Wave functions, Ket vectors and Dirac equation: why can't I use ket formulation on Dirac equation?

From non-relativistic quantum mechanics, a $$\frac{1}{2}$$- spin system can be represented by a ket vector like:

$$|\psi\rangle = a|+\rangle_{z}+b|-\rangle_{z}. \tag{1}$$

The object on $$(1)$$, is a ket vector with a fancy name: a (Pauli) spinor. This spinor object will satisfy the Pauli equation, the low-energy dynamical equation for the Dirac equation.

Now, still in non-relativistic quantum mechanics context, the Schröedinger equation can be dealt with in both ways:

$$i\hbar\frac{\partial }{\partial t}|\psi\rangle = H |\psi\rangle, \tag{2}$$

and

$$i\hbar\frac{\partial }{\partial t}\psi(\vec{r}) = H \psi(\vec{r}). \tag{3}$$

We can use Pauli spinors, in both $$(1)$$ fashion and $$(3)$$ fashion, since $$\psi(\vec{r}) := \langle \vec{r}| \psi\rangle$$.

But, in quantum field theory, we never use ket vectors! So my question is:

Why I can't study quantum field theory with a Dirac equation acting in a "Dirac four-spinor" $$|\Psi\rangle$$ as:

$$(i\gamma^{\mu}\partial_{\mu}-m)|\Psi\rangle = 0? \tag{4}$$

• Ok, but could you elaborate an answer with math? Commented Nov 27, 2022 at 16:27
• You can do this, it would basically mean you have moved to the schrodinger picture, where the states are again the objects which change in time. Commented Nov 27, 2022 at 16:28
• @doublefelix first: do you have a reference? Second: it means that every field in QFT is in position representation? Because, it seems a bit odd to sat that a vector field $|A\rangle>$ is written as $\langle x^{\mu} | A\rangle := A_{\mu}$ Commented Nov 27, 2022 at 16:31
• @doublefelix There is no position basis in QFT. $\langle 0|\phi (x^{\mu}) |v\rangle$ gives you an approximate position basis solution. The real Schrodinger equation of QFT is the one with the integral of Hamiltonian density as its Hamiltonian. This Schrodinger equation is nothing like Klein Gordon or Dirac equation Commented Nov 27, 2022 at 16:49
• There's a lot to say here. It is true that there is no position basis with satisfactory properties in QFT. Nonetheless there are kets, in the momentum basis, which not only exist but are necessary to relate the theory to experiment. You could in principle work in the schrodinger picture, in which case it is only the state (a ket) which changes in time. But the dirac equation is almost never used in that way- in normal QFT courses we work in the heisenberg or interaction picture. A spinor field is not an element of the hilbert space, nor is any field. They are operators/operator-valued distribs Commented Nov 27, 2022 at 17:04

Instead, the Dirac equation is best understood as an equation satisfied by a quantum field $$\psi(x, t)$$. It best understood as a generalization of the Heisenberg equation for the time evolution of operators. In quantum field theory, the spinor field $$\psi(x, t)$$ is a field (an operator-valued distribution), not a state.
We can and do talk about states in quantum field theory. Usually we work in the Heisenberg or interaction pictures, in which case the time evolution is carried by the operators (fields), or a mix of the operators and states. However we can work in the Schrodinger picture, in which case the state $$|\Psi\rangle$$ (which, in the field basis, is now a functional, mapping each possible field configuration to a probability amplitude) obeys the functional Schrodinger equation $$$$i \frac{\partial}{\partial t}|\Psi\rangle = H|\Psi\rangle$$$$ where $$H$$ is a functional of the fields. For example, for a free, massive spinor field $$\psi$$ (don't confuse $$\psi$$ the field with $$\Psi$$ the state!), $$H$$ would be (see, eg, Eq 5.8 here) $$$$H = \int d^3 x \bar{\psi} \left(-i \gamma^i \partial_i + m\right) \psi$$$$