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This question arose from my question on whether the vacuum energy is actually present for a free quantum scalar field

What is the right way to treat the vacuum energy?

Part of this discussion is that the canonical quantization suffers from ordering ambiguities. In essence, before promoting the classical Hamiltonian to an operator, one can add to it \begin{equation} H(x,p) \to H(x,p) +(xp-px)f(x,p). \qquad (1) \end{equation} with any $f(x,p)$. This means that upon quantization $H(x,p)$ can be turned into an arbitrary function of operators $x$ and $p$.

Some people in the comments (@Prof.Legolasov at Vacuum energy of a real Klein-Gordon field) suggested that this is a settled issue and I should ask a separate question. Thus, I guess, either the canonical quantization should be equipped with a preferred ordering scheme or there are alternative ways (path integral?) to quantize things, which are free of this problem. Can anyone expand on this or give some references?

Ultimately, I am wondering whether the vacuum energy of a free scalar field is there or it appears as a result of a wrong ordering in canonical quantization.

Discussion

Let me comment on one idea suggested below.

One suggestion was that one should require that the canonical equations of motion for $x$ and $p$ should remain the same as in the classical case. In the simplest case of a free particle, one has $$\dot x = \frac{p}{m}, \qquad \dot p =0.$$ If we insist that the same equations result from quantum commutators in the quantised case, we end up fixing $[x,H]$ and $[p,H]$. With these commutators fixed, one is only left with the freedom to add a number to the Hamiltonian -- this does not change the commutators above.

In the given case this suggestion seems very reasonable. However, I do not quite see how it can be extended to general Hamiltonians. Namely, for a general Hamiltonian the Poisson brackets $\{ x ,H\}$ and $\{p,H \}$ are some general functions of phase space variables $(x,p)$. Due to ordering ambiguities, there is no unique way how these should be promoted to operators when quantized. The reason why this logic worked for a free particle is that these brackets were ultimately simple $\{ x ,H\} \sim p$, $\{p,H \}\sim 0$: that is the first one is just one of the canonical coordinates, while the other one is zero. And, it seems natural to promote them to operators simply by $p \to \hat p$, $0\to 0$. Instead, for more general $\{ x ,H\}$ and $\{p,H \}$ there is no preferred way to promote them to operators.

In other words, I would say that the question remains open. Overall, the idea that to fix the way how functions on phase space should be promoted to commutators is to demand that their Poisson brackets are consistent with the commutators of the respective operators is problematic, as the algebra of the Poisson bracket and the Heisenberg algebra are different algebras and such a map does not exist. So, I tend to think that there is no preferable quantization. A similar discussion occurred here Physically distinct quantizations

Dr.Yoma
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    Related: What does the ordering of creation/annihilation operators mean? – Qmechanic Feb 26 '23 at 10:41
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    What about my answer to the question you link is not sufficient here? If you choose a quantization ordering prescription that ends up with the infinite piece, then you need to renormalize the infinite piece away, i.e. renormalization is the normal way in which QFT deals with these infinities. What more to you want to know? – ACuriousMind Feb 26 '23 at 11:03
  • @ACuriousMind You write there that one can regularise an infinite vacuum energy, which results from the standard ordering. But there are infinitely many other orderings. Is there any way to justify that the standard ordering is the right one? If no, then instead of regularising infinity in the standard ordering one can pick another one in which it does not appear at all. But then if one allows to play with different orderings one runs into a problem that one can replace the classical Hamiltonian with an arbitrary operator and the canonical quantization gets completely ill-defined. – Dr.Yoma Feb 26 '23 at 11:53
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    What do you mean by any ordering being the "right" one? For an $H$ produced by any ordering prescription, you can always write $H = :H: + E_0$, where $:H:$ is the standard-ordered Hamiltonian and $E_0$ is the vacuum energy (possibly zero). As you can always renormalize (not regularize, those are two different operations!) $E_0$ to whatever value you want, there is no meaningful difference between the ordering procedures, is there? – ACuriousMind Feb 26 '23 at 12:01
  • @ACuriousMind I guess, when you talk about any ordering, you mean a limited set of orderings such as the normal ordering, the symmetric ordering etc. However, in principle, one can start from $H=0$ and represent it as $0=(a a^* - a^* a)f(a,a^)$ with any $f$. Then, by promoting $a$ and $a^$ to operators, you will get the quantum Hamiltonian $\hat H=f(a,a^\dagger)$. – Dr.Yoma Feb 26 '23 at 12:18
  • @ACuriousMind Ok, there was another answer below, which says that one should put an extra requirement that the canonical variables (the associated operators) evolve in time the same way they do in the classical case. This, indeed, reduces the number of possible ambiguities in the Hamiltonian due to reordering to the addition of a constant. – Dr.Yoma Feb 26 '23 at 12:48
  • @ACuriousMind ok. Suppose that I have accepted that the only ambiguity related to the operators ordering when it concerns the Hamiltonian is just adding a number. Do you say that this number is completely unphysical? For example, it does not show up in any computations? Say, if you compute a transition amplitude between states at different times, it only produces an overall phase, which goes away once we look at the absolute value squared, which gives the probability. This makes sense. – Dr.Yoma Feb 26 '23 at 13:09
  • @ACuriousMind Probably, then, the only reason why these additive terms may be fixed is if we have symmetries and we want the associated charges at quantum level (the Hamiltonian being one of them) to commute the same way the symmetry generators commute themselves. – Dr.Yoma Feb 26 '23 at 13:11
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    No. The ordering issue cannot be resolved on its own. This discussion is quite related. – ZeroTheHero Feb 26 '23 at 18:00

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The physical content of a theory is governed by the equations of motion. In quantum theory, these are the Heisenberg equations of motion $\dot{A}(t)= i [H, A(t)]$. Replacing the original Hamiltonian $H(Q,P)$ by $H^\prime(Q,P) = H(Q,P) -i [Q,P] F(Q,P) = H(Q,P)+F(Q,P)$, with some selfadjoint operator $F(Q,P)$, would clearly change the equations of motion (and thus the physical content of the system you want to describe), unless $[F(Q,P), A]= 0$ for all observables $A$. But this implies $F(Q,P) = c \mathbf{1}$ with some arbitrary constant $c$. Thus the ordering ambiguity simply boils down to an energy shift $H \to H+c$, which you are, of course, free to do without changing the equations of motion.

Analogously, in relativistic quantum field theory, one takes advantage of the freedom to chose the origin of the energy scale by imposing the condition $P^\mu | 0 \rangle =0$ for the energy-momentum operator $P^\mu$ acting on the vacuum state $|0\rangle$. (See also the links in the comments by @Qmechanic and @ACuriousMind and the answer given by Valter Moretti in https://physics.stackexchange.com/q/751914 .)

Hyperon
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  • Probably, I am missing something. How do you know which equations of motion are correct? You have your $H$ with equations of motion $\dot A(t)=i[H,A(t)]$ and I have my $H'$ which results from a different ordering and gives a different evolution $\dot A(t)=i[H',A(t)]$ (if $H$ and $H'$ differ more substantially than just by a constant). Is there any way to tell that your evolution is better than mine? – Dr.Yoma Feb 26 '23 at 12:10
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    Suppose you want to describe the motion of a free particle, then you know that the equations of motion for $X$ and $P$ are $\dot{X}(t)=P(t)/m$ and $\dot{P}(t)=0$, which implies $H= P^2/2m + c \mathbf{1}$. If you replace $H$, for instance, by $H^\prime = H - F_0 X$, you are describing now a particle under the influence of a constant force $F_0$ with equations of motion $\dot{X}(t)= P(t)/m$ and $\dot{P}(t) =F_0$, clearly a different physical situation! – Hyperon Feb 26 '23 at 12:22
  • The reason why you want $\dot X(t)=P(t)/m$ and $\dot P(t)=0$ is that it is true classically. Right? Then, I agree with you that this fixes the quantum Hamiltonian up to a constant. – Dr.Yoma Feb 26 '23 at 12:27
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    The equations of motion $\dot{X}= P/m$ and $\dot{P}= -V^\prime(X)$ are the same in classical mechanics and quantum mechanics. – Hyperon Feb 26 '23 at 12:36
  • I changed my mind. Your trick with the requiring that the equations of motion remain the same works only because you considered Hamiltonians of a very particular type. Namely, for your Hamiltonians the right hand side of equation of motion is very simple so that there is only 1 natural way to promote them to operators. Say, 0 as a function of x and p is naturally promoted to 0 as an operator. Similarly, p as a function of x and p is naturally promoted to p as an operator. But there is no such a simple procedure for general {x,H} and {p,H}. – Dr.Yoma Mar 04 '23 at 12:59