# Why does the Dirac Lagrangian not already use operators (instead of canonical quantization)?

I've learned that in canonical quantization you take a Lagrangian, transform to a Hamiltonian and then "put the hat on" the fields (make them an operator). Then you can derive the equations of motion of the Hamiltonian.

What is the reason that you cannot already put hats in the Dirac Lagrangian? Therefore write the Lagrangian with operators and go straight to Euler-Lagrange equations. Or is there any way this has been tried?

As mentioned in the comments you'd have to make an adjustment to get a real scalar from the operator expression. Can it be done right in the Lagrangian, without artificially putting hats on after a transformation?

• Why are you asking specifically about the Dirac Lagrangian? Doesn’t your question apply to any Lagrangian? Sep 14, 2022 at 17:41
• Just for simplicity. I'm not familiar with the other parts of the general Lagrangian yet. But more general answers are fine.
– Gere
Sep 14, 2022 at 17:42
• Because the Lagrangian density gets integrated to give the action, which is supposed to be a number that you extremize. Sep 14, 2022 at 17:46
• A quantum-mechanical Hamiltonian, by contrast, has to be an operator. Sep 14, 2022 at 17:55
• @Ghoster Is being a number the only catch? What about introducing an operation which extracts a number from an operator expression and applying it to the Lagrangian? Can this work?
– Gere
Sep 14, 2022 at 18:01

In quantization the starting point is a classical theory.

1. On one hand, in a classical Hamiltonian formulation, there is a (super) Poisson bracket $$\{\cdot,\cdot\}$$, which we can formally replace with a (super) commutator $$[\cdot,\cdot]$$, e.g. using deformation quantization, cf. e.g. this related Phys.SE post. In this way the classical Hamiltonian formalism is closely related to the quantum mechanical operator formalism.

2. On the other hand, in a classical Lagrangian formulation, it is more indirect how to introduce non-commutativity and operators$$^1$$.

Of course, the classical Lagrangian and Hamiltonian formulations are formally equivalent, so the two approaches are ultimately related.

E.g. the time evolution operator $$\hat{U}_I$$ in the interaction picture is \begin{align} \hat{U}_I~=~& T \exp\left\{ -\frac{i}{\hbar}\int\!dt~H_{\rm int}(\hat{q},\hat{p})\right\}\cr ~=~& T_{\rm cov} \exp\left\{ \frac{i}{\hbar}\int\!dt~ L_{\rm int}(\hat{q},\dot{\hat{q}})\right\} ,\end{align} cf. e.g. my related Phys.SE answer here.

--

$$^1$$ Note that inside the path integral the value of the integration variables and the action itself are (super) numbers, not operators,

• It is more indirect for the Lagrangian, but can it be done? Can I put hats on things in the Lagrangian, apply an operation to get a number from my operator expression (to get a proper real number for the Lagrangian), and finally use Euler-Lagrange equations to solve all of QFT? If it can be done, how does it look like? If it cannot be done, how to show that mathematically?
– Gere
Sep 14, 2022 at 18:57

Note that the Lagrangian density, the Hamiltonian density and the equations of motion all can be deduced from each other via the Legrende transformation for fields and the Euler - Lagrange equation for fields. Regarding the Dirac equation specifically, there is no classical Lagrangian/Hamiltonian density to quantize (since it is dealing with fermions). But for scalars, you deduce the QFT Klein - Gordon equation from a (complex version of a) classical relativistic scalar Lagrangian density (i.e. $$\mathcal L= \partial_\lambda \Phi^\dagger \partial^\lambda \Phi-\mu^2 \Phi^\dagger \Phi$$ ) and the Euler - Lagrange equation for fields. (What makes something an operator is not determined by whether you start with the Hamiltonian- or Lagrangian density, but by invoking the commutation relations for fields).