Suppose an infinite square well extends from \(-L / 2\) to \(+L / 2 .\) Solve the time-independent Schrödinger's equation to find the allowed energies and stationary states of a particle with mass \(m\) that is confined to this well. Then show that these solutions can be obtained by making the coordinate transformation \(x^{\prime}=x-L / 2\) for the solutions obtained for the well extending between 0 and \(L\)

The potential function V(x) for an infinite square well is given by: \[V(x) = \begin{cases} 0, & \quad \text{ if } -\frac{L}{2} \le x \le \frac{L}{2} \\ \infty, & \quad \text{ otherwise } \end{cases} \]

The time-independent Schrödinger's equation for the confined particle with mass m is given by: \[-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi(x) + V(x)\psi(x) = E\psi(x)\] Since V(x) is zero inside the well, the Schrödinger's equation can be simplified to: \[ -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi(x) = E\psi(x) \]

Solve the second-order differential equation to obtain the wavefunction ψ(x): \[ \frac{d^2}{dx^2}\psi(x) = -\frac{2mE}{\hbar^2}\psi(x) \] The general solution for this equation is given by: \[ \psi(x) = A\sin(kx) + B\cos(kx) \] where k satisfies the equation: \[ k = \sqrt{\frac{2mE}{\hbar^2}} \]

At x = -L/2 and x = L/2, since ψ(x) is finite, we get: \[ \psi(-\frac{L}{2}) = A\sin(-\frac{kL}{2}) + B\cos(-\frac{kL}{2}) = 0 \] \[ \psi(\frac{L}{2}) = A\sin(\frac{kL}{2}) + B\cos(\frac{kL}{2}) = 0 \] But the wavefunction cannot be zero everywhere, so B = 0. The acceptable solution is: \[ \psi(x) = A\sin(kx) \] To satisfy the boundary condition at x = L/2, we need: \[ \sin(\frac{kL}{2}) = 0 \] which implies that k is quantized as: \[ k = 2n\frac{\pi}{L} \] where n is an integer (n=1,2,3,...).

With the quantized k values, we can find the allowed energies (E) using the relation: \[E_n = \frac{\hbar^2}{2m}\left(2n\frac{\pi}{L}\right)^2\] The corresponding wavefunctions are given by: \[\psi_n(x) = A_n\sin\left(2n\frac{\pi}{L} x\right)\] Now, let's find the solution by using the coordinate transformation x' = x - L/2.

Perform the coordinate transformation x' = x - L/2, so our new interval is 0 to L. The wavefunction is now given by: \[ \psi(x') = A'\sin(kx') \] The boundary conditions become: \[\psi(0) = A'\sin(0) = 0 \] \[\psi(L) = A'\sin(kL) = 0 \] It's clear that A' = 0 is not an acceptable solution, so the condition for the wavefunction to satisfy the boundary condition at x' = L is: \[ \sin(kL) = 0 \] This implies that k is quantized as: \[ k = n\frac{\pi}{L} \] Then, we can find the allowed energies (E) using the relation: \[E_n = \frac{\hbar^2}{2m}\left(n\frac{\pi}{L}\right)^2\] The corresponding wavefunctions are given by: \[\psi_n(x') = A'\sin\left(n\frac{\pi}{L} x'\right)\] In conclusion, the allowed energies and the corresponding wavefunctions are the same for both intervals due to the coordinate transformation.