# Quantum Mechanics: Eigenstates of $\hat{\mathbb{S}}_x$

1. Feb 17, 2015

### Robben

1. The problem statement, all variables and given/known data

Determine the eigenstates of $\hat{\mathbb{S}}_x$ for a spin$-1$ particle in terms of the eigenstates $|1,1\rangle, \ |1,0\rangle,$ and $|1,-1\rangle$ of $\hat{\mathbb{S}}_z.$

2. Relevant equations

3. The attempt at a solution

Not sure exactly how to set this problem correctly.

We have for the matrix representation: $S_z= \hbar\left[\begin{array}{ c c }1&0& 0 \\0 & 0 &0\\0&0&-1\end{array} \right]$ and $S_x= \frac{\hbar}{\sqrt{2}}\left[\begin{array}{ c c }0&1& 0 \\1 & 0 &1\\0&1&0\end{array} \right].$ From this it asks to determine the eigenstates of $S_x$ so do I just find the eigenvalues for $S_x$ and then determine the eigenstates from those? Not sure exactly what is being asked.

2. Feb 17, 2015

### phyzguy

The state space is a 3-dimensional vector space, so it can be spanned with any three linearly independent vectors. In particular, the three eigenvectors of Sx can span the space, or the three eigenvectors of Sz can span the space. The problem is asking you to find the three eigenvectors of Sx, and then transform those three vectors into the basis given by the three eigenvectors of Sz. Do you know how to transform a vector from one basis to another?

3. Feb 18, 2015

### Robben

So I need to find the eigenvalues of $S_x$, i.e., the $S_x$ matrix I have above? Once I find the eigenvalues I will then find the eigenvectors?

Could you elaborate on how I can do that, please?

4. Feb 18, 2015

### Idoubt

What you should understand is that when you write a matrix down for an operator such as you have done for $S_z$ and $S_x$, you have already chosen a basis to represent that operator in. In this case the basis you have written it in is $| 1 -1 \rangle$ , $| 1 0 \rangle$ and $| 11 \rangle$ . Can you see this?

The question is asking you to find the eigenvectors of $S_x$ and write it as a linear combination of these vectors.

Last edited: Feb 18, 2015
5. Feb 18, 2015

### Robben

I can't see it. Can you elaborate please?

In order to find the eigenvectors I must find the eigenvalues. Now from this, what exactly are we writing the eigenvectors as a linear combination of?

6. Feb 19, 2015

### Idoubt

For example let us take $S_z$ and systematically write down it's representation in the $\{ |1,-1\rangle, |1,0\rangle, |1,1\rangle \}$ basis. First of all we need to know the action of the operator $S_z$ on this basis. Since these states are eigenstates of the $S_z$ operator this is simple to write down as,
$S_z |1,-1\rangle = -\hbar |1 -1\rangle$
$S_z |1,0\rangle = 0$
$S_z |1,1\rangle = \hbar |1 1\rangle$

Now that we know this we can construct the matrix representation of $S_z$ in this basis by making the following assignements

$|1,1\rangle = \left( \begin{array}{c} 1\\ 0 \\ 0 \end{array} \right), |1,0\rangle = \left( \begin{array}{c} 0\\ 1 \\ 0 \end{array} \right), |1,-1\rangle = \left( \begin{array}{c} 0\\ 0 \\ 1 \end{array} \right)$
and the matrix representation of the operator is

$S_z = \left( \begin{array}{ccc} \langle 1,1 | S_z | 1,1 \rangle &\langle 1,1 | S_z |1,0\rangle &\langle 1,1| S_z | 1,-1 \rangle \\ \langle 1,0 | S_z | 1,1 \rangle &\langle 1,0 | S_z |1,0\rangle &\langle 1,0| S_z | 1,-1 \rangle \\ \langle 1,-1 | S_z | 1,1 \rangle &\langle 1,-1 | S_z |1,0\rangle &\langle 1,-1| S_z | 1,-1 \rangle \end{array} \right)$

which you can verify is the definition that you have given.
$S_x$ is also given in the same basis, so if you find the eigenvectors of $S_x$ you will get some vector

$|\phi\rangle = \left( \begin{array}{c} a\\ b \\ c \end{array} \right) = a\left( \begin{array}{c} 1\\ 0 \\ 0 \end{array} \right) + b\left( \begin{array}{c} 0\\ 1 \\ 0 \end{array} \right) + c\left( \begin{array}{c} 0\\ 0 \\ 1 \end{array} \right)$
or
$| \phi \rangle = a|1,1\rangle + b|1,0\rangle + c|1,-1\rangle$

So essentially you need to find the eigenvectors of the matrix representation of $S_x$ you have given.

7. Feb 19, 2015

### Robben

I see now, thanks to your help. Thank you very much!