Particles' Intrinsic Properties in QFT

[LarryS]

TL;DR

Intrinsic vs extrinsic properties in QFT?

In single-particle nonrelativistic QM, a particle, say an electron, has intrinsic properties like rest mass, charge and spin magnitude and extrinsic/dynamic properties like momentum and energy. Does this intrinsic/extrinsic dichotomy carry over to, say, the electron field in QFT?

Thanks in advance.

 

[Demystifier]

I don't think that intrinsic and extrinsic are the right words for what you are trying to say. Perhaps better words are invariant and non-invariant. But yes, this dichotomy carries over to QFT.

 

[LarryS]

Are the invariant properties part of the unexcited (vacuum state) state of the field?

 

[Roberto Pavani]

@Demystifier nailed it. Perhaps the simplest way to see why: energy and momentum are frame-dependent, so they can't be intrinsic to the particle.

 

[Matterwave]

Are the invariant properties part of the unexcited (vacuum state) state of the field?

Define "part of"?

As I read the question, I would say "no". The vacuum state itself does not posses charge, mass, spin, etc.

But also QFT is "not my thing" so hopefully someone more knowledgeable will chime in and correct me if I'm wrong. :)

 

[Demystifier]

The vacuum state itself does not posses charge, mass, spin, etc.

I would say it has zero charge, mass, spin, etc.

 

[gentzen]

@Demystifier nailed it. Perhaps the simplest way to see why: energy and momentum are frame-dependent, so they can't be intrinsic to the particle.

So you mean, only a scalar property can be intrinsic to the particle?
I don't see anything wrong with a Lorentz covariant vector like the four-momentum.
Of course a scalar property is frame-invariant, but I see nothing wrong with frame-covariant properties.

 

[Roberto Pavani]

You're right that the 4-momentum is covariant. But its only frame-invariant content is the norm $p_\mu p^\mu = -m^2$, which is just the mass.
The direction of $p^\mu$ (i.e., the velocity) is frame-dependent. So the intrinsic information carried by the 4-vector reduces to a scalar: $m$.

 

[gentzen]

The direction of $p^\mu$ (i.e., the velocity) is frame-dependent. So the intrinsic information carried by the 4-vector reduces to a scalar: $m$.

I see. But then it mostly boils down to how one wants to interpret the word "intrinsic" (including connotations) and how one wants to denote "frame-covariant properties" with another word like "extrinsic" with other appropriate connotations.

 

[Matterwave]

I would say it has zero charge, mass, spin, etc.

Fair enough~

 

[PeterDonis]

Are the invariant properties part of the unexcited (vacuum state) state of the field?

They're not part of any state. They're part of the mathematical description of the field--more precisely, they are properties of the group representation that is chosen to describe the field. Quantum fields are operators, not states.

The vacuum state is the state $\ket{\psi}$ that satisfies $a \ket{\psi} = 0$ for all annihilation operators $a$. So it tells you nothing about the specific properties of any particular operator.

 

[Matterwave]

Expanding a little. States in QFT live in a Hilbert space. The irreducible subspaces of that Hilbert space has certain symmetries represented by operators acting on that irreducible space. Those symmetries are representated by elements of Lie Groups (the poincare group for spacetime symmetries, unitary groups for internal "gauge" symmetries). The Lie Algebra of such a lie group gives rise to certain observables/operators, e.g. angular momentum (note: the lie algebra elements are also called generators of the Lie Group as the exponential map generates the connected identity element of the group from the algebra). A Casimir operator is an operator (technically in the universal enveloping algebra of the Lie Algebra) which commutes with all other generators and is shown by Schurs lemma to be a multiple of the identity in the irreducible space. The intrinsic properties you talk about like Mass, and Spin are eigenvalues of some Casimir operator. Charge arises from internal gauge symmetries vs. spacetime symmetry which gives rise to mass and spin.

This is as much a response I wrote for myself as for the OP. I would take it with a grain of salt as I am by no means an expert in QFT. I would appreciate corrections and comments.

 

[bhobba]

They're not part of any state. They're part of the mathematical description of the field--more precisely, they are properties of the group representation that is chosen to describe the field. Quantum fields are operators, not states.

Interestingly, in the non-relativistic limit of QFT, they still remain operators.

It is possible to formally recover the states of ordinary QM as found in QFT, The Why, What, and How by T Padmanabhan on page 39. But it is non-trivial and, as the author notes, 'is a reminder that the single particle description is not easy to incorporate into relativistic quantum theory' (direct quote). While formally true, the author also believes the physical meaning behind it is not well understood.

Thanks
Bill

 

[PeroK]

So you mean, only a scalar property can be intrinsic to the particle?
I don't see anything wrong with a Lorentz covariant vector like the four-momentum.
Of course a scalar property is frame-invariant, but I see nothing wrong with frame-covariant properties.

In the standard model how would you indicate the intrinsic four-momentum of each particle?

 

[PeterDonis]

I see nothing wrong with frame-covariant properties.

But such properties aren't properties of the particle alone; they're properties of the particle combined with a particular choice of frame. Which means that "intrinsic to the particle" doesn't seem like a good way of describing them.

 

[gentzen]

But such properties aren't properties of the particle alone; they're properties of the particle combined with a particular choice of frame.

As I said, I don't see why "frame-covariant properties" should be considered to depend on "a particular choice of frame".

Which means that "intrinsic to the particle" doesn't seem like a good way of describing them.

I already agreed to that part above:

But then it mostly boils down to how one wants to interpret the word "intrinsic" (including connotations) and how one wants to denote "frame-covariant properties" with another word like "extrinsic" with other appropriate connotations.

 

[gentzen]

In the standard model how would you indicate the intrinsic four-momentum of each particle?

What is your point? I already agreed with Roberto Pavani that one can interpret the word "intrinsic" (including connotations) in a way which excludes "frame-covariant properties".

But if I have an electron, then I do think of its four-momentum as a property of the electron.

 

[PeterDonis]

if I have an electron, then I do think of its four-momentum as a property of the electron.

The problem is that there is no single mathematical object that is referred to by "its four-momentum". Physicists, with their usual sloppiness about mathematical rigor (which I fully sympathize with, since I do it myself plenty of times), refer to an object's 4-momentum as a 4-vector and call it a single "frame-covariant" (to use your term) geometric object. But that doesn't really hold up when you try to unpack it.

Say, for example, that we assign a 4-momentum $(m, 0, 0, 0)$ to an object in its rest frame. Then we Lorentz transform to a frame in which the object is moving with speed $v$ in the $x$ direction, and we say that the object's 4-momentum gets transformed to $( \gamma m, \gamma m v, 0 ,0)$ in the new frame. But to a mathematician, those are two different vectors in the vector space "Minkowski spacetime". The Lorentz transformation induces a hyperbolic rotation on timelike vectors that maps one to the other, but that doesn't make them the same.

If you're ok with saying that "the object's 4-momentum" is not "a vector", but a whole equivalence class of vectors (all the ones that are related to $(m, 0, 0, 0)$ by Lorentz transformations), then I guess you could call that an "intrinsic frame-covariant property" of the object. But I don't think that's what most physicists would call it, because we don't measure an equivalence class of vectors. We measure components--the energy and the momentum in a given direction. And the components are straightforwardly frame-dependent; you have to pick a particular frame (the frame specified by the measuring device) to know their values. So there's no way to view them as properties of just the object alone, without specifying a frame (the measuring device and the frame it defines).