Ground state of Hamiltonian describing fermions

[Morto]

Homework Statement

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.

Homework 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)

The Attempt at a Solution

But when it comes to calculating the ground state of this Hamiltonian, I find something unusual..
$$H|0> = \sum_{k}\epsilon_k c_k^{\dag} c_k |0> - \mu \sum_k c_k^{\dag}c_k|0> \\<br /> = \sum_k \epsilon_k|1> - \mu|1>$$
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.

 

[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??)

 

[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?

 

[borgwal]

You made a mistake in the Hamiltonian.

 

[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.

 

[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?

 

[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?

 

[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.

 

[Morto]

Ahah. Thank you for your help.

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

Can I use the same method to determine the ground state? What does this Hamiltonian represent?