# Time independent Schrodinger equation and uncertainty in x

1. May 3, 2015

### wood

1. The problem statement, all variables and given/known data
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 $\Delta x$.

2. Relevant equations

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

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

3. 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

Last edited: May 3, 2015
2. May 3, 2015

### Orodruin

Staff Emeritus
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.

3. May 3, 2015

### 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

4. May 3, 2015

### vela

Staff Emeritus
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 $x$?

5. May 8, 2015

### 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 $\langle x^2\rangle$