Quantum fields and infinite tensor products

Question

As I understand it, a naive interpretation of the state space of a quantum field theory is an infinite tensor product $$\otimes_{x\in M} H_x,$$ where $x$ runs over the points of space. This corresponds to the fact that a field $\phi$ and the conjugate momentum $\pi$ can be viewed as a composite system of the array of $\phi(x)$ and $ \pi(x)$. Thus, again naively, the amplitude assigned by a quantum state $\Psi(\phi, \pi)$ to a classical initial condition $(\phi, \pi)$ is a tensor product of the amplitudes $\Psi(\phi(x), \pi(x))\in H_x$.

Of course, this doesn't quite make sense, for many reasons including the fact that the infinite tensor product is quite badly behaved. Instead, the standard way of quantising, say a scalar field satisfying the Klein-Gordon equation, is to write it in terms of Fourier modes $$\phi(x,t)=(2\pi)^{-3/2}\int [a(p)e^{i(px-\epsilon_pt)}+a^*(p)e^{-i(px-\epsilon_pt)}]\frac{d^3p}{2\epsilon_p}$$ with $\epsilon_p=\sqrt{p^2+m^2}$ (this being the KG equation). The canonical commutation relation for $\phi$ and $\pi$ become $$[a(p), a^*(p')]=2\epsilon_p\delta(p-p'); \ [a(p), a(p')]=[a^*(p), a^*(p')]=0,$$ which can individually be quantised in a Segal-Bargmann fashion to act on a Hilbert space $H_p$. To quantise all these operators as we run over all momenta, we would again require an infinite tensor product $$\otimes_p H_p.$$ This is avoided by imposing an additional condition, the existence of a vector $\Psi_0$ (interpreted as the vacuum), satisfying $$a(p)\Psi_0=0$$ for all $p$. After this, it all works out and we have a nicely quantised free field by putting the operators into the integral above. I think I sort of understand this procedure, with the level of uncertainty that I'm normally stuck with when thinking about physics.

However, I've come upon the following passage in the book by Streater and Wightman, page 86-87.

When, in fact, do non-separable Hilbert spaces appear in quantum mechanics? There are two cases which deserve mention. The first arises when one takes an infinite tensor product of Hilbert spaces... Infinite tensor products of Hilbert spaces are always non-separable. Since a Bose field can be thought of as a system composed of an infinitely of oscillators, one might think that such an infinite tensor product is the natural state space. However, it is characteristic of field theory that some of its observables involve all the oscillators at once, and it turns out that such observables can be naturally defined only on vectors belonging to a tiny separable subset of the infinite tensor product. It is the subspace spanned by such a subset that is the natural state space rather than the whole infinite tensor product itself. Thus, while it may be a matter of convenience to regard the state space as part of the infinite tensor product, it is not necessary.

My question is, how does one relate this passage to the usual quantisation procedure described above. In particular, what is the 'tiny separable subset' alluded to by Streater and Wightman?

Because the infinite tensor product picture is so intuitively compelling (this is emphasised, I think, by all authors on QFT), it would be nice to spell out the relation between it and standard quantisation with at least some level of mathematical clarity.

Answer 1

For free quantum fields I think this issue can be dealt with using this theory:

• John Baez, Irving Segal and Zhenfang Zhou, Introduction to Algebraic and Constructive Quantum Field Theory, Section 4.5: Infinite products of Hilbert spaces, Princeton U. Press, 1992, pp. 125-130. Available in pdf and dvju format here.

We describe a well-behaved notion of grounded tensor product for a possibly infinite collection of grounded Hilbert spaces: that is, Hilbert spaces $(K_\lambda)_{\lambda \in \Lambda}$ equipped with unit vectors $z_\lambda \in K_\lambda$. If each $K_\lambda$ is separable and the index set $\Lambda$ is countable, this tensor product is separable!

This doesn't help you for a tensor product of uncountably many Hilbert spaces, but it still helps you a little with your question. There is a way to reduce the Hilbert space of a free quantum field to an infinite but countable tensor product of grounded Hilbert spaces.

Namely:

When you have a free bosonic quantum field, the single-particle Hilbert space $H$ is a countable direct sum of 1-dimensional spaces $H_\lambda$. Quantizing each $H_\lambda$ is just like quantizing a harmonic oscillator: the Fock space of $H_\lambda$, say $K_\lambda$, is a Hilbert space completion of the polynomial algebra on $H_\lambda$.

Moreover, each $K_\lambda$ is grounded: there's an obvious 'vacuum vector' $z_\lambda \in H_\lambda$, namely the element 1 in the polynomial algebra. And here's the best part: the Fock space of $H$, say $K$, is the grounded tensor product of the $K_\lambda$:

$$ H = \bigoplus_{\lambda} H_\lambda \implies K = \bigotimes_{\lambda} K_\lambda $$

where, just to emphasize, the tensor product here is the grounded tensor product.

If we're dealing with a free quantum field on a spacetime $\mathbb{R} \times S$ where the spatial manifold is compact, we can do the decomposition

$$ H = \bigoplus_{\lambda} H_\lambda $$

using momentum or energy eigenstates, since the Laplacian and other elliptic operators on $S$ will have discrete spectrum.

If we're working on Minkowski spacetime, as you are, this doesn't work: your momentum $p$ takes a continuum of values. So you're trying to write $H$ not as a direct sum but as a direct integral of 1-dimensional Hilbert spaces.

So, what seem to be asking for is a generalization of the grounded tensor product to a kind of 'grounded continuous tensor product' operation that sets up an analogy

direct sum : grounded tensor product :: direct integral : grounded continuous tensor product

My hunch is that this should be doable. For one thing, physicists are implicitly using a nonrigorous version of this idea in their daily work on quantum field theory---as you pointed out. For another, it's one of those situations where the final answer you're shooting for has been made rigorous, and you're just looking for a new way of getting there.

However, I am happy enough knowing that countable tensor products of grounded Hilbert spaces work as they should. In the book, we use them to investigate the question of when a linear symplectic transformation of $H$ can be quantized to obtain a unitary operator on $K$.

Answer 2

To add to John Baez's answer, you can regard the symmetric Fock space over $L^2(X)$ as a measurable tensor product of the Hilbert spaces $l^2(\mathbb{N})$ over the index set $X$, and the antisymmetric Fock space over $L^2(X)$ as a measurable tensor product of the Hilbert spaces $\mathbb{C}^2$ over the index set $X$.

Spelling this out a little, in the symmetric Fock space $\mathbb{C} \oplus L^2(X) \oplus L^2(X)^{\otimes 2}_s \oplus \cdots$ (the direct sum of the symmetric tensor powers of $L^2(X)$), the initial $\mathbb{C}$ represents the vacuum state. An element $f \in L^2(X)$ of the second summand represents the direct integral over $x \in X$, weighted by the scalar function $f(x)$, of the state of the system which is in its first excited state at $x$ and vacuum everywhere else. Elements of $L^2(X)^{\otimes 2}_s$ are direct integrals of states of the system in which two fibers are in their first excited state or one fiber is in its second excited state, and all other fibers are vacuum. And so on. The antisymmetric case is similar. This is discussed in Section 2.5 of my book Mathematical Quantization.

Answer 3

The "infinite tensor product" picture may be useful as a sort of concrete image of the state space of a quantum field theory, but in practice is rarely used because of the technical difficulties it brings along. Moreover, the paragraph from Streater and Wightman you quoted could be read as saying that infinite tensor products of Hilbert spaces could be thought of as a bigger mathematical receptacle for the actual "natural state space" of the theory, but even that may not be tenable.

Let me unpack my last sentence above. First of all, what do Streater and Wightman mean by the "natural state space" of the theory? This must be read in the context of the subject of their book as a whole. In physical terms, they actually mean "states accessible from the vacuum state through local operations", or at least to a finite but arbitrary degree of accuracy. Implicit in this sentence is the assumption that the vacuum state is unique, which may not always be the case. Let us assume for the time being that it is, though (we will return to this point later). Mathematically speaking, this sentence summarizes the content of the Wightman(-GNS) reconstruction theorem, which tells us how to recover this Hilbert space from the vacuum expectation values of field products smeared with Schwartz test functions in space-time, seen as distributions (as mentioned by jjcale in his comment to the OP). This Hilbert space can be seen as the completion of the linear span of all smeared field products applied to the vacuum state vector.

In the case of free fields, this relates to the canonical field quantization as a continuum of oscillators corresponding to the different Fourier modes once you write the field operator smeared with a Schwartz test function in space-time in terms of these oscillators. The (Fock) Hilbert space you obtain in this fashion is the same - in fact, the Wightman(-GNS) reconstruction theorem can be seen as a generalization to possibly interacting quantum fields (which generally cannot be represented by means of creation and annihilation operators) of the Fock space construction. Streater and Wightman's sentence "some of its observables involve all the oscillators at once" can then be understood as a consequence of the uncertainty principle: if you smear your field product with test functions supported in a bounded region of space-time, representing the smearing in momentum space using the Fourier transform shows that all field oscillators contribute to the observable since the Fourier transform of these test functions vanish in no nonvoid open subset of 4-momentum space if they are not identically zero.

All seems well and good, but does the reconstruction theorem recover all physical states? The answer is unfortunately no, not even within finite but arbitrary accuracy. A typical example of a state not accessible from the vacuum state by means of local operations within arbitrary precision is a thermal equilibrium state of finite, nonzero temperature with respect to some Lorentz frame. The same can be said about any state reached from the latter state by means of local operations - they live in "disjoint" Hilbert spaces. Moreover, there are theories where the vacuum state is not unique, and with distinct vacua not locally accessible from each other in the above fashion. It is by no means obvious whether all these states fit into an infinite tensor product of Hilbert spaces such as the the one in the OP or not. If the theory is interacting, this is even less clear since (as mentioned in the previous paragraph) it may not be representable at all as a continuum of oscillators.

Of course, you could try to perform instead a "big" direct sum / integral of all these disjoint Hilbert spaces and use that as your state space receptacle, but this space (if it can be constructed at all in this fashion) is bound to be also technically extremely unwieldy to use. If you need to work with disjoint states in the above sense, is technically and conceptually better to work instead with the algebraic concept of state: as a certain kind of linear functional on the algebra of local observables of the theory and perform the GNS construction with respect to a convenient reference state (e.g. the vacuum or more generally a thermal equilibrium state with respect to some Lorentz frame) when a concrete Hilbert space is needed. All the information you need for that is already in the algebra of local observables. In other words, you no longer treat all states of the theory as belonging to a single Hilbert space but only look at subspaces of "mutually locally accessible" states, so to speak. These subspaces are usually called sectors of the theory. Under fairly general assumptions such as the ones in the Wightman-Garding formalism, sectors are always separable. Another advantage of the algebraic notion of state is that it incorporates both pure and mixed states into a single concept.

More broadly, the basic tenet of the so-called algebraic approach to quantum field theory (AQFT) by Haag and Kastler is that all the physical information of a quantum field theory is contained in how the algebras of local observables associated to each space-time region (generated in our present context by field products smeared with test functions supported in that region) embed into each other. This is the appropriate language to deal with families of disjoint physical states, in whatever context they may appear (thermal equilibria, superselection sectors and charged states, etc.).

Answer 4

There has been many answers, from many points of view on QFT: canonical quantization, algebraic QFT, etc. Let me add another perspective using the Euclidean path integral quantization, as emphasized in textbooks like "Quantum Field Theory and Critical Phenomena" by Jean Zinn-Justin.

For simplicity, I will take $M=\mathbb{R}^d$ In quantum mechanics, of say one particle in one dimension, the position is say $\phi\in\mathbb{R}$ and the momentum is say $\pi$. The physical Hilbert space is $L^2(\mathbb{R},d\phi)$ made of square integrable wave functions $\Psi(\phi)$, with respect to Lebsgue measure $d\phi$. When going to to QFT, the single number $\phi$, now becomes a classical field configuration $\phi=(\phi(x))_{x\in\mathbb{R}^d}$ and the physical Hilbert space should again be the $L^2$ space for a space of fields $\phi$ with respect to some measure $d\nu(\phi)$. Intuitively, one can think of this as a tensor product indexed by $x\in\mathbb{R}^d$ of $L^2$'s of $\mathbb{R}$, but trying to use this idea rigorously is probably counterproductive.

Suppose your theory is given by an honest Borel probability measure $\mu$ on $\mathscr{S}'(\mathbb{R}^{d+1})$, i.e., the Euclidean correlations or Schwinger functions $$ S_n(x_1,t_1;\ldots;x_n,t_n)=\frac{1}{Z}\int \phi(x_1,t_1)\cdots\phi(x_n,t_n)\ e^{-S(\phi)}D\phi $$ seen as distributions, namely, elements of $\mathscr{S}'(\mathbb{R}^{n(d+1)})$, are moments of a probability measure $\mu$. Suppose one can define a Borel-measurable restriction map $R:\mathscr{S}'(\mathbb{R}^{d+1})\rightarrow\mathscr{S}'(\mathbb{R}^{d})$ which sends the distribution $\phi(x,t)$ to $\phi(x,0)$. Then one can define a push-forward measure $\nu=R_{\ast}\mu$ and the physical Hilbert space should essentially be $$ L^2(\mathscr{S}'(\mathbb{R}^d),d\nu)\ . $$

Answer 5

The quote of Streater and Wightman is worked out in some detail in this paper by Thiemann and Winkler. In particular, they show why the restriction to this "tiny separable subset" (which is the Fock space of the fermionic or bosonic harmonic oscillators) breaks down in the context of quantum gravity.

The Infinite Tensor Product (ITP) in quantum field theory (QFT) decomposes into an uncountable direct sum of Hilbert spaces which in most applications are separable. Each of these tiny subspaces of the complete ITP are isomorphic with the usual Fock spaces of quantum field theory on Minkowski space (or some other background). The fact that one can do with separable Hilbert spaces in ordinary QFT is directly related to the fact that one fixes the background since this fixes the vacuum. The necessity to deal with the full ITP in quantum gravity could therefore be based on the fact that, in a sense, one has to consider all possible backgrounds at once. More precisely, the metric cannot be fixed to equal a given background but becomes itself a fluctuating quantum operator.

Answer 6

One way is to bypass (most) infinities altogether by going from a tensor product over an infinite number of modes to a tensor product over a finite number of modes. This is, effectively, what is done in e.g. quantum optics, though the details are not usually spelled out. One can do this, at least for quadratic Hamiltonians, in the following way:

Introduce a finite set of mode functions, $\{f_i\}_{i=1}^M$, which (a) satisfy the equations of motion and (b) are orthonormal with respect to the Klein-Gordon inner product, i.e. $(f_i,f_j)_\text{KG}=\delta_{ij}$, where $\delta_{ij}$ is the Kronecker delta-function (not the Dirac delta-function). Physically, $f$ is just a wave-packet.

One can then define the creation and annihilation operators accordingly via \begin{equation} \hat{a}_i\overset{\text{def}}{=}(f_i,\hat{\phi})_\text{KG}, \end{equation} where, e.g., $[\hat{a}_{f_i},\hat{a}_{f_j}^\dagger]=(f_i,f_j)_\text{KG}=\delta_{ij}$. We then build a basis for the Fock space, $\mathcal{F}_{\{f\}}=\bigotimes_{i=1}^M\mathcal{F}_{f_i}$ ($\mathcal{F}_{f_i}$ is the "single mode" Fock space) from the vacuum state, $|0\rangle$, (the ground state of the Hamiltonian; note the vacuum is unique in QFT in flat space-time) by successive applications of the creation operators, $\{\hat{a}_{f_i}^\dagger\}$; i.e., \begin{equation} \bigotimes_{i=1}^M|n_{f_i}\rangle\overset{\text{def}}{=}\bigotimes_{i=1}^M\frac{\hat{a}_{f_i}^{\dagger n_i}}{\sqrt{n_i!}}|0\rangle. \end{equation}

One can then show that these states form an orthonormal basis and that Hamiltonian operator is diagonal in this basis. Further, the Hamiltonian operator reduces to, \begin{equation} \hat{H}=\sum_{i=1}^{M}\hbar\omega_{f_i}\hat{n}_{f_i} + E_{gs}, \end{equation} where $\{\omega_{f_i}\}$ are the effective frequencies of the wave packets, $\hat{n}_{f_i}=\hat{a}^\dagger_{f_i}\hat{a}_{f_i}$ is the number operator for the $f_i$-mode, and $E_{gs}=\int d {k}\hbar\omega_k/2$ is the divergent vacuum (ground state) energy.

Digression: As one sees, the infinite collection of oscillators has then been reduced to a finite set of $M$ quantum harmonic oscillators. The only conundrum that remains is the divergent vacuum energy, $E_{gs}$, which is unavoidable once we go over to fields. It is possible, though, that the field description of the system is only an approximation to a large but finite set of coupled oscillators (taken "in the continuum limit"). It is also possible that the QFT breaks down altogether at some high enough energy scale (high enough $\omega$) and a better, more proper theory is required.