How to Solve Commutators Using the Jacobian?

[vela]

You need to use \hbar, not /hbar.

I'm going to use hats over the operators so you can keep track of what's going on. The first thing on the right is $\hat{p}_x \psi = -i\hbar \frac{\partial \psi}{\partial x}$, so
$$\hat{p}_x \hat{x} \hat{p}_x \psi = \hat{p}_x \hat{x} (-i\hbar)\frac{\partial \psi}{\partial x} = -i\hbar\,\hat{p}_x \hat{x} \frac{\partial \psi}{\partial x}.$$
Now, again, evaluate the what the rightmost operator does.

 

[Rorshach]

if it is supposed to be $$\frac{∂ψ}{∂x}$$, then I don't know what to do with this. If it is $$-i\hbar \frac{∂}{∂x} x$$, then it is just $$-i\hbar$$

 

[vela]

What is the operator farthest on the right in
$$-i\hbar\,\hat{p}_x \hat{x} \frac{\partial \psi}{\partial x}?$$

 

[Rorshach]

It is $\frac{∂ψ}{∂x}$

 

[vela]

I meant a hatted operator. $\psi'(x)$ is just a function.

 

[Rorshach]

Then it is $\hat{x}$, but I don't know its formula, or if it even has one besides $\hat{x}$

 

[vela]

In the coordinate basis, which is the basis you're working in, the operator $\hat{x}$ when acting on a function $f(x)$ simply multiplies that function by $x$. That is to say, $\hat{x}f(x) = xf(x)$.

 

[Rorshach]

So in this case it does nothing that would influence its right side, except for multiplication, right? Then I only see multiplication of two operators, $\hat{p}x$, which is just $-i\hbar$, and $x\hat{p}\psi$. What can I do with this?

 

[vela]

No. Say A and B are matrices and x is a vector. What you keep doing is equivalent to saying ABx = (Ax)(Bx). That's not how it works.

Just work right to left and remember that $\hat{p}$ acts on everything to its right.

 

[Rorshach]

the only thing I see is $-i\hbar\frac{∂x}{∂x}\frac{∂ψ}{∂x}$. Where am I making mistake?

 

[vela]

$\hat{p}$ acts on everything to its right.

 

[Rorshach]

$-i\hbar\frac{∂x\frac{∂\psi}{∂x}}{∂x}$?

 

[Rorshach]

Gosh, I finally understood my mistake:P I just treated the whole thing as simple standard multiplication, just like I should do in normal mathematical case, however in case of operators I should just multiply every operator by the variable on its right side:D In normal situation it seems illogical, looking like this: $(A+B)CD=(AD+BD)CD$, but in case of operators it should look like this $(-i\hbar\frac{\partial }{\partial x}x+i\hbar\frac{\partial}{\partial x})(-i\hbar\frac{\partial }{\partial x})\psi=(-i\hbar(x\frac{\partial \psi}{\partial x}+\psi\frac{\partial x}{\partial x})+i\hbar\frac{\partial \psi}{\partial x})(-i\hbar\frac{\partial \psi}{\partial x})=(-i\hbar x\frac{\partial \psi}{\partial x}-i\hbar\psi i\hbar x\frac{\partial \psi}{\partial x})(-i\hbar\frac{\partial \psi}{\partial x})=(-i\hbar\psi)(-i\hbar\frac{\partial \psi}{\partial x})=-\hbar^2\psi \frac{\partial \psi}{\partial x}=-\hbar^2\frac{\partial }{\partial x}$

 

[Rorshach]

Unfortunately, I still do not know how to solve it using Jacobi identity... I am not even sure if I interpret it correctly. I tried to use the commutator $[p^2_x,x]=[p_x p_x,x]$, use the given fact $[p_x,x]=-i\hbar$, or $[x,p_x]=i\hbar$ and substitute it to Jacobi identity formula $[p_x,[p_x,x]]+[p_x,[x,p_x]]+[x[p_x,p_x]]=0$, but those commutators zero themselves even before I sum them up. How to do this?

 

[vela]

Gosh, I finally understood my mistake:P I just treated the whole thing as simple standard multiplication, just like I should do in normal mathematical case, however in case of operators I should just multiply every operator by the variable on its right side:D In normal situation it seems illogical, looking like this: $(A+B)CD=(AD+BD)CD$, but in case of operators it should look like this $(-i\hbar\frac{\partial }{\partial x}x+i\hbar\frac{\partial}{\partial x})(-i\hbar\frac{\partial }{\partial x})\psi=(-i\hbar(x\frac{\partial \psi}{\partial x}+\psi\frac{\partial x}{\partial x})+i\hbar\frac{\partial \psi}{\partial x})(-i\hbar\frac{\partial \psi}{\partial x})$

Unfortunately, you haven't understood your mistake. You're still doing the same thing you did before. It doesn't work at all like (A+B)CD = (AD + BD)CD.

After you apply the rightmost $\hat{p}$, you have $\hat{p}\hat{x}\,\left(-i\hbar \frac{\partial\psi}{\partial x}\right)$. Now $-i\hbar\frac{\partial\psi}{\partial x}$ is just another function. Let's call it $f$. So now you have $\hat{p}\hat{x}f$. What is $\hat{x}f$ equal to? Call that result $g$. Then apply $\hat{p}$ to $g$. What do you get?

 

[Rorshach]

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

 

[vela]

Ignoring the constants, yes, that's what you get. So what's the derivative of g equal to?

 

[Rorshach]

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

 

[vela]

No. What is $g$ equal to?

 

[Rorshach]

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

 

[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 $-i\hbar(x\frac{\partial f}{\partial x}-i\hbar\frac{\partial \psi}{\partial x})$