Are fields in particle physics some wave functions, that is amplitude of probability?

Question

In particle physics, there are fields for fermions, typically labelled $\psi(x)$, and fields for scalar, typically labelled $\phi(x)$, which are functions of the space-time coordinate $x=(t, x, y, z)$.

What do represent the fields?

Are these fields in particle physics some wave functions of quantum mechanic, that is an amplitude of probability of presence of the particle at a given coordinate of space-time, as noted in quantum mechanism formalism?

(if so, I understand that a field may be invariant under a phase transformation $\phi\rightarrow e^{i\alpha}\phi$ since the measurement obtains the square of the wave function)

What is the mathematical codomain of the field : $\mathbf{R}$, $\mathbf{C}$?

Answer 1

In particle physics, there are fields for fermions, typically labelled $\psi(x)$, and fields for scalar, typically labelled $\phi(x)$, which are functions of the space-time coordinate $x=(t, x, y, z)$.

This is admittedly a confusing notation, since we also typically denote single particle wave functions in first-quantization by $\psi(x)$.

This is why sometime we are a little more clear and write the second-quantized fields with little hats like $\hat \psi(x)$ to remind ourselves that they are operator-valued functions, not complex number valued functions.

The thing denoted by $\psi(\vec x)$ in the "first quantized" theory is the wave function of a single particle. It is a complex number $\psi(\vec x)$ at every point in space $\vec x$. It is very different from the thing denoted by $\hat \psi(\vec x)$ in "second quantized" theory. (And, I do apologize for using these terms "first quantized" and "second quantized," as a vague shorthand for some much longer nuanced discussion.)

What do represent the fields?

In quantum field theory the $\hat \phi(\vec x)$ are linear operators; one for each point in space $\vec x$. So too are the $\hat \psi(\vec x)$ linear operators. (Those operators denoted by $\hat \psi(\vec x)$ often are so-denoted because they transform under a non-trivial representation of the Lorentz group, whereas the $\hat \phi(\vec x)$, notation often denotes a scalar quantum field.)

(if so, I understand that a field may be invariant under a phase transformation $\phi\rightarrow e^{i\alpha}\phi$ since the measurement obtains the square of the wave function)

The field is not invariant under that transformation. Clearly, the value changes from $\hat \phi(x,t)$ to $e^{i\alpha}\hat \phi(x,t)$. (If $\phi(\vec x)$ denotes a classical function rather than a quantum field, it is still not invariant under that transformation.)

However an operator like $$ \phi^\dagger \phi $$ is invariant under such a transformation.

What is the mathematical codomain of the field : $\mathbf{R}$, $\mathbf{C}$?

In quantum field theory the codomain of the Schrodinger picture operator-valued function $\hat\phi$ is the set of linear operators $\hat\phi(\vec x)$.

Answer 2

There seems to be confusion about what constitutes a wave function in Quantum Field Theory (QFT). Albeit QFT is not usually formulated in the wave function format (it's awkward and skibidi), let's humor OP and do a comparison of wave function of quantum mechanics (QM) vis-a-vis wave function of QFT:

In QM, the wave function is $$ \psi(\vec x, t) $$ whereas in QFT, the wave function is $$ \psi(\phi_{ x_1}, \phi_{ x_2}, \phi_{ x_3}, ..., t) $$ where we take QFT of scalar field $\phi(x)$ as an illustrative example. Note that wave function $\psi$ is complex-valued, while scalar field $\phi(x)$ could be either real or complex. Of course, given the bosonic nature of scalar field $\phi(x)$, the wave function $\psi$ is required to be symmetric under exchanging of any pair of $\phi_{ x_i}$ and $\phi_{ x_j}$.

As you can see, $\phi(x)$ is not the wave function of QFT. Rather, the whole collection of field values ($\phi_{ x_1}, \phi_{ x_2}, \phi_{ x_3}, ...$) at all the space points (${ x_1}, { x_2}, { x_3}, ...$) plays the role of space variable $\vec x$ in QM. It speaks to the fact that $\phi(x)$ (and its canonical conjugate $\pi(x)$) in QFT is promoted to quantum operator $\hat{\phi}(x)$/$\hat{\pi}(x)$, paralleling $x$ (and its canonical conjugate $p$) in QM being promoted to quantum operator $\hat{x}$/$\hat{p}$.

The QFT counterpart of QM amplitude of probability $$ \psi^\dagger(\vec x, t)\psi(\vec x, t) $$ is $$ \psi^\dagger(\phi_{ x_1}, \phi_{ x_2}, \phi_{ x_3}, ..., t)\psi(\phi_{ x_1}, \phi_{ x_2}, \phi_{ x_3}, ..., t) $$

Therefore, in QFT, you can't have an amplitude of probability interpretation for $$ \phi^\dagger \phi $$ It would be tantamount to having an amplitude of probability interpretation in QM for $$ x^\dagger x $$ which is obviously wrong.

The fun fact is that $\phi^\dagger \phi$ in QFT is indeed invariant under a phase transformation (a $U(1)$ gauge transformation in QED for complex scalar field $\phi$) of $$ \phi\rightarrow e^{i\alpha}\phi $$ It is related to the conservation of electric charge, conceptually very different from the invariance of amplitude of probability under a change of phase factor of wave function $ \psi\rightarrow e^{i\alpha}\psi $ in QM.