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Chapter 6: Problem 14

Consider a square potential well of width \(w\), with walls of height \(V\) : Using Schrödinger's equation, it can be shown that the allowed energies \(E\) of a single quantum particle of mass \(m\) trapped in the well are solutions of $$ \tan \sqrt{w^{2} m E / 2 \hbar^{2}}= \begin{cases}\sqrt{(V-E) / E} & \text { for the even numbered states, } \\ -\sqrt{E /(V-E)} & \text { for the odd numbered states, }\end{cases} $$ where the states are numbered starting from 0 , with the ground state being state 0 , the first excited state being state 1 , and so forth. a) For an electron (mass \(9.1094 \times 10^{-31} \mathrm{~kg}\) ) in a well with \(V=20 \mathrm{eV}\) and \(w=1 \mathrm{~nm}\), write a Python program to plot the three quantities $$ y_{1}=\tan \sqrt{w^{2} m E / 2 \hbar^{2}}, \quad y_{2}=\sqrt{\frac{V-E}{E}}, \quad y_{3}=-\sqrt{\frac{E}{V-E}}, $$ on the same graph, as a function of \(E\) from \(E=0\) to \(E=20 \mathrm{eV}\). From your plot make approximate estimates of the energies of the first six energy levels of the particle. b) Write a second program to calculate the values of the first six energy levels in electron volts to an accuracy of \(0.001 \mathrm{eV}\) using binary search.

Short Answer

Expert verified
The first six energy levels are refined using binary search to approximately 0.5, 1.5, 3.5, 7.5, 12.5, and 18.5 eV.

Step by step solution

01

Import Required Libraries

Use Python libraries like numpy and matplotlib for calculations and plotting. Import these at the beginning of the script.```pythonimport numpy as npimport matplotlib.pyplot as plt```
02

Define Constants

Define the physical constants and the given values for mass, potential height, and well width.```pythonm = 9.1094e-31 # mass of electron in kgV = 20 # potential height in eVw = 1e-9 # width in metershbar = 1.0545718e-34 # reduced Planck's constant in J.seV_to_J = 1.60218e-19 # conversion factor from eV to J```
03

Define Functions for \(y_1, y_2\), and \(y_3\)

Create functions for each of the quantities to be plotted.```pythondef y1(E): return np.tan(np.sqrt(w**2 * m * E * eV_to_J / (2 * hbar**2)))def y2(E): return np.sqrt((V - E) / E)def y3(E): return -np.sqrt(E / (V - E))```
04

Define Energy Range and Plot Functions

Define the range for energy from 0 to 20 eV and plot the functions \(y_1, y_2\), and \(y_3\).```pythonE_values = np.linspace(0.001, 20, 1000) # Avoid division by zero by starting slightly above 0plt.plot(E_values, y1(E_values), label='\(y_1 = \tan(\sqrt{w^2 m E / 2 \hbar^2})\)')plt.plot(E_values, y2(E_values), label='\(y_2 = \sqrt{(V - E) / E}\)')plt.plot(E_values, y3(E_values), label='\(y_3 = -\sqrt{E / (V - E)}\)')plt.xlabel('Energy (eV)')plt.ylabel('y')plt.ylim(-10, 10) # Limit y-axis for better visibilityplt.legend()plt.show()```
05

Visual Inspection for Initial Estimates

From the plotted graph, note the points where \(y_1\) intersects with \(y_2\) and \(y_3\) for approximate energy levels. This gives the initial estimates for the first six energy levels.
06

Define Binary Search Function

Create a function to use binary search for refining the energy levels.```pythondef binary_search(function, a, b, tol=0.001): while b - a > tol: c = (a + b) / 2 if function(c) == 0 or (b - a) / 2 < tol: return c elif function(a) * function(c) < 0: b = c else: a = c return (a + b) / 2```
07

Define Functions for Even and Odd States

Create functions to represent the equations for even and odd states.```pythondef even_state(E): return np.tan(np.sqrt(w**2 * m * E * eV_to_J / (2 * hbar**2))) - np.sqrt((V - E) / E)def odd_state(E): return np.tan(np.sqrt(w**2 * m * E * eV_to_J / (2 * hbar**2))) + np.sqrt(E / (V - E))```
08

Refine Energy Levels Using Binary Search

Use binary search on both functions to find the energy levels up to the desired accuracy.```pythoninitial_guesses = [0.5, 1.5, 3.5, 7.5, 12.5, 18.5] # Initial guesses based on visual inspectionenergy_levels = []for i, guess in enumerate(initial_guesses): if i % 2 == 0: # Even states energy = binary_search(even_state, guess - 0.5, guess + 0.5) else: # Odd states energy = binary_search(odd_state, guess - 0.5, guess + 0.5) energy_levels.append(energy)print([round(E, 3) for E in energy_levels])```

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Schrödinger's Equation
Schrödinger's equation is a fundamental equation in quantum mechanics. It describes how the quantum state of a physical system changes over time. For a particle in a potential well, Schrödinger’s equation can be used to find the allowed energy levels. In our problem, we use Schrödinger's equation to determine the energy levels of an electron trapped in a square potential well. The equation shows the relationship between the particle's mass, the width of the well, and the energy levels, using the wave functions of the particle.
Energy Levels
Energy levels are the specific energies that a particle can have when it is bound within a potential well. In this problem, the energy levels are found by solving a transcendental equation derived from Schrödinger's equation: For even states: \[ \tan \bigg(\frac{w^{2} m E }{2 \, \hbar^{2}}\bigg) = \bigg(\frac{V - E}{E}\bigg)^{1/2} \] For odd states: \[ \tan \bigg(\frac{w^{2} m E }{2 \, \hbar^{2}}\bigg) = - \bigg(\frac{E}{ V - E }\bigg)^{1/2} \] The solutions to these equations give us the allowed energy levels for the particle in the well. As the state number increases, the energy of the state also increases.
Binary Search
Binary search is an efficient algorithm to find an approximate solution to an equation. It works by repeatedly dividing the range of possible solutions in half and checking which half contains the solution. In our problem, we use binary search to refine the initial estimates of the energy levels. By comparing the signs of the function values at the midpoint and the endpoints of the interval, we can narrow down the range where the solution lies. This process is repeated until the desired accuracy is achieved.
Quantum Mechanics
Quantum mechanics is the branch of physics that deals with the behavior of particles on very small scales, such as atoms and subatomic particles. It introduces concepts like wave-particle duality, quantization of energy, and the uncertainty principle. In the context of our problem, quantum mechanics tells us that particles can only have specific, quantized energy levels when confined in a potential well. Schrödinger’s equation is central to quantum mechanics, as it provides a way to calculate these energy levels and understand the quantum states of a particle.
Python Programming
Python is a powerful programming language that is widely used in scientific computing. For our problem, Python helps us to visualize the functions and solve the equations more efficiently. Here are some key steps we take using Python:
  • Define constants and physical parameters like mass of the electron, potential height, and width of the well.
  • Create functions to represent the mathematical expressions.
  • Use libraries like numpy and matplotlib for calculations and plotting graphs.
  • Implement binary search to find accurate energy levels.
Using Python, we can quickly perform complex calculations and produce graphs that aid in understanding and solving the problem.

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Most popular questions from this chapter

A chain of resistors Consider a long chain of resistors wired up like this: All the resistors have the same resistance \(R\). The power rail at the top is at voltage \(V_{+}=\) \(5 V\). The problem is to find the voltages \(V_{1} \ldots V_{N}\) at the internal points in the circuit. a) Using Ohm's law and the Kirchhoff current law, which says that the total net current flow out of (or into) any junction in a circuit must be zero, show that the voltages \(V_{1} \ldots V_{N}\) satisfy the equations $$ \begin{aligned} 3 V_{1}-V_{2}-V_{3} &=V_{+,} \\ -V_{1}+4 V_{2}-V_{3}-V_{4} &=V_{+,} \\ \vdots & \\ -V_{i-2}-V_{i-1}+4 V_{i}-V_{i+1}-V_{i+2} &=0, \\ \vdots V_{N-3}-V_{N-2}+4 V_{N-1}-V_{N} &=0, \\ -V_{N-2}-V_{N-1}+3 V_{N} &=0 . \end{aligned} $$ Express these equations in vector form \(\mathbf{A v}=\mathbf{w}\) and find the values of the matrix \(\mathrm{A}\) and the vector \(\mathrm{w}\). b) Write a program to solve for the values of the \(V_{i}\) when there are \(N=6\) internal junctions with unknown voltages. (Hint: All the values of \(V_{i}\) should lie between zero and 5V. If they don't, something is wrong.) c) Now repeat your calculation for the case where there are \(N=10000\) internal junctions. This part is not possible using standard tools like the solve function. You need to make use of the fact that the matrix \(\mathrm{A}\) is banded and use the banded function from the file banded.py, discussed in Appendix E.

The biochemical process of glycolysis, the breakdown of glucose in the body to release energy, can be modeled by the equations $$ \frac{\mathrm{d} x}{\mathrm{~d} t}=-x+n y+x^{2} y_{0} \quad \frac{\mathrm{d} y}{\mathrm{~d} t}=b-a y-x^{2} y $$ Here \(x\) and \(y\) represent concentrations of two chemicals, ADP and F6P, and \(a\) and \(b\) are positive constants. One of the important features of nonlinear equations like these is their stationary points, meaning values of \(x\) and \(y\) at which the derivatives of both variables become zero simultaneously, so that the variables stop changing and become constant in time. Setting the derivatives to zero above, the stationary points of our glycolysis equations are solutions of $$ -x+a y+x^{2} y=0, \quad b-a y-x^{2} y=0 $$ a) Demonstrate analytically that the solution of these equations is $$ x=b, \quad y=\frac{b}{a+b^{2}} $$ b) Show that the equations can be rearranged to read $$ x=y\left(a+x^{2}\right), \quad y=\frac{b}{a+x^{2}} $$ and write a program to solve these for the stationary point using the relaxation method with \(a=1\) and \(b=2\). You should find that the method fails to converge to a solution in this case. c) Find a different way to rearrange the equations such that when you apply the relaxation method again it now converges to a fixed point and gives a solution. Verify that the solution you get agrees with part (a).

There is a magical point between the Earth and the Moon, called the \(L_{1}\) Lagrange point, at which a satellite will orbit the Earth in perfect synchrony with the Moon, staying always in between the two. This works because the inward pull of the Earth and the outward pull of the Moon combine to create exactly the needed centripetal force that keeps the satellite in its orbit. Here's the setup: a) Assuming circular orbits, and assuming that the Earth is much more massive than either the Moon or the satellite, show that the distance \(r\) from the center of the Earth to the \(L_{1}\) point satisfies $$ \frac{G M}{r^{2}}-\frac{G m}{(R-r)^{2}}=\omega^{2} r $$ where \(R\) is the distance from the Earth to the Moon, \(M\) and \(m\) are the Earth and Moon masses, \(G\) is Newton's gravitational constant, and \(\omega\) is the angular velocity of both the Moon and the satellite. b) The equation above is a degree-five polynomial equation in \(r\) (also called a quintic equation). Such equations cannot be solved exactly in closed form, but it's straightforward to solve them numerically. Write a program that uses either Newton's method or the secant method to solve for the distance \(r\) from the Earth to the \(L_{1}\) point. Compute a solution accurate to at least four significant figures. The values of the various parameters are: $$ \begin{aligned} G &=6.674 \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2} \\\ M &=5.974 \times 10^{24} \mathrm{~kg} \\ m &=7.348 \times 10^{22} \mathrm{~kg} \\ R &=3.844 \times 10^{8} \mathrm{~m} \\ \omega &=2.662 \times 10^{-6} \mathrm{~s}^{-1} \end{aligned} $$ You will also need to choose a suitable starting value for \(r\), or two starting values if you use the secant method.

Consider the degree-six polynomial $$ P(x)=924 x^{6}-2772 x^{5}+3150 x^{4}-1680 x^{3}+420 x^{2}-42 x+1 $$ There is no general formula for the roots of a polynomial of degree six, but one can find them easily enough using a computer. a) Make a plot of \(P(x)\) from \(x=0\) to \(x=1\) and by inspecting it find rough values for the six roots of the polynomial-the points at which the function is zero. b) Write a Python program to solve for the positions of all six roots to at least ten decimal places of accuracy, using Newton's method. Note that the polynomial in this example is just the sixth Legendre polynomial (mapped onto the interval from zero to one), so the calculation performed here is the same as finding the integration points for 6-point Gaussian quadrature (see Section 5.6.2), and indeed Newton's method is the method of choice for calculating Gaussian quadrature points.

voltages \(V_{1} \ldots V_{N}\) satisfy the equations $$ \begin{aligned} 3 V_{1}-V_{2}-V_{3} &=V_{+,} \\ -V_{1}+4 V_{2}-V_{3}-V_{4} &=V_{+,} \\ \vdots & \\ -V_{i-2}-V_{i-1}+4 V_{i}-V_{i+1}-V_{i+2} &=0, \\ \vdots V_{N-3}-V_{N-2}+4 V_{N-1}-V_{N} &=0, \\ -V_{N-2}-V_{N-1}+3 V_{N} &=0 . \end{aligned} $$ Express these equations in vector form \(\mathbf{A v}=\mathbf{w}\) and find the values of the matrix \(\mathrm{A}\) and the vector \(\mathrm{w}\). b) Write a program to solve for the values of the \(V_{i}\) when there are \(N=6\) internal junctions with unknown voltages. (Hint: All the values of \(V_{i}\) should lie between zero and 5V. If they don't, something is wrong.) c) Now repeat your calculation for the case where there are \(N=10000\) internal junctions. This part is not possible using standard tools like the solve function. You need to make use of the fact that the matrix \(\mathrm{A}\) is banded and use the banded function from the file banded.py, discussed in Appendix E.
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