# Eigenstate solution for a spin chain with Hubbard model

1. Jul 16, 2015

### Thor90

1. The problem statement, all variables and given/known data
I am trying to solve the model analitically just for 2 sites to have a comparison between computational results.
The problem is my professor keeps saying that the result should be a singlet ground state and a triplet of excited states, but when I compute it explicitally I find four different eigenvalues. Where am I wrong?

2. Relevant equations
My Hamiltonian (for 2 sites) is $H = \sigma_0^x \sigma_1^x + \lambda (\sigma_0^z + \sigma_1^z)$ where the lower index stands for a site index.

3. The attempt at a solution
I have written all my independent site states on the computational basis, that is

|00> = (0,0,0,1) , |10> = (0,0,1,0) , |01> = (0,1,0,0) , |11> = (1,0,0,0)

(I have written states as row vectors because it was simpler while they are actually column vectors on which H acts, and as usual 0 stands as a down spin on the respective site, while 1 as a spin up) in a way to reduce the problem to a simple algebric task, and then I used the matrix tensor product to evaluate esplicitally the sigma matrices product (in which a single sigma stands for a product of a sigma and an identity that acts on the other remaining site) and at the end I found my Hamiltonian operator in the form

$\begin{bmatrix} 2\lambda & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & -2\lambda \end{bmatrix}$

which has autovalues $(-1,1,-\sqrt{4\lambda^2+1},\sqrt{4\lambda^2+1})$ that are not degenerate.

Last edited: Jul 16, 2015
2. Jul 16, 2015

### TSny

Hello, Thor90. Welcome to PF!

Your work looks good to me. Perhaps "singlet" and "triplet" are not referring to the degeneracy of the eigenvalues. Instead, maybe "singlet" refers to an eigenstate of total spin 0 while the other three eigenstates each have total spin 1 forming a "triplet" of states (with different energies). So, take a look at the form of the eigenstates.

3. Jul 16, 2015

### Thor90

You mean total spin along the z-axis, which operator form should be simply $\sigma_0^z + \sigma_1^z$, am I right?
I have already done it, but I am not understanding why my states should have a definite spin polarization, since the $H_\lambda$ interaction Hamiltonian does not commute with the kinetic part $H_0$. Or at least this is what I find, that is also consistent with the fact that different sigma matrices do not commute, even if I see everywhere that for the Hubbard model with nearest-neighbour interaction the total numeber of spins up and down should be a conserved quantity.

Anyway my (not yet normalized) eigenvectors are

$(0,-1,1,0)$ $(0,1,-1,0)$ $(2\lambda-\sqrt{4\lambda^2+1},0,0,1)$ $(2\lambda+\sqrt{4\lambda^2+1},0,0,1)$

which as you can see the first two are in a superposition of |01> and |10>, while the other two are a superposition of |00> and |11>, so I see a triplet nor a singlet anywhere.

Last edited: Jul 16, 2015
4. Jul 16, 2015

### TSny

I think you have a typographical error in one of your first two eigenvectors. Any two of the eigenvectors should be orthogonal.

Singlets and triplets are defined with respect to the square of the total spin operator: $\hat{S}^2 = \hat{S}_x^2+\hat{S}_y^2+\hat{S}_z^2$, where for two electrons $\hat{S}_x =\hat{S}_{x,0} +\hat{S}_{x,1}$, etc. Here I used your subscripts "0" and "1" to label the two sites.

$\hat{S}^2$ commutes with the Hamiltonian. Eigenvalues of $\hat{S}^2$ are $S(S+1)\hbar^2$ where the quantum number $S$ has two possible values for a system of two electrons: $S = 0$ and $S = 1$. There will be one state with $S = 0$ (the singlet state) and three states with $S = 1$ (the triplet states).

Take your eigenstate (0, -1, 1, 0), for example. Write this state as a superposition of the states |10> and |01>. Here, |10> could be written symbolically as $|\uparrow \downarrow>$ while |01> = $|\downarrow \uparrow>$ .

You should then be able to recognize (0, -1, 1, 0) as singlet or triplet state.

Last edited: Jul 17, 2015
5. Jul 17, 2015

### Thor90

You are absolutely right, the fact is I have found them with wolfram alpha so I thought they were good..I think I will try do the calculations again.

so basically I have to evalute the product $\sum_\alpha \sigma_0^\alpha \sigma1^\alpha$ since $\sigma^2=1$?
Anyway thank you for your help, I will try that way.

6. Jul 17, 2015

### TSny

I think it's just a sign error in one of your entries.

Yes. You can express $\hat{S}^2$ as a 4x4 matrix in your tensor product space. Then check that your eigenvectors of $\hat{H}$ are also eigenvectors of $\hat{S}^2$.

7. Jul 22, 2015

### Thor90

Ok you were right, one of the eigenvectors is (0,-1,1,0), as you were about the $S^2$ operator; what I still do not understand is why my singlet of 0 spin is the first excited state and not the ground state, as it usually is, since the energy eigenvalue of $(2\lambda - \sqrt{4\lambda^2 +1},0,0,1)$ is $-\sqrt{4\lambda^2+1}$, while the one of (0,-1,1,0), which has spin 0, has autovalue -1. Is it a problem? Or is it right for some systems to have a ground state which has not 0 spin?

I also have that the triplet has eigenvalue of $S^2$ equal to 2, but I think I have just missed a 1/2 somewhere.

8. Jul 22, 2015

### TSny

I don't think the ground state necessarily has to have spin zero for a system of two interacting spins.

In some specific cases you can conclude that the ground state must have spin zero. For example, the ground state of the helium atom. But in this case, the spatial part of the wavefunction must be symmetric in the ground state which forces the spin part to be antisymmetric (i.e., spin 0).

The eigenvalue of the operator $\hat{S}^2$ for the triplet states is $2 \hbar^2$. So, if you are using units where $\hbar = 1$, you should get 2 for the eigenvalue of $\hat{S}^2$. But these triplet states are called "spin 1" states because the value of the quantum number S is 1.

For systems of particles of spin 1/2, the eigenvalues of $\hat{S}^2$ are $\hbar^2 S(S+1)$ where the quantum number S can be 0, 1/2, 1, 3/2, 2, ....

A single "spin 1/2" particle has S = 1/2, so the eigenvalue of $\hat{S}^2$ for a spin 1/2 particle is $\frac{3}{4} \hbar^2$.
For a system of two spin 1/2 particles, S can be 0 or 1 corresponding to eigenvalues of $\hat{S}^2$ of 0 or $2 \hbar^2$.