I am reading Jakob Schwichtenberg Physics from Symmetry where in 5.2 conjugate momentum density $\pi(x)$ is defined as generator of displacement of the field itself (1): $$ \pi(x) = −i\hbar\frac{\partial}{\partial Φ(x)}\tag{1} $$

From that definition it follows that $[Φ(x), π(y)] = i\hbar δ(x − y).$

So far so good, but in 9.1 we define (2) $$ \begin{equation} \pi(x) = \frac{\partial {\cal L}}{\partial (\partial_0 Φ(x))}\tag{2} \end{equation} $$ and still use $[Φ(x), π(y)] = i\hbar δ(x − y)$ that was derived from definition (1)

So my two questions

  1. What is the connection between two distinct definitions of $\pi(x)$?

  2. Why we can still use $[Φ(x), π(y)] = i\hbar δ(x − y)$ for the definition (2)?


1 Answer 1


Ref. 1 is possibly a bit skimpy on details at the 2 placed mentioned by OP. (OP's issue already arise in QM, and transcribes in a natural way to QFT.) Let us stress the following facts:

  • The momentum definition (2) is valid within the context of a classical Lagrangian formulation.

  • It is then implicitly understood that we next Legendre transform to a classical Hamiltonian formulation with a fundamental Poisson brackets $$\{\Phi(x),\Phi(y)\}~=~ 0, \qquad \{\Phi(x),\pi(y)\}~=~ \delta(x-y), \qquad \{\pi(x),\pi(y)\}~=~0,$$ which we in turn quantize to a quantum mechanical formulation with a CCR that OP mentions.

  • The momentum definition (1) is the Schrödinger position representation of this CCR. (There exists other representations of the CCR, such as e.g., the Schrödinger momentum representation, cf. e.g. this Phys.SE post.)

  • Moreover, the CCR implies that the momentum operator is the generator of position translations.


  1. J. Schwichtenberg, Physics from Symmetry, 2nd edition, 2018.
  • $\begingroup$ I think I get lost in what is postulated in the book (i e assumed "by definition") and what is deduced from math. It seems author of book postulated (1). From this it is deduced CCR ([Φ(𝑥),π(𝑦)]=𝑖ℏδ(𝑥−𝑦)) and then it is implicitly assumed that (2) is a equal (1) (i e generator of displacement). I also found this question from which it follows that conserved charge Q is indeed a generator of displacement. Which confuse me even more because Q != 𝜋(𝑥) in (2). $\endgroup$
    – Igor Batov
    Commented Feb 24, 2023 at 14:21
  • $\begingroup$ So, you answer is describing a standard way to do QFT - with "first quantization" and "second quantization". But what I really try to understand is a way to do QFT from symmetry - described in the OP mentioned "Physics from Symmetry". The author of the book do not use quantization nor path integral formulation to derive QFT. And so I try to unravel his approach. $\endgroup$
    – Igor Batov
    Commented Feb 25, 2023 at 14:26

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