Question:

A particle of mass $m$ is confined to one dimension within a potential well described by the function:

where $a$ is a positive constant.

(a) Determine the general form of the wavefunction $\Psi(x)$ for the particle inside the potential well.

(b) The energy $E$ of the particle is given by $E = \frac{n^2h^2}{8ma^2}$, where $n$ is a positive integer and $h$ is the Planck's constant. Calculate the value of $n$ for which $\Psi(x)$ is non-zero at $x = \frac{a}{2}$.

(c) Let's assume the particle is in the state described by $\Psi(x) = A \sin(\frac{\pi x}{a})$ for $0 \leq x \leq a$. Determine the value of the constant $A$.

Answer:

(a) Inside the potential well, the particle is bound and must satisfy the time-independent Schrödinger equation:

Since the potential energy inside the well is zero ($V(x) = 0$), the equation becomes:

Simplifying the equation further, we have:

Let's define $\kappa^2 = \frac{2mE}{\hbar^2}$. So, the differential equation becomes:

Solving this differential equation, we find that the general form of the wavefunction inside the potential well is:

where $A$ and $B$ are constants that can be determined from the boundary conditions.

(b) To determine the value of $n$ for which $\Psi(x)$ is non-zero at $x = \frac{a}{2}$, we substitute $x = \frac{a}{2}$ into the wavefunction:

Since $\sin(\frac{\kappa a}{2})$ and $\cos(\frac{\kappa a}{2})$ cannot be zero simultaneously for any value of $\kappa$, we must have $B \neq 0$ for $\Psi(x)$ to be non-zero at $x = \frac{a}{2}$.

Now, let's determine the value of $\kappa$ using the given formula for energy:

Substituting this into the expression for $\kappa$, we get:

Taking the square root of both sides, we find:

Therefore, $B \neq 0$ if $\kappa a$ is not an integral multiple of $\pi$. This implies that:

Since $n$ is a positive integer, the possible values for $n$ are $\left\lceil \frac{m}{\pi} \right\rceil,$ $\left\lceil \frac{m}{\pi} \right\rceil + 1, \left\lceil \frac{m}{\pi} \right\rceil + 2, \ldots$ where $\left\lceil x \right\rceil$ represents the smallest integer greater than or equal to $x$.

(c) The given wavefunction is $\Psi(x) = A \sin(\frac{\pi x}{a})$. To determine the value of constant $A$, we normalize the wavefunction by integrating $|\Psi(x)|^2$ over the entire domain $(0 \leq x \leq a)$ and setting it equal to 1:

Using the trigonometric identity $\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}$, we have:

The integral $\int_{0}^{a} \cos(\frac{2\pi x}{a}) dx$ evaluates to $\left[\frac{a}{2\pi} \sin(\frac{2\pi x}{a})\right]_0^a = \frac{a}{2\pi} (\sin(2\pi) - \sin(0)) = 0$, since $\sin(2\pi) = \sin(0) = 0$. Therefore, we are left with:

Solving this equation for constant $A$, we get:

Hence, the value of the constant $A$ is $\frac{2}{\sqrt{a}}$.

*Note: The answer for part (b) may vary depending on the chosen convention for the wavefunction. Here, we assumed the wavefunction to be in the form $\Psi(x) = A \sin(\kappa x) + B \cos(\kappa x)$.