# Quantum Field Theory

From a previous post, I now know that the 'qualities' of QM such as entanglement, superposition of states, decoherence, measurement problem, are still with us in QFT.

1. Is the Dirac equation the primary equation in QFT?
2. What equation governs the evolution of a macroscopic object in QFT?
3.
In quantum field theory, unlike in quantum mechanics, position is not an observable, and thus, one does not need the concept of a position-space probability density.
How does one calculate the superposition of positions for a microscopic or macroscopic object in QFT? Does that equation evolve in the same way as the Schrodinger Equation?

I found this http://en.wikipedia.org/wiki/Newton–Wigner_localization, but doesn't that only apply to single particle? Is it even an equation?

and one final question:
How do you write all the possible states of a quantum system in QFT?
e.g. I just started playing with QFT, but I can say this much: the Dirac equation isn't the fundamental equation. It describes the electron (or positron), relativistically. But it doesn't describe other particles.

You might try taking a flip through Zee's book: Quantum Field Theory in a Nutshell. Its very readable and you can get a general flavor out of it even if you're not working the problems (to a certain degree).

A. Neumaier
1. Is the Dirac equation the primary equation in QFT?
No. The primary equations are the operator version of a coupled Dirac and Maxwell equations. This defines (in some sense) QED.
http://en.wikipedia.org/wiki/Quantum_electrodynamics#Equations_of_motion
The Dirac equation is a warm-up toy describing a single electron or positron, not a quantum field.
2. What equation governs the evolution of a macroscopic object in QFT?
A macroscopic object is simply a few quantum fields with macroscopic mass density.
Condensed matter physics studies how macroscopic objects are modelled. http://en.wikipedia.org/wiki/Condensed_matter
3. How does one calculate the superposition of positions for a microscopic or macroscopic object in QFT?
Positions have no superpositions. Only states can be superimposed. The basic states in perturbative QFT (i.e., what you usually meet in textbooks) are the states in Fock space. http://en.wikipedia.org/wiki/Fock_space .A serious discussion of QFT is impossible without having understood how to work with such states. Get some practice and then ask more!

Positions have no superpositions. Only states can be superimposed. The basic states in perturbative QFT (i.e., what you usually meet in textbooks) are the states in Fock space. http://en.wikipedia.org/wiki/Fock_space .A serious discussion of QFT is impossible without having understood how to work with such states. Get some practice and then ask more!
Hmmm, I guess that is what confuses me, when in standard QM you have superposition of positions, but no longer in QFT! Or is it there are still superposition of positions, but they aren't dealt with in QFT?

A. Neumaier
Hmmm, I guess that is what confuses me, when in standard QM you have superposition of positions, but no longer in QFT! Or is it there are still superposition of positions, but they aren't dealt with in QFT?
It would be better first to get a good working knowledge of the basic QM and clear up your confusions there, before thinking about how one can do things in QFT.

Even in QM, one only has superpositions of states, not of positions.
You can have a superposition of an up-state and a down-state, but not a superposition |x>+|y> with x and y being positions, since this is not a normalizable wave function.

One can have a bilocal state of a particle, though, with psi(x) concentrated near two positions. This would be represented in QFT as the 1-particle state
$$\psi=\int dx \psi(x)|x\rangle$$

Now I'm really confused:
In quantum field theory, unlike in quantum mechanics, position is not an observable
= position is an observable in Quantum Mechanics

I thought observables are superimposed in QM?

Yes, position is no longer an operator in QFT like it was in nonrelativistic quantum mechanics. Why do you think observables are superimposed in QM? A particle might be in a superposition of states that are eigenstates of some observable, is that what you're thinking?

A. Neumaier
I thought observables are superimposed in QM?

One can have a bilocal state of a particle, though, with psi(x) concentrated near two positions. This would be represented in QFT as the 1-particle state
$$\psi=\int dx \psi(x)|x\rangle$$
What does $|x\rangle$ mean in QFT? I didn't think there was an X operator. From my understanding, the closest thing to $| x \rangle$ would be something like $\phi^\dag(x)|0\rangle$.

A. Neumaier
What does $|x\rangle$ mean in QFT? I didn't think there was an X operator. From my understanding, the closest thing to $| x \rangle$ would be something like $\phi^\dag(x)|0\rangle$.