# Eigenvectors of an operator

1. Jun 28, 2017

### LagrangeEuler

1. The problem statement, all variables and given/known data
$H=\frac{J}{4}\sum_{i=1}^2 \sigma_i^x \sigma_{i+1}^x$

2. Relevant equations
$\sigma^x$ is Pauli matrix and $J$ is number.

3. The attempt at a solution
For $i=1$ to $3$ what is dimension of eigen vector? I think it is $8$. Because it is like that I have tri sites. For example in one site spin is up in second is up and in third is up. My question is what is the easiest way to solve this. For example do I need to take Kronecker product $\sigma_1^x \otimes \sigma_{2}^x \otimes 1$ in every term in the sum to form 8x8 matrix. Thanks a lot for your answer.

Last edited: Jun 28, 2017
2. Jun 28, 2017

It looks like you have 4 sites and this is essentially an exchange type Hamiltonian or it could be. (in which case $J$ would be large and negative when the spins are aligned). I think the only concern is the spin in the x-direction. I think your states might be those of a 1 x 4 column vector where up and down is possible for each of the 4 sites. Otherwise though, this is slightly beyond my expertise in this area.

3. Jun 28, 2017

### LagrangeEuler

Yes you are right. I made a mistake. I corrected it now.

4. Jun 28, 2017

Since it is using a 2x2 spin matrix for each particle, perhaps you do need an 8 x 8 hamiltonian. Suggestion: Write out the 8x 8 Hamiltonian and try to solve the eigenvector equation.

5. Jun 28, 2017

### LagrangeEuler

Yes but could you help me how to write 8x8 hamiltonian in this case?

6. Jun 28, 2017

To isolate each matrix element of $H$, just have one state occupied on the 1 x 8 row vector and another state (single particle up or down) occupied on the 1x8 column vector. Compute the elements as if it were an $s_z$. Each combination will get a +J/4 if they are adjacent to each other and in the same direction, a $-J/4$ if opposite, and zero if they are not adjacent sites. Also a zero if it is trying to interact with itself. $\\$ Also please check your OP: Is it i=1 to 3 or i=1 to 2? $\\$ e.g. For the above, your row vector for particle 2 spin down would read: (0 0 0 1 0 0 0 0 ).

7. Jun 28, 2017

### LagrangeEuler

Thanks I understand that. But to my mind is a problem if I have 1x8 row matrix and I have product of 2 2x2 matrices $\sigma_i^x \sigma_{i+1}^x$. What is the way to form from this 8x8 matrix?

8. Jun 28, 2017

You don't need the exact form of the spin matrices. You simply know the answer: e.g. if particle 2 is spin down and particle 3 is spin up it gets a -J/4. See also my edited previous post at the bottom. Editing: I'm going to have to look this over to see if this method is correct or not...And I think the answer is yes, because you could then make both row vector and column vector each a combination of these individual states and even have them be the same and you would get the correct answer=it would give the energy of the state, assuming you normalized the selections.

9. Jun 28, 2017

And a follow-up to post #8=it takes about 5 minutes or less to write out the $H$ using this method=once you get the rows and columns sorted out for the adjacent particles, it's quite simple.

10. Jun 28, 2017

### LagrangeEuler

I will try to do it tomorrow and I will send here results.

11. Jun 28, 2017

Just an additional comment: I thought what I proposed would work, but for some reason it isn't giving the results that I expected. I expected state vectors that we can simply write down and know the energy from the configuration is what I was expecting to find as eigenvectors, but a couple of calculations using the computed $H$ and these column vectors did not give this for the result. I must conclude that it is likely there is an error somewhere in the process I proposed.

Last edited: Jun 28, 2017
12. Jun 29, 2017

### LagrangeEuler

Let me try to understand better. So for example if I have three spin up then state of the system is for example $\left[ {\begin{array}{cc} 1 \\ 0 \\ \end{array} } \right]⊗ \left[ {\begin{array}{cc} 1 \\ 0 \\ \end{array} } \right]⊗ \left[ {\begin{array}{cc} 1 \\ 0 \\ \end{array} } \right]= \left[ {\begin{array}{cc} 1 \\ 0 \\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ \end{array} } \right]$. Let me denote $\sigma^x= \left[ {\begin{array}{cc} 0 & 1\\ 1 & 0 \\ \end{array} } \right]$. Then I could define operator as $\sigma^x ⊗ \sigma^x ⊗ 1= \left[ {\begin{array}{cc} 0 & 1\\ 1 & 0 \\ \end{array} } \right]⊗ \left[ {\begin{array}{cc} 0 & 1\\ 1 & 0 \\ \end{array} } \right]⊗ \left[ {\begin{array}{cc} 1 & 0\\ 0 & 1 \\ \end{array} } \right]= \left[ {\begin{array}{cc} 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\ \end{array} } \right]$
This is some permutation matrix. Is this make sence? Could you elaborate your method. Maybe to add picture and say why you think it is incorrect.

13. Jun 29, 2017

In my method, $H_{ij}=<i| H |j>$ is found by simply putting one particle of the 4 in a given state in each row and column vector. The energy for two adjacent like states is $J/4$, but we could simply call it one unit. $\\$ In any case, e.g. suppose we put particle 2 in the down state-that makes for a row matrix: $\\$ (0 0 0 1 0 0 0 0). And suppose we put particle 3 in the down state in the column vector-that makes (0 0 0 0 0 1 0 0 ) in a column vector. Since the particles are adjacent and in the same state, this matrix element, which is found by a product of the row matrix on the left and column vector on the right with (H) in the middle, would get one positive unit (J/4 ) of energy. Anyway, the process is kind of simple, but it wasn't successful in finding the eigenvector states. $\\$ I tried the same process with a two particle system, and the method just doesn't seem to work. One eigenvector state for two particles was (1 1 -1 -1 ). I was expecting the occupation numbers of these states to be 0 or 1 or some normalized linear combination. $\\$ I would think this one has a simple solution, but I don't know what it is at the moment... $\\$ @vanhees71 Might you give us a hint or two on this one? I'm stuck. $\\$ One other idea I have is to use raising and lowering operators: $s_x=\frac{(s_++s_-)}{2}$ etc.

Last edited: Jun 29, 2017
14. Jun 29, 2017

### LagrangeEuler

If particle 2 is in down state and particles 1 and 3 is in up state then the row matrix is
(0 0 1 0 0 0 0 0). Right?

Because this is

$$\left[ {\begin{array}{cc} 1 \\ 0 \\ \end{array} } \right]\otimes \left[ {\begin{array}{cc} 0 \\ 1 \\ \end{array} } \right] \otimes \left[ {\begin{array}{cc} 1 \\ 0 \\ \end{array} } \right]= \left[ {\begin{array}{cc} 0 \\ 0 \\ 1\\ 0\\ 0\\ 0\\ 0\\ 0\\ \end{array} } \right]$$

15. Jun 29, 2017

A google of the topic shows it is necessary to use a Hilbert space of states which in this case would apparently make the $H$ 16x16. What I had proposed apparently doesn't work. I'm still at the drawing board though and haven't solved it yet. One thing that still isn't clear from the OP is if you have 3 particles or 4? I am now led to believe that you have 3 particles. That would make $H$ 8 x 8 which is what you are doing. The result might simply be that you enumerate (write out) the 8 possible states, compute the energy of each and have a diagonalized Hamiltonian matrix $H$.

Last edited: Jun 29, 2017
16. Jul 7, 2017

### Oxvillian

Seems like you guys are making this one way too hard. Why not just work in the $\sigma_x$ basis?

17. Jul 8, 2017

### Fred Wright

Because the first term in the Hamiltonian only involves $s_1$ and $s_2$ you can write,$$\sigma_x \otimes \sigma_x \otimes \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$
Because the second term in the Hamiltonian only involves $s_2$ and $s_3$ you can write,$$\begin {pmatrix} 1 & 0 \\ 0 & 1 \end {pmatrix} \otimes \sigma_x \otimes \sigma_x$$
Adding these two matricies you can read off the eigenvectors from the columns (or rows, it's symmetric). Find the eigenvalues by operating on the Hamiltonian with the eigenvectors.

18. Jul 8, 2017

### Fred Wright

Sorry, this is wrong. You have to go through the usual diagonalization procedure to compute the eigenvectors and eigenvalues.

19. Jul 8, 2017

### Fred Wright

Again I apologize. My previous posts are totally wrong! For what its worth I did, however, solve the problem using raising and lowering operators.