Exercise Problem

A particle in a box of length L is in the state \left|\psi\right\rangle such that its wavefunction is

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

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

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

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

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

The result of this integration should be 1/2 (Work out this integration).

Thus the probability that the particle is in right half of the box of length L in the state given in the problem is L/2.

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