Why is position just a label in QFT?

[LarryS]

TL;DR Summary

Supposedly position is just a label in QFT. Yet there are areas within QFT where it is treated like a real physical point in space.

At the beginning of every course in QFT we are told that, unlike in ordinary QM in which the position variable is a physical observable , the position variable in QFT is just a label.

Yet there are areas within QFT where the position variable is treated like a real physical degree of freedom, not just a label. For example, there exist a field Creation operator to create a particle at a point (a real physical point in space) and this operator is related to the momentum Creation operator via a kind of Fourier Decomposition (momentum is certainly not just a label). Another example is how the position variable is related (QED) to a Propagation Operator that moves a particle from one (real physical) point to another.

To me, the above 2 examples seem to imply that in QFT position is somewhat more than “just a label”.

Comments?

 

[A. Neumaier]

Position in QFT always denotes a point in a specified coordinate system.

''Just a label'' means ''not an operator'' and ''coordinate system dependent'', not ''nothing physical''. This is in contrast to QM, where position is a (coordinate system dependent) operator and only time is just a label.

 

[Elias1960]

To answer the question "Why is position just a label in QFT?":

Once time in QM is nothing but a label (there is no operator for time measurement and each clock has even some nonzero probability to go backward in time), and most physicists would prefer to have a manifestly covariant theory, they have no choice but to give the position a "label only" status too.

This requires also a preference for the Heisenberg formalism in comparison with the Schrödinger formalism. So you have operators $O(x,t)$. In the Schrödinger formalism, you would have a configuration defined by a classical field configuration u(x), the wave function would be a functional on this space of functions, $\Psi(u(.))$, and it would change in time $\Psi(u(.),t)$. Nothing which looks in any way relativistic.

This does not mean that there is any difference to usual QM, except that the infinite number of degrees of freedom leads to a lot of infinities. All the known problems of various interpretations, like the violation of Bell inequalities, or the measurement problem, remain unchanged. And the Schrödinger formalism is as good as in QM too. But what will be in the textbooks is the Heisenberg formalism, given that it looks more Lorentz-covariant.

 

[vanhees71]

It's one of the great achievements of theoretical physics to have clarified that relativistic QT can be constructed by investigating the unitary representations of the proper orthochronous Poincare group (Wigner 1939). That's why in relativistic QFT you start with the fundamental conserved quantities of special-relativistic space-time (Minkowski space), i.e., energy, momentum (space-time translations), angular momentum (spatial rotations), and boost generators.

It turns out that you can a posteriori construct position operators (and thus define position observables) for all kinds of massive fields. In the massless case proper position operators only exists for particles with spin $\leq 1/2$.

In the case of non-relativistic QT, it's more simple and thus also usually presented in introductory textbooks, to start with position and momentum operators and the Heisenberg algebra and from this to build up also the unitary ray representations of the Galilei group. The fact that the proper unitary representations of the Galilei don't lead to useful quantum dynamics makes the space-time-group approach a bit more cumbersome than in the case of Minkowski space. The reason is that the proper orthochronous Poincare group has no non-trivial central charges, while for the Galilei group mass occurs as a central charge. That's why mass is so special in non-relativistic physics, particularly it's also conserved though it's only indirectly derived from the unerlying Gaililei symmetry, namely as a central charge.

 

[Demystifier]

Summary: Supposedly position is just a label in QFT. Yet there are areas within QFT where it is treated like a real physical point in space.

At the beginning of every course in QFT we are told that, unlike in ordinary QM in which the position variable is a physical observable , the position variable in QFT is just a label.

Yet there are areas within QFT where the position variable is treated like a real physical degree of freedom, not just a label. For example, there exist a field Creation operator to create a particle at a point (a real physical point in space) and this operator is related to the momentum Creation operator via a kind of Fourier Decomposition (momentum is certainly not just a label). Another example is how the position variable is related (QED) to a Propagation Operator that moves a particle from one (real physical) point to another.

To me, the above 2 examples seem to imply that in QFT position is somewhat more than “just a label”.

Comments?

I think the confusion originates from the confusion about the meaning of the quantity $\phi({\bf x},t)$. It can mean two things. It can be a wave function that represents the state in the Hilbert space, or it can be a hermitian operator that acts on the Hilbert space. In the latter case ${\bf x}$ is just a label, while in the former case (especially if we talk about nonrelativistic QFT) ${\bf x}$ can be a physical position of the particle. To distinguish those, it is convenient to write the operator as $\hat{\phi}({\bf x},t)$ and the wave function as $\psi({\bf x},t)$. Then, in the simplest case, their relation is given by the formula
$$\psi({\bf x},t) = \langle 0|\hat{\phi}({\bf x},t)|\psi\rangle$$

 

[vanhees71]

In relativistic physics though $\hat{\phi}$ contains creation and annihilation operators in the free-field mode decomposition, referring to "particles" (positive-frequency term with annihilation operator) and "anti-particles" (negative-frequency term with creation operator). The above formula is thus correct if $|\psi \rangle$ is a single-particle state, but it's not a one-to-one relation between $|\psi \rangle$ in Fock space to wave functions $\psi(\vec{x},t)$. E.g., if $|\psi \rangle$ is a coherent state, you also get a single-particle wave function $\psi(\vec{x},t)$.

 

[Demystifier]

The above formula is thus correct if $|\psi \rangle$ is a single-particle state,

Yes, that's what I meant by the "simplest case".