0
$\begingroup$

I'm trying to find the expected position of a particle that has the wave function below. I understand that this should be given by integrating (x)(psi^2) dx from - infinity to infinity but I'm struggling to see how the hint given in the question is useful in this respect. Could anybody show me what operation I'm supposed to be doing to find the intgral? enter image description here

$\endgroup$
3
  • $\begingroup$ are you using $\beta$ as $\sqrt{-1}$? $\endgroup$
    – Jaywalker
    May 4, 2017 at 22:25
  • $\begingroup$ No, it's a constant $\endgroup$
    – A.Drake
    May 4, 2017 at 23:54
  • $\begingroup$ $\sqrt{-1}$ is a constant $\endgroup$
    – Jaywalker
    May 5, 2017 at 10:44

3 Answers 3

0
$\begingroup$

Expected position of a symmetric wavefunction is zero. Maybe the hint is for something else to calculate (variance?)

$\endgroup$
0
0
$\begingroup$

If you compute the expectation value of x, realise that the integrand $f(x)$ is antisymmetric, i.e. $f(x)=-f(-x)$, so that integral gives zero. If you compute the expectation value of $x^2$, that is non-zero, as the hint immediately shows.

$\endgroup$
0
$\begingroup$

The integral is straight-forward. For $k>0$, consider $$I = \int_{-\infty}^{\infty} x\mathrm e^{-kx^2}~\mathrm dx = \lim_{L \to \infty}\left(\int_{-L}^L x\mathrm e^{-kx^2}~\mathrm dx\right)$$ The substitution $u=-kx^2$ gives $x\,\mathrm dx = -\frac{1}{2k}\,\mathrm du$. If $x=-L$ then $u=-kL^2$, and if $x=L$ then $u=-kL^2$. (The $u$-limits are equal) $$I = \lim_{L \to \infty}\left(-\frac{1}{2k}\int_{-kL^2}^{-kL^2} \mathrm e^u~\mathrm du\right) = \lim_{L \to \infty}\left(0\right) = 0$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.