Position operator is it communitative

[zimbabwe]

I was asked to show how the position operator is not communitative in the Shrodinger Wave equation. I thought it was as it is simply mulitplication

[x]=integral from negative to positive infinite over f*(x,t) x f(x,t) dx

Can anyone help shed some light on this. I may have misunderstood the question
$$\hat{x}=\int^{-\infty}_{+\infty}\Psi^{*}\left(x,t\right) x \Psi\left(x,t\right) dx$$

 

[jdwood983]

In momentum space $x$ is not commutative, because there

$$\hat{x}=i\hbar\frac{\partial}{\partial p}$$

So perhaps your professor is inquiring about the wave function in momentum space?

 

[White]

Show (that is my guess based on your information, which is rather few):

$$x H \neq H x$$

Integrals do not help for not commuting operators.

 

[diazona]

I was asked to show how the position operator is not communitative in the Shrodinger Wave equation. I thought it was as it is simply mulitplication

A single operator can't be commutative or noncommutative on its own. You always have to talk about a pair of operators that commute or don't commute with each other. So the question we have is, the position operator does not commute with what? That's the extra information we need. (White assumed the missing operator was the Hamiltonian; jdwood83 assumed it was momentum...)