I've just started QFT. While I'm learning it using the path integral formalism, I think there is another way which is the canonical representation which I didn't learn yet.

Nevertheless, in Quantum Mechanics, the wave function in the position basis is a function which describes the probability of a particle at point $\vec{x}$.

In QFT, fields come into play. How would the wave function of a field be? $\psi(A_\mu)$?

There are two things that come into my mind when I pondered about this:

  1. $A_\mu(x)$ is the equivalent to $\psi(x)$.

  2. There is another field for the electron field $\phi(x)$, let's say, and $\psi(\phi(x))$ is the probability of finding the electron field a certain configuration and is equivalent to $\psi(A_\mu)$

I tend to believe 1. is wrong because how would a vector-field be a probability distribution? At the same time, I don't know how to visualize $\psi(\phi(x))$.


3 Answers 3


In QFT, fields come into play. How would the wave function of a field be? $\psi(A_\mu)$?

Somewhat analogously to the basis of position eigenstates in QM, there is a basis of field eigenstates in QFT. Each eigenstate is in correspondence to not just one value, but the configuration of the entire field, meaning all field values at all values of $x$.

I will use the real scalar field $\hat{\phi}(x)$ rather than the vector field $A_\mu$, for a simple example. The eigenstates $|\phi(x) \rangle$ are simultaneous eigenstates of every field operator $\hat{\phi}(x)$, namely for every vector $x$ we have an eigenvalue equation. Using $x=1, 3.14$ as symbolic examples (really they should be 4-vectors, so four numbers): $$\hat{\phi}(1) | \phi(x) \rangle = \phi(1) | \phi(x) \rangle $$ $$\hat{\phi}(3.14) | \phi(x) \rangle = \phi(3.14) | \phi(x) \rangle $$ $$...$$

Here, $\hat{\phi}(x)$ denotes the operator of the scalar field. Without the hat, $\phi(x)$ is the eigenvalue corresponding to the operator at $x$ for the eigenvector $|\phi(x) \rangle $. The notation can be a bit confusing here: There is one eigenvector per field configuration, so for every possible function $\phi(x)$ we have a different eigenvector. But given a single eigenvector, it is a simultaneous eigenstate of infinitely many field operators, one for each $x$.

What role do these field eigenstates play in QFT? Not the same as $x$ in QM. You use them to derive the path integral in the theory. However, I have never seen an application in which they were used to predict predict probabilities of measurement outcomes directly through an inner product.

So, if these field eigenstates are not used to directly predict probabilities in QFT, how are probabilities predicted? Fundamentally they use a postulate from Quantum Mechanics, namely that for a system in state $|\psi \rangle$, the probability of finding $n$ particles with momentum $p_1, p_2, ... p_n$ is

$$|\langle p_1 p_2 ... p_n |\psi \rangle|^2 d^4p_1 d^4p_2 ... d^4 p_n$$

It is from this postulate that the scattering cross section is derived.

Finally, then, what about probabilities for positions? Better not ask, as there isn't a good answer to that.

  • $\begingroup$ I like your answer, and I believe it's closest to 2. I have one more question since you brought path integrals. Do the eigenstates of the field $|\phi(0)\rangle$ obey the principle of least action? If so does that mean the superposition of all fields might not obey that principle? $\endgroup$
    – Habouz
    Aug 17, 2022 at 19:24
  • 1
    $\begingroup$ The eigenstates for the field corresponding to different times are given by $|\phi(t,\vec{x})\rangle = e^{i tH} |\phi(0,\vec{x})\rangle$. I don't know of an action to write down to give this solution but maybe it's possible. $\endgroup$ Aug 17, 2022 at 19:30

I will post an answer that discusses both probabilities in QFT, but also the object in QFT that is analogous to the wave-function in QM.

  1. Probabilities : Probabilities and amplitudes are formed with the help of states. For instance, the transition amplitude of an initial state $|\alpha\rangle$ to scatter off and then turn into a final state $|\beta\rangle$ is given by the S-matrix elements $$S_{\beta\alpha}=\langle\beta|S|\alpha\rangle$$ as in quantum mechanics. So, this is an example of the overlap of two states describing a probability amplitude. Its square will be proportional to the probability of the initial state $|\alpha\rangle$ to scatter off and then turn into a final state $|\beta\rangle$

  2. Wave-Functional: In QFT, there is the Schroedinger representation, in which a wave-functional describes (the possible states of) the system. Due to the latter being a function of (scalar, vector, tensor etc) fields, hence a function of functions, functional analysis is needed to explore this topic. The wave functional $\Psi[\phi]$ for a scalar field takes as its argument a scalar field and returns a scalar value. Similarly, for a vector field, the wave functional $\Psi[A_{\mu}]$ takes as an argument a vector field and returns a scalar value. Wave functionals are solutions to a Schroedinger-like equation, in which the Hamiltonian can determine the time evolution of a quantum (field-theoretic) system. I will say no more, as this is a quite broad topic. I will refer you to Brian Hatfield's book on QFT of point particles and strings, and specifically to Chapters 9 and 10 for more.

I hope this helps...


A quantum field $\phi(x)$ (as defined by the Wightman axioms) is a distribution that is operator-valued, that is, if you smear $\phi(x)$ with a smooth rapidly descreasing function $f$, then you obtain an operator on a Hilbert space $\mathcal{H}$: $$\phi(f) := \int_{\mathbb{R}^d} f(x) \phi(x) \mathrm{d} x.$$ The (normalised) elements of $\mathcal{H}$ are the states of the system, just as in quantum mechanics.

I don't think it makes so much sense to talk about the 'wave function of a quantum field'. To me it sounds a little bit like e.g. asking for a 'wave function of the momentum operator' in quantum mechanics.


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