# What are the eigenstates of quantum fields?

1. May 17, 2012

### Lapidus

I believe that in qft the particle states are eigenvectors of the Hamiltonian(which also commutes with the number operator), i.e. H|n> = E|n>. A Fock vector is a product of many particle state vectors.

But what are the eigenvectors of psi(x)?

thanks

2. May 17, 2012

### ndung200790

I guess that the eigenfunctions of field operator are wave functions in ''normal'' Quantum Mechanics.

3. May 17, 2012

### Lapidus

thanks for answering ndung!

But I found in this old thread, that quantum fields have no eigenstates! Interesting. Why are the standard textbook not mentioning this?

4. May 17, 2012

### kof9595995

I think what they meant is that the eigenstates are not in Hilbert space, just like in QM the eigenstates of position operator live in a rigged Hilbert space rather than a true Hilbert space. By the same token, if we are not limited in Hilbert space, then the field operators do have eigenstates, which turn out to be coherent states.

5. May 17, 2012

### Lapidus

Ahh, that makes sense. As states with no fixed particle number, coherent states describe the state of a quantum field.

As I am not mistaken, there is (roughly) a uncertainty relation between how precise you can measure a field vs. how accurate you can measure the number of particles that the field is made of. (If someone could that make more precise, that would be much appreciated.)

Along these lines, I digged up another old post from this thread, saying

So the "fluctuations"/ spread of the field are infinite because we take the expectation value in energy eigenstates. These are eigenstates of the Hamiltonian, which is an operator that does not commute with the quantum field operator.

So would there be a similar infinite spread when we would measure the Hamiltonian in coherent states?

thanks again

6. May 17, 2012

### Sonderval

Hey Lapidus,

I just asked two similar questions, but it seems I'm not allowed to link to them, being a newbie here. One was on phi as an observable, one was on the vacuum correlation, both from this week.

Short answer, as far as I understand now: For a real field, the eigenstate is a superposition of all different particle number states. This makes not much sense for a charged particle because this would mean that we have a superposition of different charge numbers.

My current take on "fluctuations" is this:
There is no real sense in talking about fluctuations in the sense of things changing in time or particles popping in and out of the vacuum - the vacuum is Lorentz invariant, and if you some over all this popping-in-and-out Feynman diagrams, you should recover that property.

But consider that zero-dimensional QFT is equivalent to the QM-harmonic oscillator. So at a single point in zero-dimensional QFT, you have a (gaussian) superposition of different field values phi, centered at zero (the mass term is responsible for fixing the maximum value at zero). So instead of a wave function psi(x) as in QM you now have a field value wave function psi(phi), giving a probability amplitude for each field value. This function is stationary.

If you have more than one isolated point, the spatial derivative acts like a coupling of the oscillators at neighbouring points. Still you have a stationary state of what now amounts to a wave functional. Now if you actually measure phi at point x, you'll fix a value for phi (this only works for real fields, as said above) - as in QM, the wavefunction collapses. The current field value at this point will now act like a wave packet. Since it is coupled to neighbouring points via the spatial derivative, it acts on them similar to a source term. As you can see by expanding a differential quotient
$\frac{d\phi_1 - d\phi_0}{dx}$

Hope this makes sense, if it is nonsense, please correct me.

7. May 17, 2012

### ndung200790

I think that there would not be a infinite ''fluctuation'' in Hamintonien in an ''eigenstate'' of field,because the eigenstate is not ''local'' so we have not the ''non-commutative'' considering in above.In each eigenstate the number of particle is fix,but in the superposition of them(of eigenstates) the number of particle is not fix.

8. May 17, 2012

### the_pulp

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 Im 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 Im not sure, I havent 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 Ive 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!!!

9. May 18, 2012

### Sonderval

@the_pulp
You may be right for a real (uncharged) field, but for a complex field, the field operator cannot have eigenvalues. It is a sum of particle-creator and anti-particle destructor (or vice versa). Any state is a superposition of particle-number eigenstates. If you act on any state with a definite number of particles, that cannot be an eigenstate because you increase that number. If your state is a superposition of different particle numbers, there will be a contribution from some state with the smallest number (possibly the vacuum). Applying the field operator will raise the number in that state.
So again the state cannot be an eigenstate.
For a real field it is different because the operator now is the sum of a particle creation and destruction operator - if you look at the same situation for the QM harmonic oscillator, you can see that this operator (the position-operator) is Hermitian.

10. May 18, 2012

### kof9595995

More or less true, but complex field can have eigenvalues if Grassmann numbers are introduced.

11. May 18, 2012

### the_pulp

Interesting! Any reference? or any additional explanation?
It sounds like this also happens with Dirac Fields, am I right?

Thanks!

12. May 18, 2012

### Lapidus

Does that correspond to the coherent states that kof was refering to?

But you can rewrite the Lagrangian of a complex field into a Lagrangian of two real fields! The physics remain the same.

13. May 18, 2012

### kof9595995

Sorry I was referring to Dirac field, shouldn't have said complex field. For complex field the eigenvalues exist and are just complex numbers.

14. May 18, 2012

### Sonderval

The coherent states are a special case, but in principle, any superposition of different partice numbers is an allowed state for a real scalar field. See for example the qftinfo-page by Bob Klauber, ch. 3.

But if you separate the Lagrangian, shouldn't you then also re-write the field operators in a different way? not sure about that.

15. May 20, 2012

### lugita15

So is there such a beast as a "rigged Fock space"? What would it look like?

16. Jun 4, 2012

### A. Neumaier

psi(x) is only an operator-valued distribution, hence not an operator. So talking about its eigenstates is meaningless.

17. Jun 4, 2012

### A. Neumaier

There are multiple possibilities (as with rigging any Hilbert space). In one possibility the bottom space is the space of all superpositions of multiparticle states whose wave functions are infinitely differentiable and have compact support, and whose top space is the dual, and also contains extremely singular superpositions of multiparticle distributions.

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