# Kronecker delta commutation relations for QFT

Setup: In many textbook treatments of canonical quantization (e.g., Peskin and Schroeder), one imposes canonical (equal time) Dirac delta commutation relations on the conjugate field operators. e.g., for the scalar field,

\begin{align} [\phi(\vec{x},t),\pi(\vec{y},t)]=i\hbar \delta^{(3)}(\vec{x}-\vec{y})\\ [\phi(\vec{x},t),\phi(\vec{y},t)]=0=[\pi(\vec{x},t),\pi(\vec{y},t)]. \tag{*}\label{*} \end{align}

Now, I believe I understand the formal "dictionary" that one typically employs to recast formulas for discrete finite-dimensional systems into relations valid for continuous infinite-dimensional systems: sums become integrals and, thus, Kronecker deltas become Dirac deltas, etc. Under these considerations, \eqref{*} is a natural generalization of the canonical commutation relations familiar from finite-dimensional quantum mechanics.

Hypothetical: However, suppose one instead tried to naively define a Kronecker delta over the reals---I appreciate that is precisely the role the Dirac delta is playing here---but, nevertheless, suppose that one stubbornly defines instead

\begin{align} \delta_{\vec{x},\vec{y}}\equiv \begin{cases} 1 & \text{if }\vec{x}=\vec{y}\in\mathbb{R}^3\\0 & \text{otherwise}\end{cases} \end{align}

And then attempt to replace the nontrivial commutator in \eqref{*} with

\begin{align} [\phi(\vec{x},t),\pi(\vec{y},t)]=i\hbar \delta_{\vec{x},\vec{y}} \tag{**}.\label{**} \end{align}

Question: Essentially, I am trying to figure out precisely where $\eqref{**}$ will first go off the rails. In particular, before even specifying the equations of motion for the classical $\phi(x)$, is the Kronecker delta commutation relation $\eqref{**}$ in conflict with some basic assumption we would normally want to make about the quantum field? Or does $\eqref{**}$ only run into issues when trying to simultaneously satisfy $\eqref{**}$ and the equations of motion of a particular theory--e.g., the Klein-Gordon equation.

Attempt: The first thing I notice is that while the Fourier transform of the Dirac delta commutation relations \eqref{*} should be well-defined (assuming proper fall-off behavior of the field), the Fourier transform of \eqref{**} is not well-defined because $\delta_{\vec{x},\vec{y}}$ does not integrate like the Dirac delta distribution does---I think $\delta_{\vec{x},\vec{y}}$ is a set of measure zero.

This is perhaps consequential because the first thing one typically does in Peskin and Schroeder type treatments is to assume the field can be Fourier transformed so that the Klein-Gordon equations of motion, e.g., can be recast as a linear combination of harmonic oscillator equations with "frequency" $\omega_\vec{p}^2\equiv \vec{p}^2+m^2$, where $m$ is the coupling appearing in the linear Klein-Gordon equation, $(\partial^2+m^2)\phi(x)=0$, and $\vec{p}$ is the Fourier momentum space variable.

thoughts?

• I think you basically got it; the sum over the Kronecker delta should give one, so we want the integral over the Dirac delta to give one as well, to get similar properties. The integral of your modified delta function is always zero and hence it gives no useful results. – knzhou Dec 23 '17 at 20:28
• Concerning a related comparison of Kronecker & Dirac delta normalization in a Hilbert space, see my Phys.SE answer here. – Qmechanic Dec 23 '17 at 20:28