Consider an infinite square well with wall boundaries \(\begin{array}{llllll}x=0 & \text { and } & x=L & \text { Show } & \text { that } & \text { the function }\end{array}\) \(\psi(x)=A \sin k x \quad\) is the solution to the stationary Schrödinger equation for the particle in a box only if \(k=\sqrt{2 m E} / \hbar .\) Explain why this is an acceptable wave function only if \(k\) is an integer multiple of \(\pi / L\)

For a particle trapped in a one-dimensional infinite square well, the potential energy inside the well is zero (\(V(x)=0\)), and the stationary Schrödinger equation becomes: \[-\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}=E\psi(x)\] where $$\hbar$$ is the reduced Planck's constant, $$m$$ is the mass of the particle, $$E$$ is the energy of the particle, and $$\psi(x)$$ is the wave function.

We have the given wave function $$\psi(x) = A\sin kx$$, where $$A$$ is the amplitude and $$k$$ is the wave vector. We need to compute the second derivative of the wave function with respect to $$x$$: \[\frac{d^2\psi(x)}{dx^2} = \frac{d^2}{dx^2}(A\sin kx)\] Using the chain rule, we get: \[\frac{d^2\psi(x)}{dx^2} = -A k^2\sin kx\]

Now, we substitute the second derivative of the wave function back into the Schrödinger equation: \[-\frac{\hbar^2 k^2}{2m}\cdot A\sin kx = E\cdot A\sin kx\] Since the sine function is non-zero in general, we can divide both sides of the equation by $$A\sin kx$$: \[-\frac{\hbar^2 k^2}{2m} = E\] Solving for $$k$$, we find: \[k = \sqrt{\frac{2mE}{\hbar^2}}\] This is the desired expression for $$k$$ that we wanted to show.

To see why the wave function is only acceptable if $$k$$ is an integer multiple of $$\frac{\pi}{L}$$, let's apply the boundary conditions to our wave function. We know that at the walls of the well, the wave function must vanish: 1. At $$x=0$$: $$\psi(0) = A\sin 0 = 0$$ 2. At $$x=L$$: $$\psi(L) = A\sin kL = 0$$ The first condition is already satisfied. For the second condition to be satisfied, we need to have: \[\sin kL = 0\] The sine function vanishes for integer multiples of $$\pi$$, i.e., $$kL = n\pi$$, where $$n$$ is an integer. Thus, we have: \[k = \frac{n\pi}{L}\] Using this condition, the wave function satisfies both the boundary conditions and is an acceptable solution for the one-dimensional infinite square well problem.