Commutator, operators momentum and position

[fluidistic]

Homework Statement

I must calculate [X,P].

Homework Equations

Not sure. What I've researched through the Internet suggests that $$[\hat A, \hat B]=\hat A \hat B - \hat B \hat A$$ and that $$[\hat A, \hat B]=-[\hat B, \hat A]$$.
Furthermore if the operators commute, then $$[\hat A, \hat B]=0$$ obviously from the anterior property.

The Attempt at a Solution

So I've checked out in wikipedia the definition of position and momentum operators and they seems to involve the wave function $$\Psi$$?

If I consider $$\hat X =x$$ and $$\hat P =-i\hbar \frac{\partial}{\partial x}$$, I get that $$\hat X\hat P =-i \hbar$$ but for $$\hat P \hat X$$ I have a doubt.
I get that it's worth $$x \left ( - \frac{i \hbar \partial}{\partial x} \right )$$. I'm guessing it's worth $$-i\hbar$$? So that if follows that $$[\hat X,\hat P]=0$$ and hence the position and linear momentum commute. I'm not sure I'm right on this, nor do I have any idea about some of the implications the commutativity implies.
Any insight is greatly appreciated.

 

[fzero]

To avoid confusion, compute this commutator acting on an arbitrary wavefunction:

$$[\hat{X},\hat{P}]\Psi.$$

As a differential operator, $$\hat{P}$$ acts on everything to the right, so the term

$$\hat{P}\hat{X} \Psi = \hat{P}(\hat{X} \Psi),$$

so you need to use the product rule.

 

[fluidistic]

To avoid confusion, compute this commutator acting on an arbitrary wavefunction:

$$[\hat{X},\hat{P}]\Psi.$$

As a differential operator, $$\hat{P}$$ acts on everything to the right, so the term

$$\hat{P}\hat{X} \Psi = \hat{P}(\hat{X} \Psi),$$

so you need to use the product rule.

Thank you so much. I'm so glad I've asked on this forum, I would have done meaningless work without you.
I reach, after some basic alegebra: $$[\hat X, \hat P] \Psi =i \hbar \Psi$$. So the 2 operators do not commute.
Am I right?

 

[fzero]

Thank you so much. I'm so glad I've asked on this forum, I would have done meaningless work without you.
I reach, after some basic alegebra: $$[\hat X, \hat P] \Psi =i \hbar \Psi$$. So the 2 operators do not commute.
Am I right?

Correct. Also, since the function $$\Psi$$ is arbitrary, you can turn this into an operator statement:

$$[\hat X, \hat P] =i \hbar \hat{I},$$

where $$\hat{I}$$ is the identity operator.

 

[fluidistic]

Correct. Also, since the function $$\Psi$$ is arbitrary, you can turn this into an operator statement:

$$[\hat X, \hat P] =i \hbar \hat{I},$$

where $$\hat{I}$$ is the identity operator.

Nice. Problem solved.