S_z[/itex] operator in the [itex]S_x[/itex] basis just equal to the [itex]S_x[/itex] operator in the [itex]S_z[/itex] basis? That is,

[tex]
S_z=
\frac{\hbar}{2}
\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}
[/tex]

in the [itex]S_x[/itex] basis? If so, I don't really understand how to show this in any meaningful way... I've also tried solving the eigenvalue equations for [itex]S_z[/itex] and converting the kets to the [itex]S_x[/itex] basis but that just lead me in circles. I can see that the question is a very basic one but something is just not letting me get a satisfactory answer. Any help would be much appreciated, thanks!
 

Answers and Replies

  • #2
After some thought, I've considered that maybe I can just use the vector representation of [itex]S_z[/itex] kets in the [itex]S_x[/itex] basis. That is,

[tex]
|+>_x
\doteq
\begin{pmatrix}
1\\
0
\end{pmatrix}
[/tex]

in the [itex]S_x[/itex] basis, and the same for the spin down ket, so I can just plug these vector representations into the eigenvalue equation and solve for the [itex]S_z[/itex] matrix. This will indeed give me the matrix which I asked about above (namely, the same as the [itex]S_x[/itex] the [itex]S_z[/itex] basis). I'm going to assume this is correct unless anyone tells me otherwise!
 
  • #3
Phruizler - to find the required matrix, you just have to find the four matrix elements involving the eigenvectors of [itex]S_x[/itex].

That is, you need to find [itex]\langle x,\pm \lvert S_z \lvert x,\pm \rangle[/itex].
 
  • #4
Thanks! This yields the answer I got doing it the above way but is much more satisfying. I didn't even think of solving the matrix elements individually like that for some reason. Quantum mechanical problem solving simply doesn't adhere to the same intuition as other problems in physics. I can tell it will be a while before I will have useful insight into even some of the easier problems!

Thanks again!
 

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