Second quantization and partial traces

[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!

 

[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?

 

[schieghoven]

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?

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.)

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?

Yes, I think so.

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?

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.

 

[ledamage]

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.)

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.

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.

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.