# Help me (Spin Operators)

1. Sep 12, 2003

### kakarukeys

An easy (not easy) 1st year undergraduate project:

Energy eigenvalues of a single spin-1/2 system which Hamiltonian is given by

H = - k S^z

are -/+ (1/2 k).

I got that using the spin operator S^z = 1/2 Sigma^z (let h_bar = 1), Hamiltonian is in diagonalized form:

( -1/2 k 0
0 1/2k )

So, Eigenvectors are given by (1, 0), (0, 1), spin up and spin down respectively.

How do I compute the energy eigenvalues, eigenvectors of an double identical spin-1/2 system which Hamiltonian is given by

H = - J vec(S1) dot vec(S2) - k S1^z - k S2^z

J is a const. > 0

After going some references, Griffiths, Sakurai, Merzbacher, etc,
I have no idea how to begin,

I have the following problems in mind:
(1) if |u u> represents both spins up, (1/sqrt 2)(|u d> +/- |d u>) represent one spin up, |d d> represent both spins down. |u u> and |d d> should be eigenstates of the system (because they are ground states), for |u d>, |d u> states I am not sure.

(2) I can't use same spin matrices for S1^z, S2^z. But there is only one S^z matrix namely, 1/2 Sigma^z. Could S1^z be a tensor product of 1/2 Sigma^z with an identity matrix, and S2^2 the other way round? If it is so, I have no idea how to do the maths!

(3) Do vec(S1) and vec(S2) commute?

Anyone could give me some hints?

2. Sep 13, 2003

### Sonty

Each operator acts on the part of the state vector that concerns it. If an operator belongs to the first particle it will act on the part of the state vector that belongs to the first particle. So S1 and S2 must commute.
Tensor product:
&sigma;z = &sigma;1z * I2 + I1 * &sigma;2z
Meaning:
Code (Text):

/ 1 0  0 0 \       / 1 0  0 0 \
| 0 1  0 0  |      | 0 -1  0 0 |
| 0 0 -1 0  |  +  | 0  0  1 0 |
\ 0 0  0 -1 /       \ 0 0 0 -1/

You should figure the math out of this.
There are quite a few holes in my QM as I've just finished reading Sakurai, but did it on my own and I haven't got a lot of exercise to nail down the theory. Maybe that's why that letter J and the S1S2 would take me towards Perturbation Theory if the product is small and the variational method if it's not. That's where you'll get the energy shifts from the base E0=+/-k

3. Sep 13, 2003

### Tom Mattson

Staff Emeritus
Yes, that's right.

You know the matrices &sigma;i (i=1,2,3), so you can determine the eigenvalues from those. (I am going to use subscripts to denote the components of &sigma; and superscripts in parentheses to denote the spin operators for particles 1 and 2). Let |&alpha;,&beta;> represent the spin state, where &alpha; is the state of particle 1 and &beta; is the state of particle 2. |&alpha;,&beta;> is the direct product of |&alpha;> and |beta;>, the individual spin states of particles 1 and 2, respectively. Just remember that &sigma;(1) only acts on particle 1 and &sigma;(2) only acts on particle 2. The two operators pass right through the other particle's spin states.

Yes, that's right. The only thing I would add is that (1/2)-1/2(|u,d>+|d,u>) is part of the triplet S=1, MS=0, and (1/2)-1/2(|u,d>-|d,u>) is the singlet S=0, MS=0.

Yes, you can use the same matrix for S(1)z and S(2)z. In fact, you have no choice because that is what the matrices are! Just remember that the matrix for particle 1 only acts on the state for particle 1, and likewise for particle 2.

Yes. Their Hilbert spaces are disjoint.

Let me know what else you need.

edit: fixed bold font bracket

4. Sep 13, 2003

### Tom Mattson

Staff Emeritus
You've made a mistake here. The question is asking about the Pauli matrix (&sigma;z), and you have posted the Dirac matrix (&gamma;3).

5. Sep 14, 2003

### Sonty

It might be one of the Dirac matrixes, I haven't got into relativistic QM. I'm sure though that that is exactly what I said it is: the &sigma;z for the two spin 1/2 particle system. The matrix must be part of a 2x2 tensor space. If you want to prove me wrong, please compute that tensor product and show me your results. I don't understand where you see that mistake.

6. Sep 14, 2003

### kakarukeys

Thanks for everyone's help, I really appreciate.

I figured out a way, using a "trick":

Since vec(S1), vec(S2) commute,

2 vec(S1) dot vec(S2) = [vec(S1) + vec(S2) ]^2
- vec(S1)^2 - vec(S2)^2

now ,
vec(S1)^2 = s1 (s1 + 1)
vec(S2)^2 = s2 (s2 + 1)

vec(S1) + vec(S2) = 1 for triplet, 0 for singlet

using the above, it is very easy to compute the the matrix elements of Hamiltonian, provided if

|u u>
|d d>
1/sqrt 2 (|u d> +/- |d u>)

really form a basis.

I think they should, since |u>, |d> span the single particle hilbert space, their direct products should span the double particle Hilbert space which is the direct products of two single particle Hilbert spaces.

So the Hamiltonian looks like
(
x 0 0 0
0 x x 0
0 x x 0
0 0 0 x)

x means non-zero entry.

Solving the eigenvalue equations will give the eigenvalues and correct eigenvectors.

I will figure out another way using matrices, I think that's a better way, since you can no longer use 'trick' when things get complicated.

7. Sep 15, 2003

### Tom Mattson

Staff Emeritus
I made a mistake--it is not the Dirac matrix. Writing the direct product as a 4x4 matrix is not how I usually see it done, but on closer inspection there is in fact nothing wrong with it.