Why is the result not always 0 in the product rule for derivatives?

[rayman123]

$$[x_{\alpha}, p_{\alpha}]\psi(r)=[x_{\alpha}(-i\hbar \frac{\partial}{{ \partial x_\alpha}})-(-i\hbar\frac{\partial}{\partial x_{\alpha}})x_{\alpha}]\psi(r)$$

why the result is

$$i\hbar\psi(r)$$ should not be 0?

and then the same situation
why in this case we get 0?

$$[x_{\beta}, p_{\beta}]\psi(r)=0$$ ? Can someone please explain it?

 

[Fredrik]

You need to use the product rule for derivatives when you evaluate the term on the right, and if the indices are different (you made them the same in both your examples), the extra term that comes from the product rule is zero.

 

[rayman123]

hm...i still do not know what they just don't dissapear..they have opposite signs..
Why does the first term on the right side of the equation just dissapear? and the second one remains?
you mean the rule
$$[f(x)g(x)]= f'(x)g(x)+f(x)g'(x)$$ ?

 

[Fredrik]

Yes that's the rule, but I would write the left-hand side as $$(fg)'(x)$$. You could also write it as $$\frac{d}{dx}\big(f(x)g(x)\big)$$. Show us what you got so we can tell you what you did wrong.

 

[rayman123]

this was made by our teacher on classes
$$[x_{\alpha}, p_{\alpha}]\psi(r)=[x_{\alpha}(-i\hbar \frac{\partial}{{ \partial x_\alpha}})-(-i\hbar\frac{\partial}{\partial x_{\alpha}})x_{\alpha}]\psi(r)=i\hbar(\frac{\partial}{\partial x_\alpha}}x_{\alpha})\psi(r)=i\hbar\psi(r)$$
this was the first one

the second case was
$$[x_{\alpha}, p_{\beta}]\psi(r)=[x_{\alpha}(-i\hbar \frac{\partial}{{ \partial x_\beta}})-(-i\hbar\frac{\partial}{\partial x_{\beta}})x_{\alpha}]\psi(r)=0$$

 

[Fredrik]

Those are correct, but you didn't actually include the step where you use the product rule.

 

[rayman123]

because that step does not exist in the solution, that's why i do not get it, he just jumped over this moment and i don't know how make it myself

 

[Fredrik]

At least try. You have the product rule in front of you. Do you understand what it says?

 

[rayman123]

oh i found it! so stupid i have missed it before
$$[x_{\alpha}, p_{\alpha}]\psi(r)=x_{\alpha}(-i\hbar \frac{\partial}{{ \partial x_\alpha}}\psi(r))+i\hbar\frac{\partial}{\partial x_{\alpha}})(x_{\alpha}\psi(r))=-i\hbar x\frac{\partial\psi(r)}{\partial x}+i\hbar\psi(r)+i\hbar x\frac{\partial\psi(r)}{\partial x}=i\hbar\psi(r)$$

thank you for your help

 

[Mr.Miyagi]

I'll try to pitch in, since I vividly remember being confused when I first saw this.

When you see ABf, with A and B operators and f some function, remember you first operate with B on f and then you operate with A on the result (which is a product in your case!). It's not $$Af \cdot Bf$$

edit: never mind;)