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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:

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

Qmechanic
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BasicMathGuy
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1 Answers1

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With our modern understanding of quantum field theory, the Dirac equation is not a generalization of the Schrodinger equation from non-relativistic 1-particle quantum mechanics to relativistic 1-particle quantum mechanics, despite the fact that Dirac discovered his equation by looking for an equation to describe relativistic 1-particle quantum mechanics. In fact, we now understand that it there no so consistent theory of (interacting) 1-particle relativistic quantum mechanics; by combining relativity and quantum mechanics, we are necessarily led to theories with an indefinite number of particles.

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 \begin{equation} i \frac{\partial}{\partial t}|\Psi\rangle = H|\Psi\rangle \end{equation} 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) \begin{equation} H = \int d^3 x \bar{\psi} \left(-i \gamma^i \partial_i + m\right) \psi \end{equation}

Andrew
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