Show that the position operator does not preserve H

[Moolisa]

Homework Statement

Consider a Hilbert space H that consists of all functions ψ(x) such that

$\int_{-\infty}^{\infty} |\psi(x)|^2, dx$


is finite. Show that there are functions in H for which ˆxψ(x) ≡ xψ(x) is not in H.
That is to say, there are functions in H that are taken out of H when acted upon by
the position operator (or equivalently, the position operator does not preserve H)

Relevant Equations

$\int_-\infty^\infty |ψ(x)|^2\, dx$
ˆxψ(x) ≡ xψ(x)

The attempt

$\int_{-\infty}^{\infty} |ψ^*(x)\, \hat x\,\psi(x)|\, dxˆ$

Using ˆxψ(x) ≡ xψ(x)

=$\int_{-\infty}^{\infty} |ψ^*(x)\,x\,\psi(x)|\, dxˆ$

=$\int_{-\infty}^{\infty} |ψ^*(x)\,\psi(x)\,x|\, dxˆ$

=$\int_{-\infty}^{\infty} |x\,ψ^2(x)|\, dxˆ$

I'm pretty sure this is not the correct approach. ψ(x) is a stationary right? Is it safe to attempt this using the definition of a stationary state? I apologize if I'm completely wrong, I've read the text and lecture notes but am still having a lot of trouble understanding this

 

[Abhishek11235]

If I parsed your question correctly,then think of this function:
$\psi=1/x$

Does this meets hypothesis?

 

[Moolisa]

If I parsed your question correctly,then think of this function:
$\psi=1/x$

Does this meets hypothesis?

I think so? If acted on by the position operator it would no longer be finite would it?

 

[PeroK]

If I parsed your question correctly,then think of this function:
$\psi=1/x$

Does this meets hypothesis?

That function is not defined at $x = 0$. And, the integral of $\psi^2$ does not converge in any case.

 

[PeroK]

I think so? If acted on by the position operator it would no longer be finite would it?

That $\psi$ is not square integrable.

You'll need to be a little cleverer. Maybe take a look at this for inspiration:

https://math.stackexchange.com/questions/141695/how-to-calculate-the-integral-of-sin2x-x2

 

[WWGD]

Wolfram integrator is your friend. You can use integral test to have an idea what will converge. Consider e.g. Sum $1/ n^s$. Edit: My idea is to use the comparison criteria: We know $\Sigma \frac {1}{n^s}$ converges for $s>2$. Using the integral test ( and doing something to avoid the 0 in the denominator), create an integral that converges, barely, and then multiply by $x$ to tweak it out of converging.

 

[hilbert2]

For some problem where the potential energy is singular at the origin, a wavefunction like

$\psi (x) = \frac{1}{1+\left|x\right|}$

could be a meaningful state. And even if the $\psi$ is required to be differentiable everywhere, it's possible to approximate that function with a differentiable function to any accuracy you want while having the same asymptotic behavior in the limit $\left|x\right|\rightarrow\infty$.