I Is the ground state energy of a quantum field actually zero?

[vanhees71]

Yes sure, but that doesn't make my argument wrong. Of course, you can express everything with the vacuum state itself, i.e., $hat{\rho}_{\text{vac}}=|\Omega \rangle \langle \Omega|$, but what's the point of this?

 

[ftr]

1. If you remove all fields do you still have vacuum. If fields cannot be removed does it mean that each field has its own vacuum, if yes doesn't that mean that the mentioned vacuums are not physical.
2. How does Heisenberg Uncertainty apply to vacuum since there is nothing that has "position".

 

[A. Neumaier]

If you remove all fields do you still have vacuum.

One cannot remove fields, since they are everywhere. One can only reduce their intensity in some small region of space-time.

 

[A. Neumaier]

here are no non-trivial central charges of the covering group of the proper orthochronous Lorentz group, and it's most convenient to work with the proper unitary representations, but it's not necessary.

It is necessary if you want to have a good classical limit. There a relativistic particle satisfies the mass shell relation $p^2=(mc)^2$, and this becomes invalid if you shift the energy (i.e., the positive zero component of $p$) by any nonzero amount.

Indeed, it is convenient to use the proper representation precisely because it is also necessary to match the standard meaning of the terms.

 

[vanhees71]

It's convention to call $p^0$ "the" energy of the particle. You can as well quote kinetic energy, which is $p^0-m c^2$ (it's common in the fixed-target experiments). Nothing of the physics is changed of course. To use the standard "on-shell condition" is just convenience, because then $p^{\mu}$ are four-vector components and thus transform in a simple way via a Lorentz-transformation matrix. It's really semantics we discuss about now.

What is, however, important, and that's why insist on the ray representations, is to admit that there is no absolute energy level, and that's why the problem of "dark energy" in cosmology is not solved at all by your argument.

 

[A. Neumaier]

It's convention to call $p^0$ "the" energy of the particle. You can as well quote kinetic energy, which is $p^0-m c^2$ (it's common in the fixed-target experiments). Nothing of the physics is changed of course. To use the standard "on-shell condition" is just convenience, because then $p^{\mu}$ are four-vector components and thus transform in a simple way via a Lorentz-transformation matrix. It's really semantics we discuss about now.

But it is important semantics. Every concept in theoretical physics is specified by convention, which mathematically amounts to a definition. The convention tells the usage. Moreover, conventions are usually chosen such that the formulas are nice and easy to use.

Exactly because $cp_0$ is by convention the energy of the particle in a particular frame, it is fixed in magnitude by convention, and there is no freedom to redefine it (by shifting), except by changing the convention - i.e., the definition of its meaning. That's why $cp^0-m c^2$ is called the kinetic energy, and not simply the energy - it is the contribution to the energy due to the particle motion. And $m c^2$ is called the rest energy because by the same convention, it is the energy of a particle in its rest frame. In each case, the convention defines the meaning!

What is, however, important, and that's why insist on the ray representations, is to admit that there is no absolute energy level

According to your nowhere defined alternative convention, the rest energy of a particle with mass $m$ would be arbitrary, since energy can be shifted arbitrarily. Clearly you are making your own conventions, for no good reason at all.

that's why the problem of "dark energy" in cosmology is not solved at all by your argument.

I didn't claim to solve this problem, the latter is completely unrelated. The cosmological constant is a parameter in a Lagrangian, and not the energy of a particular stationary state of a quantum field in flat space (which the present discussion is about).

 

[vanhees71]

No, I don't make my own convention anywhere. I only quote the very important fact that symmetries are represented by ray representations and not necessarily unitary representations of the corresponding groups and that due to this fact within special relativity there's no physical meaning in an absolute additive constant on energies or energy densities.

The only place in contemporary physics, where the absolute additive constant in energies is physically relevant is General Relativity, and that's why the "cosmological-constant aka. dark-energy problem" is not solved. I think it's related to the fact that we don't have a complete quantum theory of all interactions including gravity. Now we have to be content with the explanation that the energy content of the universe is not determined by any fundamental law but has to be taken from observation, implying that our contemporary theories need fine tuning of their parameters to an astonishing accuracy of $\sim 10^{60}$-$10^{120}$.

 

[A. Neumaier]

I only quote the very important fact that symmetries are represented by ray representations and not necessarily unitary representations of the corresponding groups and that due to this fact within special relativity there's no physical meaning in an absolute additive constant on energies or energy densities.

The second part does not follow from the first. No physicist ever except for you takes the possibility of defining a trivial central extension by adding such shifts as a permission to regard the standard generators with a standard physical meaning as being defined only up to a constant shift.

According to your interpretation of Weinberg's argument, the angular momentum of a ray representation in the rest frame of a particle is also defined only up to an arbitrary constant, but neither in classical nor in quantum mechanics I ever heard of the ''fact'' (that should be implied by your reasoning) that angular momentum in the rest frame is defined only up to arbitrary shifts. Particles and resonances are classified in the PDB according to their angular momentum, and it is exactly zero for proton and neutron - not an arbitrary number as allowed by a ray representation.

Thus Weinberg's argument implies nothing for shifting physical observables defined by universal conventions.

 

[ftr]

So are you claiming that the total ground energy of the harmonic oscillators is not there.

 

[A. Neumaier]

So are you claiming that the total ground energiy of the harmonic oscillators is not there.

We were discussing a relativistic quantum field theory.

In the nonrelativistic case, the energy is determined only up to a shift determined by fixing a reference with zero energy, since Hamiltonians are not constrained by a mass shell condition and the equations of motion are independent of a shift of $H$ by a constant.

Whether one takes the ground state energy of a harmonic oscillator as zero or something else therefore depends on which Hamiltonian is used to define the oscillator, $H=a^*a$ (used in quantum optics) gives a zero energy, $H=p^2/2m+kq^2$ (used in introductions to quantum mechanics) gives a positive energy for the ground state. The physics, i.e., all equations involving measurable stuff, is completely independent of this.

On the other hand, if one would introduce such a shift into the energy/Hamiltonian of a relativistic system, many equations would be seriously affected and would have to contain explicitly the shift used. This shows that something basic is wrong with such a procedure.

 

[ftr]

On the other hand, if one would introduce such a shift into the energy/Hamiltonian of a relativistic system, many equations would be seriously affected and would have to contain explicitly the shift used. This shows that something basic is wrong with such a procedure.

Ok thanks. So then why normal ordering is done if you are assuming them to be zero beforehand, isn't it to remove these infinities. Also, isn't renormalization is also done to remove the effect from charge and mass during interaction or the procedure is done for some other thing altogether.

 

[A. Neumaier]

Ok thanks. So then why normal ordering is done if you are assuming them to be zero beforehand, isn't it to remove these infinities. Also, isn't renormalization is also done to remove the effect from charge and mass during interaction or the procedure is done for some other thing altogether.

The normal ordering is necessary to get a zero vacuum energy $E=p_0c$, which is necessary because $p$ must transform like a 4-vector, This has hardly anything to do with removing infinities. The latter appear only if meaningless operations are done based on a wrong, naive understanding of the math behind operator-valued distributions.

 

[ftr]

Sorry for nagging, but on one hand you say

normal ordering is necessary to get a zero vacuum energy E=p0c,

on the other

such a shift into the energy/Hamiltonian of a relativistic system, many equations would be seriously affected

They seem to be contradictory or most likely I don't understand something.

edit: I guess I am very confused. Are you saying that Vacuum energy is zero period otherwise QFT will not work or we must remove them from the equation because they are not physical to make the equations work.
xtra edit: does relativity has forced us to use normal order or we used normal ordering because SR(in QFT) demands VE to be zero.

 

[vanhees71]

The second part does not follow from the first. No physicist ever except for you takes the possibility of defining a trivial central extension by adding such shifts as a permission to regard the standard generators with a standard physical meaning as being defined only up to a constant shift.

According to your interpretation of Weinberg's argument, the angular momentum of a ray representation in the rest frame of a particle is also defined only up to an arbitrary constant, but neither in classical nor in quantum mechanics I ever heard of the ''fact'' (that should be implied by your reasoning) that angular momentum in the rest frame is defined only up to arbitrary shifts. Particles and resonances are classified in the PDB according to their angular momentum, and it is exactly zero for proton and neutron - not an arbitrary number as allowed by a ray representation.

Thus Weinberg's argument implies nothing for shifting physical observables defined by universal conventions.

The fact that ray representations and not unitary representations represent symmetries enable the important fact that there are particles with half-integer spin. This strengthens my argument rather than contradicting it! It's of course clear that differences of additive conserved quantities are physical within special relativistic physics, which is exactly my point. So nothing I claim contradicts standard physics.

The claim you make, namely that the vacuum states's energy eigenvalue is necessarily 0 and thus the cosmological-constant problem solved, however, is simply unjustified. We obviously agree to disagree.

 

[A. Neumaier]

The fact that ray representations and not unitary representations represent symmetries enable the important fact that there are particles with half-integer spin.

No; you mix conceptually completely different things.

Weinberg's argument allows arbitrary shifts of the angular momentum in an unphysical central extension.

On the other hand, particles with half integer spin are represented in QFT already by a vector representation of the Poincare group, no central extension is necessary to do so!

 

[vanhees71]

Sigh, I think this is really a superfluous discussion.

If the proper orthochronous Poincare group in the classical sense was the very group you have to use in QT, which you must if you insist on that only unitary representations of the symmetry groups of physics are "allowed descriptions" of symmetry principles in QT, then you'd not be allowed to use the covering group of the rotation group (as a subgroup of the Poincare group) and then only integer-spin representations would be allowed. As observations in Nature, however, show there are half-integer spin realizations of the group in nature like electrons, nucleons, etc. etc.

 

[PeterDonis]

particles with half integer spin are represented in QFT already by a vector representation of the Poincare group

Don't you mean a spinor representation? A vector representation would be spin 1, not spin 1/2, correct?

 

[A. Neumaier]

No; you mix conceptually completely different things.

Weinberg's argument allows arbitrary shifts of the angular momentum in an unphysical central extension.

On the other hand, particles with half integer spin are represented in QFT already by a vector representation of the Poincare group, no central extension is necessary to do so!

Sigh, I think this is really a superfluous discussion.

If the proper orthochronous Poincare group in the classical sense was the very group you have to use in QT, which you must if you insist on that only unitary representations of the symmetry groups of physics are "allowed descriptions" of symmetry principles in QT, then you'd not be allowed to use the covering group of the rotation group (as a subgroup of the Poincare group) and then only integer-spin representations would be allowed. As observations in Nature, however, show there are half-integer spin realizations of the group in nature like electrons, nucleons, etc. etc.

Sorry, of course one has the phases in the unitary transformations that give a central extension of order 2, i.e., one has a vector representation of ISL(2,C) rather than one of ISO(1,3). Note that these have the same Lie algebra commutation relations defining the standard generators without any shift! You should therefore interpret my comments to apply to ISL(2,C) rather than the Poincare group.

Weinberg posits an arbitrary central extension of this Lie algebra (thus changing the definition and hence the meaning of the generators by positing different commutation rules) and shows that no physics can result, hence that this Lie algebra extension is physically spurious - in contrast to the Galilei group where a nontrivial central extension with mass as a central charge (the one realized in nonrelativistic physics) appears by the same kind of analysis.

In general, physical observables with a meaning in terms of symmetry are (in all cases, without exception) defined on the theoretical level by their commutation rule. If the commutation rule change, the meaning of the observables change. In the extended Lie algebra considered by Weinberg, all generators, including those for the rotation group, alter their meaning by being shifted.

You can apply exactly the same argument to SO(3) or SU(2) - which one doesn't matter since their Lie algebra is the same and Weinberg only argues with the Lie algebra. After similar calculations you end up with the same result - that there is no central charge. So if one follows your argument one should conclude that angular momentum is physically determined only up to an arbitrary shift in each component. Of course, this is nonsense, hence your argument implies no such thing - also not in the case of energy where you originally applied it.

This is completely independent of the double-valuedness of spin 1/2. Ths enters the discussion only on the group level, but Weinberg's argument is solely on the Lie algebra level! Therefore the conclusions also apply only on the Lie algebra level! This is what i had meant when saying that you mix two completely different things!

 

[A. Neumaier]

Don't you mean a spinor representation? A vector representation would be spin 1, not spin 1/2, correct?

Yes. This is equivalent to a vector representation of SU(2), which has the same Lie algebra as SO(3), generated by the components of angular momentum, without involving a central extension on the Lie algebra level. The situation discussed earlier is the same except that the group ISL(2,C) is bigger, but it too has the same Lie algebra as the Poincare algebra ISO(1,3), also without involving a central extension on the Lie algebra level.

Thus the phenomenon of half integer spin has nothing at all to do with a discussion such as Weinberg's or vanhees71's, where the Lie algebra is augmented by a (in both cases spurious) central charge.

 

[PeterDonis]

This is equivalent to a vector representation of SU(2),

Probably this is just an issue of terminology. By "vector representation of SU(2)", do you mean the fundamental representation? As described, for example, here:

https://en.wikipedia.org/wiki/Representation_theory_of_SU(2)

 

[A. Neumaier]

"vector representation of SU(2)", do you mean the fundamental representation?

No. All representations of SU(2) are vector representations, i.e., representations on a vector space - in contrast to ray representations, which are representations on a projective space.

 

[PeterDonis]

vector representations, i.e., representations on a vector space

Ah, ok, as I thought, you are using the term "vector representation" differently from the way I'm used to seeing it used. Your usage is much more general.

 

[A. Neumaier]

Ah, ok, as I thought, you are using the term "vector representation" differently from the way I'm used to seeing it used. Your usage is much more general.

"vector representation" = "linear representation" is analogous to
"ray representation" = "projective representation"; cf. https://en.wikipedia.org/wiki/Projective_representation

One can either put the emphasis on the elements on which the group acts, or on the whole collection of such elements.

 

[samalkhaiat]

Sigh, I think this is really a superfluous discussion.

If the proper orthochronous Poincare group in the classical sense was the very group you have to use in QT, which you must if you insist on that only unitary representations of the symmetry groups of physics are "allowed descriptions" of symmetry principles in QT, then you'd not be allowed to use the covering group of the rotation group (as a subgroup of the Poincare group) and then only integer-spin representations would be allowed. As observations in Nature, however, show there are half-integer spin realizations of the group in nature like electrons, nucleons, etc. etc.

You are absolutely correct. You cannot obtain spinors from the (classical) symmetry groups \mbox{SO}(3) and \mbox{SO}(1,3).