How to Solve Commutators Using the Jacobian?

[Rorshach]

Oh right, I confused just derivative with applying the operator:P

 

[Rorshach]

it came down to $-i\hbar(\frac{\partial g}{\partial x})=-i\hbar (-i\hbar\frac{\partial \psi}{\partial x})$, which is basically a half of the result, so it is correct, right?

 

[vela]

$-i\hbar x\frac{\partial x}{\partial x}$

That's not correct. Where's you get $\frac{\partial x}{\partial x}$ from?

 

[Rorshach]

I just made a mistake while writing the result in code, it should be $-i\hbar x\frac{\partial \psi}{\partial x}$

 

[vela]

OK, so what do you get when you differentiate that?

 

[Rorshach]

$-i\hbar\frac{\partial \psi}{\partial x}$

 

[vela]

How'd you get that?

 

[Rorshach]

I differentiated it by x, but from my experience I again made a mistake somewhere...

 

[vela]

What if you wrote $g=xf$? What do you get when you differentiate g?

 

[Rorshach]

oh shoot... it is $x\frac{\partial f}{\partial x}+f$... but it only complicates the matter, now I have $-i\hbar(x\frac{\partial f}{\partial x}-i\hbar\frac{\partial \psi}{\partial x})$

 

[vela]

That's only the first term (with some typos, I hope). You still have to subtract off the result from the xp term.

 

[Rorshach]

Now I'm completely lost. I repeated that calculation and got the same result, I cannot see the mistake. I thought that xp term was included in that $-i\hbar\frac{\partial g}{\partial x}$ combination?

 

[Rorshach]

my mistake again, forgot about the second term in the bracket. But I cannot use g substitution, since x is on the left side of the operator, I can only use f. What can I do with this?

 

[vela]

Try starting over from the beginning now. You began with
$$[p^2,x]\psi = (p[p,x]+[p,x]p)\psi = p[p,x]\psi + [p,x]p\psi.$$ Now redo your calculations for $p[p,x]\psi$ and $[p,x]p\psi$.

 

[Rorshach]

the result for the first term was correct, it came out as $-\hbar^2\psi\frac{\partial \psi}{\partial x}$. I just have problem with the second term, because I can't think of the solution with changed order of the operator...how to bite this?

 

[Rorshach]

I think I finally did it. Here it goes: $-i\hbar\frac{\partial }{\partial x}(-i\hbar\frac{\partial x\psi}{\partial x}+i\hbar x\frac{\partial \psi}{\partial x})+(-i\hbar\frac{\partial xf}{\partial x}+i\hbar x\frac{\partial f}{\partial x})$, where we let $f=-i\hbar\frac{\partial \psi}{\partial x}$. Then it goes on:
$-i\hbar\frac{\partial }{\partial x}(-i\hbar(x\frac{\partial \psi}{\partial x}+\psi\frac{\partial x}{\partial x})+i\hbar x\frac{\partial \psi}{\partial x})+(-i\hbar(x\frac{\partial f}{\partial x}+f\frac{\partial x}{\partial x})+i\hbar x\frac{\partial f}{\partial x})$, proceeding
$-i\hbar\frac{\partial }{\partial x}(-i\hbar x\frac{\partial \psi}{\partial x}-i\hbar\psi+i\hbar x\frac{\partial \psi}{\partial x})+(i\hbar x\frac{\partial f}{\partial x}-i\hbar f+i\hbar x\frac{\partial f}{\partial x})$, and then
$-\hbar^2\frac{\partial \psi}{\partial x}+(-i\hbar f)=-\hbar^2\frac{\partial \psi}{\partial x}-(-i\hbar\frac{\partial \psi}{\partial x})=-\hbar^2\frac{\partial \psi}{\partial x}-\hbar^2\frac{\partial \psi}{\partial x}=-2\hbar\frac{\partial \psi}{\partial x}$.
Please tell me that this time I am right.

 

[vela]

Just a few minor errors, probably typos, and you're not quite done.

I think I finally did it. Here it goes:
$$-i\hbar\frac{\partial }{\partial x}(-i\hbar\frac{\partial x\psi}{\partial x}+i\hbar x\frac{\partial \psi}{\partial x})+(-i\hbar\frac{\partial xf}{\partial x}+i\hbar x\frac{\partial f}{\partial x}),$$ where we let $f=-i\hbar\frac{\partial \psi}{\partial x}$. Then it goes on:
$$-i\hbar\frac{\partial }{\partial x}(-i\hbar(x\frac{\partial \psi}{\partial x}+\psi\frac{\partial x}{\partial x})+i\hbar x\frac{\partial \psi}{\partial x})+(-i\hbar(x\frac{\partial f}{\partial x}+f\frac{\partial x}{\partial x})+i\hbar x\frac{\partial f}{\partial x})$$
$$-i\hbar\frac{\partial }{\partial x}(-i\hbar x\frac{\partial \psi}{\partial x}-i\hbar\psi+i\hbar x\frac{\partial \psi}{\partial x})+({\color{red}-}i\hbar x\frac{\partial f}{\partial x}-i\hbar f+i\hbar x\frac{\partial f}{\partial x})$$
$$-\hbar^2\frac{\partial \psi}{\partial x}+(-i\hbar f)
= \hbar^2\frac{\partial \psi}{\partial x}-{\color{red} {(i\hbar)}}(-i\hbar\frac{\partial \psi}{\partial x})
= -\hbar^2\frac{\partial \psi}{\partial x}-\hbar^2\frac{\partial \psi}{\partial x}
= -2\hbar^{\color{red}2} \frac{\partial \psi}{\partial x}$$.
Please tell me that this time I am right.

Now use the fact that $-\hbar^2 = (-ih)^2$ and rewrite the final result in terms of $\hat{p}_x$ to show that $[\hat{p}_x^2,\hat{x}] = (-i\hbar)2\hat{p}_x$.

 

[Rorshach]

I don't quite understand what do You mean by presenting it in terms of $p_x$, I thought that my calculations (without those stupid typos) are enough to show the correct result?

 

[vela]

Well, ask yourself what you're trying to show. The problem asks you to calculate $[\hat{p}_x^2,\hat{x}]$ which is equal to $-2i\hbar \hat{p}_x$. You haven't shown that. You've shown that when you work in the position basis, the action of $[\hat{p}_x^2,\hat{x}]$ on some arbitrary function $\psi(x)$ is to differentiate it and multiply it by $-2\hbar^2$. You need to show that that's equivalent to applying $\hat{p}$ and multiplying by $-2i\hbar$.

 

[Rorshach]

So basically I should take function $\psi(x)$, act on it with $\hat{p}$ and multiply it by $-2i\hbar$?