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If I find a Hamiltonian $H = \sum_{k} \varepsilon_k a_k^{\dagger} a_k + \sum_k V_k a_k^{\dagger} a_k$ then I was wondering: As far as I know this is many body theory and so these operators act on symmetrized or antisymmetrized states respectively, but I am not sure which quantity would determine in this case what the state that we symmetrize or antisymmetrize are? So which states do these operators $a_k$ actually create and annihilate? I guess that the answer is eigenstates of the Hamiltonian, but actually the spectrum of the Hamiltonian does not have to be discrete, so I don't know this.

EDIT: It was suggested that these are the eigenstates of the kinetic part of the Hamiltonian, but then we have the problem, that the spectrum is not necessarily discrete (think of unrestricted motion in $x$ direction)

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  • $\begingroup$ $a^\dagger_k|0\rangle$ is the eigenstate of the Hamiltonian $H_0=\sum_k\epsilon_ka^\dagger_ka_k$ with continuum spectrum $\epsilon_k$. Sum is an integral for continuum $k$ $\endgroup$
    – Fedxa
    Nov 22 '14 at 15:22
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Usually in many body theory these operators create and annihilate particles. There are different annihilation and creation operators for fermions and bosons (they obey different commutation relations).

The states they act upon and the states "created" by them respect the required symmetries (antisymmetric for fermions, symmetric for bosons). The operators in your hamiltonian could act for instance on a fock state (note that the operator pairs in your Hamiltonian represent the number operator). You might want to take a look at http://en.wikipedia.org/wiki/Fock_state , especially take a look at the two sections of "Action on some specific Fock states".

NB: If you came across your Hamiltonian while dealing with harmonic oscillators, you rather want to think of $a_k^{\dagger}$ and $a_k$ in terms of ladder operators.

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  • $\begingroup$ my question was: The states in that case are determined by what? (I guess the answer is: Eigenstates of the hamiltonian, but I am not sure) $\endgroup$
    – Xin Wang
    Nov 22 '14 at 9:58
  • $\begingroup$ Your question is somewhat unclear - you can of course look at whichever states you please. But, if you look at eigenstates of the system, your Hamiltonian will, by very definition, be diagonal in the respective ladder operator basis. I presume the answer you are looking for is that you are looking at the number states corresponding to the ladder operator basis you are working in. Your Hamiltonian above is bilinear in these and only contains normal terms. Thus its diagonalization is simply a reformulation of a single-particle problem. $\endgroup$
    – ulf
    Nov 23 '14 at 3:10
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The Hilbert space of the states in this case is the Fock space. It is a linear space "constructed" by acting by the creation operators $a^\dagger_k$ on the vacuum state $|0\rangle$, which has the property $a_k|0\rangle=0$. All other states are related so that the commutation relations between $a,a^\dagger$ are satisfied.

The individual states like $a^\dagger_{k_1}a^\dagger_{k_2}|0\rangle$ are not in general eigenstates of the Hamiltonian. Usually they correspond to the eigenstates of a "free part of the Hamiltonian", while the eigenstates of the full Hamiltonian are in general linear combinations of these basic vectors.

Note, that the "free part" is not uniquely defined. One can have the same system which looks differently when written in creation and annihilation operators, and then there will be a nontrivial homomorphism between the two different Fock spaces, this is called Bogoliubov transformation.

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  • $\begingroup$ well, this sounds problematic. if you have a free hamiltonian over the whole real line( so free motion in one direction), then there are not countable many eigenstates, so you cannot write this as a sum, right? (or do you mean by free part the single particle Hamiltonian without the particle-particle interaction?) also, do you mean by usually 99% of all cases you ever encountered or do you rather mean in 3 out of 5 cases? $\endgroup$
    – Xin Wang
    Nov 22 '14 at 11:54
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    $\begingroup$ The "sums" in such discussions are always infinite. For the case of continuous "label", like momentum $k$, the sum should be replaced by an integral (not much difference, apart form proper normalization of the states, which is delta-function over $k$ in this case). Example: a one particle wavepacket would look like $\int dk \psi(k) a^\dagger_k|0\rangle$. Two particle states would look like $\int dk_1 dk_2 \psi(k_1,k_2)a^\dagger_{k_1}a^\dagger_{k2}|0\rangle$ (with proper symmetry properties for the function $\psi(k_1,k_2)$). Of course the function $\psi$ here should be properly normalized. $\endgroup$
    – Fedxa
    Nov 22 '14 at 13:22
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    $\begingroup$ "Usually" means that this is the normal convention, and you'll never find anything else. However, you really find different choices of the free part -- in may or may not contain the mass fo the particle, for example. $\endgroup$
    – Fedxa
    Nov 22 '14 at 13:27
  • $\begingroup$ thanks, is the free part just the kinetic part or would you also include an external potential that is not due to interaction of the different particles( for example a constant electric field) $\endgroup$
    – Xin Wang
    Nov 22 '14 at 13:35
  • $\begingroup$ Simple answer-just the free kinetik part. If the operators are labeled by momentum, this is virtually always the case. $\endgroup$
    – Fedxa
    Nov 22 '14 at 14:06

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