# Commutator of two functions

I want to compute the commutation relation between a function of momentum and a function of space in Quantum Mechanics. I know the commutation relation between momentum and a function of space but how can I compute the commutation relation between functions of momentum $$p$$ and space $$x$$? Some thing such as
$$[f(x),g(p)]$$

There are dozens of tricks to do this, but most of them reside in phase-space quantization, with which you are likely unfamiliar if you are asking this, so I'm staying out of it. Let's skip the hats for operators, set $$\hbar=1$$ for simplicity, non-dimensionalizing, and work in the coordinate representation, so, essentially, p is shorthand for $$p=-i\partial_x$$.

You are then seeking $$[g(p),f(x)]$$.You already said you have $$[p,f(x)]= -i f'(x),$$ from which you may compute the following $$[p^2,f(x)]=p[p,f]+ [p,f]p \\ = -ipf'(x) -i f'(x) p = -f''(x)-2if'(x)p,$$ and likewise for all monomials in p, whence all polynomials, by linearity of the commutator.

I assume you are familiar with the combinatoric tricks for monomials, which mathematicians like, but this is a bit barking up the wrong tree; most of them don't get their hands dirty in our messy world, nor have they any interest in that.

The key point is to convince your self that, for a c-number y (not operator like p), $$[e^{iy p}, f(x)] = \bigl(f(x+y) -f(x)\bigr) e^{iyp},$$ by virtue of the fundamental Lagrange shift, with which you are familiar from the translation operator.

But every function g(p) is a linear combination of such exponentials, by virtue of its Fourier transform, $$g(p) = {1\over \sqrt{2\pi}}\int\!\! dy ~~e^{iyp} \tilde g(y)~~~~\leadsto \\ \tilde g(y)= {1\over \sqrt{2\pi}}\int\!\! dp ~~e^{-iyp} g(p).$$

Consequently $$[g(p),f(x)]= {1\over \sqrt{2\pi}}\int\!\! dy ~~ \tilde g(y)~~[e^{iyp},f(x)]\\= {1\over \sqrt{2\pi}}\int\!\! dy ~\bigl (f(x+y)-f(x)\bigr )~ \tilde g(y) e^{iyp},$$ a numerical integral of functions in coordinate space when acting on a constant or translationally invariant state.

• Can you swap the roles of $g$ and $f$ here and derive a similar expression involving an integral over $\tilde{f}(y)$ and $g(p + y) - g(p)$? If so, and if these two expressions are equal up to sign, it might be an interesting statement about Fourier transforms. Mar 3 at 14:58
• Yes, mutatis mutandis: x and p are dual; they are canonical transforms of each other up to signs... same quantity proportional to $\int \!\!dk \tilde f(k) [e^{-k\partial_p},g(p)]=\int\!\! dk \tilde f(k) (g(p-k)-g(p))$... Mar 3 at 15:31
• @Michael Seifert. ...but when acting on a constant only. when acting on states, they differ, in intriguing ways. Mar 3 at 18:38
• @modest Thanks. The obvious answer is java-tablet and spellchecker mismatches, and haste! I rarely see what renders: only how I imagine my TeXcode renders after decades of habit. Aug 9 at 13:54

I don't think you can do it unless You are given what $$f(x)$$ and $$g(p)$$ are. If they are given you can use different properties for commutation relation and $$[x,p]=i\hbar$$ to compute what $$[f(x),g(p)]$$ will be.

Consider a simple example (More complicated function may not solve this way) $$[x^2,p]=x[x,p]+[x,p]x=xi\hbar+i\hbar x=2i\hbar x$$ Further generalization can be found here.