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I need to evaluate $[1/x, p]$. Note: the $p$ is the momentum operator. So far this is what i have:

$$= (1/x)(p) - (p)(1/x)$$ $$= (1/x)(-ih*d/dx)-(-ih*d/dx)(1/x)$$

Ii then factor out $-ih$ to get...


This is where im lost. Does the final answer just equal 0 because $(1/x)(d/dx) - (d/dx)(1/x) = 0$?

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Have you considered putting in a test function for the derivatives to act on? – Muphrid Mar 2 '13 at 19:11
ya, i have tried using the function x^2 (which gives me a final answer of -ih) but i also tried x^3 (this function gave me -ihx). any thoughts? @Muphrid – quantum savant Mar 2 '13 at 19:22
Don't use a specific test function; use an arbitrary one, as Nijankowski V. suggests. What will happen is that derivatives of the arbitrary test function will cancel out, so you never need to know what it actually is. – Muphrid Mar 2 '13 at 19:24
up vote 1 down vote accepted

You do the following, act with the commutator on a function of $x$

$[\frac{1}{x},p]\psi(x)=\frac{1}{x}p\psi(x)-p(\frac{\psi(x)}{x})=-\frac{i\hbar}{x} \frac{\partial\psi(x)}{\partial x}+i\hbar\frac{\partial}{\partial x}(\frac{\psi(x)}{x})=\frac{-i\hbar}{x} \frac{\partial\psi(x)}{\partial x}+\frac{i\hbar}{x} \frac{\partial\psi(x)}{\partial x}-\frac{i\hbar}{x^2}\psi(x)=-\frac{i\hbar}{x^2}\psi(x)$

From this you can see that $[\frac{1}{x},p]=-\frac{i\hbar}{x^2}$

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nevermind.. the image came up once i refreshed THANKS! – quantum savant Mar 2 '13 at 19:28
Oh ok, sure thing. – nijankowski Mar 2 '13 at 19:29

As an interesting extension of this, one can show that for any differentiable function $f= f(x)$ of the position operator $x$ one has $$ [p,f(x)] = -i\hbar f'(x) $$ This is pretty cool because it says that if we act by commutation on an operator that is a function $f$ of the position operator, then this commutation action does to the operator $f$ precisely what it would have done to an ordinary function of position!

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