Help me (Spin Operators)

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In summary: N &\left( &&sigma^{ij} &+ &h_bar^{ij}&\right)\\&= &\left( &-\frac{1}{2} &\left( &sigma^{1j} &+ &h_bar^{1j}&\right)+ &\frac{1}{2} &\left( &sigma^{2j} &+ &h_bar^{2j}&\right)\\&+ &\frac{1}{4
  • #1
kakarukeys
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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?
 
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  • #2
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:
σz = σ1z * I2 + I1 * σ2z
Meaning:
Code:
/ 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
Originally posted by 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.

Yes, that's right.

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,

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

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.

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.

(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!

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.


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

Yes. Their Hilbert spaces are disjoint.

Anyone could give me some hints?

Let me know what else you need.

edit: fixed bold font bracket
 
  • #4
Originally posted by Sonty
Tensor product:
σz = σ1z * I2 + I1 * σ2z
Meaning:
Code:
/ 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've made a mistake here. The question is asking about the Pauli matrix (σz), and you have posted the Dirac matrix (γ3).
 
  • #5
Originally posted by Tom
You've made a mistake here. The question is asking about the Pauli matrix (σz), and you have posted the Dirac matrix (γ3).

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 σz 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
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
Originally posted by Sonty
It might be one of the Dirac matrixes, I haven't got into relativistic QM.

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.
 

What is a spin operator?

A spin operator is a mathematical operator that describes the intrinsic angular momentum of a particle. It is used in quantum mechanics to represent the spin of a particle, which is a fundamental property that can have values of either 1/2 or -1/2.

How do spin operators work?

Spin operators work by representing the spin of a particle as a vector in three-dimensional space. The components of this vector correspond to the different possible spin states of the particle, and the operator acts on the wave function of the particle to determine its spin state.

What is the relationship between spin operators and spin states?

The relationship between spin operators and spin states is that the operator acts on the wave function of the particle to determine the spin state. The eigenvalues of the spin operator correspond to the different possible spin states of the particle, and the eigenvectors represent the probability amplitudes for each state.

How are spin operators used in quantum mechanics?

In quantum mechanics, spin operators are used to describe the spin of a particle and its interactions with other particles. They are an important tool for understanding the behavior of particles at the quantum level and are used in calculations and theoretical models.

Are there different types of spin operators?

Yes, there are different types of spin operators depending on the type of particle being described. For example, there are spin operators for spin-1/2 particles, spin-1 particles, and even spin-3/2 particles. Each type of spin operator has its own set of eigenvalues and eigenvectors that correspond to the possible spin states of the particle.

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