# Ground state of Hamiltonian describing fermions

1. Nov 3, 2008

### Morto

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

I have been given the Hamiltonian

$$H = \sum_{k} (\epsilon_k - \mu) c^{\dag} c_k$$

where $$c_k$$ and $$c^{\dag}_k$$ are fermion annihilation and creation operators respectively. I need to calculate the ground state, the energy of the ground state $$E_0$$ and the derivative $$\frac{\delta E_0(\mu)}{\delta \mu}$$. Apparently this last quantity is 'famous' and I should recognise it. However, I think that I am making some fundamental mistake quite early on.

2. Relevant equations

I know that
$$c^{\dag} c |1> = 1|1>$$
and
$$c^{\dag}c|0>=0|0>$$
So that
$$c^{\dag}|0> = |1>$$
and
$$c|1> = |0>$$
and
$$c|0>=0$$
and
$$c^{\dag}|1> = 0$$
(All of this is proven by writing these operators as matrices and multiplying by state vectors. These relations are confirmed in 'Quantum theory of solids' by Kittel)

3. The attempt at a solution
But when it comes to calculating the ground state of this Hamiltonian, I find something unusual..
$$H|0> = \sum_{k}\epsilon_k c_k^{\dag} c_k |0> - \mu \sum_k c_k^{\dag}c_k|0> \\ = \sum_k \epsilon_k|1> - \mu|1>$$
Using the first relation.
How do I now calculate the energy of this ground state?
$$<0|H|0> = <0|\sum_k \epsilon_k|1> - <0|\mu|1>$$

What do I do with this? Have I made some fundamental error somewhere? This doesn't look right to me.

2. Nov 3, 2008

### borgwal

For each k there are only 2 states available: |0> and |1>. For each k, write the kth term of the Hamiltonian as a 2x2 matrix, and find its eigenvalues (really easy!). The ground state corresponds to the lowest eigenvalue. The full ground state is then a tensor product of all the ground states for each k.

What you calculated is not the ground state, you just applied H to |0> and then made a mistake (why would you get a |1> in there??)

3. Nov 4, 2008

### Morto

Okay, so the matrix representation of these operators is
$$c^{\dag}c = \left(\begin{array}{cc}0 & 1 \\ 0 & 0\end{array}\right)\left(\begin{array}{cc}0 & 0 \\ 1 & 0\end{array}\right) = \left(\begin{array}{cc}1 & 0 \\ 0 & 0\end{array}\right)$$

And the Hamiltonian of the kth term will be
$$H_k = \left(\begin{array}{cc}\epsilon_k - \mu & 0 \\ 0 & \epsilin_k - \mu\end{array}\right)$$
which has only a single eigenvalue $$\lambda = \epsilon_k - \mu$$

Is this the ground state of the kth term? $$\psi_0(x) = A e^{(\epsilon_k - \mu)x}$$? What do I do with this?

4. Nov 4, 2008

### borgwal

You made a mistake in the Hamiltonian.

5. Nov 4, 2008

### Morto

Ah, yes, of course.

$$H_k = \epsilon_k\left(\begin{array}{cc}1&0\\0&0\end{array}\right)-\mu\left(\begin{array}{cc}1&0\\0&0\end{array}\right) = \left(\begin{array}{cc}\epsilon_k - \mu&0\\0&0\end{array}\right)$$

So calculating the eigenvalues
$$det\left(\begin{array}{cc}\epsilon_k - \mu-\lambda&0\\0&-\lambda\end{array}\right)=0$$
$$\left(\lambda + \mu -\epsilon_k\right)\lambda = 0$$
So this has eigenvalues $$\lambda=0, \lambda=\epsilon_k - \mu$$

For the non-zero eigenvalue to be the lowest eigenvalue, then $$\epsilon_k - \mu < 0$$. I'm not sure what this requirement means. What do I do with this?

When calculating the eigenvectors, I find that for either case then $$x$$ or $$y$$ are non-zero only when $$\mu=\epsilon_k$$, in which case the entire Hamiltonian is zero.

Last edited: Nov 4, 2008
6. Nov 4, 2008

### borgwal

Yes, now we're getting somewhere! For those k for which \epsilon_k<\mu, the ground state energy is \epsilon_k-\mu and the ground state is |1>_k. For all other k's the energy is zero, and the ground state is |0>.

So, the fermions would ideally only occupy those states with \epsilon_k<\mu. Were you given a fixed number of particles or some other relation about the number of fermions?

7. Nov 4, 2008

### Morto

Nope, I was just given that Hamiltonian and told to find the ground state, the energy of the ground state $$E_0$$ and the derivate wrt $$\mu$$, so if $$E_0 = \epsilon_k - \mu$$ then $$\frac{\partial E_0}{\partial \mu} = -1$$. (and if $$E_0 = 0$$, then obviously the derivate is zero).

Is this a 'famous' result?

8. Nov 4, 2008

### borgwal

No, that's not a famous result because that's not quite the result you'd get! In any case:
I'm not sure how much background knowledge was assumed for this question, but here is what it is really all about:

http://en.wikipedia.org/wiki/Fermi-Dirac_statistics

The \mu is the chemical potential.

9. Nov 4, 2008

### Morto

Ahah. Thank you for your help.

$$H = \sum_k \left(\epsilon_k - \mu\right) c^{\dag}_k c_k + \gamma \sum_{kp} c_k^{\dag}c_p$$