# Free particles in second quantization

1. Feb 17, 2010

### Niles

Hi guys

Today my lecturer talked about second quantization, and at the end he talked about free fermions in second quantization. He said that free electrons in second quantization satisfy that their Hamiltonian is only written as a linear combination in terms of $c^\dagger c$ (the creation and annihilation operators). I've searched the WWW, but no other source confirms this, but is the statement correct? And what is the physical reason for this?

2. Feb 17, 2010

### Physics Monkey

Hi niles,

I assume what you mean is the Hamiltonian for free fermions is quadratic in the creation and annihilation operators. This is essentially the statement that there are no interactions, because interactions involve more than a single particle at a time. For example, the interaction operator $$V = \int dx \int dy \,c^+(x) c^+(y) U(x-y) c(y) c(x)$$ may be thought of as performing the following operation: it tries to remove an electron at x and then an electron at y, and if successful, contributes $$U(x-y)$$ to the energy before putting the electrons at x and y back. This is physically an electron interaction because it depends on the presence of multiple electrons in the system. On the other hand, operators quadratic in the creation and annihilation operators are single particle operators, they check for an electron in one state, do something, and then put that one electron back in a (possibly different) state.

However, there is a subtlety in the notation. To some people, free fermions mean fermions that do not interact with each other or any external potential. However, the class of fermion Hamiltonians that are quadratic in the creation and annihilation operators includes the possibility of external potentials. However, as I said above, this class does not include the possibility of direct fermion-fermion interactions.

3. Feb 17, 2010

### Niles

Thanks. That H is quadratic in the creation/annihilation operators is the same as it being "bilinear", i.e. $a^\dagger a$ (or $a a^\dagger$)? E.g. $a a^\dagger a$ or $a^\dagger aa^\dagger$ are not quadratic, because here we have the possibility of interactions (multiple fermions)?

4. Feb 17, 2010

### xepma

In general you can write the Hamiltonian for "free" fermions as

$$H = \sum_{ij} H_{ij} c^\dag_i c_j$$

which is always diagonalizable. You call this bilinear in a similar sense explained here: http://en.wikipedia.org/wiki/Bilinear_form

Terms that involve three creation/annihilation operators indeed correspond to interactions, since the involve multiple fermions. However, do note that terms with an odd number of creation/annihilation operators change the number of fermions. For example, the term $c_i^\dag c_j^\dag c_k$ destroys one fermion, and creates two in return. Such terms are usually absent since it breaks not only particle number conservation, but also charge conservation.

Interactions between two fermions are indeed given by a product of 2 creation and 2 annihilation operators.

Final remark: using the commutation relations you can always put the Hamiltonian into a normal ordered form. This comes down to putting all annihilation operators "to the right" and creation operators "to the left".

5. Feb 17, 2010

### Niles

Why is it necessarily always diagonalizable?

Just to be clear: So "bilinear" is equivalent to "quadratic", when talking about free fermions?

And thanks, I appreciate it very much.

6. Feb 17, 2010

### xepma

It is diagonalizable because the Hamiltonian H is hermitian, and so a basis of eigenvectors exists.

The diagonalized version is
$$\sum_k \epsilon_k d^{\dag}_k d_k$$

where the d's are linear combinations of the c's.

You can interpet the $$\epsilon_k$$ as the energy carried by the excitation created by d.

In the simple case of fermions in the absence of an external potential the $d^\dag_k$ creates a fermion with momentum k, and its energy $\epsilon_k$ is simply $k^2/2m$. In a harmonic oscillator or an infinite potential well the $d^\dag_k$ creates a fermion in the k'th energy level, etc.

Last edited: Feb 18, 2010
7. Feb 18, 2010

### Niles

Thanks guys, you are so awesome. This means alot to me.

8. Feb 18, 2010

### Niles

By the way, are there any restrictions regarding time-dependence? Personally I think yes (it has to be time-independent) - am I correct?

9. Feb 18, 2010

### muppet

In this framework you create and destroy particles at spacetime points; it's implicit in the above notation that x=(ct,x,y,z) is a 4-vector.

10. Feb 18, 2010

### Physics Monkey

The Hamiltonian can be time dependent and still be quadratic. This describes fermions interacting with some external time dependent potential but still without direct fermion-fermion interactions.

11. Feb 18, 2010

### peteratcam

Just to add, all hamiltonians are 'diagonalizable' in the sense that they are hermitian operators so there exists a basis of orthonormal eigenvectors.
Hamiltonians which are quadratic in creation and annihilation operators are easy to diagonalise: solving their spectrum is no harder than solving a single particle problem. But interacting many body quantum mechanics is hard!

12. Feb 18, 2010

### Niles

Thanks guys; I have one more question related to free fermions, and I am just gonna ask it here instead of creating a new thread. In some notes I have they write the Hamiltonian for free fermions as

$$H = \sum_{\mathbf{k}, \sigma}{\epsilon_{\mathbf{k}, \sigma}c^{\dagger}_{\mathbf{k}, \sigma}c_{\mathbf{k}, \sigma}},$$

i.e. the Hamiltonian is diagonal in the quantum numbers k and σ (σ denotes spin). When they write the Hamiltonian this way, are they writing it in k-space?

Also what is the connection between the Hamiltonian in the notes and the one we have been discussing? They are both for free fermions, but when we diagonalize "ours", is it written in the same space as the one in the notes?

Last edited: Feb 19, 2010