# What is the meaning of a state in QFT?

I guess this may be more of a mathematical than a physics question, but it comes down to physical interpretations, so I'm posting it here.

In classical Quantum Mechanics, we can define a state $\left| \psi \right\rangle$ to represent some probability amplitude over all of space. Specifically, it corresponds to a square-integrable function $\psi:\mathbb{R}^3 \rightarrow \mathbb{C}$. This state may vary in time, or one may take the view that the state remains constant and operators on the Hilbert space vary in time. (Schrodinger vs. Heisenberg)

To do a classical real scalar field, a state $\left| \Psi \right\rangle$ represents a functional probability amplitude over the possible field configurations: specifically, it corresponds to a functional $\Psi: \mathbb{R}^{\mathbb{R}^3} \rightarrow \mathbb{C}$. We could again take either the Schrodinger or Heisenberg picture here. (is this right so far?)

Most QFT introductions jump straight into fields over Minkowski space $\mathbb{M}^4$. This is where I get confused. It seems our field states still correspond to fields over spatial coordinates in $\mathbb{R}^3$ that vary in time. In $\mathbb{M}^4$, this would be saying that, given coordinates $(t,\mathbf{x})$, each constant-time slice $t=t_0$ has a field state associated with it. It seems to me, though, that choosing the slices and setting up the Hilbert spaces on each slice to get the states breaks Lorentz covariance. The alternative to me is to treat a state as being a functional $\Psi: \mathbb{R}^{\mathbb{M}^4} \rightarrow \mathbb{C}$, which is a probability amplitude over possible field configurations on the entire $\mathbb{M}^4$. However, this doesn't fit the math as far as I can tell.

What am I doing wrong? Or am I way off the mark? And what books/references can I find to better understand the formal/mathematical underpinnings of states in QFT?

• Think of the $\mathbb{R}^3$ in your quantum mechanical system not as a point in space but the generalised coordinates for your degrees of freedom. Then, you view a scalar field as providing one degree of freedom per spatial point. That might help you in understanding QFT better. – suresh Sep 28 '14 at 5:16
• FrancisFlute: I may write this up as a full answer later, but I would strongly recommend you learn about "second quantization" as it is the crucial element in answering your question for real. See, for example, my answer to this question: physics.stackexchange.com/questions/122570/…. Field theory differs from intro quantum mechanics because in QFT you have many particles. This forces you to seriously re-interpret what quantum state vectors mean, and you find that wave functions are far less useful. – DanielSank Sep 28 '14 at 6:24
• BTW, we needn't necessarily talk of relativistic theories. QFT is perfectly applicable in non-relativistic contexts as well, eg: cond-mat. – Siva Nov 2 '14 at 9:55
• I feel compelled to link this collaborative project: https://github.com/tobiasosborne/What-is-a-quantum-field-state- It should be an introduction to the same question from a quantum information point of view! – justmyfault Dec 27 '16 at 18:15

As correctly pointed out by Daniel Sank in the comment section, the key to understanding the state space in quantum field theory is the realization that it contains information about the excitations of operator-valued functions (quantum fields) of spacetime. The latter consists of one time and three spatial coordinates (at least in the context of the standard model). Note that unlike in non-relativistic quantum mechanics, there is no longer a position operator, hence space and time are treated on an equal footing, as can be understood from the fact that we seek to deal Poincaré invariant theories. In order to achieve this, there is also the possibility of instead promoting time to an operator, as is done in string theory, but this is a different story.

As mentioned above, states in quantum field theory encode information about the excitations of quantum fields. By quantizing these fields, one introduces ladder operators acting on the ground state at each point in spacetime. The ground state is usually denoted as $|0\rangle$ and corresponds to the vacuum, while excited states represent particles. In the case of a non-interacting theory, the Hilbert space is simply a Fock space, while in the interacting case, the construction of the state space is a highly non-trivial problem (for more detail see the answers to this question by Arnold Neumaier and me). The problem of choosing slices in spacetime and therefore breaking Lorentz invariance does not arise, the formalism can be written down in a completely covariant way.

The amplitudes one looks at are specified by the particle content of the in- and out-states. A typical example is that of particle decay: one has to calculate the amplitude for an in-state with one field excitation and an out-state with several excitations, not necessarily of the same field.

• I don't agree with Frederic completely, it seems to me that whenever one uses the Hilbert space/canonical quantization formalism of quantum mechanics (as opposed to the path integral) one is explicitly breaking Lorentz invariance, since one needs to specify a particular time direction on your manifold. States/operators in the Schroedinger/Heisenberg picture are then evolved in time using the Hamiltonian operator, which itself is not a Lorentz invariant quantity. See conformal field theory by di Francesco et al., chapter 6. – Meer Ashwinkumar Feb 5 '15 at 18:17

To put it simply, a QFT state is a linear superposition of the possibilities of one-particle wavefunction, two-particle wave function, three-particle wavefunction, ad infinitum. Each of those wavefunctions is a map from $\mathbb{R}^3 \rightarrow \mathbb{C}$ just like what you said.

BTW, the time evolution of that spatial configuration "follows" once you specify a Hamiltonian.

The linear combination over particle number is because the number of particles is not a fixed quantity in QFT (a la statistical mechanics in the Grand Canonical Ensemble). Refer to @DanielSank's comment on second quantization.

You can identify states with fields, which are maps from Minkosky space-time to an operators space. By definition of fields you want this maps to satisfy some invariance proprieties under suitable transformation.

• You can't in general identify states with fields. Any operator -- including a field operator -- creates a state. But it's not the case that two different operators must create different states. – user1504 Nov 1 '14 at 0:34

States in QFT are not so relevant, because what you are calling a "state" doesn't contain the physical information of the system in QFT. The relevant quantities in QFT are the correlation functions, which allow us to calculate explicitly transition probabilites between configurations of the same or different number of particles (the latter is the main goal of QFT's: a quantum mechanical state only describes the physics of a fixed number of particles). Q.M. states are called "fields" in QFT. For example, a classical field is the Klein-Gordon field. However, if you canonically quantize the Klein-Gordon field (via a formalism that relies on canonical conmutation relations) you promote the classical scalar field $\phi(x, t)$ to an operator $\hat{\phi}$ acting on a Fock space, which is the direct sum of (anti)symmetrized Hilbert spaces of fixed number of particles, $$\mathcal{F} = \bigoplus_i^\infty S_{\pm} \left(H_1\otimes H_2\otimes \cdots \otimes H_i \right)$$ so in canonical quantization we keep using quantum mechanical states, though these are not the fundamental quantities.

• QFT is a quantum theory. It has states just as any quantum theory has states. Why do you say that states don't make sense in QFT? – user1504 Nov 1 '14 at 0:14
• Yes, of course it has states. I mean that the word "state" in Q.M. refers to an object that contain all the physical information of the theory. In QFT the objects we are calculating are vacuum expectation values of time ordered products of fields, which contain the information we can measure experimentally. – Alvaro Nov 1 '14 at 0:21
• Sure, but those VEVs are amplitudes between certain states. It's really not any different from QM. – user1504 Nov 1 '14 at 0:28
• Yes, I've edited the answer. Thanks for the comment! – Alvaro Nov 1 '14 at 0:46
• This answer is full of misconceptions. While it is true that often people in particle physics and condensed matter tend to prefer correlation functions, this is a property of quantum field theorists, not of quantum field theory. The reason for this preference is that the full quantum state of an interacting QFT is horrendously complicated. Nevertheless, any time you calculate a transition amplitude or a correlation function, you are computing one "piece" of the quantum state. The statement "Q.M. states are called "fields" in QFT" is just totally incorrect. – Mark Mitchison Feb 5 '15 at 19:35