# Homework Help: Diagonalization of a Hamiltonian for two fermions

1. Jan 21, 2014

### hmdkdl

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

I want to diagonalize the Hamiltonian:

2. Relevant equations

$H=\phi a^{\dagger}b + \phi^{*} b^{\dagger}a$

a and b are fermionic annihilation operators and $\phi$ is some complex number.

3. The attempt at a solution

Should I use bogoliubov tranformations? I have put the following:

$a = u c + v d^{\dagger}$
$b = w c + z d^{\dagger}$

(which c and d are some other fermionic annihilation operators and u, v, w, z complex numbers) into hamiltonian but I cannot find the final answer (it does not seem straightforward).. Is it correct? and if so how can I find the final answer.

2. Jan 21, 2014

### strangerep

It's difficult to help you if you don't show the details of what you've actually tried.

3. Jan 22, 2014

### hmdkdl

My try

I supposed that the transformations are like this (stars on u,v,w,z means complex conjugate and on a,b,c,d means dagger):

a = uc + vd* --> a*=u*a + v*d
b = wc + zd* --> b*=w*c + z*d

we have fermions, so:

{a,a*}=1 --> |u|^2 + |v|^2=1
{b,b*}=1 --> |w|^2 + |z|^2=1

Now the hamiltonian becomes:

H=( øu*w + ø*uw* )c*c + (øv*z + ø*vz*)dd* + (øu*z + ø*v w*)c*d* + (øv*w + ø*u*z)dc

We want the hamiltonian to be diagonalized so:

øu*z + ø*v w* = 0

Now we have 8 unknown parameters and 4 equations.

4. Jan 22, 2014

### strangerep

You seem not to have used the other anticommutation relations: $\{a,b\} = 0 = \{a, b^\dagger\}$, etc.

5. Jan 23, 2014

### hmdkdl

Yes, you're right. {a,b}=0 gives nothing but {a,b*}=0 gives two other equations that can be used to find the final answer.
By the way, my final answer is:
$H = \pm |\phi| (c^{\dagger}c + d^{\dagger}d - 1 )$

This hamiltonian has three eigenvalues: $\epsilon=\pm |\phi|, 0$
However, my reference says that we have just two: $\epsilon=\pm |\phi|$. any ideas?

Anyway, Thanks a lot strangerep.

6. Jan 23, 2014

### strangerep

What reference is this from?

What are your values for the Bogoliubov coefficients in the transformation?

7. Jan 24, 2014

### hmdkdl

It is from a paper about graphene ( PRL 101, 026805 (2008) ). This hamiltonian with sum over wave vector k and $\phi = \phi(k)$, is the hamiltonian of graphene in the tight bonding approximation, and I think it is a very standard form of hamiltonian for graphene.

if $\phi = |\phi| e^{i \alpha}$:
$u = \frac{ \sqrt 2 }{2} e^{i(\alpha+\theta_{w} + n\pi)} \\ v = \mp \frac{ \sqrt 2 }{2} e^{i(\alpha+\theta_{z} +n\pi)} \\ w =\pm \frac{ \sqrt 2 }{2} e^{i\theta_{w}} \\ z = \frac{ \sqrt 2 }{2} e^{i\theta_{z}} \\$

$\theta_w$ and $\theta_z$ are two free parameters. There are other answers with different signs but they all give the same answer.

Last edited: Jan 24, 2014
8. Jan 24, 2014

### strangerep

OK, I see what's gone wrong. I had blithely assumed your initial ansatz was correct, but it wasn't. I should have realized this earlier since you mixed annihilation and creation operators, which usually means the B-transformed Fock space is unitarily inequivalent to the original. But it's not.

The diagonalization should be dead easy for this case. Suppressing your scalar coefficients, one can write the Hamiltonian as:
$$H ~=~ a^\dagger b + b^\dagger a ~=~ \pmatrix{a^\dagger &b^\dagger} \pmatrix{0 & 1 \\ 1 & 0} \pmatrix{a \\ b}$$Then find a diagonalizing matrix P for the matrix in the middle of the above (by finding its eigenvectors). The diagonalizing matrix P in this case turns out to be
$$P ~=~ \frac{1}{\sqrt{2}} \pmatrix{1 & 1 \\ 1 & -1}$$
Apply $P$ to the column vector of $a,b$ to get $c,d$. Or rather, apply its inverse to get $a,b$ in terms of $c,d$. I get
$$\pmatrix{a \\ b} ~=~ \frac{1}{\sqrt{2}} \, \pmatrix{c+d \\ c-d} ~.$$
When you substitute for $a,b,$ in the original $H$ it becomes diagonal -- without the pesky constant term. (The latter was happening because you'd mixed annihilation and creation operators, and unwittingly transformed to a new vacuum, with different ground state energy from the old.)

Sorry for not recognizing all this sooner. Hopefuly you can generalize the above to your specific case.

9. Jan 25, 2014

### hmdkdl

Your method is more easy and more straightforward. Thanks.
Now, is it true that we have three eigenvalues: $\epsilon = \pm |\phi|,0$ ? with eigenstates:

|1,0> , |0,1> and a linear combiantion of |0,0> and |1,1>

That's true. But why shouldn't I? In other words, when can I mix them? The example in Wikipedia and an example in the book "Quantum theory of solids" by Charles Kittel (for bosons) uses the mixed combination of them.

10. Jan 25, 2014

### strangerep

I'm not sure what you're doing. Show me how you're getting those 3 eigenvalues, and what the values in the kets mean. (I could guess, but I'd rather not.)

I don't have that book, but I'm guessing it's for cases when the lowest energy state of the system doesn't correspond to the free vacuum. One introduces "quasi-particles" via such a transformation, and (as a consequence) a new vacuum, being the ground state of that system.

So it depends on what one is trying to achieve.

Edit: This is really an exercise in condensed matter, and hence outside my comfort zone. This wasn't obvious from the original post, else I wouldn't have tried to answer. I'll try to get someone better qualified in that area to take over. BTW, for other readers, the PRL paper mentioned above (Localized Magnetic States in Graphene) can be downloaded here.

Last edited: Jan 26, 2014
11. Jan 26, 2014

### hmdkdl

$|n_c,n_d>$ is the fock state with $n_c (n_d)$ particle in state c (d), which can be 0 or 1. I think my question has a very simple the answer: if we consider multiparticle states we have three eigenvalues, but if we are looking for state of one paticle, we have just two.

Thanks a lot for all the help strangerep.