# Second quantization and partial traces

1. Apr 27, 2009

### ledamage

Hi!

Is there a common way to write a fermionic Fock space (finite dimensional) as a tensor product such that it is possible to do a partial trace over one particle type? Sorry, if this is an obvious question, but I just can't see it.

Thanks!!

2. Apr 28, 2009

### ledamage

Let me refine the question. In bosonic Fock space, due to the commutation relations of the creation and annihilation operators, it is possible to write every state vector as a tensor product

$$|\psi\rangle = |n_1\rangle \otimes |n_2\rangle \otimes |n_3\rangle \otimes \hdots$$

where $n_1=0,1,2,3,\hdots$ etc., right?

Since creators and annihilators commute, it is possible to write an operator, e.g., $Q=a_1^\dagger a_3^\dagger a_2 a_1 a_2^\dagger$ as $Q=a_1^\dagger a_1 \otimes a_2 a_2^\dagger \otimes a_3^\dagger \otimes 1 \otimes 1 \otimes \hdots$ as long as the ordering within the same species is maintained, right?

This makes it particularly easy to take partial traces over modes, just by taking the expectation value of, say, $|n_2\rangle$ and taking the sum $\sum_{n_2=0}^\infty$. In fermionic Fock space, however, I always get minus signs when commuting since even different species don't commute trivially, so it seems suspicious to do something like above. Is there a common treatment for this? How do I take a partial trace over, say, one species in a fermionic Fock space?

3. May 1, 2009

### schieghoven

No, not every state is of this form. Firstly, a state vector $|\psi\rangle$ for one particle species, in general, is a superposition of number eigenstates $|n\rangle$, where $n=0,1,2,3,\hdots$. Secondly, it's possible to entangle particles of different species; so you need to include finite linear sums of states of this form. (That is, the states you mention are a basis for the state space: state space is the closure of the set of finite linear sums of basis states. Reed and Simon, Methods of Modern Math Physics is a great reference.)

Yes, I think so.

Good question. I'm not sure. I never really thought about it before but my first inclination was that creation/annihilation operators of different species commute, even for fermions.

4. May 1, 2009

### ledamage

Oh sure! Actually, this is what I meant - a possible basis for the Fock space. But this is sufficient for the present purposes if I choose this basis for my trace.

Unfortunately, I noted that I implicitly assumed this after a quite tedious calculation. But we have $\{ a_i^{(\dagger)},a_j^{(\dagger)} \} = 0$, don't we? Indeed, when expressing basis states of the fermionic Fock space in terms of occupation numbers, a determined order of the creators has to be specified, e.g.,

$$|n_1, n_2, \hdots \rangle := (a_1^\dagger)^{n_1} (a_2^\dagger)^{n_2} \hdots |0\rangle \neq (a_2^\dagger)^{n_1} (a_1^\dagger)^{n_2} \hdots |0\rangle \ .$$

(Actually, for three of four cases, the $\neq$ is a $=$, but not in general.) In another thread, I found an interesting suggestion: Write the basis states of the fermionic Fock space as

$$|\psi\rangle = |n_1\rangle \otimes |n_2\rangle \otimes \hdots \ , \qquad (n_i=0,1)$$

and represent the operators by

$$a_1^{(\dagger)} = a_1^{(\dagger)} \otimes 1 \otimes 1 \otimes 1 \otimes \hdots ,$$
$$a_2^{(\dagger)} = (-1)^{a_1^\dagger a_1} \otimes a_2^{(\dagger)} \otimes 1 \otimes 1 \otimes \hdots ,$$
$$a_3^{(\dagger)} = (-1)^{a_1^\dagger a_1} \otimes (-1)^{a_2^\dagger a_2} \otimes a_3^{(\dagger)} \otimes 1 \otimes \hdots$$
etc.

The $(-1)^{a_i^\dagger a_i}$ stands for $(1-2 a_i^\dagger a_i)$ and accounts for the additional minus signs due to anticommuting when necessary.