# Expectation value of spin operators.

1. Apr 26, 2010

### hhhmortal

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

If an electron is in an eigen state with eigen vector :

[1]
[0]

what are the expectation values of the operators $$S_{x}, and S_{z}$$

Interpret answer in terms of the Stern-Gerlach experiment.

3. The attempt at a solution

Im not too sure how to calculate the expectation value of the spin operators. Do you get rid off the integral in this case, when I did this I got :

[0]
[-1] ħ/2

Thanks.

2. Apr 26, 2010

### nickjer

There is no integral, it is just matrix multiplication. First, what are the spin matrices for $S_x$, $S_y$, and $S_z$. Once you have them in matrix form, you can operate on it with your spin vector from the left and right.

3. Apr 27, 2010

### hhhmortal

Im not exactly sure. Is it the spin matrix multiplied by the eigen vector multiplied by complex conjugate of eigen vector?

4. Apr 27, 2010

### nickjer

Yes that is right. But the complex conjugate is a row vector that you multiply on the left.

5. Apr 27, 2010

### hhhmortal

Thanks very much for your help.

I got the expectation values to be:

$<S_x>$ = 0

$<S_z>$ = -ħ/2

$<S^2_z>$ = ħ²/4

6. Apr 27, 2010

### nickjer

Hmmm... You got a negative value? I thought your eigenvector was a spin up vector. That should have given you a positive answer.

7. Apr 27, 2010

### hhhmortal

Not so sure. I think this eigen vector is a spin-down. It was represented as β_z in the question. I think spin-up is the α_z eigen vector.

How come when I square the spin matrix of $$S_{x}$$ and calculate the expectation value of it, it's no longer zero? instead I get ħ²/4 .

8. Apr 27, 2010

### nickjer

In your first post you said the eigenvector was (1,0) which is spin up.

Also, the spin up eigenvector isn't an eigenvector of $S_x$, in fact it is a superposition of the $S_x$ eigenvectors. So just because $\langle S_x\rangle = 0$, it says nothing about what $\langle S_x^2\rangle$ is.

Or you can think of it this way:

$$\langle S^2\rangle = \hbar^2 \frac{1}{2}(\frac{1}{2}+1)= \frac{3\hbar^2}{4}$$

And you also know:

$$\langle S_z^2\rangle= \frac{\hbar^2}{4}$$

and...

$$S^2 = S_x^2 + S_y^2 + S_z^2$$

So taking the expectation value of the above operator...
$$\frac{3\hbar^2}{4} = \langle S_x^2\rangle + \langle S_y^2\rangle + \frac{\hbar^2}{4}$$

You see that you can't have the first two expectation values equal to 0 for that equality to hold true.

9. Apr 28, 2010

### hhhmortal

Oh! yes I meant spin-down which is (0,1) sorry about that.

That makes perfect sense. Thanks for your help!

10. May 1, 2010

### hhhmortal

Hello again nickjer! I was going to create another post on spin operators but I thought I might aswell post the question here, seeing I already have another post around !

(Q) Use the three angular momentum commutation relations [$S_x$,$S_y$] = iħ$S_z$ and its cyclic permutations in x, y and z to determine the matrix forms of the operators $S_x$ and $S_y$ if we know the matrix forms of the operators $S_z$ and $S^2$

I attempted the question by saying:

$S^2_z$ + $S^2_x$ + $S^2_y$ = $S^2$

I know two of the operators by I need to find out the other two..

This is where I got stuck, I cant see how I can find out what the spin operators are if I have two unknowns and only two spin operators that are known. Is there any method to tackle this question?

Thanks!

11. May 1, 2010

### nickjer

Looks like a real messy algebra problem. You will just have to solve for the values of the spin matrices using those conditions you listed above. You will be able to solve it since they have been solved for before. But looks a bit messy.

12. May 2, 2010

### hhhmortal

I dont see how I can work it out using the commutation relations, I dont know where to start..

13. May 2, 2010

### nickjer

Just set up a matrix with unknown values. And try to solve for those values.