# B Is there a tension between QM and QFT?

#### ftr

There seems to be some -at least- conceptual difference between particles in QFT which is just a point -eventually- in the field AND the particle in QM which is described by a wavefunction which is extended in space. As if QFT somehow "collapses" the wavefunction.

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#### mikeyork

There seems to be some -at least- conceptual difference between particles in QFT which is just a point -eventually- in the field AND the particle in QM which is described by a wavefunction which is extended in space. As if QFT somehow "collapses" the wavefunction.
A particle in QM is described purely by a set of intrinsic quantum numbers, none of which have anything to do with space-time. Space-time enters the picture only when we attempt to describe the behavior of the particle in the observer's space-time frame. The difference with QFT is that the concept of a field has meaning only in a space-time context.

#### ftr

A particle in QM is described purely by a set of intrinsic quantum numbers, none of which have anything to do with space-time. Space-time enters the picture only when we attempt to describe the behavior of the particle in the observer's space-time frame. The difference with QFT is that the concept of a field has meaning only in a space-time context.
the point I am trying to make is that the particle is a point in QFT in a particular place and that's that, and presumably they still have the same intrinsic quantum numbers.

#### Orodruin

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the point I am trying to make is that the particle is a point in QFT in a particular place and that's that
Where did you read this? This is not what field theory is about, nor quantum field theory.

#### atyy

There seems to be some -at least- conceptual difference between particles in QFT which is just a point -eventually- in the field AND the particle in QM which is described by a wavefunction which is extended in space. As if QFT somehow "collapses" the wavefunction.
There are many types of particles in QFT. A particle is basically a quantized excitation of the field. There are many ways to describe field excitations, so there also correspondingly many types of particles in QFT. Some particles are localized, and others are spread out, like the eigenfunctions of the free particle of non-relativistic QM.

You can find a discussion an analogous discussion of different types of photons in https://books.google.com.sg/books?id=l-l0L8YInA0C&vq=photon&source=gbs_navlinks_s (section 4.6: Quantized radiation states and photons; Complement 4C: Photons in modes other than travelling planes waves; Section 5.4.3: "They might be referred to as quasi-particles states, because they are the quantum states whose properties most closely resemble those of an isolated particle propagating at the speed of light").

Roughly, a particle corresponds to the state created by a creation operator. However, fermion creation operators are usually not Hermitian, so single fermion operators are usually not observables.

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#### ftr

spread out, like the eigenfunctions of the free particle of non-relativistic QM.
but these are generally do not discussed in the established textbooks correct. I have most of those, they typically rehash the same things.

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#### bhobba

Mentor
The difference is quite simple, and there is no tension.

In QFT you can put it in a form similar to the second quantisiation interpretation of ordinary QM - it's one of a number of equivalent formulations:

The difference is the number of particles is not fixed like in ordinary QM - but can itself be in a superposition.

It's of practical importance because you need it to explain things like spontaneous emission which cant be explained otherwise:
http://www.physics.usu.edu/torre/3700_Spring_2015/What_is_a_photon.pdf

Thanks
Bill

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#### vanhees71

Gold Member
the point I am trying to make is that the particle is a point in QFT in a particular place and that's that, and presumably they still have the same intrinsic quantum numbers.
To the contrary! In relativistic QT the QFT formulation is so much more approriate than the 1st-quantization approach, because you cannot localize particles in a more strict sense than already in non-relativistic QM. The reason is that to resolve a particles position you need other particles to scatter with sufficiently large momenta to have the wanted resolution in position. In relativistic QT an ever higher momentum to scatter particles to localize other particles doesn't lead to a better position resolution, because one creates new particles rather then get better position resolution.

In the formalism of relativistic QT this occurs in the known difficulties to interprete Poincare covariant wave equations in a 1st-quantization way: It simply doesn't make sense for interacting particles (interacting of, e.g., charged with an external em. field is already enough!) to work in a one-particle picture, because with some probability new particles are created or initially present particles are destroyed. So the natural way to formulate relativistic QT is in terms of a QFT, working with a Hilbert space of indefinite particle number.

Another formal hint about the problematic issue of position is that relativistic QT admits the possibility of massless particles (which is not so in non-relativistic physics, where massless particles just don't have a sensible dynamics), and massless particles with a spin $\geq 1$ don't admit the construction of a position operator in the strict sense, i.e., for a massless particle like the photon you cannot even define a position observable to begin with!

Another point is that within non-relativistic QT in the cases, where you deal with a fixed number of particles like in atomic physics, where you have a fixed number of electrons around a nucleus, the first-quantization formulation and the second-quantization formulation (the latter is just non-relativistic QFT) are equivalent. It's just writing the same theory in different mathematical terms, but the physics is completely the same.

"Is there a tension between QM and QFT?"

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