# Spin operators

1. Sep 28, 2014

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

2. Relevant equations

See question

3. The attempt at a solution

Ok 1a) is dead simple. But I am struggeling with part b. I plug in all the variables. But cannot seem to get a and b..... I do this:

(-b)(S_x)(a,b) = E(a,b) and then get

-b^2h/2= Ea and ah/2=E

I dont know.. Can you guys help me.

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2. Sep 28, 2014

### Orodruin

Staff Emeritus
Note that the b in the Hamiltonian is not the same b as that in the wave function ...

3. Sep 28, 2014

Well that is a big help thanks. haha. Although I am sure still not sure how to proceed to find the Energy states. Because when I multiply the X matrix with the S_x matrix I get the transpose of X...

4. Sep 28, 2014

### Orodruin

Staff Emeritus
Well, it is not really a transpose - you simply switch the elements (a transpose would take a column matrix to a row one). The energy states are simply the states for which $H\chi = E\chi$, where $E$ is a number, i.e., the eigenstates of the Hamiltonian.

5. Sep 28, 2014

Ya I understand that. But how can I get back the orginal X to find E. Since, as you say the elements switch places of X.

6. Sep 28, 2014

### Orodruin

Staff Emeritus
So, what did you obtain for the eigenstates? If you have found the energy eigenstates, all you need to do is to check what the constant multiplying $\chi$ is after you apply the Hamiltonian to it.

7. Sep 28, 2014

$$\frac{-b\hbar}{2}\begin{pmatrix} 0&1\\ 1&0\\ \end{pmatrix} \begin{pmatrix} a\\ b\\ \end{pmatrix}= \frac{-b\hbar}{2} \begin{pmatrix} b\\ a\\ \end{pmatrix}$$

Now what?

8. Sep 28, 2014

### Orodruin

Staff Emeritus
Well, to start I would call the $b$ in the $\chi$ $c$ instead to separate it from the $b$ in the Hamiltonian (or rename the Hamiltonian $b$ to $B$ ...). Now you need to solve the eigenvalue equation that states that this should be equal to $E \chi$.

9. Sep 28, 2014

Which is preciously where I'm stuck. I get a relationship for a and b but don't know how to solve for them.

10. Sep 28, 2014

### Orodruin

Staff Emeritus
I suggest starting by solving for the energy eigenvalue E.

11. Sep 28, 2014

Then I just get what I had over X.... Now sure how that helps. E should be a value and not contain matrices . How can I eliminate them?

12. Sep 28, 2014

### Orodruin

Staff Emeritus
For the matrix equation to hold, it must hold for every element.

13. Sep 28, 2014

$$\frac{-B\hbar}{2}\begin{pmatrix} 0&1\\ 1&0\\ \end{pmatrix} \begin{pmatrix} a\\ b\\ \end{pmatrix}= \frac{-B\hbar}{2} \begin{pmatrix} b\\ a\\ \end{pmatrix}=E \begin{pmatrix} a\\ b\\ \end{pmatrix}$$
So that implies:

$$\frac{-B\hbar a}{2b} = E$$ and $$\frac{-B\hbar b}{2a} = E$$

Equating the:

$$\frac{-B\hbar b}{2a} = \frac{-B\hbar a}{2b}$$

so
$$a^2=b^2$$

Which implies:

$$a=\pm b$$

But we have the relationship:

$$|a|^2+|b|^2=1$$

So plugging it in I get:

$$a=\frac{1}{\sqrt{2}} b=\pm \frac{1}{\sqrt{2}}$$

So how does this give me the ground state and the excited states? or the value for E.

Like I said I am not sure how to solve this problem. Please some more guidance.

14. Sep 29, 2014

### Orodruin

Staff Emeritus
You are essentially done. You have two expressions for E in terms of a and b that you now know - so just insert them. The $\pm$ gives you two possible values of E - the lower one corresponding to the ground state and the other the excited.

15. Sep 29, 2014

$$\frac{-B\hbar}{2} = E$$
$$\frac{B\hbar}{2} = E$$