# Commutation Relations in Second Quantization

I understand that if I have the field operators $\psi(r)$ and $\psi^\dagger(r)$, then I have the canonical commutation relation (in the boson case) $$[ \psi(r) , \psi^\dagger(r')]=\delta(r-r').$$ My question is that if during manipulation of an equation I want to use the commutator, but both field operators are evaluated at zero, then I have $$[ \psi(0) , \psi^\dagger(0)]=\delta(0).$$ But $\delta(0)$ does not seem to be well defined even under the integral. I know I'm missing something really simple here.

I suppose my question then is what explicitly is the value of the commutator $[ \psi(0),\psi^\dagger(0)]$?

## 2 Answers

You are not missing anything: the commutator $[\psi(0),\psi^\dagger(0)]$ is ill defined. This is related to the fact that operators are actually distributions, not functions of $x$, so taking $x=0$ is meaningless.

• So then I am only allowed to use the commutation relations if $\psi$ is left unevaluated? what then is the meaning of the symbol $\psi(0)$? – Paul Malinowski Apr 18 '16 at 19:57
• The context of this is that I am dealing with the function $C(x) = <n(x)n(0)>$, where $n(x) = \psi\dagger(x)\psi(x)$, and the brackets denote an expectation value. – Paul Malinowski Apr 18 '16 at 19:59
• 1) yes, so to speak. 2) $\psi(0)$ can only be used in formal expressions when it is not multiplied by another factor of $\psi$. For example, $\langle 0|\psi(0)|0\rangle$ is well defined, but $\langle 0|\psi(0)^2|0\rangle$ is not. This is because there is not a well defined notion of multiplication for distributions: $\psi(0)^2$ doesn't make sense. Functions can be multiplied, but distributions cannot. 3) Are you sure that $C(x)=\langle n(x)n(0)\rangle$? shouldn't the fields be normal-ordered? (normal ordering enables us to multiply distributions in a well defined way). – AccidentalFourierTransform Apr 18 '16 at 20:03
• This was how the problem was posed to me. The context is that I am dealing with noninteracting spinless fermions in one dimension. – Paul Malinowski Apr 18 '16 at 20:05
• @PaulMalinowski then I guess the fields should be time-ordered at least (right?). My best bet is that you are supposed to expand $\psi$ in creation/annihilation operators, where time-ordering is to some extent equivalent to time-ordering because of Wick's theorem. I'd suggest you to expand $\psi(x)$ in $a,a^\dagger$ and see what happens - but if there is no ordering prescription the expression is ill-defined. – AccidentalFourierTransform Apr 18 '16 at 20:10

When you write $n(x)=\psi^\dagger(x)\psi(x)$, you basically claim that $n(x)$ is a composite operator. Such naively defined composite operators in QFT suffer from UV-divergences, and this is essentially what you observe. In order to have a well-defined $n(x)$, you need to renormalize it, and the standard approach for free theories is to take a normal-ordered product. For bilinears like $n(x)$ this is basically equivalent to shifting $n(x)$ by its (infinite) vacuum expectation value, $$n_R(x)=n(x)-\langle n(x) \rangle.$$ For more general operators you need to use more complicated formulas (you will know which when you know what is normal-ordering). In interacting theories the definition of composite operators is more intricate and generally subject to ambiguities.