# Given this formula, is this formula true by symmetry? (1 Viewer)

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#### bjnartowt

1. The problem statement, all variables and given/known data

I know that,
$$\left[ {f(x),p} \right] = {\bf{i}}\hbar \frac{{df}}{{dx}}$$

By symmetry, is it also true that,
$$\left[ {f(p),x} \right] = {\bf{i}}\hbar \frac{{df}}{{dp}}$$

...since x and p are just symbols?

#### Mindscrape

No. These are commutators, and the symbols aren't just placeholders, they actually mean something. Almost everything in quantum mechanics has a meaning that can change based on how you associate that meaning to something else. That is why operators that commute have such a high importance, the overall meaning of what you are doing doesn't change between the two operators that commute.

$$[f(x),p]g=-i*h*\frac{d}{dx}(f*g)+ihf*dg/dx = -ih(gdf/dx+fdg/dx-fdg/dx)= ihgdf/dx$$

The g's go away and you get your answer. (btw, the stars aren't convolutions, just multiplication)

So then work out the next one in a jiffy.

$$[f(p),x]g=f*x*g-g*f*x=0$$

#### diazona

Homework Helper
Well... according to my calculations, it is true that
$$\left[f(p),x\right] = i\hbar\frac{\mathrm{d}f}{\mathrm{d}p}$$
but not just because of symmetry. I'll second what Mindscrape wrote about the symbols not being just placeholders; they do have particular meanings.

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