Hi there, I had some months ago the same doubt and here I found a lot of help. As an amateur of these topics, perhaps my answer is not absolutely correct, but perhaps you would find it more accesible than the others.
The psi (x) operator creates a particle in "x" position. That means, it takes the original state, perhaps defined as a superposition of states (phii) that has each one of them "i" particles in "x" position, to a state which is the same superposition, but now of the phii+1 states. Perhaps there is a state, defined as a superposition of states (phii) defined each one of them as having '"i" particles in "x" position', that when you add a particle in all those phii states, you arrive as a result in an end state which is equal as the original multiplied by a complex number. In that case, that state is in fact an eigenstate of psi(x) and that complex number is an eigenvalue.
I am pretty sure (but perhaps I am wrong) that the eigenvalues of that operator (for every "x") will be complex numbers with unit norm. That guess comes from the following idea:
1) That operator is like an operator that produces a migration from one kind of state ("n" particle state in "x" position) to another state ("n+1" particle state in "x" position).
2) In a way it is sort of similar to an "n" dimensional permutation matrix (look it up in wikipedia). For example, the matrix that, when you insert the (1,0,0,0...) vector gives back the (0,1,0,0,...) vector, when you insert the (0,1,0,0,0,...) gives (0,0,1,0,...) and when you put (0,0,0...0,1) gives back (1,0,0,0,0) (just in this last point the analogy is not exact).
3) to make the analogy you have to think that the first vector mentioned is the 0 particle state, the second one is the 1 particle and so on.
4) at first sight is difficult to think that there is an eigenvector of this matrix (operator from now on). But if you start thinking in complex numbers(lets suppose that we are working with a 5d mattrix), you can take the number "exp(i*2*pi/5)" and make the following vector with it (exp(1*i*2*pi/5);exp(2*i*2*pi/5);exp(3*i*2*pi/5);exp(4*i*2*pi/5);exp(5*i*2*pi/5))
5) If you apply this permutation matrix you will get that every element will flip one position to the right (only the last element will flip back to the first position). And the result will be equal to multiply the original vector by exp(-i*2*pi/5). So the eigenvector is the original vector and the eigenvalue will be this complex number.
6) every other eigenvalue is exp(-i*2*pi/5)^(1 or 2 or 3 or 4 or 5).
7) every other eigenvector is similar to what is described in 4 but ordered in different ways
8) for an "n" dimensional mattrix the eigenvalues are exp(-i*2*pi/n)^(1 or 2 or 3 or ... or n) and the eigenvectors have similar form as what is mentioned in 4) and 7).
So, I think that if you have to calculate the eigenvalues of psi(x), where you would have infinite ocupation states, you will have that the eigenvalues are something not very diferent than what is mentioned in 8). Perhaps something like exp(-i*2*pi)^(every rational number), but I am not sure, I haven't thought this last point carefully.
Finally we can extrapolate this idea for every "x" and think about the state |phi> where, for every "x", we have that if we apply psi(x) to phi> (ie psi(x)|phi>) we obtain as a result f(x)*phi> (where f(x) is a function that, because of what I've told, I think can take only complex values of unit norm). In that case, it is used to say that we are in the state f(x)>.
What is the interpretation of that state? Nothing. Just another basis of the same space. Just as you can, in 2d, represent everything with the vectors (1;0) and (0;1) or with the vectors (0,5^0,5 ; 0,5^0,5) and (0,5^0,5 ; 0,5^0,5), you can in infinite d represent every state as a superposition of f(x)> states or as a superposition of "n particles"> states.
Hope this helps you, and hope my lack of precision does not make anyone of you throw your laptops through the window!
