# QFT confusion

1. Sep 2, 2011

### Hymne

Hello!
Im trying to do QFT on my own and its going fine.. except one confusion now.

We have our operator fields corresponding to our observables, and our state which is a function of space and time.

But doing the second quantization we get the creation and destruction operators which now take the place in the Klein-Gordon solution that our coefficients a(k) and a^(+)(k) for the planewaves use to have.
So the Klein-Gordon solution is not our state but an operator field?

Edit: Maybe I got it now. generalizing a(k) and a^(+)(k) to operators the KG solution gives the general form of a operator that when acting on |0> gives a wavefuntion solution?

Last edited: Sep 2, 2011
2. Sep 2, 2011

### Polyrhythmic

You're right. The physical state is given by the creation/annihilation operators acting on the vacuum.

3. Sep 2, 2011

### vanhees71

The confusion comes from learning quantum mechanics from an approach which is dominated by the wave-function representation (i.e., the position-space representation). This is a bad starting point for learning quantum-field theory.

You should first become familiar with the representation-free formulation of quantum mechanics. This is slightly more abstract than thinking in terms of "wave mechanics", but at the same time it clarifies the mathematical structure of quantum theory and finally the picture of nature quantum theory paints of the physical world.

A (pure) state of a quantum system is represented by vectors $|\psi \rangle$ in a Hilbert space and the observables by self-adjoint operators acting on these state vectors. The outcome of a (precise) measurement of an observable is always an eigenvalue of the operator representing this observable.

The physical interpretation of the state vectors is then given by Born's probability postulate, i.e., that for a system prepared in a (pure) state, represented by the state vector, $|\psi \rangle$, the probability to find the value $\alpha$ when measuring an observable is given by $|\langle \alpha|\psi \rangle|^2$, where $|\alpha \rangle$ is the eigenvector of the observable's representing operator for the eigenvalue, $\alpha$. Here I simplified things a bit and assumed that for any eigenvalue of the operator there exists only one linearly independent eigenvector (non-degeneracy of the observable). If there is degeneracy for some eigenvalues of the operator you have to measure more than one observable, and to have a complete measurement you have to measure a complete set of compatible observables. Such a set is defined as being represented by a set of selfadjoint operators commuting with each other, but this is not so important to understand the principle structure of QT.

Now, in QFT you look primarily at many-body systems, where the number of particles is not specified. Stated differently, you look at all quantum systems consisting of an arbitrary not specified number of identical particles (e.g., electrons) (or several types of such identical particles, e.g., electrons and protons). Thus the Hilbert space of states consists of all vectors describing "the vacuum" (no particles present), one particle, two particles, and so on.

Now the indistinguishability of particles leads to the idea of fermions and bosons, i.e., you can build a basis in the $N$-particle space by a complete set of antisymmetrized for fermions (symmetrized for bosons) product vectors of $N$ single-particle states. This is much easier achieved by defining anticommuting (fermions) or commuting (bosons) creation and annihilation operators (also simply called field operators) with respect to the chosen single-particle basis (e.g., single-particle momentum eigenstates), because then you can incorporate the antisymmetry/symmetry of the N-particle basis states within these creation and annihilation operators.

The trick is now, that you can build the operators representing observables of your N-particle systems in terms of these field operators, and this gives you quantum-field theory as a convenient description of many-body systems.

It of course applies also to the case of a few particles, as in relativistic quantum field theory when describing the collision of elementary particles. But here the qft description is more or less even a necessity since in relativistic high-energy-particle physics you always may create new particles and/or destroy the original particles colliding with each other.

4. Sep 5, 2011

### Hymne

I´m sorry I am not quite sure I still get it. Can we summarize qft concepts maybe?

In QFT...

.. a state is still a wavefunction, complex or scalar, defined over spacetime.
However when solving the eqn of motion we get an operator field(/quantum field) that needs to act on our vacuum state to actually give us the wavesolution.

.. an observable is represented by a operator field.

In QM an operator acting on a statevector gives a new statevector. Should a operator field acting on a state (for example the vacuumstate) yield a new state?

I feel like theres a big intuition gap here that the books I´m using doesnt cover.

5. Sep 5, 2011

### vanhees71

Again, states are not necessarily represented by wave functions. In QFT the Hilbert space is the Fock space, and it's even quite unnatural (at least to me) to represent them as sets of wave functions. You should first get familiar with the abstract Hilbert-space formalism (or Dirac's bra-ket formalism which is the most convenient notation) in quantum mechanics before learning QFT.

6. Sep 5, 2011

### Hymne

I am familiar with all you wrote earlier and the bra-ket notation. I have taken a course in advanced QM which included Wigner-Eckart thm, crossection, 2nd quantization and all in between..

maybe I am mixing up PI-approach and Canonical quantization approach to QTF, I dont konw.. but when we author is talking about states it seem that i can be both a quantum field or a state in the QM sense.

I need a summarize about what each concept mathematically corresponds to in each approach.

7. Sep 5, 2011

### DrDu

"We have our operator fields corresponding to our observables, and our state which is a function of space and time. "

Before second quantization you have a classical wave which fulfills the KG equation and the observables aren't operators.
Doing the quantization there are many states corresponding to a KG wave, which contain a varying amount of quanta.

8. Sep 5, 2011

### vanhees71

Ok, then perhaps it helps you, if I say that what is called "second quantization" in non-relativistic quantum theory is nothing else than non-relativistic quantum-field theory. Of course "second quantization" is a misnomer from the early days of modern quantum theory. There's only one quantum theory after all, and the socalled "first quantization" formalism is only a special case for situations, where the particle number is fixed. In that case first and second quantization formalism are equivalent. In the case of creation and destruction of particles the first-quantization formalism is simply inapplicable.

9. Sep 5, 2011

### Hymne

double post

10. Sep 5, 2011

### Hymne

Maybe I missunderstood all of classical field theory, but the observables are operators before, what about the energy-momentum tensor? or angularmomentum tensor?

11. Sep 5, 2011

### DrDu

They are functions of the field, but not operators.

12. Sep 5, 2011

### Hymne

Ah right, functionals that is.

So in classical field theory we have:

Observable - Functional

State - Real, scalar or even vector field defined over spacetime​

After 2nd quantization we get Quantum Field Theory in which we have:

Observable - Quantum Field / Operator Field / "Field"

State - Real, scalar or vectorfield that can be constucted by an operator field acting on a state, lets say the vacuum state?​

13. Sep 5, 2011

### Hymne

I thought I knew how this worked.. second quantization just appeared as a trick to create and destroy particles with help of operators that can be formulated by or express our usual operators, energy, momentum etc.

But then the author says that these creation and destruction operators also are our coefficients in our state for the Klein Gordon solution - which means that 2nd quantization made our states from wavefunctions to quantum fields. We are integrating over opertors..

14. Sep 6, 2011

### DrDu

Just one point: In the KG case, the field itself is observable, not only the quantities energy, momentum etc you mention.

15. Sep 6, 2011

### kith

No. In QFT, the fields itself are observables. So the Dirac equation is an operator equation. The states are vectors in Hilber space, just like in ordinary QM. The evolution of states is still governed by the Schrödinger equation and H can be expressed via creation and annihilation operators like every other operator.

I think some of the confusion originates from the fact that in relativistic wave mechanics (not QFT), the Schrödinger equation for the wavefunction is replaced by the Dirac equation and therefor describes the evolution of wavefunctions in this formalism. But relativistic wave mechanics is not a consistent theory, we need QFT to really include relativity in quantum mechanics.

16. Sep 6, 2011

### The_Duck

No. Forget about states as wave functions of spacetime. States are vectors in Hilbert space. The physics is determined by the Hamiltonian operator. For both a classical and a quantum scalar field theory, the Hamiltonian looks something like

$$H = \int d^3x \frac{1}{2}(\dot{\phi}^2 + (\vec{\nabla}\phi)^2 + m^2 \phi^2)$$

H is an operator in the quantum case, therefore so is phi. In the Heisenberg picture, phi is a function of time as well as space. Plugging phi into the Heisenberg equation of motion for operators we find that phi obeys the classical equation of motion. And of course, in the Heisenberg picture the state (vector) is static and unchanging in time. The other operators we care about, like the momentum and angular momentum, are functions of phi just like H is.

You can use Fourier expansion to write phi as something like
$$\int \frac{d^3k}{(2 \pi)^3}(a(k)e^{ikx} + a^\dagger(k)e^{-ikx})$$
where a(k) is some operator function of k. *Any* Hermitian operator can be written like this for some a(k). If you like you can invert this expression and write a(k) explicitly in terms of phi(x). Defining these a(k) is useful because it turns out that you can rewrite the Hamiltonian in terms of a(k) as something like
$$\int \frac{d^3k}{(2 \pi)^3}\sqrt{k^2 + m^2}(a^\dagger(k)a(k) + \frac{1}{2})$$
which is an integral over a bunch of harmonic oscillator Hamiltonians. As a consequence we can define a state |0> which is annihilated by all the a(k)'s and therefore has minimum energy. Call this the vacuum state. From the above form of H, any state $$a^\dagger(k)|0>$$ formed by hitting the vacuum with an annihilation operator is a state of definite energy, and has the energy expected of a particle of mass m and momentum k. We can confirm that this state has momentum k by hitting it with the momentum operator, which is found via Noether's theorem. Like H, the momentum operator has a particularly simple form in terms of a.

There's no wave function. Hitting the vacuum with the field operator phi(x) gives a very specific state. Perhaps you can convince yourself that at time x_0 this state is localized at the spatial position (x_1, x_2, x_3).

17. Sep 7, 2011

### vanhees71

More precisely stated: It does not generate a state-representing vector in Hilbert space but a generalized state. A true state vector must be normalizable to 1, which is not the case for the generalized ket $\hat{\phi}(x) |\Omega \rangle$.

18. Sep 7, 2011

### kof9595995

Fields may not be observables, e.g. the fundamental field of Dirac is not even Hermitian. I think it's more appropriate to say fields are convenient tools of constructing other operators.

19. Sep 7, 2011

### kith

Of course you're right!

20. Sep 7, 2011

### DrDu

Bosonic fields are often observable. If a field generator or anihilator is not observable, this is usually due to the field carrying some kind of charge which induces a superselection rule.