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

[samalkhaiat]

@samalkhaiat is there any book/paper that covers issues you raised in your last post?

1) P. N. Hoffman and J. F. Humphreys, “Projective representations of the symmetric groups”. Oxford Mathematical Monographs. The Clarendon Press, Oxford University Press, New York, 1992.
2) Alexander Kleshchev, “Linear and projective representations of symmetric groups”, volume 163 of Cambridge Tracts in Mathematics. Camb. Uni. Press, 2005.
3) Jose A. De Azcarraga & Jose M. Izquierdo, “Lie groups, Lie algebras, cohomology and some applications in physics”, Camb. Monographs on Mathematical Physics, Camb. Uni. Press, 1998.
4) V. Ovsienko, S. Tabachnikov, “Projective Differential Geometry, Old and New: From the Schwarzian Derivative to the Cohomology of Diffeomorphism Groups”, Camb. Uni. Press, Camb.Tracts in Mathematics, 2004.
5) G. Tuynman, W. Wiegerinck, “Central extensions and physics”, J. Geom. Phys. 4(2), 207–258 (1987).

 

[weirdoguy]

@samalkhaiat thank you very much

 

[ftr]

So, what does group representation say about vacuum energy?

 

[vanhees71]

The proper unitary representations of the Poincare group tells you the vacuum energy should be 0. This doesn't however imply that this is necessarily the only allowed value since in QT symmetries are not only realized by unitary transformations but more generally by unitary ray representations, and this leaves the freedom to choose any value for the vacuum energy you like. Only energy differences are observable within special (sic!) relativity (as well as in Newtonian physics), in classical as well as quantum theory.

 

[ftr]

Thanks. Is vacuum looked upon as an extended object. Is it physical or mathematical.

 

[vanhees71]

Thanks. Is vacuum looked upon as an extended object. Is it physical or mathematical.

Vacuum is simply the state where no particles are present. It's physical but pretty much impossible to realize in the lab.

 

[ftr]

Vacuum is simply the state where no particles are present. It's physical but pretty much impossible to realize in the lab.

What equation/function denotes "vacuum quantum fluctuation".

 

[vanhees71]

There are no vacuum flucutations. It's a misconception of popular-science writings overcomplicating the true issue. Most popular-science writings paradoxically make things more complicated than they are, because it's everything than trivial to correctly describe what physics is about if you are not allowed to use the only adequate language to talk about it, which is mathematics. So popular-science writing is much more complicated if you like to get it right than writing a textbook.

 

[Demystifier]

There are no vacuum flucutations.

Of course there are. Perhaps you wanted to say that there are no virtual particles that pop out and disappear in the vacuum? But you cannot deny that in the vacuum we have
$$\langle 0|\phi(x)|0\rangle=0, \;\; \langle 0|\phi(x)\phi(x)|0\rangle\neq 0 $$
and the technical name for this fact is - vacuum fluctuations of the field $\phi(x)$.

 

[vanhees71]

What's called "vacuum fluctuations" are in fact radiative corrections with particles/fields present. The vacuum is the only thing which doesn't change over time. Nothing "pops in and out of existence" as most textbooks claim.

E.g., what's usually quoted as "proof" for vacuum fluctuations is the Casimir effect, applied to two uncharged plates. Of course, that's no vacuum at all since the plates consists of a humongous amount of charges, and the Casimir effect is due to quantum fluctuations of the quantized electromagnetic field due to the presence of these charges.

It's not possible to observe the vacuum at all since to observer something you need a measurement apparatus (and be it simply your own eyes to observe light), and then it's no more vacuum.

 

[Demystifier]

The vacuum is the only thing which doesn't change over time.

Not the only. Any Hamiltonian eigenstate has this property. E.g. a two-particle state in the free theory.

Nothing "pops in and out of existence" as most textbooks claim.

You mean popular science books, not textbooks.

 

[vanhees71]

Well, even many textbooks write such nonsense in their "heuristic introductions" ;-)). Of course you are right concerning the eigenstates of the Hamiltonian.

 

[ftr]

Of course there are. Perhaps you wanted to say that there are no virtual particles that pop out and disappear in the vacuum? But you cannot deny that in the vacuum we have
$$\langle 0|\phi(x)|0\rangle=0, \;\; \langle 0|\phi(x)\phi(x)|0\rangle\neq 0 $$
and the technical name for this fact is - vacuum fluctuations of the field $\phi(x)$.

Two issues. I thought VP is associated with force between two charged particles. Why is it associated with vacuum.
Why is the result you gave does not equal to zero. What is the expression for phi(x). THANKS

 

[Demystifier]

Two issues. I thought VP is associated with force between two charged particles. Why is it associated with vacuum.
Why is the result you gave does not equal to zero. What is the expression for phi(x). THANKS

Are you familiar with quantum mechanics of a harmonic oscillator?

 

[ftr]

Are you familiar with quantum mechanics of a harmonic oscillator?

yes

 

[ftr]

I am familiar with all the topics as in this as an example.
https://www.youtube.com/user/DrPhysicsA
And I have read many book on QM/QFT but still confused about the vacuum issue.
even more so when I read threads like this
https://physics.stackexchange.com/questions/75834/the-vacuum-in-quantum-field-theories-what-is-it

 

[Demystifier]

yes

So, do you know what does it mean that in the ground state of the harmonic oscillator we have
$$\langle 0|x|0\rangle =0, \;\; \langle 0|x^2|0\rangle \neq 0 \; ?$$

 

[samalkhaiat]

the state is Poincare invariant
$$\exp(\mathrm{i} \alpha_G \hat{G}) |\Omega \rangle=\exp(\mathrm{i} \alpha_G g) |\Omega \rangle,$$
where $G$ can be chosen as the 10 basic generators of the Poincare group (i.e., four-momentum and four-angular-momentum). The corresponding eigenvalue $g$ of the vacuum vector is arbitrary. If you choose $g \neq 0$ you have to consider the general unitary ray representations of the Poincare group to make it consistent with the Poincare Lie algebra built by the $\hat{G}$.

There is a nicer way of saying almost the same thing: Consider the element $U( A , x) \in \mbox{U}(\mathcal{H})$, with $(A , x) \in \mbox{SL}(2 , \mathbb{C}) \ltimes \mathbb{R}^{(1,3)}$, and use the group law to write it as follows
$$U(A , x) = U(1 , x) \ U(A , 0) = U(A , 0) \ U(1 , A^{-1}x) .$$ Now, translation invariance of the vacuum means that $$U(1, y) \ \Omega = U(1, A^{-1}x) \ \Omega = \Omega .$$ Thus
$$U(A , x) \ \Omega = U(1, x) \big( U(A , 0) \ \Omega \big) = U(A , 0) \ \Omega . \ \ \ \ \ (1)$$ But, the middle and the RHS of (1) means that $U(A , 0) \Omega$ is another translational invariant vector. Then, the uniqueness of the vacuum (up to constant) implies that $$U(A , 0) \ \Omega = C_{A} \ \Omega , \ \ \ \ \lvert C_{A} \rvert = 1.$$
This means that $A \mapsto C_{A}$ is a one-dimensional representation of the “Lorentz” group $\mbox{SL}(2 , \mathbb{C})$. However, as a perfect* group, the Lorentz group has no non-trivial 1-dimensional representation. Thus, we must have $C_{A} = 1$, and
$$U(A , x) \Omega = U(A , 0)\Omega = \Omega .$$

*A group $G$ is called perfect, if its Abelianization $\big($ i.e., $\frac{G}{[G ,G]} \cong \mbox{H}^{1}(G , \mathbb{Z})$ $\big)$ is a trivial group.

 

[vanhees71]

But you argue again with the proper representation! That's my whole point! $C_A \neq 1$ is "allowed" here, because kets that differ only by a phase factor represent the same state. That's why one has to consider unitary ray representations rather than unitary representations to draw the correct conclusion about the eigenvalues of the symmetry generators of the vacuum state, and in fact they are arbitrary as in classical relativistic physics. There's no argument to define absolute additive values for energy, momentum of the vacuum state, and there's no necessity (within SRT not GRT!) to do so because additive constants to the additive conserved quantities are unobservable.

Of course, you are right with the math. There 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 represantations, but it's not necessary. That would be great, because then there'd be no "cosmological-constant problem" within the Standard Model, because then the energy of the vacuum would be necessarily 0 by a symmetry, and there'd be no fine-tuning problem for "dark energy", i.e., one issue considered as a problem for decades of the Standard Model.

 

[samalkhaiat]

But you argue again with the proper representation!

So did you my friend! Or, to be precise, that is the meaning of the equation you wrote:
1) Your operator $e^{i \alpha \cdot G} \equiv U(\alpha)$ is an element of $\mbox{U}(\mathcal{H})$, the (topological) group of unitary operators on the (separable) Hilbert space $\mathcal{H}$, i.e., it is not an element of $\mbox{PU}(\mathcal{H}) \cong \mbox{U}(\mathbb{P}\mathcal{H})$, the projective unitary group of $\mathcal{H}$.
2) You applied $U(\alpha) \in \mbox{U}(\mathcal{H})$ on the vacuum vector $\Omega \in \mathcal{H}$, which is a choice of normalised representative of the vacuum state $[\Omega] = \mathbb{C}\Omega \in \mathbb{P}\mathcal{H}$, the distinguished state in the quantum space of the states $\mathbb{P}\mathcal{H}$ also called the projective Hilbert space.
3) Therefore, the equation you wrote simply means that $$U(\alpha) : \mathcal{H} \to \mathcal{H}.$$ That is to say that $$U : \mbox{SL}(2 , \mathbb{C}) \ltimes \mathbb{R}^{(1,3)} \to \mbox{U}(\mathcal{H})$$ is a unitary representation of the simply connected group (also called the quantum Poincare' group) $\mbox{SL}(2 , \mathbb{C}) \ltimes \mathbb{R}^{(1,3)}$ in $\mathcal{H}$.
So, I was improving on your post. This is why I said “There is a nicer way of saying almost the same thing”.

 

[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!