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I have a question:

If I have to find the commutator $[x, p^2]$ (with $p= {h\over i}{d \over dx} $) the right answer is:

$[x,p^2]=x p^2 - p^2x = x p^2 -pxp + pxp - p^2x = [x,p]p + p[x,p] = 2hip$

But why can't I say:

$[x,p^2]=x p^2 - p^2x = - x h^2{d^2 \over dx^2} + h^2 {d^2 \over dx^2}x = 0$ ?

Thank you for your reply.

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How do you derive '0' in the end? The point is that the 'multiply by $x$' operator and the 'differentiate with respect to $x$' operator don't commute, so you can't just blithely say that $xh^2\frac{d^2}{dx^2}$ = $h^2\frac{d^2}{dx^2}x$. –  Steven Stadnicki Sep 25 '13 at 19:11
    
Of course, you can say that. But, it's is more involved. –  Felix Marin Sep 25 '13 at 20:25

2 Answers 2

What you describe is a quite common situation which pops up when dealing with commutators of operators. On an appropriate space of functions $\mathcal D$ (like an $L^2$-space or the Schwartz space etc...), the operators $x$ and $p$ are given by

$$x(f)(x):=xf(x), $$ $$p(f)(x):=\frac{h}{i}\frac{df}{dx}, $$

for all $f\in \mathcal D$ and $x$ in the domain of $f$. In other words, $x(f)$ and $p(f)$ are elements in $\mathcal D$, i.e. functions. In particular $\frac{df}{dx}$ is the derivative of $f$ w.r.t. $x$ at the point $x$, by convention.

The commutator

$$[x,p]$$

is the operator that, evaluated at any $f$, gives the function $[x,p](f)$ s.t.

$$[x,p](f)(x):=\frac{h}{i}x\frac{df}{dx}-\frac{h}{i}\frac{d}{dx}(xf)= \frac{h}{i}x\frac{df}{dx}-\frac{h}{i}x\frac{d}{dx}(f)-\frac{h}{i}f(x)= -\frac{h}{i}f(x), $$

or $[x,p](f)=-\frac{h}{i}f$.

Similarly, $$[x,p^2]$$ is the operator that, once evaluated at any $f\in \mathcal D$, gives the function $[x,p^2](f)$, with

$$[x,p^2](f)(x)=-h^2x\frac{d^2f}{dx^2}+h^2\frac{d}{dx}\left(\frac{d}{dx} (xf) \right)= -h^2x\frac{d^2f}{dx^2}+h^2\frac{d}{dx}\left(f+x\frac{df}{dx} \right)= -h^2x\frac{d^2f}{dx^2}+h^2\frac{df}{dx}+ h^2\frac{d}{dx}\left(x\frac{df}{dx}\right)=\\ -h^2x\frac{d^2f}{dx^2}+h^2\frac{df}{dx}+ h^2\frac{df}{dx}+ h^2x\frac{d^2f}{dx^2}=2h^2\frac{df}{dx}.$$

Equivalently

$$[x,p^2](f)(x)=2hip(f)(x)$$

or

$$[x,p^2](f)=2hip(f)$$

as expected.

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1  
Of course, there are many more catches lurking behind the scenes too: for instance, both operators can easily take a function out of the $L^2$ space (this is a big part of why Schwartz spaces are so convenient - though of course many functions we're interested in don't belong to them). –  Steven Stadnicki Sep 26 '13 at 21:59
    
yop, that is right: but Schwartz spaces are not so bad, after all :-) –  Avitus Sep 29 '13 at 8:35

You treat $x$ like a scalar when in fact it is an operator. Let's write it as $\hat{x}$ instead. Then $[\hat{x}, p^2]$ is also an operator, so let's see what happens when we apply an arbitrary function $\psi$ to it. \begin{align} [\hat{x},p^2]\psi&=\left(\hat{x}p^2-p^2\hat{x}\right)\psi \\ &=-xh^2\frac{\mathrm{d}^2}{\mathrm{d}x^2}\psi+h^2\frac{\mathrm{d}^2}{\mathrm{d}x^2}(x\psi) \\ &=-xh^2\frac{\mathrm{d}^2}{\mathrm{d}x^2}\psi+h^2x\frac{\mathrm{d}^2}{\mathrm{d}x^2}\psi+2h^2\frac{\mathrm{d}}{\mathrm{d}x}\psi \\ &=2h^2\frac{\mathrm{d}}{\mathrm{d}x}\psi \\ &=2hi\frac{h}{i}\frac{\mathrm{d}}{\mathrm{d}x}\psi=2hip\psi \end{align} so that $[\hat{x},p^2]=2hip$.

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+1 for the Physics notation! –  Avitus Sep 25 '13 at 19:40
1  
Thank you everyone for your replies. My problem was that i viewed 0 like a function that I could derive, (with derivate = 0) but it's not the case. I have to view $[x,p^2]$ also like an operator that operate on a "test function" $f$ –  Benzio Sep 25 '13 at 19:52
    
Yes, this is the point: to consider operators acting on functions. Same happens with the $a$ and $a^{\dagger}$ operators... ;-) –  Avitus Sep 25 '13 at 20:14

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