Time independent Schrodinger equation and uncertainty in x

[wood]

Homework Statement

Is the gaussian
$$\sqrt{\frac{\pi}{2\alpha}}e^{-\alpha x^{2}}$$
an eigenfunction of ## \widehat{T} = \frac{\hat{p}^{2}}{2m}## ? If so, what is the corresponding eigenvalue? If not, find a P.E. operator ##\widehat{U} = U(\hat{x}) ## which gives rise to a Hamiltonian ##\widehat{H}## for which this Gaussian is an energy eigenfunction. What physical system are we talking about?
Find the uncertainty in ##\Delta x##.

Homework Equations

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

time independent Schrodinger equation
##\{-\frac{\hbar^{2}}{2m} \frac{\partial^{2}}{\partial x^{2}}+U(x)\}\psi_n(x)=E_n\psi_n(x)##

The Attempt at a Solution

I have worked out the relative derivatives and determined that it isn't an eigenfunction.
##\widehat{T}\psi(x)= -\frac{\hbar}{2m} 2\alpha e^{-\alpha x^{2}}(2\alpha x^{2} -1)##
Then I plug all that into the TISE and solve for ##\widehat{U}## and get

##\widehat{U}=\frac {\hbar^{2} + 4\alpha^{2}x^{2}}{2m}##

Hopefully I am correct up to here.

Now I am asked to find ##\Delta x ##

I think I need

## \Delta x= \sqrt{ \langle x^{2} \rangle - \langle x \rangle^{2}}##

where

## \langle x \rangle = \int_{-\infty}^\infty x|\psi|^{2} \,\mathrm{d}x##

which gives me 0

but I get confused here where I think I need

## \langle x^{2} \rangle = \int_{-\infty}^\infty x^{2}|\psi|^{2} \,\mathrm{d}x##

is the x² term just the square of my function? So in principal I am just multiplying the square of my function by the mod of the function squared? THen taking the integral?

Thanks

edit :- sorted formatting of equation

 

[Orodruin]

but I get confused here where I think I need

## \langle x^{2} \rangle = \int_{-\infty}^\infty x^|\psi|^{2} \,\mathrm{d}x##

is the x² term just the square of my function? So in principal I am just multiplying the square of my function by the mod of the function squared? THen taking the integral?

If you intended to write
## \langle x^{2} \rangle = \int_{-\infty}^\infty x^2 |\psi|^{2} \,\mathrm{d}x##
then yes, this is the expectation value of ##x^2##. Your wording makes it unclear what you are actually having trouble with.

 

[wood]

That is what the equation was meant to look like.

My problem is do I square the function ##\psi## and multiply that by the mod squared of the function or multiply the function by x². In short i am not sure what x² is in the above formula

thanks

 

[vela]

When you calculated ##\langle x \rangle##, how did you treat the factor of ##x##? Why do you think it's different in the case of ##\langle x^2 \rangle## in how you deal with ##x##?

 

[wood]

Thanks very much I think I am getting the hang of what is going on. I don't have to treat ##\langle x^2 \rangle## any different to how I dealt with ##x## i.e.I just use ##x^{2}## when workingout ##\langle x^2\rangle ##