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.