For a particle trapped in an infinite square well of length \(L\), what happens to the probability that the particle is found between 0 and \(L / 2\) as the particle's energy increases?

The probability of finding the particle between \(x=0\) and \(x=L/2\) can be calculated using the Born rule, which states that the probability is given by the integral of the wavefunction squared over the region of interest: $$P_{0 \to L/2}(n) = \int_{0}^{L/2} |\psi_n(x)|^2 dx.$$ Substitute the wavefunction into the integral: $$P_{0 \to L/2}(n)= \int_{0}^{L/2} \left(\sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right)\right)^2 dx.$$ #Step 3: Simplify the expression and find the general result for any n#

We know that the energy of the particle is given by the formula \(E_n = \frac{n^2\pi^2\hbar^2}{2mL^2}\), where \(m\) is the mass of the particle and \(\hbar\) is the reduced Planck constant. Therefore, as energy increases, the quantum number \(n\) also increases. Now let's analyze the probability expression as \(n\) increases: $$P_{0 \to L/2}(n) = \frac{1}{L}\left(\frac{L}{2} - \frac{L}{2n\pi} \sin\left(n\pi\right)\right).$$ As we can observe from the above expression, the term \(\sin(n\pi)\) is always equal to 0 for any positive integer value of \(n\). Thus, the probability simplifies as: $$P_{0 \to L/2}(n) = \frac{1}{L}\left(\frac{L}{2}\right) = \frac{1}{2}.$$ This result implies that regardless of the energy (or the value of \(n\)), the probability of finding the particle between 0 and \(L/2\) remains constant at 1/2 or 50%.