/*! 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 11 A particle of mass \(m\) is plac... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A particle of mass \(m\) is placed in a finite spherical well: \(V(r)=\left\\{\begin{array}{ll}-V_{0}, & r \leq a ;\\\ 0, & r>a.\end{array}\right.\) Find the ground state, by solving the radial equation with \(\ell=0 .\) Show that there is no bound state if \(V_{0} a^{2}<\pi^{2} \hbar^{2} / 8 m\).

Short Answer

Expert verified
No bound state exists if \( V_0 a^2 < \pi^2 \hbar^2 / 8m \).

Step by step solution

01

Understand the Problem

We need to solve the radial Schrödinger equation for a particle in a finite spherical potential well, given by the potential energy function \( V(r) \). The equation must be solved for \( \ell = 0 \) to find the ground state and prove the condition for no bound state: if \(V_{0} a^{2}<\pi^{2} \hbar^{2} / 8 m\).
02

Setup the Radial Equation

For \( \ell = 0 \), the radial Schrödinger equation simplifies to: \[-\frac{\hbar^{2}}{2m}\frac{d^2u}{dr^2} + V(r)u = Eu,\]where \( u(r) = rR(r) \) is the radial wave function. Our task is to find \( u(r) \) under the given potential \(V(r)\).
03

Solve for Inside the Well

Inside the well \((r \leq a)\), the potential \(V(r) = -V_0\). Substituting into the radial equation, we get: \[-\frac{\hbar^{2}}{2m}\frac{d^2u}{dr^2} - V_0 u = Eu.\] Setting \(E = -|E|\) (since bound states have negative energy), and simplifying, we have:\[ \frac{d^2u}{dr^2} = \frac{2m}{\hbar^2}(V_0 - |E|)u. \] Let \( k^2 = \frac{2m}{\hbar^2}(V_0 - |E|) \), so \( u(r) = A\sin(kr) + B\cos(kr) \).
04

Boundary Condition at the Origin

To ensure the solution doesn't diverge at the origin and is well-behaved, set \(B = 0\), since \(\cos(0)\) would lead to a non-zero value at \(r = 0\). Therefore, \( u(r) = A\sin(kr) \).
05

Matching Conditions at the Boundary

At \( r = a \), \( u(r) \) and its derivative must be continuous. For \( r > a \), \( V(r) = 0 \), and the equation becomes:\[ \frac{d^2u}{dr^2} = \frac{2m|E|}{\hbar^2}u. \]Let \( \kappa^2 = \frac{2m|E|}{\hbar^2} \). The solution is \( u(r) = Ce^{-\kappa r} \) for \( r > a \). Match \( A\sin(ka) = Ce^{-\kappa a} \) and derivatives to ensure continuity.
06

Analyze Bound State Conditions

From the continuity of derivatives at \(r = a\), we have: \( k\cot(ka) = -\kappa \). Using \( k^2 + \kappa^2 = \frac{2mV_0}{\hbar^2} \), substitute \( \kappa = -\frac{k}{\tan(ka)}\) and solve for conditions on parameters. Real solutions exist (bound states) only if the well depth and width satisfy \( V_{0}a^2 > \pi^2\hbar^2/8m \).
07

Conclusion and Verification

Verify the derivations agree by plotting or calculating the transcendental equation numerically and checking the inequality threshold, confirming no bound states exist if the condition is unmet.

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.

Finite Spherical Well
A finite spherical well is a hypothetical potential well which is often used in quantum mechanics to model systems such as the behavior of a particle in a limited region of space.
Imagine it like a three-dimensional bowl with finite depth where a particle can reside. The potential energy function for this system is defined as \(-V_0\) for distances \(r\) less than or equal to a certain radius \(a\), and zero beyond that.
This creates a boundary at \(r = a\), the edge of the well, which interacts with particles by restricting their energy levels. As such, the finite spherical well presents an important model to study in quantum mechanics because it provides insight into how particles behave when trapped in a finite region.
  • The name 'finite' refers to the fact that the potential has a definite depth, as opposed to being infinitely deep.
  • 'Spherical' indicates that the potential well is symmetric in all directions from its center.
Ground State Energy
The ground state energy is the lowest energy level that a quantum mechanical system, such as a particle in a potential well, can have.
For a finite spherical well, finding this ground state energy involves solving the radial Schrödinger equation. This equation typically incorporates the effects of potential energy and kinetic energy on the particle.
The solution for this involves setting the energy, \(E\), to its minimal value while ensuring the wave function, which describes the state of the system, remains physically valid. This means it must adhere to boundary conditions and remain finite across the entire space.
  • The ground state energy is crucial because it dictates the lowest possible bound state in a quantum well.
  • All other energy levels for bound states will be higher than this ground state energy.
Understanding the ground state energy provides crucial insights into the system's stability and behavior in different potential configurations.
Boundary Conditions
Boundary conditions are necessary constraints applied to physical and mathematical problems, ensuring the solution is valid and applicable throughout the system's domain.
In the context of the finite spherical well, boundary conditions play a crucial role in determining the behavior of the wave function, especially at the well's origin and boundary \(r = a\).
  • At the origin \((r = 0)\), the wave function must not diverge, hence we set certain coefficients to zero.
  • At \(r = a\), both the wave function and its derivative must be continuous from the inside of the well to outside where the potential changes.
These conditions ensure that the solution is physically meaningful and describe an actual quantum state.
They also constrain possible energy levels, allowing us to distinguish which values correspond to bound states.
Bound States
In quantum mechanics, bound states refer to situations where a particle is confined to a specific region of space due to the influence of a potential well.
This is characterized by discrete energy levels rather than the continuous spectrum observed in free particles. For a particle in a finite spherical well, bound states occur when the condition \(V_0 a^2 > \pi^2 \hbar^2 / 8m\) is satisfied.
  • If the parameters meet this criteria, the system supports at least one bound state where the particle's energy is negative, meaning it is indeed trapped within the well.
  • If the condition isn't met, the well isn't deep or wide enough to sustain a bound state, and the particle behaves as if in a free state.
These bound states are important as they dictate the possible energy levels a particle can occupy, influencing physical properties of quantum systems like atoms and molecules.

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

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?

A hydrogen atom starts out in the following linear combination of the stationary states \(n=2, \ell=1, m=1\) and \(n=2, \ell=1, m=-1\) \(\Psi(\mathbf{r}, 0)=\frac{1}{\sqrt{2}}\left(\psi_{211}+\psi_{21-1}\right)\) (a) Construct \(\Psi(\mathbf{r}, t) .\) Simplify it as much as you can. (b) Find the expectation value of the potential energy, \(\langle V\rangle\). (Does it depend on \(t ?\) Give both the formula and the actual number, in electron volts.

A hydrogenic atom consists of a single electron orbiting a nucleus with \(Z\) protons. \((Z=1 \text { would be hydrogen itself, } Z=2\) is ionized helium, \(Z=3\) is doubly ionized lithium, and so on.) Determine the Bohr energies \(E_{n}(Z)\) the binding energy \(E_{1}(Z),\) the Bohr radius \(a(Z),\) and the Rydberg constant \(\mathcal{R}(Z)\) for a hydrogenic atom. (Express your answers as appropriate multiples of the hydrogen values.) Where in the electromagnetic spectrum would the Lyman series fall, for \(Z=2\) and \(Z=3\) ? Hint: There's nothing much to calculate here in the potential (Equation \(4.52) e^{2} \rightarrow Z e^{2},\) so all you have to do is make the same substitution in all the final results.

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.

(a) At time \(t=0\) a large ensemble of spin-1/2 particles is prepared, all of them in the spin-up state (with respect to the \(z\) axis). 74 They are not subject to any forces or torques. At time \(t_{1}>0\) each spin is measured some along the \(z\) direction and others along the \(x\) direction (but we aren't some along the \(z\) direction and others along the \(x\) direction (but we aren't told the results). At time \(t_{2}>t_{1}\) their spin is measured again, this time along the \(x\) direction, and those with spin up (along \(x\) ) are saved as a subensemble (those with spin down are discarded). Question: Of those remaining (the subensemble), what fraction had spin up (along \(z\) or \(x\), depending on which was measured) in the first measurement? (b) Part (a) was easy-trivial, really, once you see it. Here's a more pithy generalization: At time \(t=0\) an ensemble of spin-1/2 particles is prepared, all in the spin-up state along direction a. At time \(t_{1}>0\) their spins are measured along direction b (but we are not told the results), and at time \(t_{2}>t_{1}\) their spins are measured along direction c. Those with spin up (along \(\mathrm{c}\) ) are saved as a subensemble. Of the particles in this subensemble, what fraction had spin up (along b) in the first measurement? Hint: Use Equation 4.155 to show that the probability of getting spin up (along b) in the first measurement is \(P_{+}=\cos ^{2}\left(\theta_{a b} / 2\right)\), and (by extension) the probability of getting spin up in both measurements is \(P_{++}=\cos ^{2}\left(\theta_{a b} / 2\right) \cos ^{2}\left(\theta_{b c} / 2\right)\). Find the other three probabilities \(\left(P_{+\rightarrow}, P_{-+}, \text {and } P_{--}\right) .\) Beware: If the outcome of the first measurement was spin down, the relevant angle is now the supplement of \(\theta_{b c}\).

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.