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I have a quantum mechanics homework problem, where I have a particle in one dimension harmonic potential.

Particle's wavefunction is defined as:

$$ \psi(x)=\left\{\begin{matrix} A(b^2-x^2), |x|<a\\ B/x^2, |x|>a \end{matrix}\right. $$

From normalization, continuity($\psi(+a)=\psi(-a)$) and smoothness($\psi'(+a)=\psi'(-a)$) conditions I got that $b=2a^2$, $B=Aa^4$, $A=\frac{15}{16a^5(30a^4-5a^2+1}$, where $a=\sqrt{\frac{\hbar}{m\omega}}$.

Now I need to find a probability to find a particle in region from -a to +a.

Therefore

$$ \int_{-a}^{a}(A(b^2-x^2))^2dx=A^2(32a^9-\frac{16a^7}{3}+\frac{2a^5}{5})=\frac{240a^4-40a^2+3}{240a^4-40a^2+8} $$

But an expression like this is quite meaningless, since it should be a number. However, should I just write a limit that a is approaching 0 or have I made mistake in my calculations somewhere (however, I have checked all of them twice with "WolframAlpha" and haven't managed to find a mistake)?

Thank you very much for any kind of help!

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    $\begingroup$ What do you mean this expression is meaningless? It is a probability (thank God $<1$) which depends on the width $a$. $\endgroup$
    – ZeroTheHero
    Commented Dec 14, 2019 at 14:19
  • $\begingroup$ @ZeroTheHero, thank you very much for the comment! Yes, now that I think about it, it makes so much sense that it depends on $a$, I was just probably very strongly anticipating to get some kind of exact numeric probability value as in previous exercises this semester. $\endgroup$
    – aerospace
    Commented Dec 14, 2019 at 16:24

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I'm assuming you meant to apply the continuity conditions at the point $a$, i.e. examining $\psi(\pm a)$ and $\psi^{\prime}(\pm a)$ rather than at $x$?

As for your probability calculation, if someone asks for the probability inside a region of width $2a$ it is no surprise that your answer depends upon that width; it will only be a number as you expect once you are given the numerical value of $a$.

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  • $\begingroup$ Thank you very much for the answer! Yes, I calculated the continuity at the point a, fixed that in the initial question. And yes, it makes so much sense that it depends on the value of $a$, I was probably just strongly expecting to get the exact number at the end (as it was in previous exercises that we have done this semester) $\endgroup$
    – aerospace
    Commented Dec 14, 2019 at 16:20

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