If the wavefunction of the electron in a box of length $L$ is $$\psi{(x)}=\sqrt{\frac{2}{L}}\cdot\sin{\frac{n\pi x}{L}}$$ What would be the probability of finding the electron between $0.25L$ and $0.75L$?.

From postulate 1 of quantum mechanics I know that $$1=\int_{0}^{L}{\psi(x) \psi^{*}(x)\,\mathrm{d}x}.$$

Trying out a similar reasoning leads me to think that the required probability is the integral $$ \int_{0.25L}^{0.75L}{\psi(x) \psi^{*}(x)\,\mathrm{d}x}$$ which gives the answer as $0.5$. But the book gives the answer as $0.82$.

What is the problem with my method and can anyone explain me why the answer is $0.82$?

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    $\begingroup$ Have you tried to calculate the integral by finding the Antiderivative of $\psi(x) \psi^{*} (x)$ and then substituting the integration limits in? $\endgroup$ – Philipp Sep 27 '13 at 5:05

You have the reasoning correct, so I suspect you made a mistake in your integral calculation. The probably of finding the particle between $a$ and $b$, for the $n$-th state, is indeed given by:

$$P_n(a,b) = \int_a^b \left|\psi_n(x)\right|^2\, \mathrm{d}x $$

but if I do the calculation for the ground state ($n = 1$), I find the same results as your textbook:

$$ P_1\left(\frac{L}{4},\frac{3L}{4}\right) = \frac{1}{2} + \frac{1}{\pi} \approx 0.818 $$


Here is a more complete solution enter image description here

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    $\begingroup$ Welcome to chemistry.se! If you have questions about how to beautify your posts, have a look at the help center. Do you want to know more about this site, please take the tour. While this answer might be correct, it would certainly profit from some more explanation. Also it would be better if it was typeset with the tools available on this site. $\endgroup$ – Martin - マーチン Feb 26 '15 at 3:44

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