# How to calculate the probability of finding an electron in a box between 0.25L and 0.75L?

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 $0.82$?

• Have you tried to calculate the integral by finding the Antiderivative of $\psi(x) \psi^{*} (x)$ and then substituting the integration limits in? – 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:
$$P_1\left(\frac{L}{4},\frac{3L}{4}\right) = \frac{1}{2} + \frac{1}{\pi} \approx 0.818$$