- #1

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## Homework Statement

I know that,

[tex]\left[ {f(x),p} \right] = {\bf{i}}\hbar \frac{{df}}{{dx}}[/tex]

By symmetry, is it also true that,

[tex]\left[ {f(p),x} \right] = {\bf{i}}\hbar \frac{{df}}{{dp}}[/tex]

...since x and p are just symbols?

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- Thread starter bjnartowt
- Start date

- #1

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I know that,

[tex]\left[ {f(x),p} \right] = {\bf{i}}\hbar \frac{{df}}{{dx}}[/tex]

By symmetry, is it also true that,

[tex]\left[ {f(p),x} \right] = {\bf{i}}\hbar \frac{{df}}{{dp}}[/tex]

...since x and p are just symbols?

- #2

- 1,860

- 0

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

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.

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

- #3

diazona

Homework Helper

- 2,175

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[tex]\left[f(p),x\right] = i\hbar\frac{\mathrm{d}f}{\mathrm{d}p}[/tex]

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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