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

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A. Neumaier
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.

vanhees71
Gold Member
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 enery 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).

A. Neumaier
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.

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vanhees71
Gold Member
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.

A. Neumaier
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.

vanhees71
Gold Member
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.

A. Neumaier
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.

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A. Neumaier
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!

vanhees71
Gold Member
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).

vanhees71
Gold Member
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.

A. Neumaier
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.
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.
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.

A. Neumaier
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.

A. Neumaier
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.

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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samalkhaiat
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 $$T^{\mu\nu} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\varphi_{r})}\partial^{\nu}\varphi_{r} - \eta^{\mu\nu}\mathcal{L} \ ,$$ and a conserved Lorentz current given by the moment tensor $$M^{\rho\mu\nu} = x^{\mu}T^{\rho\nu} -x^{\nu}T^{\rho\mu} + S^{\rho\mu\nu} \ ,$$ $$S^{\rho\mu\nu}(x) = - i \frac{\partial \mathcal{L}}{\partial (\partial_{\rho}\varphi_{r})} \left( \Sigma^{\mu\nu}\right)_{r}{}^{s}\varphi_{s}(x) .$$ And the corresponding Noether charges are given by $$P^{\mu} = \int d \sigma_{\rho}(x) \ T^{\rho\mu}(x) = \int d^{3}x \ T^{0 \mu}(x) \ , \ \ \ \ \ \ \ \ \ (A)$$$$J^{\mu\nu} = \int d \sigma_{\rho}(x) \ M^{\rho\mu\nu}(x) = \int d^{3}x \ M^{0 \mu\nu} (x) . \ \ \ \ \ \ \ (B)$$ In a classical field theory, where we can almost always ignore surface terms, we can show that $( P^{\mu} , J^{\mu\nu})$ are time-independent Lorentz vector and Lorentz tensor respectively. So, in a classical field theory, this means that $( P^{\mu} , J^{\mu\nu})$ generate the proper Poincare’ algebra (i.e., without central charges). In other words, one can show that the classical $( P^{\mu} , J^{\mu\nu})$ are unique Poincare generators. Clearly this cannot be true in a QFT because 1) surface integrals may not vanish, and 2) the uniqueness of $( P^{\mu} , J^{\mu\nu})$ 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 $( P^{\mu} , J^{\mu\nu})$, it is an easy exercise to show that $$[iP^{\mu} , \varphi (x)] = \partial^{\mu}\varphi (x) \ , \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$$$[iJ^{\mu\nu}, \varphi (x)] = \left( x^{\mu}\partial^{\nu}- x^{\nu}\partial^{\mu} - i \Sigma^{\mu\nu} \right) \varphi (x) . \ \ \ \ \ (2)$$

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

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

$$\big[ [iP^{\mu} , P^{\nu}] , \varphi (x) \big] = 0 ,$$
$$\big[ [iP^{\sigma} , J^{\mu \nu}] - \eta^{\sigma \nu}P^{\mu} + \eta^{\sigma \mu}P^{\nu} , \varphi (x) \big] = 0 ,$$
$$\big[ [iJ^{\mu \nu} , J^{\rho \sigma}] - \eta^{\mu \rho} J^{\nu \sigma} - \eta^{\mu \sigma} J^{\rho \nu} - \eta^{\nu \rho} J^{\mu \sigma} - \eta^{\nu \sigma} J^{\rho \mu} , \varphi (x) \big] = 0 .$$
The most general solutions of these equations are given by
$$\big[ iP^{\mu} , P^{\nu}\big] = C^{\mu , \nu} ,$$
$$\big[ iP^{\sigma} , J^{\mu \nu}\big] = \eta^{\sigma \nu} P^{\mu} - \eta^{\sigma \mu} P^{\nu} + C^{\sigma , \mu \nu} ,$$
$$\big[ iJ^{\mu \nu} , J^{\rho \sigma} \big] = \eta^{\mu \rho} J^{\nu\sigma} + \cdots + C^{\mu \nu , \rho \sigma} ,$$ where all the $C$’s are constants. If these constants are not all zero, we conclude that the Noether charges $\big( P^{\mu} , J^{\mu\nu})$ 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
$$\big[ J^{\mu \nu} , [ P^{\rho}, P^{\sigma}] \big] + \big[ P^{\sigma} , [ J^{\mu \nu}, P^{\rho}] \big] + \big[ P^{\rho} , [ P^{\sigma}, J^{\mu \nu}] \big] = 0 ,$$ leads to $$C^{\mu , \nu} = 0 .$$ Another useful Jacobi identity is
$$\big[ J^{\lambda \tau} , [ J^{\mu \nu} , P^{\rho}] \big] + \cdots \ = 0 .$$ This allows us to express $C^{\mu , \rho \sigma}$ as
$$C^{\mu , \lambda \tau} = \eta^{\mu \tau} \left( \frac{1}{3} \eta_{\rho \nu} C^{\rho , \lambda \nu}\right) - \eta^{\mu \lambda} \left( \frac{1}{3} \eta_{\rho \nu} C^{\rho , \tau \nu}\right),$$ which may be used to define the constant $$C^{\mu} \equiv \frac{1}{3} \eta_{\rho \nu} C^{\rho , \mu \nu} .$$ And the last Jacobi identity is between 3 $J$’s
$$\big[ J^{\lambda \tau} , [ J^{\mu \nu}, J^{\rho \sigma}] \big] + \cdots \ = 0 .$$ This allows us to define yet another constant in terms of $C^{\mu \rho , \nu \sigma}$ $$C^{\mu \nu} \equiv \frac{1}{2} \eta_{\rho \sigma} C^{\mu \rho , \nu \sigma} .$$ Now, it is an easy exercise to show that the shifted Noether charges $$\bar{P}^{\mu} = P^{\mu} + C^{\mu},$$$$\bar{J}^{\mu \nu} = J^{\mu \nu} + C^{\mu \nu} ,$$ 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 $(C^{\mu} , C^{\mu \nu})$ in QFT. So, we will do better than the old man by showing that $$C^{\mu} = - \langle 0 | P^{\mu}| 0 \rangle \ ,$$$$C^{\mu \nu} = - \langle 0 | J^{\mu \nu} | 0 \rangle \ .$$

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

Similar, but more complicated calculation, leads to $$\bar{J}^{\mu\nu} = J^{\mu \nu} - \langle 0 | J^{\mu \nu} | 0 \rangle .$$

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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'. :-) :-) :-)

bhobba
Mentor
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 .

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

Thanks
Bill

A. Neumaier
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.
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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samalkhaiat
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".

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".
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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vanhees71
Gold Member
However, Weinberg does not explain the important meaning of $(C^{\mu} , C^{\mu \nu})$ in QFT. So, we will do better than the old man by showing that $$C^{\mu} = - \langle 0 | P^{\mu}| 0 \rangle \ ,$$$$C^{\mu \nu} = - \langle 0 | J^{\mu \nu} | 0 \rangle \ .$$
[/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

ftr
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.

Demystifier
2018 Award
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.

samalkhaiat
The “crucial part in my lengthy derivation” is the fact that $(P^{\mu}, J^{\mu\nu})$ generate a Projective Unitary Representation, while the $(\bar{P}^{\mu}, \bar{J}^{\mu\nu})$ 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 $G_{1}^{A} = (P_{1}^{\mu} , J_{1}^{\mu\nu})$ and $G_{2}^{A}= (P_{2}^{\mu} ,J_{2}^{\mu\nu})$ both satisfy the ordinary Poincare’ algebra, then one can prove that $G_{1}^{A} = G_{2}^{A}$, i.e., the vacuum subtraction is not possible.
In #41, this uniqueness theorem applies to the $(\bar{P}^{\mu}, \bar{J}^{\mu\nu})$ but not to the $(P^{\mu}, J^{\mu\nu})$, because the algebra of the latter contains central charges, i.e., not an ordinary Poincare’ algebra.