# How to interpret this construction of the states in QFT?

## Non-Relativistic Quantum Mechanics

To make this question clear it might be useful to contrast with non-relativistic quantum mechanics.

In any quantum theory, the states of a system are unit rays in a Hilbert space and hence can be described by unit vectors $$\Psi$$.

In the non-relativistic quantum mechanics of a particle, the appropriate Hilbert space is $$\mathscr{H}=L^2(\mathbb{R},dx)$$. The position operator $$X$$ is simply $$X\Psi(x)=x\Psi(x)$$. This allows to interpret $$|\Psi(x)|^2$$ as a probability density for position.

In any quantum theory, the time evolution is given by a unitary $$U(t,t_0)$$ which is generated by the Hamiltonian.

Once some state $$\Psi_0(x)$$ is given, one evolves it to $$\Psi(t,x)=U(t,t_0)\Psi_0(x)$$. So our conclusions are:

1. A state at one fixed time instant is a map $$\Psi_0 : \mathbb{R}\to \mathbb{C}$$;

2. The time-evolution of the system is a path of states $$\Psi : \mathbb{R}\times \mathbb{R}\to \mathbb{C}$$ so that for each $$t$$ fixed we have one such state as in (1) which is $$\Psi(t,\cdot)$$.

## Free QFT in Minkowski Spacetime

Here we consider a free scalar QFT in Minkowski spacetime. This already is able to show the essence of the doubt.

One way to construct the space of states is outlined in Wald's GR book. The question is about this specific construct because it can be generalized to other spacetimes easier than the one based on representations of the Poincare group.

We define the one-particle Hilbert space, $$\mathscr{H}$$, to be the vector space composed of positive frequency solutions of the Klein-Gordon equation whose Klein-Gordon norm is finite, with inner product on $$\mathscr{H}$$ defined by

$$(\alpha,\beta)_{KG}=i\int_{\Sigma}(\bar{\alpha}\nabla_a\beta -\beta \nabla_a \bar{\alpha})n^a d\Sigma$$

The Hilbert space of all possible states of the Klein-Gordon scalar field is taken to be the symmetric Fock space, $$\mathscr{F}_S(\mathscr{H})$$, constructed from $$\mathscr{H}$$.

So focus on the one-particle Hilbert space $$\mathscr{H}$$ so constructed. Its elements are the states of one particle. But now these are maps $$\Psi : M\to \mathbb{C}$$ where $$M$$ is Minkowski spacetime. In coordinates such a map is $$\Psi(t,\mathbf{x})$$.

But wait a minute. Now this is confusing. Compare to non-relativistic quantum mechanics. There the state has no time-dependence. The curve of states representing the time-evolution which has time-dependence.

Here the state itself has time-dependence.

So the evolution would be something like $$\Psi(t')(t,\mathbf{x})$$ with "two time parameters"? This is extremely weird.

Or these states already carry some idea of time evolution into them?

It is also hard to interpret such a state $$\Psi(t,\mathbf{x})$$. There is no position operator in QFT so it is not possible to say that $$|\Psi(t,\mathbf{x})|^2$$ is the probability density for the particle to be at $$\mathbf{x}$$ at time $$t$$ as in the non-relativistic case.

So how do we interpret this construction of states? How such a $$\Psi(t,\mathbf{x})$$ is to be interpreted and what justifies the interpretation?

• Time-dependent states aren't weird, they appear even in undergraduate nonrelativistic quantum mechanics under the name of Heisenberg picture. The time-dependence captures, well, the time-dependence of the state. To recover a time-independent state you can take a timeslice, if applicable in your spacetime. – knzhou Feb 20 at 1:37
• @knzhou I think I'm missing something obvious. In the Heisenberg picture we fix the state as the initial state and the observables evolve right? In other words, given $\Psi_0$ the initial state, it remains fixed and observables evolve according to $A(t)=U^\dagger(t,t_0)A U(t,t_0)$. In that case $\Psi_0$ doesn't depend on time, the observables do. What I'm missing here? – user1620696 Feb 20 at 1:44
• It depends on how you set it up, I suppose. For example, I've seen the state $|x, t \rangle$ defined "in Heisenberg picture" to mean the eigenvector of $\hat{x}_H(t)$ with eigenvalue $x$. In that sense, the states extend across time. But in practice it's nothing to worry about because the calculations will look exactly the same up to minor bookkeeping swaps. – knzhou Feb 20 at 1:55
• For example, the amplitude to propagate from $(x, t)$ to $(y, t')$ would be written as $\langle y, t' | x, t \rangle$ in this notation, while in standard notation it would be $\langle y | U(t', t) | x \rangle$. But you can see how they're really the same thing. – knzhou Feb 20 at 1:56
• But this is a different thing IMHO. In your example we have a time-dependent state. So for each $t$ you have one state $|x,t\rangle$. That's fine. We have a set of states $\mathscr{H}$ and for each $t$ we pick one element labelled by it. In the construction I've outlined that's not the case. The states themselves are functions of time. So the function $\Psi$ - which is a state - defined on spacetime (and which hence depends on time) is one single element of $\mathscr{H}$. Not many parametrized by time. That's the thing I'm having a hard time to reconcile with non-relativistic QM. – user1620696 Feb 20 at 2:35

I think it's probably best to drop the quantum and just understand what's going on here in the classical setting.

In classical mechanics, we usually take a state to be specified by initial value data for the equations of motion, e.g., a point $$(x,p) \in T^*\mathbb{R}^3$$ for a single particle. But, because we work with equations of motion for which the initial value problem is well-posed, we've got an isomorphism

$$\{\mbox{solutions to equations of motion}\} \simeq \{\mbox{initial value data}\},$$

so we could equally well define the set of states to be the set of solutions to the equations of motion. This is an equally good way of specifying the state of the system. This is what Wald is doing. The only difference is that he's in the quantum setting, so he's considering superpositions of classical states.

$$\def\bp{{\bf p}} \def\bx{{\bf x}}$$ I'd suggest an equivalent approach, which perhaps you'll find more palatable.

Instead of working with

positive frequency solutions of the Klein-Gordon equation whose Klein-Gordon norm is finite

as Wald does, you could deal with $$L^2(V^+,d^3\bp/p^0)$$ where $$V^+$$ is the positive mass shell in 4-momentum space. This is in Fourier 3D transform relationship with Wald's $$\mathscr H$$: $$\Psi(t,\bx) = \int_{V^+}\!\exp(i p_\mu x^\mu)\,\Phi(\bp)\>{d^3\bp \over p^0}$$ Note that $$p_0 = +\sqrt{\bp^2 + m^2}$$ and $$\Psi$$ is defined as a function of $$t$$ because of $$\exp\ldots$$ which contains time evolution. On the other hand $$\Psi(t,\bx) = U(t)\,\Psi(0,\bx)$$ as you would like.

• Let's see if I got it. The space $L^2(V^+,d^3\mathbf{p}/p^0)$ has the advantage that a state $\Phi(\mathbf{p})$ has a direct interpretation as an amplitude for momentum. Now the translation operator acts exactly as $$U(a)\Phi(\mathbf{p})=e^{i p_\mu a^\mu}\Phi(\mathbf{p})$$. In particular the free time evolution is encoded as a pure translation in time. So when we Fourier transform to get Wald's space, we actually end up getting $\Psi(t,\mathbf{x})$ which is the free time evolution of a state $\Psi(0,\mathbf{x})$. In that case, states can be seen as $\Psi(0,\mathbf{x})$. Is that your point? – user1620696 Feb 20 at 16:49

In relativistic theories, space and time should be treated on equal footing as coordinates, and this applies also to relativistic quantum theories.

Therefore, the quantum fields (i.e. the fundamental observables in relativistic quantum theories) depend on both space and time as coordinates, and thus a spacetime vector $$x_\mu = (x_0,\vec{x})$$ substitutes the space coordinate vector $$\vec{x}$$ used in non-relativistic theories.

This also translates to some extent to quantum states, that are (noncommutative) probability distributions acting on observables (that are formed from spacetime fields). The distinction "to some extent" is due to the fact that in relativistic quantum mechanics, since particles can be created and annihilated, the fields rather than (the quantum version of) coordinates and momenta are the fundamental observables of the theory, on which states act upon.

The concept of time evolution in relativistic quantum mechanics thus becomes a bit more subtle than in the non-relativistic case. This is due to the fact that it would be unwise (more properly, non necessarily covariant) to characterize the evolution with respect to the time of a fixed frame.

A relativistic quantum system in Minkowski spacetime is specified by two things: a representation of the Poincaré group, defined somewhat abstractly on the algebra of observables; and a pure state that is invariant under the (adjoint) action of the Poincaré group. Such state, the so-called vacuum state, determines the representation of states as Hilbert space vectors, and determines uniquely the considered theory. The generator, in such representation, of the time-translations of the Poincaré group is the Hamiltonian of the system. Finding the vacuum for interacting theories is one of the most difficult and still open problems in mathematical and theoretical physics. For free theories, the vacuum is the Fock vacuum, and thus this justifies the construction mentioned by the OP.

In curved spacetime, the situation is even more complicated, because there is no more a symmetry group for the spacetime, and therefore one should find a suitable analogue of the vacuum state. For free theories, the so-called Hadamard states are used.

• Thanks @yuggib. But you see, by Wald's construction, a state is a function of space and time $\phi(t,\mathbf{x})$ which solves the KG equation and is of positive frequency to the inertial observer. So at least we have a natural vacuum construction here. Still, how do we interpret such state? I don't see how to justify saying that $|\phi(t,\mathbf{x})|^2$ is a probability density for position as in non-relativistic QM since we have no position observables. – user1620696 Feb 20 at 13:50
• @user1620696: You are forgetting an important thing mentioned by Wald which is to take the Fock space $\mathcal{F}$ over the space $\mathcal{H}$ which contains your classical KG solution $\phi(t,\mathbf{x})$. The thing whose modulus squared has a probability interpretation belongs to $\mathcal{F}$ not $\mathcal{H}$. – Abdelmalek Abdesselam Feb 20 at 18:56

A function is a function, not a state. It is said that states are in one-to-one correspondence with such functions. It is not said that a state at time $$t$$ is described by the function at time $$t$$. Imagine you are working in flat space, with usual creation-annihilation operators. There is a one-particle state for every creation operator. There is also a field mode, which is a function of $$x$$ and $$t$$, for every creation operator. This mode is the function that Wald is taking about, if I understand correctly. It doesn't mean that creation operators depend on time, and so it doesn't mean that the states they create depend on time. Everything is in Heisenberg picture.