In summary, quantum field theories have a unique ground state, the vacuum state, which is defined as a Poincare invariant state with a 4-momentum of zero. This results in a ground state energy of zero due to Lorentz covariance. The issue of the vacuum energy being arbitrary can be resolved through normal ordering or by setting an arbitrary ultraviolet cutoff.
  • #36
vanhees71 said:
it's this arbitrary choice which fixes ##\hat{H}## absolutely
It is not an arbitrary choice but the only invariant choice. It makes the mass shells be ##p^2=m^2## rather than ##(p_0-E)^2-p_1^2-p_2^2-p_3^2=m^2##. It is also needed for many other formulas that require ##P## to be covariant.
vanhees71 said:
it's an argument to use "normal ordering" for free-field
This postulated relation between normal ordering and energy shifts through introducing a central charge is nonexistent, at least it is not in Weinberg's book.

Weinberg (on whom you rely for your central charge argument) doesn't use this argument. He introduces normal ordering on p.175 as a normal form in which to represent arbitrary field operators, which is indeed the natural thing to do. He later gives a complete discussion of the free field without any reference to normal ordering. The next mention of the term is on p.200, where he uses it to normally order the interaction term in the classical action. Not in order to make the vacuum energy zero, which he obtains on p.65 (case (f)) as an automatic consequence of his definitions (2.4.18-24) of the commutation relations.
vanhees71 said:
the starred sections in Weinberg's books are not "hardly relevant" but "not so relevant at a first read".
I meant, hardly relevant for the subject of his book, relativistic QFT.
 
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  • #37
vanhees71 said:
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.
it is as good from a purely mathematical point of view. But not from a physics point of view since all symbols used for the generators no longer have the standard meaning. Instead energy, momentum, angular momentum and boost differ from it by an arbitrary and physically irrelevant constant shift.
 
  • #40
ftr said:
https://en.wikipedia.org/wiki/Zero-point_energy
especially the section on The quantum electrodynamic vacuum
This wikipedia article is a digest of what one can read about the topic in the popular literature (and in carelessly written scientific literature) - but little of it is founded in substance. For example, the tiny initial section on terminology is already confused:
wikipedia said:
A vacuum can be viewed not as empty space but as the combination of all zero-point fields. In quantum field theory this combination of fields is called the vacuum state, its associated zero-point energy is called the vacuum energy and the average energy value is called the vacuum expectation value (VEV) also called its condensate.
In QFT, fields are quantum fields, and there is no notion of a zero point field. A state (and therefore also the vacuum state) is not a combination of fields but an assignment of expectation values to fields and their products. A condensate is a nonvanishing vacuum expectation value of a field, and not an average energy.
In the section you mention, already the first formula (the commutator) is wrong - it describes the relations of a single harmonic oscillator, not of a free quantum field.
wikipedia said:
In a process in which a photon is annihilated (absorbed), we can think of the photon as making a transition into the vacuum state. Similarly, when a photon is created (emitted), it is occasionally useful to imagine that the photon has made a transition out of the vacuum state.
If you think of it this way, you are mislead. A photon cannot transit into or be created from a vacuum state. This is forbidden by the structure of the S-matrix. Instead, photons are absorbed and created by matter only, and this is a transition of a non-vacuum state to a non-vacuum state. You can imagine whatever you like, but imagination is not physics!

This shows the level of quality of the whole text. You can use it to inform yourself about the buzzwords, phrases and references, but not about proper physical content.

wikipedia said:
any point in space that contains energy can be thought of as having mass to create particles. Virtual particles spontaneously flash into existence at every point in space due to the energy of quantum fluctuations caused by the uncertainty principle.
This is an assertion of the content of the vacuum fluctuation myth that I critically discussed here. It is the way physicists (even famous ones) report their abstract findings to the general public, painting mythical stories that substitute for the unintuitive abstract reality of their actual findings. It gives a vague intuition about the real QFT, but if you want to learn the subject properly you need to unlearn all that mythical stuff.

ftr said:
This book by Milonni does real physics but also appeals to the imagination without giving a valid physical reasoning for it. E.g., p.48; ''we can imagine them to be fluctuating about their mean values of zero''. And then he states without proof the wrong statement in Wikipedia (which probably copied it from Milonni) about transitions of photons to the vacuum state.

So read and understand his formulas and proper calculations. But doubt his imagination and interpretations, which are not substantiated by the formulas!
 
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  • #41
bhobba said:
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.
At last on PF, I see a question worth answering. It is, indeed, a very good question. I am glad that you brought this up, for I am not aware of any textbook or paper that tackles this issue. So, bellow you will see (probably for the first time) how the vacuum-subtraction (or normal ordering) arises almost naturally in QFT.

All good and bad features in QFT have their origin in the process of “integrating by parts and ignoring surface terms”. A process which almost all authors use as if it is taken for granted that surface terms do vanish. So, I will avoid this process by using expressions that can be derived with no reference to the behaviour at the boundary.

Let us consider a field theory whose action integral is invariant under the Poincare’ group. Then, by Noether theorem, we have a conserved translation current given by the energy-momentum tensor [tex]T^{\mu\nu} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\varphi_{r})}\partial^{\nu}\varphi_{r} - \eta^{\mu\nu}\mathcal{L} \ ,[/tex] and a conserved Lorentz current given by the moment tensor [tex]M^{\rho\mu\nu} = x^{\mu}T^{\rho\nu} -x^{\nu}T^{\rho\mu} + S^{\rho\mu\nu} \ ,[/tex] [tex]S^{\rho\mu\nu}(x) = - i \frac{\partial \mathcal{L}}{\partial (\partial_{\rho}\varphi_{r})} \left( \Sigma^{\mu\nu}\right)_{r}{}^{s}\varphi_{s}(x) .[/tex] And the corresponding Noether charges are given by [tex]P^{\mu} = \int d \sigma_{\rho}(x) \ T^{\rho\mu}(x) = \int d^{3}x \ T^{0 \mu}(x) \ , \ \ \ \ \ \ \ \ \ (A)[/tex][tex]J^{\mu\nu} = \int d \sigma_{\rho}(x) \ M^{\rho\mu\nu}(x) = \int d^{3}x \ M^{0 \mu\nu} (x) . \ \ \ \ \ \ \ (B)[/tex] In a classical field theory, where we can almost always ignore surface terms, we can show that [itex]( P^{\mu} , J^{\mu\nu})[/itex] are time-independent Lorentz vector and Lorentz tensor respectively. So, in a classical field theory, this means that [itex]( P^{\mu} , J^{\mu\nu})[/itex] generate the proper Poincare’ algebra (i.e., without central charges). In other words, one can show that the classical [itex]( P^{\mu} , J^{\mu\nu})[/itex] are unique Poincare generators. Clearly this cannot be true in a QFT because 1) surface integrals may not vanish, and 2) the uniqueness of [itex]( P^{\mu} , J^{\mu\nu})[/itex] does not permit the possibility of vacuum subtractions.

Okay, let us start. People call Noether theorem “the beautiful theorem”. However, one property of Noether charge is more beautiful that the entire theorem. Indeed we can show, with no reference to 1) symmetry considerations (i.e., conservation law), 2) dynamical consideration (i.e., equation of motion) and/or 3) the behaviour at the boundary, that the Noether charge generates the correct infinitesimal transformation on local operators. In fact, using only the canonical equal-time commutation relations and the expressions for [itex]( P^{\mu} , J^{\mu\nu})[/itex], it is an easy exercise to show that [tex][iP^{\mu} , \varphi (x)] = \partial^{\mu}\varphi (x) \ , \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)[/tex][tex][iJ^{\mu\nu}, \varphi (x)] = \left( x^{\mu}\partial^{\nu}- x^{\nu}\partial^{\mu} - i \Sigma^{\mu\nu} \right) \varphi (x) . \ \ \ \ \ (2)[/tex]

So, these are the equations that we can start working with. Now, we use (1) and (2) to evaluate the RH sides of the following Jacobi identities

[tex]\big[ [P^{\mu},P^{\nu}] , \varphi (x) \big] = \big[ P^{\mu}, [ P^{\nu} ,\varphi (x)] \big] - \big[ P^{\nu}, [ P^{\mu} ,\varphi (x) ] \big] ,[/tex]
[tex]\big[ [P^{\sigma} , J^{\mu \nu}] , \varphi (x) \big] = \big[ P^{\sigma} , [ J^{\mu \nu}, \varphi (x)] \big] - \big[ J^{\mu \nu} , [ P^{\sigma}, \varphi (x) ] \big],[/tex]
[tex]\big[ [J^{\mu \nu} , J^{\rho \sigma}] , \varphi (x) \big] = \big[ J^{\mu \nu} , [J^{\rho \sigma}, \varphi(x) ] \big] - \big[ J^{\rho \sigma}, [J^{\mu \nu}, \varphi (x) ] \big] .[/tex] After couple of pages of easy algebra, we obtain the following relations

[tex]\big[ [iP^{\mu} , P^{\nu}] , \varphi (x) \big] = 0 ,[/tex]
[tex]\big[ [iP^{\sigma} , J^{\mu \nu}] - \eta^{\sigma \nu}P^{\mu} + \eta^{\sigma \mu}P^{\nu} , \varphi (x) \big] = 0 ,[/tex]
[tex]\big[ [iJ^{\mu \nu} , J^{\rho \sigma}] - \eta^{\mu \rho} J^{\nu \sigma} - \eta^{\mu \sigma} J^{\rho \nu} - \eta^{\nu \rho} J^{\mu \sigma} - \eta^{\nu \sigma} J^{\rho \mu} , \varphi (x) \big] = 0 .[/tex]
The most general solutions of these equations are given by
[tex]\big[ iP^{\mu} , P^{\nu}\big] = C^{\mu , \nu} , [/tex]
[tex]\big[ iP^{\sigma} , J^{\mu \nu}\big] = \eta^{\sigma \nu} P^{\mu} - \eta^{\sigma \mu} P^{\nu} + C^{\sigma , \mu \nu} ,[/tex]
[tex]\big[ iJ^{\mu \nu} , J^{\rho \sigma} \big] = \eta^{\mu \rho} J^{\nu\sigma} + \cdots + C^{\mu \nu , \rho \sigma} ,[/tex] where all the [itex]C[/itex]’s are constants. If these constants are not all zero, we conclude that the Noether charges [itex]\big( P^{\mu} , J^{\mu\nu})[/itex] generate a centrally extended Poincare’ algebra. Next, we may consider various Jacobi identities to establish certain algebraic relations for the central charges. For example, the Jacobi identity
[tex]\big[ J^{\mu \nu} , [ P^{\rho}, P^{\sigma}] \big] + \big[ P^{\sigma} , [ J^{\mu \nu}, P^{\rho}] \big] + \big[ P^{\rho} , [ P^{\sigma}, J^{\mu \nu}] \big] = 0 ,[/tex] leads to [tex]C^{\mu , \nu} = 0 .[/tex] Another useful Jacobi identity is
[tex]\big[ J^{\lambda \tau} , [ J^{\mu \nu} , P^{\rho}] \big] + \cdots \ = 0 .[/tex] This allows us to express [itex]C^{\mu , \rho \sigma}[/itex] as
[tex]C^{\mu , \lambda \tau} = \eta^{\mu \tau} \left( \frac{1}{3} \eta_{\rho \nu} C^{\rho , \lambda \nu}\right) - \eta^{\mu \lambda} \left( \frac{1}{3} \eta_{\rho \nu} C^{\rho , \tau \nu}\right),[/tex] which may be used to define the constant [tex]C^{\mu} \equiv \frac{1}{3} \eta_{\rho \nu} C^{\rho , \mu \nu} .[/tex] And the last Jacobi identity is between 3 [itex]J[/itex]’s
[tex]\big[ J^{\lambda \tau} , [ J^{\mu \nu}, J^{\rho \sigma}] \big] + \cdots \ = 0 .[/tex] This allows us to define yet another constant in terms of [itex]C^{\mu \rho , \nu \sigma}[/itex] [tex]C^{\mu \nu} \equiv \frac{1}{2} \eta_{\rho \sigma} C^{\mu \rho , \nu \sigma} .[/tex] Now, it is an easy exercise to show that the shifted Noether charges [tex]\bar{P}^{\mu} = P^{\mu} + C^{\mu},[/tex][tex]\bar{J}^{\mu \nu} = J^{\mu \nu} + C^{\mu \nu} ,[/tex] form an ordinary representation of the Poincare’ algebra, i.e., with no central charges. The boring algebraic details of all this can be found in Weinberg’s book QFT, Vol.1, P(84-86). However, Weinberg does not explain the important meaning of [itex](C^{\mu} , C^{\mu \nu})[/itex] in QFT. So, we will do better than the old man by showing that [tex]C^{\mu} = - \langle 0 | P^{\mu}| 0 \rangle \ ,[/tex][tex]C^{\mu \nu} = - \langle 0 | J^{\mu \nu} | 0 \rangle \ .[/tex]

Okay, let us commute the Noether charge [itex]P^{\sigma}[/itex] with the moment tensor [itex]M^{\rho \mu \nu}(x)[/itex]. By the linearity of the bracket, we get
[tex]\big[ i P^{\sigma} , M^{\rho\mu\nu}(x) \big] = x^{\mu} \big[iP^{\sigma} , T^{\rho \nu}(x) \big] - x^{\nu} \big[ iP^{\sigma} , T^{\rho \mu}(x) \big] + \big[iP^{\sigma} , S^{\rho \mu \nu}(x) \big] .[/tex] Using the fact that Eq(1) holds for any local operator, we get [tex]\big[ iP^{\sigma} , M^{\rho \mu \nu}(x) \big] = x^{\mu}\partial^{\sigma}T^{\rho \nu} - x^{\nu} \partial^{\sigma}T^{\rho \mu} + \partial^{\sigma}S^{\rho \mu \nu} .[/tex] We rewrite this as [tex]\big[ iP^{\sigma} , M^{\rho \mu \nu}(x) \big] = \partial^{\sigma}M^{\rho \mu \nu} - \eta^{\sigma \mu}T^{\rho \nu} + \eta^{\sigma \nu}T^{\rho \mu} .[/tex] Integrating this over the hyper-surface [itex]\int d \sigma_{\rho}(x)[/itex], and using the definitions of the Noether charges [Eq’s (A) and (B)], we obtain [tex]\big[ iP^{\sigma} , J^{\mu\nu} \big] = \eta^{\sigma \nu}P^{\mu} - \eta^{\sigma \mu}P^{\nu} + \int d \sigma_{\rho}(x) \ \partial^{\sigma}M^{\rho \mu \nu}(x) .[/tex] Now, we rewrite this in terms of the shifted Noether charges [itex](\bar{P}^{\mu} , \bar{J}^{\mu\nu})[/itex]. Using [itex]P^{\mu} = \bar{P}^{\mu} - C^{\mu}[/itex] and [itex]J^{\mu\nu} = \bar{J}^{\mu\nu} - C^{\mu\nu}[/itex], we find [tex]\big[ i\bar{P}^{\sigma} , \bar{J}^{\mu\nu} \big] = \eta^{\sigma \nu}\bar{P}^{\mu} - \eta^{\sigma \mu}\bar{P}^{\nu} + \left( \eta^{\sigma \mu}C^{\nu} - \eta^{\sigma \nu}C^{\mu} + \int d \sigma_{\rho}(x) \ \partial^{\sigma}M^{\rho \mu \nu}(x) \right) .[/tex]
Since the shifted Noether charges satisfy the ordinary Poincare’ algebra (no central charges), we must have
[tex]\eta^{\sigma \nu}C^{\mu} - \eta^{\sigma \mu}C^{\nu} = \int d \sigma_{\rho} \ \partial^{\sigma}M^{\rho \mu \nu}(x) . \ \ \ \ \ \ (3)[/tex]
Of course, if [itex]M^{\rho \mu \nu}(x) \to 0[/itex] as [itex]|\vec{x}| \to \infty[/itex], then we can apply the Schwinger identity [itex]\int d \sigma^{\rho}(x) \ \partial^{\sigma}F(x) = \int d \sigma^{\sigma}(x) \ \partial^{\rho}F(x)[/itex] on the RHS of (3) and obtain [tex]\eta^{\sigma \nu}C^{\mu} - \eta^{\sigma \mu}C^{\nu} = \int d \sigma^{\sigma}(x) \ \partial_{\rho}M^{\rho \mu \nu}(x) = 0,[/tex] because of the conservation law [itex]\partial_{\rho}M^{\rho \mu \nu} = 0[/itex]. This then leads to [itex]C^{\mu} = 0[/itex]. However, in QFT it is not always true that the operator [itex]M^{\rho \mu \nu}[/itex] vanishes at infinity. So, we cannot always use the Schwinger identity. Instead, we will stay away form the behaviour at infinity and try to determine the constant [itex]C^{\mu}[/itex] from Eq(3). Contracting Eq(3) with [itex]\eta_{\sigma \nu}[/itex] and doing the differentiation on the moment tensor, leads us to
[tex]3C^{\mu} = \eta_{\sigma \nu} \int d \sigma_{\rho} \left( x^{\mu} \partial^{\sigma}T^{\rho \nu} - x^{\nu}\partial^{\sigma}T^{\rho \mu} + \partial^{\sigma}S^{\rho \mu \nu} \right) - 3P^{\mu}.[/tex] Now, if we take the vacuum expectation value, the integrand vanishes by translation invariance of the vacuum: [itex]\langle 0 | \partial \mathcal{O}(x) | 0 \rangle = \partial \langle 0 | \mathcal{O}(0) | 0 \rangle = 0[/itex]. Thus, we obtain [tex]C^{\mu} = - \langle 0 | P^{\mu}| 0 \rangle ,[/tex] and the vacuum subtraction is, therefore, justified [tex]\bar{P}^{\mu} = P^{\mu} - \langle 0 | P^{\mu} | 0 \rangle .[/tex]

Similar, but more complicated calculation, leads to [tex]\bar{J}^{\mu\nu} = J^{\mu \nu} - \langle 0 | J^{\mu \nu} | 0 \rangle .[/tex]
 
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  • #42
samalkhaiat said:
At last on PF, I see a question worth answering. It is, indeed, a very good question.
That's a hell of a long-winded way to say 'yes'. :-) :-) :-)
 
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  • #43
samalkhaiat said:
I am glad that you brought this up, for I am not aware of any textbook or paper that tackles this issue. So, bellow you will see (probably for the first time) how the vacuum-subtraction (or normal ordering) arises almost naturally in QFT.
.
WOW - amazing :woot::woot::woot::woot::woot:.

And yes, no text I have seen has anything like it.

Thanks
Bill
 
  • #44
samalkhaiat said:
the vacuum subtraction is, therefore, justified
The subtraction of vacuum expectation values is just the normal ordering with respect to the vacuum state. The C's vanish automatically when one begins directly with a meaningful operator expression for the generators; the unordered expression is mathematically ill-defined even in the free case. In the interacting case, the expressions need further renormalization beyond normal ordering to be meaningful, except in 1+1 dimensions.
samalkhaiat said:
Weinberg does not explain the important meaning of [the C's] in QFT.
This follows directly from his exposition and the fact that his vacuum expectation values vanish, so after a shift the C's become (negative) vacuum expectation values.
 
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  • #45
Almighty BOB said:
That's a hell of a long-winded way to say 'yes'. :-) :-) :-)
No, that was a way of saying: I don't do Alice, and I certainly don't do Bob. Therefore, I don't do "Alice & Bob". :wink:
 
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  • #46
samalkhaiat said:
No, that was a way of saying: I don't do Alice, and I certainly don't do Bob. Therefore, I don't do "Alice & Bob". :wink:
A teacher friend of mine once said "a good teacher is not someone who can explain, a good teacher is someone who can explain TO THE PEOPLE AT THE BACK.". :-) :-) :-)

Edit... he also used to say "Hey ho, we're bounded below..."
 
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  • #47
samalkhaiat said:
However, Weinberg does not explain the important meaning of [itex](C^{\mu} , C^{\mu \nu})[/itex] in QFT. So, we will do better than the old man by showing that [tex]C^{\mu} = - \langle 0 | P^{\mu}| 0 \rangle \ ,[/tex][tex]C^{\mu \nu} = - \langle 0 | J^{\mu \nu} | 0 \rangle \ .[/tex]
[/tex]
Thanks a lot for this very clear and nice derivation, but that's indeed all in Weinberg's book, although he doesn't make the statement so explicitly as in your posting.

I think this confirms my argument that in SRT (including relativistic QFT) there's no absolute zero of energy (and the other conserved quantities).

Maybe you can also clarify another question, which is not clear even to many among us practitioners of relativistic (many-body) QFT, and that's the question about the densities (or currents) of the conserved Noether charges. Obviously Noether's theorem does not uniquely determine these currents. E.g., the naive canonical energy-momentum tensor of electromagnetics (QED in the quantized version) is neither symmetric nor gauge invariant. One can of course always construct a symmetric and gauge invariant em tensor which is equivalent (in a naive sense as detailed below) to the canonical one, because it leads to the same total energy and momentum.

However, and that's why I'm asking, this statement is indeed "modulo surface terms". Particularly I don't see any necessity for the energy-momentum tensor to be symmetric in the realm of special relativity. For gauge fields, of course you can argue with gauge invariance, but that aside only GR demands for a symmetric energy-momentum tensor.

Now, related with the question about the symmetry or asymmetry of the energy-momentum tensor is also the issue of angular momentum, particularly the possibility to split it into "orbital and spin" angular momentum. In my opinion, there's neither uniqueness in this split nor is there a really well defined treatment of spin in relativistic hydrodynamics, and that's an issue which is indeed of high interest today in the heavy-ion community in regard to a possible manifestation of the "chiral magnetic effect" and the relativistic rotation of the "fireball" produced in heavy-ion collisions, related to the polarization of particles. See, e.g.,

https://www.nature.com/nature/journal/v548/n7665/full/nature23004.html
https://arxiv.org/abs/1701.06657

https://www.nature.com/articles/548034a
 
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  • #48
samalkhaiat said:
So, bellow you will see (probably for the first time) how the vacuum-subtraction (or normal ordering) arises almost naturally in QFT

Can you be more direct , is there vacuum energy or not? or is it the case that whatever is there can be "handled" mathematically.
 
  • #49
samalkhaiat said:
Now, if we take the vacuum expectation value, the integrand vanishes by translation invariance of the vacuum:
I would say that's the crucial part of your lengthy derivation: It is assumed (or postulated) that the vacuum is translation invariant. It's important to emphasize it because some states that are also called "vacuum" in the literature are not translation invariant. Examples are Casimir "vacuum" and "vacuum" in curved spacetime. That's why in such examples the "vacuum" energy should not be simply thrown away.
 
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  • #50
Demystifier said:
I would say that's the crucial part of your lengthy derivation: It is assumed (or postulated) that the vacuum is translation invariant.
Of course, the postulates of relativistic QFT are given! Equations (1) and (2) are derived from the commutation relations of QFT.
The “crucial part in my lengthy derivation” is the fact that [itex](P^{\mu}, J^{\mu\nu})[/itex] generate a Projective Unitary Representation, while the [itex](\bar{P}^{\mu}, \bar{J}^{\mu\nu})[/itex] generate what we need,i.e., an ordinary Unitary Representation. So, #41 is the answer to the following question: Given (the postulates of) relativistic QFT, how would one mathematically justify the vacuum subtraction? This is not trivial because the Poincare’ generators in an ordinary representation are unique. In other words, if [itex]G_{1}^{A} = (P_{1}^{\mu} , J_{1}^{\mu\nu})[/itex] and [itex]G_{2}^{A}= (P_{2}^{\mu} ,J_{2}^{\mu\nu})[/itex] both satisfy the ordinary Poincare’ algebra, then one can prove that [itex]G_{1}^{A} = G_{2}^{A}[/itex], i.e., the vacuum subtraction is not possible.
In #41, this uniqueness theorem applies to the [itex](\bar{P}^{\mu}, \bar{J}^{\mu\nu})[/itex] but not to the [itex](P^{\mu}, J^{\mu\nu})[/itex], because the algebra of the latter contains central charges, i.e., not an ordinary Poincare’ algebra.
 
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  • #51
samalkhaiat said:
Of course, the postulates of relativistic QFT are given! Equations (1) and (2) are derived from the commutation relations of QFT.
The “crucial part in my lengthy derivation” is the fact that [itex](P^{\mu}, J^{\mu\nu})[/itex] generate a Projective Unitary Representation, while the [itex](\bar{P}^{\mu}, \bar{J}^{\mu\nu})[/itex] generate what we need,i.e., an ordinary Unitary Representation. So, #41 is the answer to the following question: Given (the postulates of) relativistic QFT, how would one mathematically justify the vacuum subtraction? This is not trivial because the Poincare’ generators in an ordinary representation are unique. In other words, if [itex]G_{1}^{A} = (P_{1}^{\mu} , J_{1}^{\mu\nu})[/itex] and [itex]G_{2}^{A}= (P_{2}^{\mu} ,J_{2}^{\mu\nu})[/itex] both satisfy the ordinary Poincare’ algebra, then one can prove that [itex]G_{1}^{A} = G_{2}^{A}[/itex], i.e., the vacuum subtraction is not possible.
In #41, this uniqueness theorem applies to the [itex](\bar{P}^{\mu}, \bar{J}^{\mu\nu})[/itex] but not to the [itex](P^{\mu}, J^{\mu\nu})[/itex], because the algebra of the latter contains central charges, i.e., not an ordinary Poincare’ algebra.
Your math is deep, but I am always more interested in physical implications. Can this math help to resolve the cosmological constant problem?
 
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  • #52
The physics implication is, and that's why we have this lengthy discussion, that in special relativity there's no possibility to define the absolute scale of energy (density) (and other conserved quantities), and this holds true also in relativistic QT. Contrary to A. Neumaier's claim it cannot be argued with the proper unitary representations of the Poincare group to this effect since in QT all ray representations is what counts. As @samalkait has confirmed (you can find this also in Weinberg's book, as argued by me above), the overall additive constant of the total conserved quantities if arbitrary and thus are their vacuum expectation values.

It is of course true that all central charges of the Poincare algebra (contrary to the issue in case of the Galileo symmetry, where mass occurs as the one crucial non-trivial central charge) are trivial, i.e., you can always lift any ray representation to a proper unitary transformation, and that's why this restriction leads to all possible relativistic QT models, i.e., the standard treatment in textbooks that are less detailed and less careful than Weinberg is correct.

The physics, which now has gone lost in all the math, is the important finding that we cannot argue from SR alone to give a zero value to all vacuum expectation values of the conserved quantities, and thus also the corresponding densities have no additive absolute zero. Now, the only place in physics, where this absolute zero is physically relevant is GR, and it has to do with the still not fully understood finetuning of the very small cosmological constant (compared to expectations from renormalization of the vacuum energy in the standard model, particularly the contribution from the Higgs field as a scalar field) of our universe.
 
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  • #53
vanhees71 said:
ontrary to A. Neumaier's claim it cannot be argued with the proper unitary representations of the Poincare group to this effect since in QT all ray representations is what counts. As @samalkait has confirmed (you can find this also in Weinberg's book, as argued by me above), the overall additive constant of the total conserved quantities if arbitrary and thus are their vacuum expectation values.
No. By your argument, one can also shift the vacuum momentum to any arbitrary value. But this is a completely unphysical spurious effect. The vacuum momentum must be zero in any Lorentz frame, and this forces the vacuum energy to be zero. Energy plays no preferred role in relativistic physics!
 
  • #54
We agree to disagree. If your claim would be true, there'd be no problem about the cosmological constant/dark energy in connection with the Standard Model, but there is a huge problem, depending on which scale you look between a factor of ##10^{60}## to ##10^{100}## or so!

It's also clear from the standard treatment in textbooks. If your claim would be true that the vacuum expectation value of energy and momentum is guaranteed to be 0 from the ray-representation theory of the proper orthochronous Lorentz group alone, you'd never get infinities.

I think Weinberg and @samalkhaiat have it right!
 
  • #55
vanhees71 said:
I think Weinberg and @samalkhaiat have it right!
So you say that physical momentum is defined only up to an arbitrary shift? This would be news to everyone in the world!

vanhees71 said:
If your claim would be true that the vacuum expectation value of energy and momentum is guaranteed to be 0 from the ray-representation theory of the proper orthochronous Lorentz group alone, you'd never get infinities.
It is the formulas that give infinities that are wrong, not my claim. Correct formulas cannot give infinite values!
 
  • #56
vanhees71 said:
there'd be no problem about the cosmological constant/dark energy in connection with the Standard Model
There is no cosmological constant problem in the standard model, which assumes Poincare invariance.

Only quantum gravity, which doesn't assume it, has this problem but is not yet a valid theory, and Samalkhaiat's argument does not even apply.
 
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  • #57
A. Neumaier said:
So you say that physical momentum is defined only up to an arbitrary shift? This would be news to everyone in the world!It is the formulas that give infinities that are wrong, not my claim. Correct formulas cannot give infinite values!
Why should this be news to the world? It's simply the special principle of relativity. The total momentum of a system depends on the reference frame, and you can thus always add an arbitrary constant momentum without changing any physics. It's just the change from one inertial frame of reference to another.

Of course, correct formulas occur via some regularization procedure of illdefined quantities, e.g., a momentum cutoff or more sophisticated procedures to keep symmetries valid in the regularized theory. All quantities are finite, and you can "renormalize" everything observable up to constants that stay finite when the physical limit is taken (momentum cutoff to infinity). That different choices are also irrelevant for the physics content of the theory and leads to the notion of the various renormalization-group equations.
 
  • #58
A. Neumaier said:
There is no cosmological constant problem in the standard model, which assumes Poincare invariance.

Only quantum gravity, which doesn't assume it, has this problem but is not yet a valid theory, and Samalkhaiat's argument does not even apply.
Well, there cannot be a cosmological-constant problem in the standard model since there's no cosmological constant in SRT, but I think we all mean to which notorious problem I refer too (see Weinberg's famous RMP article of 1989). It also has nothing to do with quantum gravity but just with classical GR cosmology.
 
  • #59
vanhees71 said:
I think we all mean to which notorious problem I refer too (see Weinberg's famous RMP article of 1989). It also has nothing to do with quantum gravity but just with classical GR cosmology.
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?
vanhees71 said:
The total momentum of a system depends on the reference frame, and you can thus always add an arbitrary constant momentum without changing any physics. It's just the change from one inertial frame of reference to another.
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?
 
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  • #60
vanhees71 said:
The total momentum of a system depends on the reference frame

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.
 
  • #61
PeterDonis said:
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.
 
  • #62
Demystifier said:
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 [itex](P^{\mu} , J^{\mu\nu})[/itex] satisfy the ordinary Poincare’ algebra (no central charges). Thus, the uniqueness theorem applies to the generators [itex](P^{\mu} , J^{\mu\nu})[/itex] and does not permit the possibility of a vacuum subtraction].

Can this math help to resolve the cosmological constant problem?
Very un-likely, because [itex]T^{\mu\nu}[/itex] is symmetric in GR. The manipulations in #41 cannot be performed on the symmetrized energy-momentum tensor.
 
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  • #63
A. Neumaier said:
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.
 
  • #64
PeterDonis said:
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.
 
  • #65
samalkhaiat said:
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?
 
  • #66
samalkhaiat said:
The manipulations in #41 cannot be performed on the symmetrized energy-momentum tensor.
Precisely what step in #41 cannot be performed?
 
  • #67
Demystifier said:
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.
 
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  • #68
vanhees71 said:
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 [itex]G[/itex] and its Lie algebra [itex]\mathfrak{g}[/itex], then every projective unitary representation [tex]\rho : G \to \mbox{U}(\mathbb{P}\mathcal{H}) ,[/tex] lifts to a unique unitary representation [tex]U : G \to \mbox{U}(\mathcal{H}) ,[/tex] if the following two conditions hold: 1) [itex]G[/itex] is simply connected, and 2) the second cohomology group of [itex]\mathfrak{g}[/itex] is trivial, i.e., [itex]\mbox{H}^{2}(\mathfrak{g}, \mathbb{R}) = 0[/itex].
The Poincare’ group [itex]\mathbb{R}^{(1,3)} \rtimes \mbox{SO}(1, 3)[/itex] 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 [tex]\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) ,[/tex] is also a homomorphism whose kernel [itex]\{ (0 , \pm I)\}[/itex] coincide with the centre of [itex]\mathbb{R}^{(1,3)} \rtimes \mbox{SL}(2 , \mathbb{C})[/itex]. In other words, we have the following short exact sequence of groups and homomorphisms [tex]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 .[/tex] This simply means that the group in the middle (the universal covering group) is the central extension of the Poincare’ group [itex]\mathbb{R}^{(1,3)} \rtimes \mbox{SO}(1, 3)[/itex] by the group [itex]\{ (0 , \pm I )\}[/itex]. Now, any (irreducible) projective unitary representation [tex]\pi : \mathbb{R}^{(1,3)} \rtimes \mbox{SO}(1, 3) \to \mbox{U}(\mathbb{P}\mathcal{H}) ,[/tex] induces (an irreducible) projective unitary representation of the universal covering group given by the following composition of homomorphisms [tex]\pi \circ \varphi = \hat{\pi} : \mathbb{R}^{(1,3)} \rtimes \mbox{SL}(2, \mathbb{C}) \to \mbox{U}(\mathbb{P}\mathcal{H}) .[/tex] This, in turn, lifts to (an irreducible) unitary representation [tex]U : \mathbb{R}^{(1,3)} \rtimes \mbox{SL}(2, \mathbb{C}) \to \mbox{U}(\mathcal{H}) \ ,[/tex] because [itex]\mathbb{R}^{(1,3)} \rtimes \mbox{SL}(2, \mathbb{C})[/itex] 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 [itex]\mathbb{R}^{(1,3)} \rtimes \mbox{SO}(1, 3)[/itex] and the (irreducible) continuous unitary representation of the simply connected covering group [itex]\mathbb{R}^{(1,3)} \rtimes \mbox{SL}(2, \mathbb{C})[/itex].

Finally, recall that the quotient (or canonical projection) map [tex]p : \mbox{U}( \mathcal{H}) \to \mbox{U}( \mathbb{P}\mathcal{H}) \ \left( \cong \frac{\mbox{U}( \mathcal{H})}{ \mbox{U}(1)} \right),[/tex] with the centre [itex]\mathcal{Z}\left(\mbox{U}(\mathcal{H})\right) = \big\{ \lambda \ \mbox{id}_{\mathcal{H}}; | \lambda | = 1 \big\}[/itex] identified with [itex]\mbox{U}(1)[/itex], gives us the following short exact sequence of groups and homomorphisms [tex]1 \rightarrow \mbox{U}(1) \overset{i}{\hookrightarrow} \mbox{U}(\mathcal{H}) \overset{p}{\rightarrow} \mbox{U}(\mathbb{P}\mathcal{H}) \rightarrow 1 .[/tex] This means that the unitary group of [itex]\mathcal{H}[/itex] [itex]\big([/itex]i.e., [itex]\mbox{U}(\mathcal{H})[/itex][itex]\big)[/itex] is the central extension of the projective unitary group [itex]\mbox{U}(\mathbb{P}\mathcal{H})[/itex] by the group [itex]\mbox{U}(1)[/itex] [itex]\big([/itex] it is, at the same time, a locally trivial principal [itex]\mbox{U}(1)[/itex]-bundle over [itex]\mbox{U}(\mathbb{P}\mathcal{H})[/itex][itex]\big)[/itex]. 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 [itex]\hat{\pi}[/itex] factors according to [tex]\pi \circ \varphi = \hat{\pi} = p \circ U .[/tex]
 
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  • #69
@samalkhaiat is there any book/paper that covers issues you raised in your last post?
 
  • #70
samalkhaiat said:
Technically, this is not correct. Given a Lie group [itex]G[/itex] and its Lie algebra [itex]\mathfrak{g}[/itex], then every projective unitary representation [tex]\rho : G \to \mbox{U}(\mathbb{P}\mathcal{H}) ,[/tex] lifts to a unique unitary representation [tex]U : G \to \mbox{U}(\mathcal{H}) ,[/tex] if the following two conditions hold: 1) [itex]G[/itex] is simply connected, and 2) the second cohomology group of [itex]\mathfrak{g}[/itex] is trivial, i.e., [itex]\mbox{H}^{2}(\mathfrak{g}, \mathbb{R}) = 0[/itex].
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.
 
<h2>1. What is the ground state energy of a quantum field?</h2><p>The ground state energy of a quantum field is the lowest possible energy state that a quantum field can have. It is also known as the vacuum energy or zero-point energy.</p><h2>2. Why is there a debate about whether the ground state energy of a quantum field is actually zero?</h2><p>There is a debate because according to classical physics, the ground state energy of a system should be zero. However, in quantum mechanics, there is a concept of zero-point energy which suggests that even in the lowest energy state, there is still some energy present.</p><h2>3. How is the ground state energy of a quantum field calculated?</h2><p>The ground state energy of a quantum field is calculated using quantum field theory, which combines principles of quantum mechanics and special relativity. It involves solving complex equations and taking into account various factors such as particle interactions and the effects of virtual particles.</p><h2>4. What evidence is there to support the idea that the ground state energy of a quantum field is not actually zero?</h2><p>One of the main pieces of evidence is the Casimir effect, which is the attraction between two uncharged metal plates in a vacuum. This effect is caused by the zero-point energy of the quantum field between the plates. Additionally, various experiments and observations in quantum mechanics have also provided evidence for the existence of zero-point energy.</p><h2>5. Why is the ground state energy of a quantum field important?</h2><p>The ground state energy of a quantum field is important because it plays a role in various physical phenomena, such as the stability of atoms and the behavior of particles in a vacuum. It also has implications for our understanding of the fundamental nature of the universe and the concept of empty space.</p>

1. What is the ground state energy of a quantum field?

The ground state energy of a quantum field is the lowest possible energy state that a quantum field can have. It is also known as the vacuum energy or zero-point energy.

2. Why is there a debate about whether the ground state energy of a quantum field is actually zero?

There is a debate because according to classical physics, the ground state energy of a system should be zero. However, in quantum mechanics, there is a concept of zero-point energy which suggests that even in the lowest energy state, there is still some energy present.

3. How is the ground state energy of a quantum field calculated?

The ground state energy of a quantum field is calculated using quantum field theory, which combines principles of quantum mechanics and special relativity. It involves solving complex equations and taking into account various factors such as particle interactions and the effects of virtual particles.

4. What evidence is there to support the idea that the ground state energy of a quantum field is not actually zero?

One of the main pieces of evidence is the Casimir effect, which is the attraction between two uncharged metal plates in a vacuum. This effect is caused by the zero-point energy of the quantum field between the plates. Additionally, various experiments and observations in quantum mechanics have also provided evidence for the existence of zero-point energy.

5. Why is the ground state energy of a quantum field important?

The ground state energy of a quantum field is important because it plays a role in various physical phenomena, such as the stability of atoms and the behavior of particles in a vacuum. It also has implications for our understanding of the fundamental nature of the universe and the concept of empty space.

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