On commuting of position and momentum operators

[dingo_d]

Homework Statement

I have to proove:

[\hat{y},\hat{p}_y]=[\hat{z},\hat{p}_z]=i\hbar\hat{I}

Homework Equations

[\hat{A},\hat{B}]=\hat{A}\hat{B}-\hat{B}\hat{A}

The Attempt at a Solution

Ok so I know that

[\hat{y},\hat{p}_y]=\hat{y}\hat{p}_y-\hat{p}_y\hat{y}=y\left(-i\hbar\frac{\partial}{\partial y}\right)-(-i\hbar\frac{\partial y}{\partial y})=i\hbar

Analogus for z component. But how to show that it's i\hbar\hat{I}?

Since they're operators they can be expressed in component form - they correspond to some kind of matrix, right?

I don't see how to get that \hat{I} :\

 

[MathematicalPhysicist]

A few hints:

1. Remember that [,] is an operator and thus it should act on an arbitrary function, in QM its the wave function.

2. Remember that (d/dy)y f= f+ydf/dy (why?).

 

[dingo_d]

Ok so I can set a wave function \psi (x) and say: \psi (x)=\langle x|\psi\rangle. And I can use product rule.

So [\hat{y},\hat{p}_y]\psi(x)=i\hbar\psi(x), but how to relate that with identity operator? :\ I mean, acting with identity operator won't change my wave function...

 

[fzero]

Ok so I can set a wave function \psi (x) and say: \psi (x)=\langle x|\psi\rangle. And I can use product rule.

So [\hat{y},\hat{p}_y]\psi(x)=i\hbar\psi(x), but how to relate that with identity operator? :\ I mean, acting with identity operator won't change my wave function...

When you write

[\hat{y},\hat{p}_y]\psi(x)=i\hbar\psi(x)

that makes mathematical sense because the left and right-hand sides are both wavefunctions. If we want to write

[\hat{y},\hat{p}_y]=?

we have an operator on the left-hand side, so we should have an operator on the right-hand side. Since you already derived the result for the first equation, just note that it is equivalent to

[\hat{y},\hat{p}_y]\psi(x)=i\hbar \hat{I}\psi(x)

and you will have the correct value for the operator statement.

 

[dingo_d]

I see, it's like when I have \vec{F}=m\vec{a} - if I have a vector on the one side I must have vector on the other side...

(we said sth about it on my group theory class, but as a side note...)