# Origin of the canonical equal time commutation relations

I stumbled upon this QFT exercise:

And I must say I'm somewhat baffled. I always thought that the commutation relations are something that is postulated in making the shift from classical physics into quantum; they replace the Poisson bracket and so on. This exercise seems to imply that the commutation relations can be derived from the Lagrangian of the theory. How does one make sense of that?

• I think you are correct that they want you to show that the Poisson bracket in the classical theory is 1. – Bobak Hashemi May 26 '17 at 18:47
• Which reference? – Qmechanic May 26 '17 at 19:07

With some technical assumption, in fact, CCR arises from requiring compatibility of Lagrangian/Hamiltonian formalism and standard formulation of Quantum Theory. I will not be completely rigorous below and I only indicate the way to follow to achieve the wanted result.

Consider the Hamiltonian of the theory, it reads $$H(t_0) = \frac{1}{2} \int_{\mathbb{R}^3, t=t_0} \Pi_\mu \Pi^\mu + \vec{\nabla} A_\mu \cdot \vec{\nabla} A_\mu d^3x\tag{1}$$ Form Heisenberg evolution of operators we are committed to assume that $$\partial_t A_\mu(t,x) = i[H(t), A_\mu(t,x)]\tag{2}\:.$$ On the other hand, from Hamilton equations, or directly form the definition of conjugated momentum in Lagrangian formulation, we have $$\Pi_\mu(t,x) = \partial_t A_\mu(t,x)\tag{3}\:.$$ So, putting (1) and (2) together, it must be $$\int [\Pi_\mu(t,x) \Pi^\mu(t,x), A^\nu(t,y) ] d^3x+ \int [\vec{\nabla} A_\mu(t,x) \cdot \vec{\nabla} A_\mu(t,x), A^\nu(t,y)] d^3x = -2i\partial_t A^\nu(t,y)$$ If we now assume that

H1. measurements at different positions at the same fixed time of generally different components of $A_\mu$ are compatible in quantum sense,

i.e., $[A_\mu(t,x), A^\nu(t,y)]=0$,

it remains $$\int [\Pi_\mu(t,x) \Pi^\mu(t,x), A^\nu(t,y) ] d^3x= -2i\partial_t A^\nu(t,y)$$ that is $$\int \Pi_\mu(t,x)\:[\Pi^\mu(t,x), A^\nu(t,y) ] d^3x + \int [\Pi_\mu(t,x), A^\nu(t,y) ]\: \Pi^\mu(t,x) d^3x= -2i\partial_t A^\nu(t,y)$$ If now we further assume that

H2. $[\Pi_\mu(t,x), A^\nu(t,y) ]$ is a number $[\Pi_\mu(t,x), A^\nu(t,y) ]= c(t,x,y)I$ so that it commutes with operators,

we have $$\int \Pi_\mu(t,x)c(t,x,y) d^3x = -i\partial_t A^\nu(t,y)\:.$$ From (3) $$\int \Pi_\mu(t,x)c(t,x,y) d^3x = -i\Pi_\nu(t,y)\:.$$ This means $$\int \Pi_\mu(t,x)\left(c(t,x,y) + i\delta(x,y) \delta^\mu_\nu \right)d^3x =0\:.\tag{5}$$ This identity must be interpreted within the fixed-time smearing procedure (also the previous lines should be interpreted this way, but here I make explicit the formalism since a crucial further hypothesis needs which is stated with this formalism): The operators $A(t,x)$ and $\Pi(t,x)$ have to be smeared with smooth compactly supported functions $f : \mathbb{R}^3 \to \mathbb{R}$ giving rise to the smeared field operators, the ones with mathematical sense. $$A(t,f) := \int A(t,x) f(x) d^3x\:,\quad \Pi(t,f) := \int \Pi(t,x) f(x) d^3x$$ For instance $$[A(t,x), \Pi(t,y)] = i \delta(x-y)I$$ has to be interpreted as a short way to write $$[A(t,f), \Pi(t,g)] = i \int f(x)g(x) d^3x \:I$$ This way (5) actually means that, for every smooth compactly supported function $f : \mathbb{R}^3 \to \mathbb{R}$, $$\int d^3y \int \Pi_\mu(t,x)\left(c(t,x,y)f(y) + i\delta(x,y) f(y)\delta^\mu_\nu \right)d^3x =0$$ In other words $$\Pi_\mu\left(t, \int c(t,\cdot,y)f(y) d^3y +i \delta^\mu_\nu f\right) =0\tag{6}$$ The last hypothesis I demand is that

H3. the operator valued distribution $f \mapsto \Pi_\mu(t,f)$ vanishes if and only if $f=0$.

Assuming this, (6) implies $$\int c(t,x,y)f(y) d^3y +i \delta^\mu_\nu f(x) =0$$ for every said function $f$, which means $$c(t,x,y) = -i\delta^\mu_\nu \delta(x-y)\:,$$ as wanted.