Expectation value of kinetic energy (QM)

[fluidistic]

Homework Statement

In a QM problem I must calculate the expectation value of the kinetic energy, namely $\langle \Psi _c ,\frac{ \hat { \vec p } ^2}{2m} \Psi _c \rangle$. Where $\Psi _c=ce^{-\alpha x^2}$.

Homework Equations

$\int _{-\infty}^\infty \exp \{ -a x^2 +bx \} dx = \sqrt{\frac{\pi}{a}}\exp \{ -\frac{b^2}{4a} \}$.

The Attempt at a Solution

$\langle \Psi _c ,\frac{ \hat { \vec p } ^2}{2m} \Psi _c \rangle = \int _{-\infty}^\infty c e^{-\alpha x^2} \frac{\hbar ^2}{2m} \frac{d^2}{dx^2}(ce^{-\alpha x^2})dx$$=...=-\frac{\hbar ^2}{m}\alpha c^2 \underbrace{\int _{-\infty}^\infty e^{-2\alpha x^2}dx}_{=I}+\frac{2\hbar ^2}{m} \alpha c^2 \underbrace{\int _{\infty}^\infty x^2 e^{-2\alpha x^2}dx}_{=J}$.
After some algebra, I reached that $I=\sqrt{\frac{\pi}{2\alpha}}$ while $J=\frac{\sqrt \pi}{2^{5/2}\alpha ^{3/2}}$.
From which I reached that $\langle \Psi _c ,\frac{ \hat { \vec p } ^2}{2m} \Psi _c \rangle = \frac{\hbar ^2 c^2 \pi ^{1/2}}{2^{3/2}m} \left ( \frac{1}{\alpha ^{1/2}} -\alpha ^{1/2} \right )$.
But I'm having a hard time to swallow the fact that this number is negative if alpha is greater than 1. What do you think? I made some mistake somewhere?! Right?

 

[TSny]

$\frac{\hbar ^2 c^2 \pi ^{1/2}}{2^{3/2}m} \left ( \frac{1}{\alpha ^{1/2}} -\alpha ^{1/2} \right )$.

Note that the two terms inside the parentheses are inconsistent with dimensions. Check to make sure you got the correct number of factors of $\alpha$ in front of the J integral.

EDIT: Also, don't forget that the momentum operator has a factor of $i$.

 

[fluidistic]

Note that the two terms inside the parentheses are inconsistent with dimensions. Check to make sure you got the correct number of factors of $\alpha$ in front of the J integral.

EDIT: Also, don't forget that the momentum operator has a factor of $i$.

I see. I've just checked out my 2 integrals with Wolfram Alpha and I didn't make any mistake there. So indeed, it should be in the factor(s) outside the integral(s).

$\hat {\vec p}^2=-\hbar ^2 \frac{d^2}{dx^2}$. So already I'm off a sign. I forgot the minus sign in my calculations. I'll recheck the rest and post tomorrow (too late here).
Thank you very much so far.

 

[fluidistic]

I've redone the whole algebra but the 2 integrals and I reached that the factor in front of I is $2c^2\hbar ^2 \alpha$ and the one in front of J is $-4c^2 \hbar ^2 \alpha ^2$ which makes the result for the expectation value of the momentum: $c^2\hbar ^2 \sqrt {\frac{\pi \alpha }{2}}$ which is always positive. However I'm not sure the units make sense.
If I'm not wrong, the units of c^2 is density of probability, i.e. 1/m (1 over meter). The units of alpha are 1/m^2 because the exponent must be adimensional. And the units of hbar squared are (J*s)^2, joule per second squared. So I get that the units of expectation of the momentum are J^2s^2/m^2 instead of J.Nevermind I forgot to divide all my algebra by 2m (2 masses). By doing so I get a result for the expectation value of $\frac{c^2\hbar^2 \sqrt{\pi \alpha }}{2^{3/2}m}$. The units are joule, a unit of energy which is a good sign.

 

[TSny]

I think that's the right answer.