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

  • #1
jcap
170
12
I start by outlining the little I know about the basics of quantum field theory.

The simplest relativistic field theory is described by the Klein-Gordon equation of motion for a scalar field ##\large \phi(\vec{x},t)##:
$$\large \frac{\partial^2\phi}{\partial t^2}-\nabla^2\phi+m^2\phi=0.$$
We can decouple the degrees of freedom from each other by taking the Fourier transform:
$$\large \phi(\vec{x},t)=\int \frac{d^3p}{(2\pi)^3}e^{i\vec{p}\cdot \vec{x}}\phi(\vec{p},t).$$
Substituting back into the Klein-Gordon equation we find that ##\large \phi(\vec{p},t)## satisfies the simple harmonic equation of motion
$$\large \frac{\partial^2\phi}{\partial t^2}=-(\vec{p}^2+m^2)\phi.$$
Therefore, for each value of ##\large \vec{p}##, ##\large \phi(\vec{p},t)## solves the equation of a harmonic oscillator vibrating at frequency
$$\large \omega_\vec{p}=+\sqrt{\vec{p}^2+m^2}.$$
Thus the general solution to the Klein-Gordon equation is a linear superposition of simple harmonic oscillators with frequency ##\large \omega_\vec{p}##. When these harmonic oscillators are quantized we find that each has a set of discrete positive energy levels given by
$$\large E^p_n=\hbar\omega_\vec{p}(n+\frac{1}{2})$$
for ##\large n=0,1,2\ldots## where ##\large n## is interpreted as the number of particles with momentum ##\large \vec{p}##.

My question is what about the harmonic oscillator solutions that vibrate at negative frequency
$$\large \bar{\omega}_\vec{p}=-\sqrt{\vec{p}^2+m^2}?$$

When these harmonic oscillators are quantized we get a set of discrete negative energy levels given by
$$\large \bar{E}^p_n=\hbar\bar{\omega}_\vec{p}(n+\frac{1}{2})$$
for ##\large n=0,1,2\ldots## where ##\large n## can now be interpreted as the number of antiparticles with momentum ##\large \vec{p}##.

If this is correct then the total energy of the ground state, per momentum ##\large \vec{p}##, is given by
\begin{eqnarray*}
\large T^p_0 &=& \large E^p_0+\bar{E}^p_0\\
&=& \large \frac{\hbar\sqrt{\vec{p}^2+m^2}}{2} + \frac{-\hbar\sqrt{\vec{p}^2+m^2}}{2}\\
&=& \large 0.
\end{eqnarray*}

Thus the total ground state energy, ##\large T_0##, is zero; there is no zero-point energy.

Does this interpretation of the negative frequency solutions make sense?
 
Last edited:

Answers and Replies

  • #2
king vitamin
Science Advisor
Gold Member
486
243
The solutions to the Klein-Gordon equation with negative frequency do not have negative energy! You need to go back to the Hamiltonian for the Klein-Gordon theory,
[tex]
H = \int d^3 x \left[ \frac{1}{2} \Pi^2 + \frac{1}{2} \left( \nabla \phi\right)^2 + \frac{m^2}{2} \phi^2 \right],
[/tex]
and consider the energy for a field
[tex]
\phi(\vec{x},t) = \int \frac{d^3 k}{(2 \pi)^3} \left( a(k) e^{i \vec{k} \cdot \vec{x} - i \omega t} + b(k) e^{i \vec{k} \cdot \vec{x} + i \omega t} \right).
[/tex]
Here you can see that we're including both the positive and negative frequency solutions to the Klein-Gordon equation. The coefficients in the Fourier expansion are operators. Now do the usual trick of writing these coefficients as ladder operators, and do the same trick for the canonical momentum [itex]\Pi(\vec{x},t)[/itex] such that [itex][\phi(\vec{x},t),\Pi(0,t)] = i \delta^3(\vec{x})[/itex] is satisfied, and then calculate [itex]H[/itex]. You should only find a unique ground state, and all other states have higher energy than this.

In fact, this is true even without quantum mechanics. Negative frequency solutions to the classical Klein-Gordon equation still result in a positive energy in the classical Klein-Gordon Hamiltonian (show this!). In contrast, a "classical" Dirac Hamiltonian has negative energy solutions which signal that it is sick.
 
Last edited:
  • Like
Likes Demystifier, jcap, atyy and 2 others
  • #3
10,151
3,282
To answer the title of the question look into something called normal ordering. The energy of the vacuum is a big fat zero.

That one fooled me to until l read a proper book on QFT a few years ago now.

Thanks
Bill
 
  • #4
A. Neumaier
Science Advisor
Insights Author
8,230
4,113
Is the ground state energy of a quantum field actually zero?

In the relativistic case, yes, by Lorentz covariance.

In the nonrelativistic case, it is a matter of convention since there energies are determined only up to an arbitrary constant shift.
 
  • #6
Demystifier
Science Advisor
Insights Author
Gold Member
13,139
5,439
To answer the title of the question look into something called normal ordering. The energy of the vacuum is a big fat zero.

That one fooled me to until l read a proper book on QFT a few years ago now.
Are you saying that normal ordering is the only correct ordering? And which book was that?

If you read e.g. Bjorken and Drell, you will see that vacuum energy can be any number, the effect of which is to modify the phase of the scattering amplitude without changing any measurable quantity.
 
  • #7
A. Neumaier
Science Advisor
Insights Author
8,230
4,113
Is the ground state energy of a quantum field actually zero?

In the relativistic case, yes, by Lorentz covariance.
It's not that simple
It is that simple, in spite of your reference.

The ground state of a relativistic quantum field theory is the vacuum state, defined as a Poincare invariant state. Hence the 4-momentum ##p## must be Lorentz invariant, which is only possible if ##p=0##. The energy is the 0-component, hence vanishes.
 
Last edited:
  • #8
10,151
3,282
Are you saying that normal ordering is the only correct ordering? And which book was that? If you read e.g. Bjorken and Drell, you will see that vacuum energy can be any number, the effect of which is to modify the phase of the scattering amplitude without changing any measurable quantity.

No I am not. What I am saying is the issue can be rectified. I read some overview books on QFT that said its arbitrary the energy you call zero because you only measure differences anyway. So you just set the infinite energy as zero. I thought it totally silly and turned me right off. The only solution was to get a proper book on QFT and I chose An Introduction to Quantum Field Theory by George Sterman as my first book. He explained it on page 44. But overall its not my favorite book - I preferred others like QFT For The Gifted Amateur - which does the same thing. I am reading Strednicki right now and it resolves it by an arbitrary ultraviolet cutoff - see page 24 - I suppose in preparation for the modern effective field theory view. I hope so because that is something I want to understand better.

Thanks
Bill
 
Last edited:
  • #9
Demystifier
Science Advisor
Insights Author
Gold Member
13,139
5,439
It is that simple, in spite of your reference.

The vacuum is defined as a Poincare invariant state. Hence the 4-momentum ##p## must be Lorentz invariant, which is only possible if ##p=0##. The energy is the 0-component, hence vanishes.
So do you claim that there is no cosmological constant problem? Another Lorentz invariant value for energy is infinity. Indeed, the vacuum energy-momentum tensor of the vacuum with cosmological constant ##\lambda## is
$$T_{\mu\nu}=\lambda g_{\mu\nu}$$
which is Lorentz-invariant as long as ##g_{\mu\nu}=\eta_{\mu\nu}## is Lorentz invariant. The 4-momentum is then
$$P^{\mu}=\int d^3x T^{\mu}_{0}=\lambda \delta^{\mu}_{0} \int d^3x$$
where ## \int d^3x=\infty##.
 
  • #10
A. Neumaier
Science Advisor
Insights Author
8,230
4,113
So do you claim that there is no cosmological constant problem?
The universe is not in a vacuum/ground state; so your question has nothing to do with the topic of the thread.
Another Lorentz invariant value for momentum is infinity.
In this case, all states must have infinite energy and all states would be ground states! Thus the notion of a vacuum/ground state makes no longer sense.
 
  • #11
Demystifier
Science Advisor
Insights Author
Gold Member
13,139
5,439
The universe is not in a vacuum/ground state; so your question has nothing to do with the topic of the thread.

In this case, all states must have infinite energy and all states would be ground states! Thus the notion of a vacuum/ground state makes no longer sense.
Are you saying that the cosmological constant problem has nothing to do with the vacuum energy?
 
  • #13
Demystifier
Science Advisor
Insights Author
Gold Member
13,139
5,439
There is no well-defined notion of vacuum energy in cosmology. Its reality status is similar to that of virtual particles popping in and out of existence for a very short time.
So are you saying that all the papers about the cosmological constant problem are misleading? Including the famous paper by Weinberg?
https://repositories.lib.utexas.edu/bitstream/handle/2152/61094/Weinberg_1989.pdf?sequence=1
 
  • #14
A. Neumaier
Science Advisor
Insights Author
8,230
4,113
So are you saying that all the papers about the cosmological constant problem are misleading?
Only the discussion in terms of vacuum energy, which is similar to discussions of QFT in terms of virtual particles. Its value is nil, apart from making it seemingly less abstract.

The cosmological constant is a property of the state of our observable universe, which surely isn't in a vacuum state. Thus it can have nothing to do with the properties of the vacuum state.
 
  • #15
Demystifier
Science Advisor
Insights Author
Gold Member
13,139
5,439
Only the discussion in terms of vacuum energy, which is similar to discussions of QFT in terms of virtual particles. Its value is nil, apart from making it seemingly less abstract.

The cosmological constant is a property of the state of our observable universe, which surely isn't in a vacuum state. Thus it can have nothing to do with the properties of the vacuum state.
If it's true, then it's new and very important. If you are convinced that you are right, then you should publish it.
 
  • Like
Likes mfb and king vitamin
  • #16
10,151
3,282
So do you claim that there is no cosmological constant problem?

Indeed there is IMHO. Normal ordering is just a way of handling the issue in a way that makes sense - but skirts the main issue - why do we have to resort to it in the first place. Like I said I am hopeful a better understanding of Effective Field Theory on my part will help - at least me anyway. We will see.

Thanks
Bill
 
  • #17
A. Neumaier
Science Advisor
Insights Author
8,230
4,113
Including the famous paper by Weinberg?
https://repositories.lib.utexas.edu/bitstream/handle/2152/61094/Weinberg_1989.pdf?sequence=1
Well, he also talks (on p.3 of his paper) about the ''demonstration in the Casimir effect of the reality of zero-point energies'' and ''the gravitational force between the particles in the vacuum fluctuations'', which is virtual particle nonsense.

Don't take verbal talk involving bare, virtual stuff too serious - the meat is always only in the (renormalized) formulas. For lack of a good renormalization prescription for quantum gravity we can say very little definite. Weinberg's paper (like much in quantum gravity) is just speculation because we don't have anything better.
 
  • #18
Demystifier
Science Advisor
Insights Author
Gold Member
13,139
5,439
Well, he also talks (on p.3 of his paper) about the ''demonstration in the Casimir effect of the reality of zero-point energies'' and ''the gravitational force between the particles in the vacuum fluctuations'', which is virtual particle nonsense.
Vacuum fluctuations are not nonsense, even if virtual particles are.

Don't take verbal talk involving bare, virtual stuff too serious - the meat is always only in the (renormalized) formulas. For lack of a good renormalization prescription for quantum gravity we can say very little definite. Weinberg's paper (like much in quantum gravity) is just speculation because we don't have anything better.
Are you going to teach Weinberg renormalization? :wideeyed:
 
  • #19
A. Neumaier
Science Advisor
Insights Author
8,230
4,113
Vacuum fluctuations are not nonsense
But he talked about ''particles in the vacuum fluctuations'', which is nonsense. The vacuum contains zero particles at all times. This shows that his discussion must be taken with large amounts of grains of salt. When he talks about summing up the zero point energies of modes, if his argument were stringent, it would also apply to QED, where we know that energies are not horrendously large.

Are you going to teach Weinberg renormalization?
No, but I know that he doesn't know how to renormalize gravity. Currently nobody knows!
 
  • #20
vanhees71
Science Advisor
Insights Author
Gold Member
2021 Award
21,594
12,414
It is that simple, in spite of your reference.

The ground state of a relativistic quantum field theory is the vacuum state, defined as a Poincare invariant state. Hence the 4-momentum ##p## must be Lorentz invariant, which is only possible if ##p=0##. The energy is the 0-component, hence vanishes.
I've never understood this argument of yours. You can simply add a constant ##E_0 \hat{1}## to the usually used normal-ordered ##\hat{H}##, and the ground-state energy is ##E_0##, where ##E_0## can take any real value. The ground state is given by
$$\hat{\rho}=|\Omega \rangle \langle \Omega|,$$
and that's Poincare invariant, particularly it's invariant under temporal translations, no matter which value of ##E_0## you choose
$$\exp(-\mathrm{i} \hat{H} t) \hat{\rho} \exp(+\mathrm{i} \hat{H} t)=\exp(-\mathrm{i} E_0 t) \hat{\rho} \exp(+\mathrm{i} E_0 t)=\hat{\rho}.$$
I don't see, where there's anything forbidden in this argument when used in relativistic QFT. Of course, it holds either in non-relativistic QT.
 
  • Like
Likes bhobba, atyy and mfb
  • #21
A. Neumaier
Science Advisor
Insights Author
8,230
4,113
I don't see, where there's anything forbidden in this argument when used in relativistic QFT.
Your vacuum state is Poincare invariant, but your Poincare generators are not. That spoils covariance.
You can simply add a constant ##E_0 \hat{1}## to the usually used normal-ordered ##\hat{H}##.
You can do this in the nonrelativistic case since the symmetry group there is elementary abelian (or a Galilei group), and this property is preserved under an energy shift. But you cannot add such a shift to a Poincare group generator, since the Poincare generators have to transform according to the adjoint representation.

Therefore relativistic energies have an absolute meaning, while nonrelativistic energies have a relative meaning only.
 
Last edited:
  • #22
Demystifier
Science Advisor
Insights Author
Gold Member
13,139
5,439
Your vacuum state is Poincare invariant, but your Poincare generators are not. That spoils covariance.

You can do this in the nonrelativistic case since the symmetry group there is elementary abelian (or a Galilei group), and this property is preserved under an energy shift. But you cannot add such a shift to a Poincare group generator, since the Poincare generators have to transform according to the adjoint representation.

Therefore relativistic energies have an absolute meaning, while nonrelativistic energies have a relative meaning only.
What about the spatial 3-momentum? Is it absolute too?
 
  • #23
vanhees71
Science Advisor
Insights Author
Gold Member
2021 Award
21,594
12,414
Your vacuum state is Poincare invariant, but your Poincare generators are not. That spoils covariance.

You can do this in the nonrelativistic case since the symmetry group there is elementary abelian (or a Galilei group), and this property is preserved under an energy shift. But you cannot add such a shift to a Poincare group generator, since the Poincare generators have to transform according to the adjoint representation.

Therefore relativistic energies have an absolute meaning, while nonrelativistic energies have a relative meaning only.
I guess you mean the main difference between the Poincare and Galileo group is that in only the latter the boosts form an Abelian subgroup, while in the Poincare group they are non-abelian and don't form a subgroup at all (i.e., only the full orthochronous Lorentz group is a subgroup, i.e., the group generated by both boosts and rotations).

Let me see, whether I understand this argument: In the usual construction of the unitary ray representations you start with showing that all possible central charges are trivial in the sense that one can redefine the representation in such a way that you get a true unitary representation which is equivalent to the ray representation you started with, and then you investigate only the proper unitary representations, which fixes the absolute value of the energy of the ground state to be 0? This would make sense to me.
 
  • #24
A. Neumaier
Science Advisor
Insights Author
8,230
4,113
What about the spatial 3-momentum? Is it absolute too?
In the relativistic case, for the vacuum, yes, as it remains a vacuum state in every frame.
I guess you mean the main difference between the Poincare and Galileo group is that in only the latter the boosts form an Abelian subgroup, while in the Poincare group they are non-abelian and don't form a subgroup at all
This difference is immaterial.

The point is that in an irreducible representation of the Poincare group, the possible values of 4-momentum form an orbit under the Lorentz group. Thus there are only a few possibilities, namely the orbits of ##(0 ,0 ,0, 0), (\pm m, 0, 0, 0), (0, 0 ,0 ,x), (\pm E, 0 ,0 ,E)##. The first orbit is a single point, the next two form hyperboloids, the final one half cones. Since the vacuum state is invariant it must correspond to the first case. Thus its momentum is zero. If you would shift its energy component you would give the vacuum a mass.

In contrast, in a massless representation of the Galilei group, any ##(E ,0 ,0 ,0)## is a singleton orbit, and shifting ##E## creates an isomorphic representation.
 
Last edited:
  • Like
Likes dextercioby
  • #25
vanhees71
Science Advisor
Insights Author
Gold Member
2021 Award
21,594
12,414
But invariance of the vacuum state only means that any representing state vector changes by a phase factor under the Poincare group. This phase factor needs not to be simply 1. Of course you can always define this phase factors in the unitary ray representations of the Poincare group away (as shown by Weinberg in Vol. 1 of his QT of Fields), but that's not necessary. Thus you can always add "central charges" to the generators of the Poincare group. Then you get a ray representation. The point is that in the case of the Poincare group from any unitary ray representation you can eliminate the central charges, i.e., you can lift any unitary ray representation to an equivalent unitary representation. If you only admit these unitary representations you get of course that the vacuum state has 0 energy, momentum, and angular momentum. Equivalently you can however assign any arbitrary value to these quantities. The only difference is you get a ray rather than a proper unitary representation of the Poincare group.

The Galileo group is different in the sense that there are ray representations that are not liftable to an equivalent unitary representation, which admits the introduction of non-zero mass into non-relativistic QT as a central charge of the Galileo algebra. This is crucial, because the proper unitary representations of the Galileo group do not lead to any satisfactory non-relativistic quantum dynamics.

Of course, your statement concerning the physical relevant representations with the standard momentum ##p^{\mu}=0## (##\mu \in \{0,1,2,3\}##) is correct: The little group is the entire (proper orthochronous) Lorentz group, which has no non-trivial unitary representations. So the only unitary representation induced in the usual way a la Frobenius (Wigner) is to represent the Poincare group trivially. Still since you need only unitary ray representations the four-momentum of the Vacuum state can be chosen arbitrarily.
 
  • #26
A. Neumaier
Science Advisor
Insights Author
8,230
4,113
Still since you need only unitary ray representations the four-momentum of the Vacuum state can be chosen arbitrarily.
But this is not the only relevant criterion. Only the zero choice gives a Poincare invariant Lie algebra of generators. The 4-momentum is generally defined as the vector of translation generators in such a Lie algebra, and the energy as its 0-component.
 
  • #27
vanhees71
Science Advisor
Insights Author
Gold Member
2021 Award
21,594
12,414
Ok, I have to do the calculation explicitly to see, whether there is a contradiction with the Poincare Lie algebra (including central charges!) if I assume that the vacuum is a eigenstate for different values for energy and momentum than 0. I doubt it. The complete Poincare Lie algebra, including central charges, is given in Weinberg, QT of Fields Vol. 1 in Eqs. (2.7.13-2.7.16). The four-momentum commutation relation reads
$$\mathrm{i} [\hat{p}^{\mu},\hat{p}^{\nu}]=C^{\rho, \mu} \hat{1}.$$
This is verbatim Weinberg except that I added a unit operator which Weinberg leaves out in the usual physicist's notation that writing a usual number for an operator it's understood to be that number times the unit operator.

The usual choice to make the vacuum energy and momentum 0 is of course the most convenient, because than you can work with the proper unitary representation rather than with the more complicated unitary ray representations. The Poincare group allows you to do that, because any unitary ray representation can be equivalently lifted to a unitary representation (that's not so for the Galileo group of non-relativistic spacetime, and that's why the quantum Galileo group is different from the classical one, with mass as a non-trivial central charge).
 
  • #28
A. Neumaier
Science Advisor
Insights Author
8,230
4,113
whether there is a contradiction with the Poincare Lie algebra
The place where things go wrong is with the commutation relation ##[K_j,P_j]=-iH## [Weinberg (2.4.22)] between the boost generator ##K_j ## and the corresponding momentum generator ##P_j##. You cannot consistently shift ##H## by a constant. In contrast, the commutation relations for the central extension of the Galiliei algebra [Weinberg p.62] does not involve the non-mass energy ##W## on the right hand side of a commutation relation, hence a shift of ##W## by a constant gives something isomorphic. But the mass ##M## appears on the right hand side, and hence cannot be shifted, though it is a central charge.

Of course, one can shift all generators of an arbitrary Lie algebra by arbitrary numbers and correct the commutation relations in such a way that one gets an isomorphic representation with additional central charges. For example, one can do this for the Lie algebra of ##SO(3)## and would have to conclude (by analogy to your argument for the Poincare group) that angular momentum is only defined up to an arbitrary shift. But I never heard such a claim, neither by mathematicians nor by physicists.

Indeed, this is very unnatural and is never done in practice since it introduces unnecessary and unphysical charges. It is done for no group at all without explicitly renaming the algebra or group. In particular, one always distinguishes the Galilei algebra and its central extension, also called the Bargmann algebra. (Weinberg also makes the distinction, without mentioning the term central extension.) They are different algebras with a different number of independent generators.

The same holds for the Poincare group: The Poincare algebra and its standard generators (including the Hamiltonian) are defined by the standard commutation relations given by Weinberg in Section 2.4, and not by the more general relations in Section 2.7. The latter don't define the Poincare algebra itself but a physically spurious and mathematically trivial central extension of it.
 
Last edited:
  • Like
Likes dextercioby
  • #29
vanhees71
Science Advisor
Insights Author
Gold Member
2021 Award
21,594
12,414
We agree about the math of both the Poincare and Galileo group. Of course the quantum Galileo group is the one possible non-trivial central extension, where the corresponding algebra is extended by the mass as a central charge. I didn't know that it's called Bargmann algebra.

I still don't understand, why the vacuum state of (say a massive) free-field representation must be necessarily an eigenstate of four-momentum with eigenvalues 0. If you choose the full general equations (2.7.13-2.7.16), there should be no contradiction. Note that there are central charges in all commutation relations. Of course in the Poincare group all central charges are "trivial" in the sense that you can always lift any unitary ray representation to a proper unitary representation of the proper orthochronous Poincare group, and I admit it's pretty much more convenient to do so and then impose 0 eigenvalues of the four-momentum operators for the vacuum state.

On the other hand, if you realize the algebra in the usual way using local field operators, you do not get this for free but you have to introduce normal ordering. In the (perturbative treatment) of the interacting theory you have to renormalize the action to give 0 four-momentum for the vacuum state. It's not a "protected property" by symmetries.
 
  • #30
A. Neumaier
Science Advisor
Insights Author
8,230
4,113
I still don't understand, why the vacuum state of (say a massive) free-field representation must be necessarily an eigenstate of four-momentum with eigenvalues 0. If you choose the full general equations (2.7.13-2.7.16), there should be no contradiction.
(2.7.13-2.7.16) does not describe the Poincare algebra, but a trivial (direct product) central extension of it.

If you do the thing analogous of (2.7.13-2.7.16) for the SO(3) Lie algebra, you would have to conclude that there is no reason why a centrally symmetric state of the hydrogen aton must necessarily be an eigenstate of ##J_3## with eigenvalue 0. This is a very strange claim. Why should your argument hold for one group but not for another one?

if you realize the algebra in the usual way using local field operators, you do not get this for free but you have to introduce normal ordering.
Surely (2.7.13-2.7.16) are invalid for infinite values of the shifts - thus this argument is besides the point.
Normal ordering is already needed to make sense at all of products and hence of commutators of the field operators.
 
  • #31
vanhees71
Science Advisor
Insights Author
Gold Member
2021 Award
21,594
12,414
As I said, I don't understand this arguments. In QT you need only ray representations and not necessarily proper representations of the Lie group or Lie algebra of a continuous symmetry. Of course, these trivial constant additions to the eigenvalues of the corresponding conserved quantities are physically irrelevant, only differences are measurable. That's already so in Newtonian classical mechanics, where a shift of the total energy of a system by an arbitrary constant doesn't make any difference in the physics.

Of course this holds for the rotation-group algebra su(2) (there's a (trivial) central charge also in the corresponding commutation relation, given by (2.7.13), which of course in this Minkowski four-vector notation, includes also the boosts, i.e., it refers to a ray representation of the entire Lorentz group as a subgroup of the Poincare group). As Weinberg shows in the said Section 2.7 of course you don't loose anything by considering only the usual commutation relations without central charges, because one can get rid of them by corresponding shifts of the generators by adding operators proportional to the unit operator.

It (thus) also doesn't make a difference to add arbitrary finite values to the renormalized additive conserved quantities. Normal ordering is only one convenient choice, you are always free to add aribtrary constants. The physics doesn't depend on them. Of course, it would be great if this was not so, because then Poincare symmetry would imply that there's no fine-tuning problem concerning the cosmological constant/"dark energy" in the Standard Model.
 
  • #32
A. Neumaier
Science Advisor
Insights Author
8,230
4,113
In QT you need only ray representations
Of course, but the generators satisfying (2.7.13-16) do not generally deserve the interpretation they have in the special case where the ##C##s vanish.

Once you allow a shift in energy ##P_0## because of (2.7.x) you must also allow a shift in all other generators of the Poincare group. The angular momentum commutator relations now read ##[J_i,J_j]=i\epsilon_{ijk}J_k + C_{ij}## by (2.7.13). Allowing this means that the angular momentum component ##J_3## of the trivial (=vacuum = ground state) representation can be made to have any value we like.

But this is not what physicists mean by angular momentum
. Instead, everyone assumes that the generators that deserve to be called angular momentum components, must have to satisfy the commutation rule ##[J_i,J_j]=i\epsilon_{ijk}J_k##, as given by Weinberg in (2.4.18).

Similarly, everyone defines the meaning of the Poincare algebra generators (and hence of ##H##) by (2.4.18-24), and no one by (2.7.x) - even Weinberg uses the subjunctive when he discusses the latter. (He also discusses it only in a starred section, i.e., as hardly relevant material.) Because of (2.4.22), this fixes ##H## absolutely, and leaves no room for a shift.
 
Last edited:
  • #33
A. Neumaier
Science Advisor
Insights Author
8,230
4,113
in Newtonian classical mechanics, where a shift of the total energy of a system by an arbitrary constant doesn't make any difference in the physics.
As I has explained, this is due to a special property of the representation [Weinberg, p.62] defining the nonrelativistic observables in terms of the central extension of the Galilei algebra, namely that the non-mass energy ##W## does not occur on the right hand side of the defining commutation relations. Thus shifting ##W## defines an isomorphism. All other observables are fixed by these relations and cannot be shifted without changing the physics!

In contrast, in case of the Poincare algebra, the relations (2.4.18-24) defining the meaning of the observables contain no such generator. All observables are fixed by these relations and cannot be shifted without changing the physics!
 
  • #34
vanhees71
Science Advisor
Insights Author
Gold Member
2021 Award
21,594
12,414
Sure, we agree about this: It's a standard choice to define all the values with respect to the vacuum state, and thus you can use the proper rather than the ray representations of the Poincare group, but it's this arbitrary choice which fixes ##\hat{H}## absolutely and it's an argument to use "normal ordering" for free-field (or fields in the interaction representation) to start with when arguing within "canonical quantization", which is the most straight-forward way to start learning QFT. It's not the best way to understand the foundations, for which you need to go through the analysis of the Poincare group/Lie algebra as Weinberg does in Chpt. 2 of his book. In any case, no matter how you approach the subject starting from free fields, you need it for renormalizing the total energy of an interacting system, and to see that a change of the renormalization description doesn't have any physically relevant consequences.

Besides, the starred sections in Weinberg's books are not "hardly relevant" but "not so relevant at a first read". Weinberg has very good reasons for including this important discussion, because a naive treatment of the Galileo group fails in the case of non-relativistic QM. There you'd come to the conclusion that there is no non-relativistic QM that describes nature if you'd restrict yourself only to the proper unitary representations of the classical Galileo group. There you need the ray representations, and the difference to the Poincare group is that it admits a non-trivial central extension with the mass as the (unique) non-trivial central charge of the Lie algebra of the corresponding central extension of the Galileo group (the Bargmann group as you told me before).
 
  • #35
vanhees71
Science Advisor
Insights Author
Gold Member
2021 Award
21,594
12,414
As I has explained, this is due to a special property of the representation [Weinberg, p.62] defining the nonrelativistic observables in terms of the central extension of the Galilei algebra, namely that the non-mass energy ##W## does not occur on the right hand side of the defining commutation relations. Thus shifting ##W## defines an isomorphism. All other observables are fixed by these relations and cannot be shifted without changing the physics!

In contrast, in case of the Poincare algebra, the relations (2.4.18-24) defining the meaning of the observables contain no such generator. All observables are fixed by these relations and cannot be shifted without changing the physics!
No, the ray representations are as good as the proper ones in the case of the Poincare group, because the central charges are all trivial in this case. The Lie algebra with the trivial central charges and the one without are physically equivalent.

Of course, by doing a contraction of the Poincare Lie algebra to the Galileo algebra starting from the Poincare Lie algebra without trivial central charges, guided by (classical) non-relativistic mechanics you end up with the quantum Galileo Lie algebra (or Bargmann Lie algebra) including the mass as non-trivial central charge of this algebra, as it should be.
 

Suggested for: Is the ground state energy of a quantum field actually zero?

Replies
4
Views
397
Replies
11
Views
518
  • Last Post
Replies
33
Views
1K
Replies
2
Views
434
Replies
36
Views
1K
  • Last Post
Replies
2
Views
305
  • Last Post
Replies
12
Views
656
  • Last Post
Replies
2
Views
355
  • Last Post
Replies
3
Views
359
Replies
1
Views
278
Top