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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... $$=-ih(\frac{1}{x}\frac{d}{dx}-\frac{d}{dx}\frac{1}{x})$$ 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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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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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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