Why can the probability function for a particle in an infinite square well be larger than 1? For a particle in a one dimensional infinite potential well of width $L$ the probability function is:
$$P_n(x)=\left(\frac{2}{L}\right)\sin^2\left(\frac{n\pi x}{L}\right)$$
for $0\leq x\leq L$.
The probability function reaches maximum when the squared sine reaches 1, that is for
$$x = \frac{L}{2n} \, .$$
For these value of x $P = 2 / L$.
$L$ is typically very small, so that would make $2/L \gg 1$. However, $P$ is a probability and thus $0\leq P\leq 1$. How can $P$ be much larger than 1?
 A: It must not be greater than 1.
To find the probability function you must integrate the probability density, $\psi^* \psi$, over the region in which you want to calculate the probability:
$$P_n(x_1,x_2)=\int_{x_1}^{x_2} \psi^*(x)\psi(x)\,dx \, .$$
A: Your integration is wrong. The probability density function measures the probability of finding the particle between $x$ and $x+dx$. If you integrate over $0 \leq x \leq L$ you don't get a function of $x$ but a number instead (one for this interval). If you state that $P_n$ is the integrated density, then it should depend on the interval in which you did the integration. If it is around the maximum density point then the probability would actually be the difference between this $P_n$ function, which would be small. 
If it is the infinite well as you state, the solution is
$\Psi_n(x) = \sqrt{\frac{2}{L}}\sin\left( \frac {n \pi x}{L}\right)$
which would have probabilities given by
$P_n = \int_{x_1}^{x_2} \Psi_n^2 (x)dx =\frac{2}{L}\int_{x_1}^{x_2} \sin^2\left( \frac {n \pi x}{L}\right)dx$
which is normalized, since
$\frac{2}{L}\int_{0}^{L} \sin^2\left( \frac {n \pi x}{L}\right)dx = \frac{2}{L} \int_{0}^{L} \frac{1}{2}\left(1 - \cos(2\frac {n \pi x}{L})\right)dx = 1 $
