prove: $[p^2,f] = 2 \frac{\hbar}{i}\frac{df}{dx}p - \hbar^2 \frac{d^2f}{dx^2}$

I need to prove the commutation relation,

$$[p^2,f] = 2 \frac{\hbar}{i}\frac{\partial f}{\partial x} p - \hbar^2 \frac{\partial^2 f}{\partial x^2}$$

where $f \equiv f(\vec{r})$ and $\vec{p} = p_x \vec{i}$

I know

$$[AB,C] = A[B,C] + [A,C]B$$

Applying this, I get

$$[p^2,f] = p[p,f] + [p,f]p$$

where $p = \frac{\hbar}{i}\frac{\partial}{\partial x}$, and $[p,f] \equiv pf - fp$

using a trial function, $g(x)$, I get

$$[p^2,f] = p[p,f]g + [p,f]pg$$

$$= \frac{\hbar}{i}\frac{\partial}{\partial x}\left[\frac{\hbar}{i}\frac{\partial fg}{\partial x} - f\frac{\hbar}{i}\frac{\partial g}{\partial x}\right] + \left[\frac{\hbar}{i}\frac{\partial fg}{\partial x} - f\frac{\hbar}{i}\frac{\partial g}{\partial x}\right] \frac{\hbar}{i}\frac{\partial}{\partial x}$$

using the product rule

$$= -\hbar^2 \frac{\partial}{\partial x}\left[g \frac{\partial f}{\partial x} + f \frac{\partial g}{\partial x} - f \frac{\partial g}{\partial x} \right] + \left[g \frac{\partial f}{\partial x} + f \frac{\partial g}{\partial x} - f \frac{\partial g}{\partial x} \right] \frac{\hbar}{i}^2 \frac{\partial}{\partial x}$$

cancelling the like terms in the brackets gives

$$= -\hbar^2 \frac{\partial}{\partial x} \left[g \frac{\partial f}{\partial x}\right] - \left[g \frac{\partial f}{\partial x}\right]\hbar^2 \frac{\partial}{\partial x}$$

using the product rule again gives

$$= -\hbar^2 \left[\frac{\partial g}{\partial x} \frac{\partial f}{\partial x} + g \frac{\partial^2 f}{\partial x^2} \right] - \left[g \frac{\partial f}{\partial x}\right] \hbar^2 \frac{\partial}{\partial x}$$

$\frac{\partial g}{\partial x} = 0$, so

$$= -\hbar^2 g \frac{\partial^2}{\partial x^2} - \hbar^2 g \frac{\partial f}{\partial x} \frac{\partial}{\partial x}$$

Substituting the momentum operator back in gives

$$= -\hbar^2 g \frac{\partial^2}{\partial x^2} - \frac{\hbar}{i} g \frac{\partial f}{\partial x} p$$

The trial function, $g$, can now be dropped,

$$= -\hbar^2 \frac{\partial^2}{\partial x^2} - \frac{\hbar}{i}\frac{\partial f}{\partial x}p$$

But this is not what I was supposed to arrive at. Where did I go wrong?

• You really need to $\LaTeX\text{ify}$ this. – Brandon Enright Mar 4 '14 at 4:19
• – Kyle Kanos Mar 4 '14 at 13:45

I didn't read your answer, but let's think about just computing the operator $\partial_x^2 f$. First we need to compute the operator $\partial_x f$. Now I am saying "the operator" because we are viewing $\partial_x f$ as a composition of first multiplying by $f$ and then taking the derivative. By the product rule, we know $\partial_x f = (\partial_x f) + f \partial_x$, were by $(\partial_x f)$, I really do just mean multiplication by the derivative of $f$.
Now lets try to compute $\partial_x^2 f$. It is $\partial_x [(\partial_x f) + f \partial_x] = (\partial_x^2 f) + (\partial_x f) \partial_x + (\partial_x f) \partial_x + f \partial_x^2 = (\partial_x^2 f) +2(\partial_x f) \partial_x + f \partial_x^2$.
Then $[\partial_x^2,f] = \partial_x^2f-f \partial_x^2 = (\partial_x^2 f) +2(\partial_x f) \partial_x$. If you understand this then you should get the right answer. You just need to put in the appropriate $i$'s and $\hbar$'s.
• I am not sure what your question is exactly. I probably messed this up the first time I did it too. So I guess the answer is experience, but if you are careful enough to really think about it, you should be able to reason it out. For example, would you expect $\hat{P} \hat{X} |\psi \rangle = -i \hbar |\psi \rangle$? – Brian Moths Mar 7 '14 at 4:16
• if $\hat{P}\hat{X}|\psi \rangle=−i\hbar |\psi \rangle$. Then the inverse of $\hat{X}$ is basically $\hat{P}$ and so $\hat{X}$ and $\hat{P}$ commute, but quantum mechanics depends very crucially on $\hat{X}$ and $\hat{P}$ not commuting. – Brian Moths Mar 7 '14 at 15:57