# Canonical field momentum in quantum field theory

In the context of the second quantization and the use of fields in the canonical quantization, the canonical momentum of the field is defined as the derivative of the field by the time coordinate. But if we're talking relativistically, shouldn't it be the derivative of the field by the proper time? What am I missing?

Thanks

• You're missing that it is a definition, and so it can be whatever the author intends to. Anyway, the proper time of what? Mar 5, 2016 at 17:30
• Isn't there a reason for such a definition? The use of proper time in the question might have been misplaced, but what I wanted to express is my doubt in relation to the definition. Why not define too a canonical momentum for each of the coordinates, why is the time derivative different? Mar 5, 2016 at 17:40
• Yes, the reason the same as in classical mechanics: it is the natural definition that arises when going from the lagrangian to the hamiltonian formalism (see Field theory: equivalence between Hamiltonian and Lagrangian formulation). In Hamiltonian mechanics time is different from the other coordinates. Some people suggested that we should use cannonical momentum for all the coordinates (check the wikipedia article on the De Donder-Weyl theory). Mar 5, 2016 at 18:19
• The question (v1) seems to be essentially a duplicate of physics.stackexchange.com/q/38286/2451 Mar 5, 2016 at 19:50
• In the covariant formulation of the Lagrangian time and space are put on an even footing by using an additional parameter; thus time loses is special place as the temporal parameter of the positions coordinates: they all share the new parameter. For some discussion, see en.wikipedia.org/wiki/… Mar 5, 2016 at 21:05

The second quantisation you mentioned is an equal time quantisation so it is specific to the frame one starts with, and for this reason time and spatial indices are not treated equally. For details, one can see ,for example, the canonical quantisation of scalar field from Srednicki's QFT Chapter 3. However, we do need to check that the canonical quantisation is compatible with Lorentzian transformations, and one way to do this is to check the following diagram commutes: (for notational convenience I only write one particle state. But this should be checked for $n$ particle states )
$$\require{AMScd} \begin{CD} p^u @>{Quantization }>> |p^u >;\\ @V{\Lambda}VV @V{U(\Lambda)}VV \\ \Lambda^u{}_vp^v @>{Quantization}>>|\Lambda^u{}_v p^v > ; \end{CD}$$
where $U(\Lambda)$ is according to Srednicki's notation the representation of Lorentzian transformation.