A1  ^?? Find the units of $\varkappa_0$.

A2  ^?? Using the one-dimensional Schrödinger equation
\[ -\frac{\hbar^2}{2m} \Psi'' + U(x) \Psi = E \Psi,\]
obtain solutions for the wave function $\Psi_-$ at $x<0$ and $\Psi_+$ at $x>0$ for the bound state. Remember that

• The probability density $\rho(x)$ of finding a particle at point $x$ is equal to $\Psi^*(x) \Psi (x)$ and therefore \[1 = \int\limits_{-\infty}^0 \Psi_-^* \Psi_- \, dx + \int\limits_0^{+\infty} \Psi_+^* \Psi_+ \, dx\]
• The wave function is continuous $\Psi_-(0)=\Psi_+(0)$
• The wave function is defined up to phase factor $e^{i \varphi}$, where $\varphi$ is a real number

Sketch the graph showing the dependence of $|\Psi|$ on $x$.

Schrödinger equation
\[ -\frac{\hbar^2}{2m} \Psi'‘ + 0 \cdot \Psi = -\frac{\hbar^2 \varkappa_0^2}{m} \Psi \quad \Rightarrow \quad \Psi’' - 2 \varkappa_0^2 \Psi = 0\]The wave function cannot go to infinity at infinite distance from zero, so only the decaying exponent should be left
\[\Psi_- = A e^{\sqrt{2} \varkappa_0 x}, \quad \Psi_+ = A^{-\sqrt{2} \varkappa_0 x}.\]The constants for $\Psi_+$ and $\Psi_-$ are equal to ensure that $\Psi_+(0) = \Psi_-(0)$.

We can find the constant $A$ from the normalization condition, replacing two integrals with one for symmetry reasons.
\[1 = 2AA^* \int\limits_0^{\infty}e^{-2\sqrt{2} \varkappa_0 x} \, dx = \frac{AA^*}{\sqrt{2} \varkappa_0}\]From this expression, we can determine $A = e^{i \varphi}\sqrt{\sqrt{2} \varkappa_0}$ with accuracy to the phase factor.

A3  ^?? Calculate the probability $p(b)$ that the particle will be found at the distance less than $b$ from the center of the well.

Verify that $p(0)=0$ and $p(\infty)=1$.

The probability $p(b)$ can be found by direct integration
\[p(b) = \int\limits_{-b}^{b} \Psi^* \Psi \, dx = 2 \int\limits_0^{b} \Psi^* \Psi \, dx = 1 - e^{-2\sqrt{2} \varkappa_0 b}\]