1
$\begingroup$

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 $[ \psi(0),\psi^\dagger(0)]$?

|cite|improve this question
$\endgroup$
2
$\begingroup$

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.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ So then I am only allowed to use the commutation relations if $\psi$ is left unevaluated? what then is the meaning of the symbol $\psi(0)$? $\endgroup$ – Paul Malinowski Apr 18 '16 at 19:57
  • $\begingroup$ 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. $\endgroup$ – Paul Malinowski Apr 18 '16 at 19:59
  • $\begingroup$ 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). $\endgroup$ – AccidentalFourierTransform Apr 18 '16 at 20:03
  • $\begingroup$ This was how the problem was posed to me. The context is that I am dealing with noninteracting spinless fermions in one dimension. $\endgroup$ – Paul Malinowski Apr 18 '16 at 20:05
  • $\begingroup$ @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. $\endgroup$ – AccidentalFourierTransform Apr 18 '16 at 20:10
1
$\begingroup$

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.

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.