# What are the states in QFT?

1. Mar 21, 2010

### Azrael84

Hello,

this is quite a basic question I know, but something I'm not sure I've fully got my head around. In classical particle mechanics the dynamical variable is the position vector x, and in classical field theory the dynamical variable becomes the field $$\phi(x)$$, with x being relagated to just a label to each of the infinitity of these dynamical variables $$\phi$$ at each point in space if you like.

Then in regualar QM, the things that were dynamical variables in classical particle mechanics (position, momentum etc...) get promoted to operators, with the actual states being abstract vectors in the hilbert space of these operators. In QFT the dynamical variable from classical field theory, the field itself, gets promoted to an operator. But what are the analogues of the eigenvalues of position/momentum/angular momentum etc from regualar QM?

Also I noticed that in QM the solutions of the Schrodinger equation are the actual states themselves, but in QFT the solutions of the KG eqn or the Dirac equation say, are the field operators not the states. So what are the actual states? are they somehow abstract vectors of the field operators?

2. Mar 21, 2010

### meopemuk

QFT is not fundamentally different from quantum mechanics. In QFT we also have a Hilbert space (it is called the Fock space) and states of the system are represented by vectors in this Hilbert space. The most significant difference is that the number of particles is not fixed in QFT. So the Fock space is a direct sum of subspaces with 0 particles (vacuum), 1 particle, 2 particles, etc. Thus in the Fock space we can describe processes in which the number of particles can change (radiation, decays, etc.). So, basically QFT is the same as "QM with variable number of particles".

Quantum fields have nothing to do with wave functions or states. Quantum fields are just certain operators in the Fock space which are convenient "building blocks" for construction of other more physical operators. For example, interaction terms in the Hamiltonian are usually constructed as products of several field operators.

Eugene.

3. Mar 22, 2010

### Fredrik

Staff Emeritus
The Hilbert space of one-particle states of a QFT without interactions can be constructed explicitly by taking the vectors to be equivalence classes of square integrable positive-frequency solutions of the classical field equation. But it doesn't really matter what the Hilbert space is, since all separable infinite-dimensional Hilbert spaces are isomorphic to each other. I suppose the point of the explicit construction is that it makes it easy to explicitly construct the generators of (an irreducible representation of) the Poincaré group (i.e. the (four-)momentum, angular momentum and boost operators) from the quantum field.

The Hilbert space of many-particle states is constructed as the Fock space of the Hilbert space of one-particle states.

When there are interactions in the theory, things get really complicated, and I don't understand this well enough to comment. (The only person here who really seems to understand these things is DarMM).

4. Mar 22, 2010

### meopemuk

You are right, QFT gets really complicated (physical particles are not the same as bare particles; the need for renormalization, etc) if interaction in the Hamiltonian has terms which create additional particles in 0-particle and 1-particle states. In the creation/annihilation operator notation these troublesome interaction terms may look like a*a*a or a*a*a*.

There is no good reason why such bad terms should be present in realistic Hamiltonians. One can build successfult QFT theories without these bad interactions. For example, there is a "dressed particle" version of QED, in which the simplest interaction term has the form a*a*aa. In this theory particles cannot appear spontaneously out of vacuum or 1-particle states. There is no difference between physical particles and bare particles. There is no need for renormalization. The entire theory is not more complicated than standard quantum mechanics.

Eugene.

5. Mar 22, 2010

### samalkhaiat

This is also true in QFT. They are called wavefunctional, $\Psi[\psi] = <\psi|\Psi(t)>$ and the Schrodinger equation of QFT is the following functional differential equation

$$i\partial_{t}\Psi[\psi] = \int d^{3}x \mathcal{H}(\psi , \frac{\delta}{\delta \psi(x)})\Psi[\psi]$$

Solving this equation can reproduce all QFT results. However, that is a very hard job.

There is a deep mathematical reason for that. The KG "field" and Dirac spinor can not be regarded as wavefunctions in Hilbert space but fields on Minkowski spacetime. The reason for this is the NON-COMPACT nature of the Poincare' group. All FINITE-DIMENSIONAL
irreducible representations of any non-compact group are NOT UNITARY. Therefore these IR representations can not be carried by functions on Hilbert space, but by functions on spacetime where non-unitarty does not cause any problem.

regards

sam

6. Mar 23, 2010

### meopemuk

samalkhaiat,

From what you've said I guess that the wavefunctional $\Psi[\psi] = <\psi|\Psi(t)>$ cannot be considered as a probability amplitude. Quantum field $\Psi(t)$ transfrorms by a non-unitary representation of the Poincare group. Therefore the total probability (which must be equal 1 in all reference frames) is not necessarily preserved in this representation. Is that correct?

Eugene.

7. Mar 23, 2010

### hamster143

Isomorphic does not mean homeomorphic. Just like R is isomorphic to R^2, but not homeomorphic to it. Hilbert space isomorphisms are useless if they are not homeomorphisms.

8. Mar 23, 2010

### hamster143

$\Psi[\psi(x)] = <\psi(x)|\Psi(x,t)>$ denotes an amplitude to find the wavefunctional in the state $\psi(x)$ everywhere on a hypersurface t=const. It is inherently Lorentz non-invariant. Since the hypersurface is not invariant under general Poincare transform, I'm not sure if the concept of representation even applies here.

If you work with a subset of Poincare group that preserves the hypersurface (it consists of translations and SO(3) rotations), the probability is certainly conserved

9. Mar 23, 2010

### Azrael84

So how do these wavefunctionals compare to the usual $$\mid 0\rangle$$, $$\mid k_1\rangle$$ etc etc, type of states that one normally sees when learning QFT? are they the same thing? (is it just like in QM, where one has the wavefunction as the position representation of the abstract vector, here the wavefunctional would be the "$$\phi$$ representation"?).

Why do these wavefunctionals obey the Schroedinger equation too? and not a Lorentz invariant equation like KG? I don't understand how the non-relativistic Schroedinger can have a role in QFT.

10. Mar 23, 2010

### Fredrik

Staff Emeritus
It does when we're talking about Hilbert space isomorphisms. They preserve both the vector space structure and the inner product. The latter requirement implies that they're bounded, and that implies that they must be continuous. Their inverses must be continuous too. So Hilbert space isomorphisms are homeomorphisms.

R and R^2 is a strange example, since they clearly aren't vector space isomorphic. You must have meant that they're isomorphic sets, i.e. that there's a bijection from one of the sets into the other, but that's not the kind of "isomorphism" we're interested in.

Last edited: Mar 23, 2010
11. Mar 23, 2010

### Fredrik

Staff Emeritus
There must exist a time translation operator which satisfies $U(t_1+t_2)=U(t_1)U(t_2)$ and this gives us $$U(t)=e^{-iHt}$$. This last equation is the definition of the Hamiltonian in both special relativistic and non-relativistic QM. States satisfy the Schrödinger equation because the time evolution operator does:

$$i\frac{d}{dt}|\psi(t)\rangle=i\frac{d}{dt}\Big(e^{-iHt}|\psi\rangle\Big)=He^{-iHt}|\psi\rangle=H|\psi(t)\rangle$$

The difference between special relativistic and non-relativistic QM is that we have to consider representations of the Poincaré group instead of the Galilei group, but the time translation group is a subgroup of both.

12. Mar 23, 2010

### Azrael84

Thanks.

13. Mar 23, 2010

### Azrael84

I was already aware of the Fock space being the analogue of the Hilbert space in QFT when I originally posted. I guess what I was really wondering is what are the states in the "non-abstract sense"? e.g. in QM you could choose to work in the position rep, express the operators of your Hilbert space in their position rep, and have wavefunctions that are dependent on space. So what I'm wondering is can you do something equivalent in QFT....after so called "second quantization", the operators P, X from QM have been relagated to labels, and the things satisfying the Schroedinger equation (or the KG or Dirac equation), that were our states in QM (Wavefunctions or more abstract Hilbert vectors) have been promoted to operators. But the new states of these field operators, are at the moment just abstract Fock vectors in my mind at the moment, I can't quite see how to visualize them, and am still left with the question what actually are they?

They are not configurations of the field, since the field is the operator, but then what are they....

Perhaps then they are these "wavefunctionals", things are essentially functions of the field config which is itself a function of spacetime. But I'm having difficultyl reconciling that view with the fock kets notion.

14. Mar 23, 2010

### meopemuk

Since boosts do not preserve "hypersurfaces", does that mean that for a moving observer the total probability of events does not necessarily add up to 1? Does that mean that probability is not longer a physically relevant quantity in QFT?

Eugene.

15. Mar 23, 2010

### meopemuk

Azrael84,

You can visualize QFT states and wavefunctions just as easily as you can do that in usual quantum mechanics.

The Fock space of QFT is a direct sum of subspaces (or sectors) with different number of particles: 0, 1, 2, 3,... Take for example the 2-particle sector. In this sector you can define operators of observables (position, momentum, spin, etc.) of both particles exactly as you do that in any 2-particle problem in QM. You can also define eigenvectors/eigenvalues of these operators. So, you can form corresponding orthonormal bases and define wavefunctions in any convenient representation (position, momentum, etc.).

The same things can be done in any N-particle sector of the Fock space. This sector looks the same as the Hilbert space of N-particle quantum mechanics. The only important difference between QFT QM is that in the QFT Fock space you can now consider states which do not have a definite number of particles. Such states can have non-zero projections on sectors with different N. Wave functions of such a states can be written as collections of N-particle wave functions with different N.

Eugene.

16. Mar 23, 2010

### hamster143

That means that states do not transform in a trivial way.

In translations, the transformation law is simply $\Psi'[\psi(x)] = \Psi[\psi(x+\Delta)]$. (Since psi is an arbitrary function, what we have here is an infinite-dimensional representation of the symmetry group. Most people would be scared at this point, but we'll move on.)

In rotations, we have to consider the possibility that psi has spin and different components go into one another under rotation. That's still fairly easy.

To compute the outcome after a boost, we have to compute the wavefunction on the "new" hypersurface from the wavefunction on the "old" hypersurface. To do that, we need dynamics of the field and the exact Hamiltonian.

To correct my earlier remark, I think that the transformation law is still a unitary representation, but it's not like any usual representations we normally see.

17. Mar 23, 2010

### Azrael84

Thanks. You say "eingenvectors/eigenvalues of these operators" but what operators? the usual momentum/position operators of the fixed dimensional Hilbert space of QM, have been turned into labels. The only operators I'm aware of the field itself $$\phi$$, and I have no idea what eigenvalues eigenvectors of this actually are, or how to view them in a particular representation.
How do you define the momentum operator in say the 3-particle sector of the Fock space?

18. Mar 23, 2010

### meopemuk

Let us start with the 1-particle sector. This subspace carries a unitary irreducible representation of the Poincare group (see Weinberg, vol. 1 chapter 2). This representation allows us to define all relevant 1-particle operators (momentum, spin, position, mass, etc.) there.

The 3-particle sector is a (appropriately symmetrized or antisymmetrized) tensor product of 3 one-particle Hilbert spaces. We've already build 1-particle observables in each 1-particle Hilbert space. These observables transfer to the 3-particle sector according to the mapping associated with the tensor product construction. So, there is no difficulty in defining 1-particle momentum operators p1, p2, and p3. These operators commute with each other. So, one can find their common eigenvectors and define wave functions in the momentum representation. This is not different from ordinary quantum mechanics.

Eugene.

19. Mar 23, 2010

### Azrael84

I don't have access to Weinberg right now, but I will check it out when I get chance. For now could you tell me then what the explicit form of the, say momentum, operator is then in the 1-particle sector, in say, position rep? Because it surely isnt just $$\hat{P}=-i\partial_x$$ like in ordinary QM position rep. Also for that matter what is the explicit form of the state $$\mid p \rangle$$ in the position rep, in the 1-particle sector.

20. Mar 23, 2010

### meopemuk

Why not? The momentum operator will be exactly as you wrote it.

It will be the usual plane wave.

For more details you might find useful chapter 5 in http://www.arxiv.org/abs/physics/0504062

Eugene.