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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

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  • $\begingroup$ are you using $\beta$ as $\sqrt{-1}$? $\endgroup$ – Jaywalker May 4 '17 at 22:25
  • $\begingroup$ No, it's a constant $\endgroup$ – A.Drake May 4 '17 at 23:54
  • $\begingroup$ $\sqrt{-1}$ is a constant $\endgroup$ – Jaywalker May 5 '17 at 10:44
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Expected position of a symmetric wavefunction is zero. Maybe the hint is for something else to calculate (variance?)

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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.

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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) $$I = \lim_{L \to \infty}\left(-\frac{1}{2k}\int_{-kL^2}^{-kL^2} \mathrm e^u~\mathrm du\right) = \lim_{L \to \infty}\left(0\right) = 0$$

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