RaySix

Post

Quantum Particle in Piecewise Potential Well

Created by @nathanedwards

at October 31st 2023, 3:23:17 pm.

AP_Physics_2-Questions

#potential-energy

#quantum-mechanics

#ap-physics-2

AP Physics 2 Exam Question

Consider a particle in a one-dimensional quantum system with a potential energy function given by:

V(x) = \left\{ \begin{array}{ll} 0 & \text{for } 0 \leq x < a \\ V_0 & \text{for } x \geq a \\ \end{array} \right.

where $a$ is a positive constant and $V_0$ is a positive constant greater than zero.

a) Sketch the potential energy function $V(x)$ . Make sure to label the axis appropriately.

b) Determine the energy levels and corresponding wave functions for this system. Provide a qualitative sketch for the lowest energy level.

c) Calculate the probability of finding the particle in the region $0 \leq x \leq a$ when it is in the lowest energy level. Express your answer in terms of the constants provided.

Assume the particle has mass $m$ and the total energy in the system is less than $V_0$ .

Answer:

a) The potential energy function $V(x)$ is given by:

V(x) = \left\{ \begin{array}{ll} 0 & \text{for } 0 \leq x < a \\ V_0 & \text{for } x \geq a \\ \end{array} \right.

The potential is zero when $0 \leq x < a$ and $V_0$ when $x \geq a$ . Below is a sketch of the potential energy function:

      | V0
      |
      |                   --
      |                 --
      |               --
      |            --
      |---------
      |  0    a

b) To determine the energy levels and corresponding wave functions for this system, we need to solve the time-independent Schrödinger equation:

-\frac{{\hbar^2}}{{2m}} \frac{{d^2\psi}}{{dx^2}} + V(x)\psi = E\psi

Inside the region $0 \leq x < a$ , the potential energy is zero, so the Schrödinger equation becomes:

-\frac{{\hbar^2}}{{2m}} \frac{{d^2\psi}}{{dx^2}} = E\psi

Solving this differential equation, we get:

\frac{{d^2\psi}}{{dx^2}} = - \frac{{2mE}}{{\hbar^2}} \psi

The general solution to this equation is:

\psi(x) = A\sin(kx) + B\cos(kx)

where $k = \frac{{\sqrt{2mE}}}{{\hbar}}$ and $A$ and $B$ are constants.

Inside the region $x \geq a$ , the potential energy is $V_0$ . So the Schrödinger equation becomes:

-\frac{{\hbar^2}}{{2m}} \frac{{d^2\psi}}{{dx^2}} + V_0\psi = E\psi

Solving this differential equation, we get:

\psi(x) = Ce^{-\alpha x} + De^{\alpha x}

where $\alpha = \frac{{\sqrt{2m(V_0 - E)}}}{{\hbar}}$ and $C$ and $D$ are constants.

To determine the energy levels and wave functions, we need to apply boundary conditions. The wave function must be continuous at $x = a$ and its derivative must be continuous as well:

\psi(a^-) = \psi(a^+)

\psi'(a^-) = \psi'(a^+)

From these boundary conditions, we can determine the energy levels and corresponding wave functions.

c) Given that the particle is in the lowest energy level, its wave function takes the form:

\psi(x) = A\sin(kx)

where $A$ is a constant.

The probability of finding the particle in the region $0 \leq x \leq a$ is given by:

P = \int_{0}^{a} |\psi(x)|^2 dx

Substituting the wave function, we get:

P = \int_{0}^{a} |A\sin(kx)|^2 dx

P = \int_{0}^{a} A^2\sin^2(kx) dx

Since $|\sin(kx)|^2 = \sin^2(kx)$ , we can simplify the integral:

P = A^2 \int_{0}^{a} \sin^2(kx) dx

Using the trigonometric identity $\sin^2(x) = \frac{1}{2} - \frac{1}{2}\cos(2x)$ , we further simplify the integral:

P = A^2 \int_{0}^{a} \left(\frac{1}{2} - \frac{1}{2}\cos(2kx)\right) dx

P = \frac{A^2}{2} \left( x - \frac{\sin(2kx)}{2k} \right)\bigg|_{0}^{a}

P = \frac{A^2}{2} \left( a - \frac{\sin(2ka)}{2k} \right)

Thus, the probability of finding the particle in the region $0 \leq x \leq a$ is $\frac{A^2}{2} \left( a - \frac{\sin(2ka)}{2k} \right)$ .