Interpretation of the nonuniqueness of vacuum of QFT in flat spacetime for a given inertial observer; No Lorentz transformation; No accelerated motion

Question

Consider an inertial observer in flat spacetime with a choice of coordinates $(t,{\vec x})$. This observer can expand a quantum field $\hat{\phi}$ in more than one complete set of orthonormal modes. Let us consider two such choices made by the observer in question.
In terms of modes $\{f_i\}$, the quantum field is expanded as $$\hat{\phi}(t,{\vec x})=\sum_i\left(\hat{a}_if_i+\hat{a}_i^\dagger f_i^*\right)$$

$$[a_j,a_j^\dagger]=\delta_{ij}, [a_i,a_j]=[a_i^\dagger, a_j^\dagger]=0.$$

Similarly, in terms of modes $\{g_i\}$, the same field $\hat{\phi}(t,{\vec x})$ is expanded as $$\hat{\phi}(t,{\vec x})=\sum_i\left(\hat{b}_ig_i+\hat{b}_i^\dagger g_i^*\right)$$

$$[b_j,b_j^\dagger]=\delta_{ij}, [b_i,b_j]=[b_i^\dagger,b_j^\dagger]=0.$$

It can be shown that the basis modes are related by the Bogoliubov transformation.
Further, it can be shown that the vacuum defined by $a_i|0_f\rangle=0$ is different from the vacuum defined by $b_i|0_g\rangle=0$. In particular, though $$\langle 0_f|a_i^\dagger a_i|0_f\rangle=0,$$ $$\langle 0_f|b_i^\dagger b_i|0_f\rangle\neq 0.$$
Does it mean that even for a given inertial observer the notion of vacuum is ambiguous? If more than one choice of annihilation operators (e.g. $a_i$ and $b_i$, above) is possible, how can we define a unique vacuum? Is there at all a preferred choice of basis modes (and thus, a preferred choice of creation and annihilation operators) and a preferred vacuum?

Please note that I am (i) neither talking about changing from one inertial frame to another (Lorentz transformation), (ii) nor about curved spacetime or accelerated observers.

I asked a related question here, but I did not receive a satisfactory answer probably because the question was not focussed and poorly worded.

Re "the vacuum of defined by" (seems somewhat incomprehensible): Do you mean "the vacuum defined by"? Or something else? — Peter Mortensen, May 05 '21 at 19:46

Chiral Anomaly · Accepted Answer · 2021-10-13T14:09:20.773

Valter Moretti gave a great answer from the perspective of rigorous mathematics. As a supplement, I'll add an answer from the less rigorous perspective of for-all-practical-purposes (FAPP) physics. The context of the question is flat spacetime with a single observer, but generality is one of the keys to insight, so I'll start with a broader perspective and then apply it to the question.

The FAPP approach

The energy operator is the operator that generates time-translations.
In quantum field theory, the vacuum state is the state of lowest energy. It can be non-unique due to things like spontaneous symmetry breaking, but that's a different subject. In many models, including models of free fields, it's unique — after we specify the energy operator.
Proper time is observer-dependent, and it's a local concept (applicable only in a neighborhood of that observer), not a global concept.

People often point out that the definition 1 only makes sense in a spacetime that has time-translation symmetry, like flat spacetime. But here's the FAPP theme: In the neighborhood of any given observer, we can always define an effective Hamiltonian $H=\int_R T^{00}$, where $T^{ab}$ is the energy-momentum tensor, $R$ is a spatial neighborhood of the observer, and the coordinate system is such that the "time" coordinate (index $0$) agrees with the observer's proper time. The operator $H$ works just fine as a generator of time-translations in a neighborhood of that observer, so we can use it in the definition 2: any state $|0\rangle$ that minimizes the expectation value of $H$ qualifies as a vacuum state locally, for that observer.

(Warning: We cannot let the local integration region $R$ be too small — not microscopic — because the energy density $T^{00}$ is not bounded below in relativistic QFT. I cited some references about this in an answer to another question. In this answer, I'm assuming that the spacetime curvature and the observer's acceleration are both mild enough so that $R$ can be large enough to make that effect unimportant.)

Particles are defined with respect to the vacuum state. But the vacuum state is the state that minimizes the energy, and the energy operator is observer-dependent (and typically only defined locally, as the generator of that observer's proper time translations locally), so particles are observer-dependent.

By the way, the expression $\int_R T^{00}$ for the (local) energy operator illustrates the important idea of a splittable symmetry — a symmetry that can be applied locally, to only part of a system. I described the significance of splittable symmetries in another answer. This is the kind of "time translation symmetry" that matters when we're talking about a given observer, because observers are localized.

Application to inertial observers in flat spacetime

In flat spacetime, thanks to time-translation symmetry, the vacuum state can be defined globally. And thanks to Poincaré symmetry, the vacuum state is the same for all inertial observers, even though different inertial observers have different energy operators. Therefore, all inertial observers in flat spacetime agree about the definition of "particle."

The question asks about a single inertial observer in flat spacetime. The answer is that physical particles are defined with respect to the lowest-energy state, and the energy operator is uniquely defined, so the lowest-energy state is also uniquely defined — ignoring things like spontaneous symmetry breaking, which is beside the point here.

Accelerating observer in flat spacetime: the Unruh effect

By the way, we can apply this same approach to the famous Unruh effect. The Unruh effect considers a uniformly accelerating observer in flat spacetime. For a uniformly accelerating observer, we can use the operator that generates boosts as the operator $K$ that generates translations of that observer's proper time. (The boost operator is globally defined, but that's not important. The important thing is that it's defined in a neighborhood of the observer, where it generates translations of that observer's proper time.) Locally, we can define a set of states that minimizes the expectation value of a local version of $K$. All of these states look the same locally, to that observer, and none of these states agree with the Minkowski vacuum state. That's why a uniformly accelerating observer "sees" particles in the Minkowski vacuum, and conversely why inertial observers "see" particles in the accelerating observer's vacuum. This is just a straightforward application of the definitions 1,2,3, applied locally FAPP.

The Hawking effect

We can also apply this approach to the even-more-famous Hawking effect. Hopefully that's pretty obvious now, so I won't ramble on about it.

A useless definition

Every once in a while, you might encounter a book/paper that defines the vacuum state as the state annihilated by a set of mutually commuting annihilation operators — operators $a_n$ whose commutator with their adoints is $[a_n,a_m^\dagger]=\delta_{nm}$. Authors are free to make whatever definitions they want, because language is arbitrary, but it's important to understand that the annihilation-operator-based definition is generally completely different than the lowest-energy definition that is standard in the quantum field theory literature. To see just how arbitrary the annihilation-operator-based definition is, suppose that $|\varnothing\rangle$ is the state satisfying $a_n|\varnothing\rangle=0$ for all $a_n$, and let $|\psi\rangle$ be absolutely any other state. A unitary operator $U$ satisfying $U|\varnothing\rangle=|\psi\rangle$ always exists, and the operators $b_n\equiv U^{-1}a_n U$ satisfy the same type of commutation relations as the original operators: $[b_n,b_m^\dagger]=\delta_{nm}$. Therefore, according to the annihilation-operator-based definition, every state in the Hilbert space would be a vacuum state. That's not a very useful definition, but it's not illegal, either. It's just different... and useless.

A similar but more useful definition

Given a preferred time coordinate, such as one that agrees locally with a given user's proper time, we can define positive frequency and negative frequency. And given any time-dependent operator in the Heisenberg picture, we can define its positive- and negative-frequency parts. In free field theories, the standard set of creation and annihilation operators are defined to be the negative- and posititive-frequency (respectively) parts of the field operators. Those annihilation operators annihilate the lowest-energy state, because the positive-frequency part of any time-dependent operator acts as an energy-lowering operator. When applied to a state that already has the lowest possible energy, an energy-lowering operator annihilates it.

So if we define the vacuum state as the state to be the state annihilated by the negative-energy parts of all time-dependent operators, then this is equivalent to defining the vacuum state as the state of lowest energy. In free field theory, these operators are called annihilation operators, but notice that these are not just any old set of annihilation operators. They're defined with respect to a given energy operator (generator of time translations), and that's the key to answering the question.

What do you mean by "the energy operator is uniquely defined"? Isn't $ H_U = U H U^{-1}$ also a valid Hamiltonian for the system? If so, given the ground state of $H$ as $ \vert E_0\rangle $, then $ U \vert E_0 \rangle$ would be the ground state of $H_U$. The definition of the vacuum as the least energetic state seems as ambiguous and "useless" as the $\hat{a_n}\vert0\rangle =0$ one. Both are tied down to the basis you choose. The later is tied to the choice of $\hat{a_n}$, while the former is tied to the choice of $H$. — Lucas Baldo, May 05 '21 at 19:22
@LucasBaldo Consider flat spacetime. The hamiltonian, the generator of time translations in a given coordinate system, is part of the model's definition. You can use a unitary transformation to change how the model is represented on the Hilbert space, which of course changes which vector represents the lowest energy state. But just because operators satisfy commutation relations that look like the canonical ones doesn't mean they're related to energy or time translations in any meaningful way. — Chiral Anomaly, May 05 '21 at 20:04
It seems to me that you are facing several fictitious (in a sense) problems due to the Hilbert space formulation...In this case, the algebraic approach really helps — Valter Moretti, May 05 '21 at 20:10
@ValterMoretti I'm familiar with the algebraic appraoch, at least the basic principles of the haag-kastler and haag-araki versions, and my perspective has been strongly influenced by it. Can you clarify what you mean by fictitious problems? — Chiral Anomaly, May 05 '21 at 21:00
All problems (not of yours) related to equivalent unitary representations for instance...I think the problem is here: how many quasi free algebraic pure states do exist which are ground states with respect to a given one parameter group of automorphisms of the * algebra of the bosonic field operators? Answer, at most one. — Valter Moretti, May 05 '21 at 21:02
@Chiral Anomaly, I didn't understand the connection between the first (up to "...lowest energy state") and last part of your response ("But just because..."). Could you clarify further? — Lucas Baldo, May 05 '21 at 21:06
@ValterMoretti Right, for automorphisms of the whole algebra. But a local observer only has access to local observables in their own neighborhood, which is why I described an effective Hamiltonian that only integrates the stress-energy tensor over some bounded local region. A state that (appoximately) minimizes the expectation value of that operator might as well be empty as far as that observer is concerned, but it's highly nonunique because it may still contain material objects elsewhere, far away . — Chiral Anomaly, May 05 '21 at 21:52
@ValterMoretti Of course, we can't let the local integration region be too small (not microscopic), because the energy density is not bounded below in QFT. — Chiral Anomaly, May 05 '21 at 21:53
@LucasBaldo I didn't say it very well. Consider a given model, with a given Hamiltonian, and a given Hilbert space representation. For any state vector in that Hilbert space, we can construct a set of operators which annihilate that state and whose commutators with their adjoins look like the usual creation/annihilation relations. Therefore, that can't be a useful definition of the model's vacuum state. Tying the definition to an appropriate integral of the stress energy tensor (e.g Hamiltonian) makes it useful. — Chiral Anomaly, May 05 '21 at 22:02
@Chiral Anomaly. Ah, thank you for explaining. I think I see your point now, and how it clashes with what I had in mind. I believe that what makes useful the definition as a state that is annihilated by a set of operators is that those are not any set of operators, but the set of operators that diagonalize the Hamiltonian: $\hat{H} = \sum{E_n \hat{a}_n^{\dagger} \hat{a}_n}$. If you change to a different basis of operators $\hat{b}_n= U\hat{a}_nU^{-1}$, you also have to transform the Hamiltonian in the same way. — Lucas Baldo, May 05 '21 at 23:33
This way, the new ground state is still the vacuum in both senses (minimum of energy and annihilated by all operators). I hope this clarifies why I think both definitions are sort of equivalent. — Lucas Baldo, May 05 '21 at 23:34
@LucasBaldo Yes, that makes perfect sense. You said it well. I tried (poorly) to address the different-annihilation-operators issue because the question comes from the fact that a generic curved spacetime doesn't have (global) time-translation symmetry, so there is no (global) unambiguous energy operator in the usual sense. Different observers can define their own local energy operators, but the corresponding vacuum states are different. What looks like the vacuum to one observer may look like a non-vacuum to another. So it's observer-dependent, but still tied to the energy operator(s). — Chiral Anomaly, May 06 '21 at 00:14
I see. Makes sense. QFT is such a complex topic and I still have much to learn. It's nice to see other people's takes on subjects like these. Thanks for the discussion. — Lucas Baldo, May 06 '21 at 00:38
@Chiral Anomaly. I thought the issue concerned global objects as inertial reference frames are global in SR, however, yes it can be also interpreted in a local sense as you did. — Valter Moretti, May 06 '21 at 06:16
@ChiralAnomaly I require some more clarifications from you. Can you help? Sorry for asking this so late. In the question, I defined two types of modes ${f_i}$ and ${g_i}$ used by the same inertial observer. Then I wrote that the operators $a_i$ and $b_i$ are related by Bogoliubov transformation. Now I wonder if this is really true. Any comments? Can't remember why I said that. I know that this is true for two different observers in curved spacetime. But the scenario I consider is much simpler: just one inertial observer using two different bases ${f_i}$ and ${g_i}$. — Solidification, Oct 11 '21 at 15:14
@ChiralAnomaly If you kindly comment on this I'll highly appreciate it. — Solidification, Oct 11 '21 at 15:14
@mithusengupta123 By "modes... used by the same intertial observer [in flat spacetime]," I assume you mean they have well-defined freq's with respect to the time coordinate that we usually associate with that observer. Then the Bogoliubov transf'n that relates the two sets of modes ${f_i}$ and ${g_i}$ is trivial in the sense that it doesn't mix positive and negative frequencies with each other, so the same vacuum state works equally well for both. As an example, ${f_i}$ could be plane-waves and ${g_i}$ could be spherical waves, both defined with respect to the same time coordinate. — Chiral Anomaly, Oct 12 '21 at 02:49
@ChiralAnomaly About the section "a useless definition". Is it true that any two states in the Hilbert space are related by a Unitary operator U? Doesn't seem so. For example, the ground state and the first excited state of the harmonic oscillator are related by $a^\dagger$ which is not a unitary operator. — Solidification, Oct 12 '21 at 18:04
@mithusengupta123 It is true. Let $|0\rangle$ and $|1\rangle$ be any two states with unit norm. Even if $|1\rangle\propto a^\dagger|0\rangle$ for a non-unitary operator $a^\dagger$, we always also have $|1\rangle=U|0\rangle$ for a unitary operator $U$. Proof: define $U$ by the condition that $U|\psi\rangle=|\psi\rangle$ for all states $|\psi\rangle$ that are orthogonal to both $|1\rangle$ and $|0\rangle$, and by the conditions $U|0\rangle=|1\rangle$ and $U|1\rangle=a|0\rangle+b|1\rangle$ with $b=\langle 0|1\rangle - a\langle 1|0\rangle$. — Chiral Anomaly, Oct 13 '21 at 00:31

Valter Moretti · Answer 2 · 2021-05-05T20:04:16.897

First of all, a state of the type you consider is said to be Gaussian or quasi-free. These states are completely defined by their two-point function $\omega(t_x,x,t_y,y)$ -- which is supposed to define a positive bi-distribution and whose antisymmetric part is fixed by the standard CCR -- using Wick's rule to compute the even $n$-functions (the odd ones being $0$). I stress that the two-point function defines a Gaussian state intrinsically without referring to an explicit Fock space structure that is an a posteriori structure defined up to unitary transformations. In the case of the two-point function of a Gaussian state,

(a) the Hilbert space is a Fock space,

(b) the operatorial representation of $\phi$ is the standard one,

(c) $\omega(t_x,x,t_y,y)$ is exactly “expectation value” of the vacuum vector of the Fock space as the notation suggest $$\omega(t_x,x,t_y,y) = \langle 0| \phi(t_x,x)\phi(t_y,y)\rangle |0\rangle\:.$$

A uniqueness result arises as soon as one requires that the two-point function (and thus the vacuum) has good properties with respect to the time translation referred to the considered reference frame and a pair of other physically meaningful requirements.

(1) The two point function is invariant under time displacements $$\omega(t_x+T,x,t_y+T,y) = \omega(t_x,x,t_y,y) \quad \mbox{for every $T\in \mathbb{R}$}$$

(2) In the Fock space the time translation is implemented by an Hamiltonian operator $H$ such that its spectrum is non-negative and $|0\rangle$ is the unique zero eigenvector of $H$ up to phases.

(3) The representation of the algebra of fields is irreducible.

There is only one two-point function satisfying the said hypotheses.

Actually (1) and a physically natural continuity requirement imply the existence of the generator $H$, so that (2) matters only concerning the spectrum and the eigenvalue property.

In Minkowski spacetime, when switching on the Poincaré invariance, this state must coincide with the analog vacuum state of every other inertial frame.

It is worth stressing that, dropping the request of dealing with Gaussian states, the picture dramatically changes and the result is no longer valid in general.

The appropriate setup to prove the statement above is the algebraic formulation of quantum field theory. The mentioned result was established by B.Kay [1] in a quite general fashion, e.g., in curved spacetime and other contexts (we used it for instance to define ground states on the light-like boundary of asymptotically flat spacetimes [2] and in the construction of the Unruh state in the Kruskal spacetime [3]),

[1] B. S. Kay, A uniqueness result in the Segal-Weinless approach to linear Bose fields,J. Math. Phys.20, 1712 (1979)

[2] V.Moretti: Uniqueness theorems for BMS-invariant states of scalar QFT on the null boundary of asymptotically flat spacetimes and bulk-boundary observable algebra correspondence Commun. Math. Phys. 268, 727 (2006).

[3] C. Dappiaggi, V.Moretti, and N. Pinamonti: Rigorous construction and Hadamard property of the Unruh state in Schwarzschild spacetime. Adv. Theor. Math. Phys. 15, vol 2, 355-448 (2011)

score 4 · Answer 3 · answered May 05 '21 at 23:48

Others have already given a deep discussion of the details of the problem, but I wanted to add a simpler approach to your doubt.

The issue with using the definition of vacuum as the state which is destroyed by all annihilation operators is that this is not true for any set of operators, but specifically for the set of operators which diagonalizes the Hamiltonian.

In the diagonal form $\hat{H}$ can be written as \begin{align} \hat{H} = \sum_n E_n \hat{a}_n^{\dagger} \hat{a}_n, \end{align} where $E_n>0$.

It's straightforward to check that state $\vert \psi\rangle$ which satisfies $\hat{a}_n \vert \psi\rangle= 0$ for all $\hat{a}_n$ is the state with minimum energy (which is the fundamental property of the vacuum).

Note that because of this there is no ambiguity for the vacuum state because the set of operators which diagonalize the Hamiltonian is unique (given no degeneracy is present), apart from a global phase.