A particle in a box of length $latex L&s=2$ is in the state $latex \left|\psi\right\rangle&s=2$ such that its wavefunction is

$latex \psi(x)=\left\langle x|\psi\right\rangle= \sqrt{\frac{2}{L}}\sin\left(\frac{\pi x}{L} \right)&s=2$

What is the probability that the particle is in right half of the box?

### Solution

The probability density associated with the given wave function is

$latex \mathrm{p}(x) = |\psi(x)|^2= \frac 2 L \sin^2\left(\frac{\pi x}{L}\right)&s=2$

The probability that the particle is in right half of the box is the integration of this probability density from $latex x=L/2&s=2$ to $latex x=L&s=2$. Thus the required probability is

$latex \mathrm{P}= \int_{L/2}^L \mathrm{p}(x) dx &s=2= \frac 2 L\int_{L/2}^L\sin^2\left(\frac{\pi x}{L}\right)dx&s=2$

The result of this integration should be $latex 1/2&s=2$ (Work out this integration).

Thus the probability that the particle is in right half of the box of length $latex L&s=2$ in the state given in the problem is $latex L/2&s=2$.

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