For a real scalar field, the equivalence $\pi^{0} = \dot{\phi}$ makes the concept of conjugate momentum intuitive at the physical level: it is the rate of change in time of the field.
This is true when the Lagrangian has a $\frac{1}{2}\dot \phi^2$ term and no other $\dot\phi$ dependence. However, in general, regardless of the $\dot \phi$ dependence of the Langrangian, we define the canonical momentum as:
$$
p_i = \frac{\partial L}{\partial \dot q}\;.
$$
Or, in field theory:
$$
\pi(\vec x) = \frac{\delta L}{\delta \dot \phi(\vec x)}\;.
$$
Or, more precisely, we always use the above definitions in classical mechanics and classical field theory. And we always look to these definitions for heuristic guidance when we perform canonical quantization (or "anti-canonical" quantization for fermions.)
- $\pi^{0}$ has lost all sort of intuition. I don't know what it represents anymore other than simply repeating like a parrot "it's $i \psi^\dagger$". But there must be a definition for $\pi^{0}$,
Yes, there is still a definition:
$$
\pi(\vec x) = \frac{\delta L}{\delta \dot \phi(\vec x)}\;,
$$
which can still be used regardless of whether the functional differentiation variable is denoted $\phi$ or $\psi$.
There is, however, a problem with the quantization of the Dirac field. Namely, we can not perform "canonical" quantization in the strict sense of implementing a canonical commutator:
$$
[\phi(\vec x),\pi(\vec y)] = [\psi(\vec x), i\psi(\vec y)^\dagger] \stackrel{?!}{=} i\delta(\vec x - \vec y)\;.
$$
Rather, we have to implement an anti-commutator:
$$
\{\phi(\vec x),\pi(\vec y)\} = \{\psi(\vec x), i\psi(\vec y)^\dagger\} {=} i\delta(\vec x - \vec y)\;.
$$
as in "$\pi^{0}$ is the variable that would allow us to know how the field behaves under translations in..." or something similar,
In bosonic quantum field theories, it is straight-forward to see that the $\pi(\vec y)$ field still generates translations of the $\phi(\vec x)$ field, in the analogous sense as it does in classical mechanics of $N$ particles. (I.e., $\phi(\vec x)\to\phi(\vec x)+c\delta(\vec x - \vec y)$.)
However, there are problems with a straightforward analog for fermionic fields, due to the anticommutation relations.
To see how this is straightforward for a bosonic field recall, in classical mechanics of $N$ particles, we have
$$
p_i = \frac{\partial L}{\partial \dot q_i}\;.
$$
In classical mechanics, we also recall, a canonical transformation can be implemented to change variables $(q,p)\to (\tilde q, \tilde p)$ without changing the form of Hamilton's equations of motion.
And, we also recall, that a special type of canonical transformation called an infinitesimal canonical transformation can be implemented by considering a generating function:
$$
F = \vec q\cdot \vec {\tilde p} + \alpha G\;,
$$
where $\alpha$ is a "small" parameter. And where, somewhat confusingly, we also call the function $G$ the "generator" of the infinitesimal translation.
In classical mechanics of $N$ particles, we can consider the generator $G$ to be the momentum $p_j$ itself. In which case, by the usual rules of classical mechanics, we have (with $G=p_j$):
$$
p_i = \frac{\partial F}{\partial q} = \tilde p_i + \alpha\frac{\partial G}{\partial q_i} = \tilde p_i
$$
and
$$
\tilde q_i = \frac{\partial F}{\partial \tilde p_i}= q_i + \alpha\frac{\partial G}{\partial \tilde p_i} = q_i + \alpha\frac{\partial G}{\partial p_i} + O(\alpha^2) = q_i + \alpha\delta_{ij}\;,
$$
where the final equality ignores higher order terms in $\alpha$. In other words, the momentum $p_i$ generates shifts of its conjugate partner $q_i$.
In classical field theory (and bosonic field theory) we similarly have canonical transformations and we can similarly set the generator to $G=\pi(\vec x)$, in which case we have:
$$
\pi(\vec x) = \tilde \pi(\vec x)
$$
and
$$
\tilde \phi(\vec x) = \phi(\vec x) + \alpha\delta(\vec x - \vec y)\;.
$$
that is valid for all types of fields at that can be applied to both the scalar and the spinor cases, allowing us to at least infer something about $\pi^{0}$ even before looking at $i \psi^\dagger$.
My question is: given a field that transforms as an arbitrary representation of the Lorentz group, what is the universally valid meaning of the conjugate momentum?
To glean some kind of "universally valid" meaning for the quantum field other than the definition:
$$
\pi = \frac{\delta L}{\delta \dot\phi}
$$
we can look at the momentum operator itself.
Recall, the momentum operator for a real scalar field is:
$$
\vec P = -\int d^3x \pi \vec \nabla \phi\;.\tag{A}
$$
The definition in Eq. (A) also holds for fermionic fields:
$$
\vec P = -\int d^3x (i\psi^\dagger)\vec \nabla \psi=\int d^3x \psi^\dagger (-i\vec \nabla\psi)\;,
$$
so, $\pi$ has the same meaning when convolved with $\nabla\phi$ in both bosonic and fermionic quantum field theories.
However, we can not straightforwardly write a direct analog of:
$$
U_j = e^{i p_j \delta q}
$$
when our field is fermionic, since in this case the analog of $p_i$ is not Hermitian, so the $U$ operator would not be unitary.
