Suppose I have the following commutator relation for an operator $a[x,\omega]$ which depends on position $x$ and frequency $\omega$

$$ \left[a[x,\omega],a^{\dagger}[x^{\prime},\omega^{\prime}]\right]=\delta(x-x^{\prime})\delta(\omega-\omega^{\prime}). $$

This is an unequal position and unequal frequency commutator. Based on the two Dirac delta functions, the unit for $a[x,\omega]$ must be $$ \text{Dim}[a[x,\omega]] = \frac{1}{\sqrt{\text{length}\times\text{frequency}}}. $$

Suppose now that the position is fixed, namely, I'm looking at an equal-space commutator. Hence $x=x^{\prime}$ and the commutator simplifies to $$ \left[a[x,\omega],a^{\dagger}[x,\omega^{\prime}]\right]=\delta(x-x)\delta(\omega-\omega^{\prime})=\delta(0)\delta(\omega-\omega^{\prime}) =\delta(\omega-\omega^{\prime}), $$ where now, the unit of $a[x,\omega]$ becomes $1/\sqrt{\text{frequency}}$.

My question is, how is this possible? Surely fixing the position can't possibly change the dimension of the operator? What am I missing here and what are some good resources to read up on something like this?

Edit: Another analogy that I can think of are simply the raising and lowering operators of the harmonic oscillator. In general, they are time-dependent (but dimensionless) such that $$ [a(t),a^{\dagger}(t)]=1. $$ But it is possible for them to be evaluated at unequal times. In which case $$ [a(t),a^{\dagger}(t^{\prime})]=\delta(t-t^{\prime}) $$ However, the Dirac delta here implies that the unit of $a(t)$ must be $1/\sqrt{\text{time}}$. How can this be?

Edit 2: I noticed I wrote $\delta(0) = 1$ which is clearly not true. I think I meant to say $\int\delta(0)dx = 1$. This suggests that some integration might be involved in making the dimensions consistent when fixing one of the variables.

  • 1
    $\begingroup$ What is $\delta(0)$...? $\endgroup$ Mar 22 at 20:03
  • $\begingroup$ @JasonFunderberker I see your point. $\delta (0)$ should be infinity. I think I just confused Kronecker delta with Dirac delta. In either case, what is the correct way of approaching the collapse of the delta function of one variable? $\endgroup$
    – kowalski
    Mar 22 at 20:11
  • 1
    $\begingroup$ The correct way is to not approach it at all. The delta function is not a true function, and evaluating it at specific points will lead to trouble. $\endgroup$
    – Javier
    Mar 22 at 21:23


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