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

[Demystifier]

If the vacuum is supposed to be Poincare invariant, then it must have zero momentum (and energy) in every frame, correct? Otherwise the vacuum would pick out a particular frame as its "rest frame" (the frame in which it has zero momentum), and would not be Poincare invariant.

As I already mentioned before, the zero momentum is not the only Poincare invariant value. Another Poincare invariant value is infinity. That's exactly the reason why "naive" QFT with default (not normal) ordering gives the infinite values.

 

[samalkhaiat]

I am always more interested in physical implications.

Okay. In order to obtain any physically sensible quantity from QFT, you need to normal-order the point-wise product of fields in the interaction Lagrangian. So, the question is: Is normal-ordering allowed or ad-hoc in QFT? Careful treatment (as in #41) shows that normal-ordering is an allowed procedure in QFT. On the other hand, if you follow the “usual treatment” of text-books (including chapter 7 of Weinberg’s), i.e., if you neglect surface terms, then normal-ordering becomes an ad-hoc procedure in QFT [By neglecting surface integrals, you can show that the Noether charges $(P^{\mu} , J^{\mu\nu})$ satisfy the ordinary Poincare’ algebra (no central charges). Thus, the uniqueness theorem applies to the generators $(P^{\mu} , J^{\mu\nu})$ and does not permit the possibility of a vacuum subtraction].

Can this math help to resolve the cosmological constant problem?

Very un-likely, because $T^{\mu\nu}$ is symmetric in GR. The manipulations in #41 cannot be performed on the symmetrized energy-momentum tensor.

 

[vanhees71]

But then why do you discuss this in the context of Weinberg's book and Samalkhaiat's argument, which both rely on a Poincare invariance vacuum?

Only in the nonrelativistic case.

But your proposal causes bad causality problems in the relativistic case. Any massive particle moves with a timelike 4-momentum, and you cannot change its energy or momentum by an arbitrary shift! This does not even preserve the time-likeness since you can change the energy to something negative or its momentum to something huge! Thus your alleged freedom violates basic principles of relativity!

On the other hand, if you cannot change the energy in case of a massive particle why do you insist on allowing the arbitrary shift for the vacuum?

No, as I've cited early on in this discussion Weinberg discusses the complete realization of the Poincare symmetry in terms of ray representations. What @samalkhaiat did in his posting was to show the fact that the vacuum state is an eigenstate for 0 eigenvalues only for the special case where you set all central charges to 0 and how to realize this standard choice with local fields. Using general ray representations, which are however all "equivalent" (i.e., can be lifted) to the unitary representations.

From this point of view it might occur as a minor issue, because as long as you discuss just special-relativistic QFT it indeed doesn't matter, but it's of great relevance for the still unsolved problem of QFT in curved spacetime and cosmology, let alone the even less understood question whether there is a consistent quantum description of gravity and whether one needs one at all.

 

[vanhees71]

If the vacuum is supposed to be Poincare invariant, then it must have zero momentum (and energy) in every frame, correct? Otherwise the vacuum would pick out a particular frame as its "rest frame" (the frame in which it has zero momentum), and would not be Poincare invariant.

The vacuum state for the Wigner-Weyl case (no spontaneous symmetry breaking) is given by the Statistical operator $|\Omega \rangle \langle \Omega|$ and not just $|\Omega \rangle$. Thus, 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 $\hat{G}$.

For details, see Weinberg, Quantum Theory of Fields Vol. I.

 

[Demystifier]

Careful treatment (as in #41) shows that normal-ordering is an allowed procedure in QFT.

Just to be sure about language, by "allowed" you don't mean mandatory, am I right?

 

[Demystifier]

The manipulations in #41 cannot be performed on the symmetrized energy-momentum tensor.

Precisely what step in #41 cannot be performed?

 

[samalkhaiat]

Just to be sure about language, by "allowed" you don't mean mandatory, am I right?

There is absolutely no language ambiguity in my posts. I’ve already told you that sensible results can only be obtained from normal-ordered Lagrangians.

Precisely what step in #41 cannot be performed?

Every single step, because the equations (A), (B), (1) and (2) get screwed up by surface terms and EOM terms.

 

[samalkhaiat]

The Poincare group allows you to do that, because any unitary ray representation can be equivalently lifted to a unitary representation

Technically, this is not correct. Given a Lie group $G$ and its Lie algebra $\mathfrak{g}$, then every projective unitary representation $$\rho : G \to \mbox{U}(\mathbb{P}\mathcal{H}) ,$$ lifts to a unique unitary representation $$U : G \to \mbox{U}(\mathcal{H}) ,$$ if the following two conditions hold: 1) $G$ is simply connected, and 2) the second cohomology group of $\mathfrak{g}$ is trivial, i.e., $\mbox{H}^{2}(\mathfrak{g}, \mathbb{R}) = 0$.
The Poincare’ group $\mathbb{R}^{(1,3)} \rtimes \mbox{SO}(1, 3)$ satisfies the second condition (this is why we were able to eliminate the central charges from algebra) but not the first (it is connected but not simply connected). However, the 2 to 1 covering map $$\varphi : \mathbb{R}^{(1,3)} \rtimes \mbox{SL}(2, \mathbb{C}) \to \mathbb{R}^{(1,3)} \rtimes \mbox{SO}(1, 3) \ \left( \cong \frac{ \mathbb{R}^{(1,3)} \rtimes \mbox{SL}(2, \mathbb{C})}{\{ (0 , \pm I)\}}\right) ,$$ is also a homomorphism whose kernel $\{ (0 , \pm I)\}$ coincide with the centre of $\mathbb{R}^{(1,3)} \rtimes \mbox{SL}(2 , \mathbb{C})$. In other words, we have the following short exact sequence of groups and homomorphisms $$1 \rightarrow \{ (0 , \pm I ) \} \overset {i}{\hookrightarrow} \mathbb{R}^{(1,3)} \rtimes \mbox{SL}(2, \mathbb{C}) \overset {\varphi}{\rightarrow} \mathbb{R}^{(1,3)} \rtimes \mbox{SO}(1, 3) \rightarrow 1 .$$ This simply means that the group in the middle (the universal covering group) is the central extension of the Poincare’ group $\mathbb{R}^{(1,3)} \rtimes \mbox{SO}(1, 3)$ by the group $\{ (0 , \pm I )\}$. Now, any (irreducible) projective unitary representation $$\pi : \mathbb{R}^{(1,3)} \rtimes \mbox{SO}(1, 3) \to \mbox{U}(\mathbb{P}\mathcal{H}) ,$$ induces (an irreducible) projective unitary representation of the universal covering group given by the following composition of homomorphisms $$\pi \circ \varphi = \hat{\pi} : \mathbb{R}^{(1,3)} \rtimes \mbox{SL}(2, \mathbb{C}) \to \mbox{U}(\mathbb{P}\mathcal{H}) .$$ This, in turn, lifts to (an irreducible) unitary representation $$U : \mathbb{R}^{(1,3)} \rtimes \mbox{SL}(2, \mathbb{C}) \to \mbox{U}(\mathcal{H}) \ ,$$ because $\mathbb{R}^{(1,3)} \rtimes \mbox{SL}(2, \mathbb{C})$ is simply connected and the second cohomology group of its Lie algebra is trivial. Indeed, there is a bijective correspondence between the (irreducible) continuous projective unitary representations of the connected Poincare’ group $\mathbb{R}^{(1,3)} \rtimes \mbox{SO}(1, 3)$ and the (irreducible) continuous unitary representation of the simply connected covering group $\mathbb{R}^{(1,3)} \rtimes \mbox{SL}(2, \mathbb{C})$.

Finally, recall that the quotient (or canonical projection) map $$p : \mbox{U}( \mathcal{H}) \to \mbox{U}( \mathbb{P}\mathcal{H}) \ \left( \cong \frac{\mbox{U}( \mathcal{H})}{ \mbox{U}(1)} \right),$$ with the centre $\mathcal{Z}\left(\mbox{U}(\mathcal{H})\right) = \big\{ \lambda \ \mbox{id}_{\mathcal{H}}; | \lambda | = 1 \big\}$ identified with $\mbox{U}(1)$, gives us the following short exact sequence of groups and homomorphisms $$1 \rightarrow \mbox{U}(1) \overset{i}{\hookrightarrow} \mbox{U}(\mathcal{H}) \overset{p}{\rightarrow} \mbox{U}(\mathbb{P}\mathcal{H}) \rightarrow 1 .$$ This means that the unitary group of $\mathcal{H}$ $\big($i.e., $\mbox{U}(\mathcal{H})$$\big)$ is the central extension of the projective unitary group $\mbox{U}(\mathbb{P}\mathcal{H})$ by the group $\mbox{U}(1)$ $\big($ it is, at the same time, a locally trivial principal $\mbox{U}(1)$-bundle over $\mbox{U}(\mathbb{P}\mathcal{H})$$\big)$. Now, if you put the above two exact sequences on top of each other, you obtain a commutative diagram (which I don’t know the correct command for it on here), because one can show that the projective representation $\hat{\pi}$ factors according to $$\pi \circ \varphi = \hat{\pi} = p \circ U .$$

 

[weirdoguy]

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

 

[vanhees71]

Technically, this is not correct. Given a Lie group $G$ and its Lie algebra $\mathfrak{g}$, then every projective unitary representation $$\rho : G \to \mbox{U}(\mathbb{P}\mathcal{H}) ,$$ lifts to a unique unitary representation $$U : G \to \mbox{U}(\mathcal{H}) ,$$ if the following two conditions hold: 1) $G$ is simply connected, and 2) the second cohomology group of $\mathfrak{g}$ is trivial, i.e., $\mbox{H}^{2}(\mathfrak{g}, \mathbb{R}) = 0$.

Thanks for the correction. Of course, that's why we use the central extension for the Poincare and the Galilei groups in special relativistic and non-relativistic QT, respectively. The point is that for the Galilei group there's one "non-trivial" central charge, which physically is the mass or the system. Of course, also for the Galilei group we use its central extension, using the covering group SU(2) instead of the "classical" rotation group SO(3).

What I was arguing about is that the finite eigenvalues of energy, momentum, and angular momentum in any representation of the Poincare group are not determined by the group, i.e., the Minkowski space-time model since in QT the most general realizations of symmetries are unitary ray representations rather than proper unitary representations. That's also the very reason why we are able to make physics sense for the use of ray representations of the central extensions of the "classical" groups rather than the classical groups themselves, and indeed obviously that's what's realized in nature since there are definitely half-integer spin representations realized in Nature.

 

[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”.