/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 Use separation of variables in c... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use separation of variables in cartesian coordinates to solve the infinite cubical well (or "particle in a box"): \(V(x, y, z)=\left\\{\begin{array}{ll}0, & x, y, z \text { all between } 0 \text { and } a; \\ \infty, & \text { otherwise .}\end{array}\right.\) (a) Find the stationary states, and the corresponding energies. (b) Call the distinct energies \(E_{1}, E_{2}, E_{3}, \ldots,\) in order of increasing energy. Find \(E_{1}, E_{2}, E_{3}, E_{4}, E_{5},\) and \(E_{6} .\) Determine their degeneracies (that is, the number of different states that share the same energy). Comment: In one dimension degenerate bound states do not occur (see Problem 2.44 ), but in three dimensions they are very common. (c) What is the degeneracy of \(E_{14}\), and why is this case interesting?

Short Answer

Expert verified
The stationary states are product solutions with energies indexed by integers. Energies increase with non-degenerate and degenerate cases due to state combinations. The degeneracy of \(E_{14}\) reflects these combinations.

Step by step solution

01

Define the Problem

The problem is to solve the Schrödinger equation for a particle confined in an infinite cubical well. The potential is zero inside a cube of side length \(a\), and infinite outside. We need to find the stationary states and corresponding energies.
02

Use Separation of Variables

Assume the wave function can be separated as \(\psi(x, y, z) = X(x)Y(y)Z(z)\). The time-independent Schrödinger equation inside the box is \(-\frac{\hbar^2}{2m}abla^2 \psi = E \psi\). This separates into separate equations for \(X(x)\), \(Y(y)\), and \(Z(z)\).
03

Solve the One-Dimensional Equation

Each separated equation resembles a one-dimensional particle in a box problem: \(\frac{d^2F}{dx^2} = -k^2 F(x)\), where \(k^2 = \frac{2mE_x}{\hbar^2}\). Boundary conditions \(F(0)=0\) and \(F(a)=0\) yield \(F(x) = A\sin\left(\frac{n\pi}{a}x\right)\) with \(n=1,2,3,...\), and energies \(E_x = \frac{n^2\pi^2\hbar^2}{2ma^2}\). Similarly for \(Y(y)\) and \(Z(z)\).
04

Write the Solutions for Three Dimensions

The total wave function is \(\psi_{n_x,n_y,n_z}(x,y,z) = A \sin\left(\frac{n_x\pi}{a}x\right) \sin\left(\frac{n_y\pi}{a}y\right) \sin\left(\frac{n_z\pi}{a}z\right)\). Energies are given by \(E_{n_x,n_y,n_z} = \frac{\hbar^2 \pi^2}{2ma^2} (n_x^2 + n_y^2 + n_z^2)\), where \(n_x, n_y, n_z\) are positive integers.
05

Calculate Distinct Energies

Order energies \(E_{n_x,n_y,n_z}\) by increasing value. Calculate \(E_1 = 3\frac{\hbar^2 \pi^2}{2ma^2}\), \(E_2 = 6\frac{\hbar^2 \pi^2}{2ma^2}\), and so on for \(E_3, E_4, E_5,\) and \(E_6\).
06

Determine Degeneracy of Energies

For each distinct energy, determine how many combinations \((n_x, n_y, n_z)\) produce that energy. For example, \(E_1 = 3\frac{\hbar^2 \pi^2}{2ma^2}\) is non-degenerate as \((n_x,n_y,n_z)=(1,1,1)\). Express degeneracies for \(E_2, E_3, E_4, E_5,\) and \(E_6\).
07

Assess Degeneracy of Higher States

Calculate the degeneracy for \(E_{14}\) by considering all combinations of \((n_x,n_y,n_z)\) that satisfy the equation for that energy. Highlight why there is interest, such as a unique property or pattern of degeneracy.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Schrödinger equation
The Schrödinger equation is a fundamental part of quantum mechanics. It describes how the quantum state of a particle changes over time. In the "particle in a box" problem, we use the time-independent form of the Schrödinger equation. This describes the stationary states of particles confined within a potential box. The equation looks like this:
  • \[-\frac{\hbar^2}{2m}abla^2 \psi = E \psi\]
Here, \(\psi\) represents the wave function of the particle, \(E\) is the energy of the particle, and \(abla^2\) is the Laplacian operator, which represents the spatial part of the wave function within the well. The boundary conditions for a cubical box entail that outside the box, the potential is infinite, hence the wave function must be zero outside. Inside the box, the potential is zero which allows for separation of variables, resulting in solvable one-dimensional problems. The core idea is that the equation allows quantum mechanics to determine possible states and corresponding energies of a confined particle.
quantum mechanics
Quantum mechanics is the branch of physics that explores the behavior of matter on very small scales, such as atoms and subatomic particles. In quantum mechanics, the position and momentum of particles cannot be precisely determined. Instead, we use probabilities to describe where a particle is likely to be found. For a particle in a box, quantum mechanics comes into play to describe energy levels that are quantized. This means a particle can only exist in certain discrete energy states. Some key aspects are:
  • Wave functions: They describe the probability distribution of a particle's location.
  • Quantization: Only specific, discrete energy levels are allowed.
  • Uncertainty Principle: Exact positions and velocities cannot be known simultaneously.
In the infinite cubical well or box scenario, quantum mechanics dictates specific patterns and rules for how the particle behaves and the allowed energy levels that stand out distinctively as being quantized.
degeneracy of energy levels
In quantum mechanics, degeneracy refers to the number of quantum states that have the same energy. When dealing with a particle in a three-dimensional box, degeneracy becomes a crucial concept. For example, if multiple combinations of quantum numbers \((n_x, n_y, n_z)\) result in the same energy calculation \(E_{n_x,n_y,n_z}\), those energy levels are degenerate. Degeneracy is significant because:
  • It describes symmetry in the system. If more states share the same energy, it indicates a high level of symmetry.
  • Leads to complex behavior as different states with the same energy can mix.
In our exercise, for energies like \(E_1\), where \(n_x = n_y = n_z = 1\), there is no degeneracy. Meanwhile, higher energy levels might show degeneracy due to combinations like \((n_x = 2, n_y = 1, n_z = 1)\), \((n_x = 1, n_y = 2, n_z = 1)\), and \((n_x = 1, n_y = 1, n_z = 2)\) leading to the same energy state, indicating a degenerate energy level.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Quarks carry spin \(1 / 2 .\) Three quarks bind together to make a baryon (such as the proton or neutron); two quarks (or more precisely a quark and an antiquark) bind together to make a meson (such as the pion or the kaon). Assume the quarks are in the ground state (so the orbital angular momentum is zero). (a) What spins are possible for baryons? (b) What spins are possible for mesons?

Construct the matrix \(S_{r}\) representing the component of spin angular momentum along an arbitrary direction \(\hat{r} .\) Use spherical coordinates, for which \(\hat{r}=\sin \theta \cos \phi \hat{\imath}+\sin \theta \sin \phi \hat{\jmath}+\cos \theta \hat{k}\). Find the eigenvalues and (normalized) eigenspinors of \(\mathrm{S}_{r} .\) Answer: \(\chi_{+}^{(r)}=\left(\begin{array}{c}\cos (\theta / 2) \\ e^{i \phi} \sin (\theta / 2)\end{array}\right) ; \quad \chi_{-}^{(r)}=\left(\begin{array}{c}e^{-i \phi} \sin (\theta / 2) \\ -\cos (\theta / 2)\end{array}\right)\). Note: You're always free to multiply by an arbitrary phase factor-say, \(e^{i \phi-\text { so }}\) your answer may not look exactly the same as mine.

Consider the three-dimensional harmonic oscillator, for which the potential is \(V(r)=\frac{1}{2} m \omega^{2} r^{2}\). (a) Show that separation of variables in cartesian coordinates turns this into three one-dimensional oscillators, and exploit your knowledge of the latter to determine the allowed energies. (b) Determine the degeneracy \(d(n)\) of \(E_{n}\).

The electron in a hydrogen atom occupies the combined spin and position state \(R_{21}\left(\sqrt{1 / 3} Y_{1}^{0} \chi_{+}+\sqrt{2 / 3} Y_{1}^{1} \chi_{-}\right)\). (a) If you measured the orbital angular momentum squared \(\left(L^{2}\right),\) what values might you get, and what is the probability of each? (b) Same for the \(z\) component of orbital angular momentum \(\left(L_{z}\right)\). (c) Same for the spin angular momentum squared \(\left(S^{2}\right)\). (d) Same for the \(z\) component of spin angular momentum \(\left(S_{z}\right)\). Let \(\mathbf{J} \equiv \mathbf{L}+\mathbf{S}\) be the total angular momentum. (e) If you measured \(J^{2}\), what values might you get, and what is the probability of each? (f) Same for \(J_{z}\). (g) If you measured the position of the particle, what is the probability density for finding it at \(r, \theta, \phi\) ? (h) If you measured both the \(z\) component of the spin and the distance from the origin (note that these are compatible observables), what is the probability per unit \(r\) for finding the particle with spin up and at radius \(r\) ?

Starting from the Rodrigues formula, derive the orthonormality condition for Legendre polynomials: \(\int_{-1}^{1} P_{\ell}(x) P_{\ell^{\prime}}(x) d x=\left(\frac{2}{2 \ell+1}\right) \delta_{\ell \ell}\). Hint: Use integration by parts.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.