# Interpreting operators in second quantization

1. Apr 17, 2010

### Niles

Hi guys

When working with operators in second quantization, I always imagine

$$c^\dagger_ic_j$$

as denoting the "good old" matrix element $\left\langle {i} \mathrel{\left | {\vphantom {i j}} \right. \kern-\nulldelimiterspace} {j} \right\rangle$. But how should I interpret an operator given by e.g.

$$c^\dagger_ic_j^\dagger?$$

2. Apr 17, 2010

### sokrates

Wouldn't it be the good old product of two bras?

3. Apr 19, 2010

In single-particle QM, one can write the Hamiltonian as

$$H=\sum H_{ij} |i \rangle \langle j|$$.

In second quantization, the single-particle operator is

$$H'=\sum H_{ij} c_i^{\dagger} c_j,$$

so one can make the identification $$c_j \leftrightarrow \langle j|$$, as long as the Hilbert space is simply the space of single particles. Both Hamiltonians H and H' have, after all, the same matrix elements in this space.

The operator $$c^\dagger_i c_j^\dagger$$, on the other hand, does not even conserve the number of particles, so in my opinion, it makes no sense to identify the operator as any combination of bras and kets.

4. Apr 19, 2010

### Physics Monkey

Hi Niles,

It wouldn't make sense to interpret $$c_i^+ c_j$$ as the matrix element $$\langle i | j \rangle$$ since one is an operator and the other is a number. Indeed, if the single particle states i and j are orthogonal then the matrix element is just zero, but the operator you wrote is definitely not zero. The meaning of $$c_i^+ c_j$$ is that is turns the state $$| j \rangle$$ into the state $$| i \rangle$$. It destroys a particle in the single particle state labeled by j and creates a particle in the single particle state labeled by i. Similarly, $$c^+_i c_j^+$$ destroys no particles and creates a particle in state i and a particle in state j.

These kinds of operators are relevant in superconductors, for example, where the particle number can fluctuate coherently.

Last edited: Apr 19, 2010
5. Apr 19, 2010

### George Jones

Staff Emeritus
I'm a little confused. Have you treated the c's as a's?

6. Apr 19, 2010

### Physics Monkey

Sorry, what are the "a's"?

7. Apr 19, 2010

### SpectraCat

By convention, the "c's" are the creation/annihilation operators for *fermions*. The "a's" are the analogous operators for bosons. The c's change occupation numbers of fermions in a Fock state vector ... due to the PEP, those occupation numbers can only ever be 0 or 1.

The a's change occupation numbers for bosons, which do not have the PEP restrictions, and so can assume arbitrarily large values (e.g. high-photon density in an intense EM field).

EDIT: So, I guess George Jones' confusion came from the fact that if you apply a c+ creation operator twice to the same state, it simply returns that state to its original configuration with no net effect. Therefore, you cannot say (as you did) that applying the ci+ operator will create a particle in the ith state, because if that state is initially occupied, then it will actually annihilate the particle in that state.

Last edited: Apr 19, 2010
8. Apr 19, 2010

### Physics Monkey

I suppose this is a fairly common convention, but it wasn't clear to me that the OP was interested only in fermions.

Perhaps this is what George was talking about. Of course you're right that if I apply the fermionic creation operator to an already filled state then I don't add a fermion. However, I think you meant to say that applying the same fermionic creation operator twice annihilates the state (rather than returning the state to its original form) I was implicitly thinking about adding particles to the vacuum, but thanks for clarifying this point.

9. Apr 19, 2010

### SpectraCat

Yes, you are correct ... I was thinking that the application of the ci+ operator just destroyed the fermion in the ith-state, but of course it must annihilate the entire Fock state according to the PEP. Thanks for catching that! (it's been a while since I had to think about this stuff, and I am a bit rusty.)