Ground state of Hamiltonian describing fermions

In summary, the Hamiltonian represents a system of particles in which each particle has an associated energy. The ground state is the lowest energy state of the system.
  • #1
Morto
12
0

Homework Statement



I have been given the Hamiltonian

[tex]H = \sum_{k} (\epsilon_k - \mu) c^{\dag} c_k[/tex]

where [tex]c_k[/tex] and [tex]c^{\dag}_k[/tex] are fermion annihilation and creation operators respectively. I need to calculate the ground state, the energy of the ground state [tex]E_0[/tex] and the derivative [tex]\frac{\delta E_0(\mu)}{\delta \mu}[/tex]. 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
[tex]c^{\dag} c |1> = 1|1>[/tex]
and
[tex]c^{\dag}c|0>=0|0>[/tex]
So that
[tex]c^{\dag}|0> = |1>[/tex]
and
[tex]c|1> = |0>[/tex]
and
[tex]c|0>=0[/tex]
and
[tex]c^{\dag}|1> = 0[/tex]
(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..
[tex]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>
[/tex]
Using the first relation.
How do I now calculate the energy of this ground state?
[tex]<0|H|0> = <0|\sum_k \epsilon_k|1> - <0|\mu|1>[/tex]

What do I do with this? Have I made some fundamental error somewhere? This doesn't look right to me.
 
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  • #2
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
Okay, so the matrix representation of these operators is
[tex]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)[/tex]

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

Is this the ground state of the kth term? [tex]\psi_0(x) = A e^{(\epsilon_k - \mu)x}[/tex]? What do I do with this?
 
  • #4
You made a mistake in the Hamiltonian.
 
  • #5
Ah, yes, of course.

[tex]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)[/tex]

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

For the non-zero eigenvalue to be the lowest eigenvalue, then [tex]\epsilon_k - \mu < 0[/tex]. 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 [tex]x[/tex] or [tex]y[/tex] are non-zero only when [tex]\mu=\epsilon_k[/tex], in which case the entire Hamiltonian is zero.
 
Last edited:
  • #6
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
Nope, I was just given that Hamiltonian and told to find the ground state, the energy of the ground state [tex]E_0[/tex] and the derivate wrt [tex]\mu[/tex], so if [tex]E_0 = \epsilon_k - \mu[/tex] then [tex]\frac{\partial E_0}{\partial \mu} = -1[/tex]. (and if [tex]E_0 = 0[/tex], then obviously the derivate is zero).

Is this a 'famous' result?
 
  • #8
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
Ahah. Thank you for your help.

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

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

1. What is the ground state of a Hamiltonian describing fermions?

The ground state of a Hamiltonian describing fermions is the lowest energy state that the system can be in. It is the state in which all the fermions are in their lowest possible energy levels, following the Pauli exclusion principle.

2. How is the ground state of a Hamiltonian describing fermions determined?

The ground state is determined by solving the Hamiltonian equation, which describes the total energy of the system. This involves finding the eigenvalues and eigenvectors of the Hamiltonian matrix.

3. Can the ground state of a Hamiltonian describing fermions change?

Yes, the ground state can change if the Hamiltonian is perturbed by an external force or if the temperature of the system is changed. In these cases, the fermions may rearrange themselves to reach a new ground state with a lower energy.

4. How does the ground state of a Hamiltonian describing fermions relate to quantum mechanics?

The ground state of a Hamiltonian describing fermions is a fundamental concept in quantum mechanics. It is one of the key components in understanding the behavior and properties of fermionic systems, such as atoms, nuclei, and electrons.

5. What is the significance of the ground state of a Hamiltonian describing fermions?

The ground state of a Hamiltonian describing fermions is important because it represents the most stable and lowest energy state of the system. It also serves as a reference point for understanding the excited states and dynamics of the system. Additionally, the ground state can provide insight into the properties and behavior of the system as a whole.

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