/*! 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 27 Consider a particle of mass \(m\... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider a particle of mass \(m\) subject to a one-dimensional potential of the following form: $$ V=\left\\{\begin{array}{lll} \frac{1}{2} k x^{2} & \text { for } & x>0 \\ \infty & \text { for } & x<0 \end{array}\right. $$ a. What is the ground-state energy? b. What is the expectation value \(\left\langle x^{2}\right\rangle\) for the ground state?

Short Answer

Expert verified
The ground-state energy is \(\frac{3}{2} \hbar \omega\), and \(\langle x^2 \rangle = \frac{3 \hbar}{2m\omega}\).

Step by step solution

01

Analyze the Potential

The potential is a harmonic oscillator potential for \(x > 0\) and infinite for \(x < 0\). This means the particle is confined to \(x \geq 0\) similar to a harmonic oscillator in a half-space.
02

Write the Ground-State Energy for a Harmonic Oscillator

The ground-state energy for a one-dimensional harmonic oscillator in full space is given by \(E_0 = \frac{1}{2} \hbar \omega\), where \( \omega = \sqrt{\frac{k}{m}} \) is the angular frequency.
03

Modify for Infinite Barrier at x < 0

Since there is an infinite barrier at \(x < 0\), only the symmetric wave functions that remain entirely in the \(x > 0\) region are physically allowed. This turns the problem into a particle in a half-space, thus the lowest allowed state corresponds the wave function resembling the half of the ground-state harmonic wave function.
04

Find the Ground-State Energy for Half-Space

The ground-state energy in this half-space is equivalent to the energy of the first excited state of the full harmonic oscillator space, which is \(E_1 = \frac{3}{2} \hbar \omega\).
05

Expectation Value of \(x^2\) for Half-Space

In the full harmonic oscillator space, \(\left\langle x^2 \right\rangle = \frac{\hbar}{2m\omega}\). For half-space, the calculation gets adjusted. By symmetry, the wave function is effectively \(\psi(x) = A\psi_{1}(x)\), leading to \(\langle x^2 \rangle\) by following same form as adjusted harmonic oscillator half-space: \(\left\langle x^2 \right\rangle = \frac{3 \hbar}{2m\omega}\).

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Key Concepts

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

Ground-State Energy
The concept of ground-state energy is crucial in quantum mechanics as it refers to the lowest energy state of a quantum system. For a quantum harmonic oscillator, this is typically denoted by \(E_0 = \frac{1}{2} \hbar \omega\) in a complete space, where \(\omega = \sqrt{\frac{k}{m}}\) is known as the angular frequency.
However, in our specific scenario, we deal with a particle constrained to one side of a space due to the infinite barrier at \(x < 0\). This modifies the physical space to only include \(x > 0\), effectively splitting the ground-state wave function.
  • Rather than the familiar symmetric wavefunctions, only those which fit entirely within the permitted region are allowed.
  • This half-space constraint modifies the ground-state energy to correspond with what would be the first excited state of a conventional harmonic oscillator system, shown mathematically as \(E_1 = \frac{3}{2} \hbar \omega\).
The barrier effectively raises the energy level because the particle is squeezed into a smaller space, forcing it to adopt a higher energy configuration.
Expectation Value
The expectation value is a fundamental concept in quantum mechanics that gives the expected average result of a measurement of a system's property. For position-related variables, like \(x^2\), it tells us the average position squared that we would find if we could measure many instances of the system.
For a harmonic oscillator in a full space, the expectation value \(\langle x^2 \rangle\) is known to be \(\frac{\hbar}{2m\omega}\). However, our setup is different due to the infinite barrier at \(x < 0\).
  • We have to adjust calculations considering only the acceptable wave functions in the \(x > 0\) region.
  • As such, for a particle in a half-space harmonic oscillator, the formula becomes \(\langle x^2 \rangle = \frac{3 \hbar}{2m \omega}\).
This difference arises from the fact that the allowed wave functions in the half-space shift the average position squared to align with the energy levels of the system, hence balancing out the probabilities across the confined region.
Particle in a Half-Space
A particle in a half-space is a quantum system where the particle is confined to move only within a certain region due to constraints like potential barriers. In the given scenario, the particle cannot penetrate the infinite barrier at \(x < 0\), confining it to the half-space of \(x \geq 0\).
This adjustment introduces a unique challenge:
  • Instead of symmetric wave functions, the valid functions are only those which consolidate entirely in the \(x > 0\) domain, reflecting a real physical restriction.
  • This influences the energy levels and behavior of the system since the barrier prevents the wave functions from extending into the negative space.
The idea of confining a particle in a half-space helps in understanding alterations in energy levels compared to full space scenarios and highlights the impact of spatial constraints on quantum states.
Infinite Potential Barrier
An infinite potential barrier is a boundary in quantum mechanics through which a particle cannot pass or penetrate, effectively sectioning off space into permitted and forbidden regions.
In mathematical terms, this is represented by an infinite potential value in certain areas, such as \(V(x)= \infty\) for \(x < 0\) in our case.
  • Such a barrier forces any particle wave function to be zero in the forbidden region since no probability of finding the particle exists there.
  • This confinement affects how the particle's wave function behaves in the allowed region, altering the usual state spacing of permissible energy levels.
The infinite potential barrier is a crucial construct in potential problems, neatly illustrating how spatial constraints might enforce physical rules on quantum particles, fundamentally changing the nature of their quantum states.

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Most popular questions from this chapter

Consider an electron confined to the interior of a hollow cylindrical shell whose axis coincides with the \(z\)-axis. The wave function is required to vanish on the inner and outer walls, \(\rho=\rho_{a}\) and \(\rho_{b}\), and also at the top and bottom, \(z=0\) and \(L\). a. Find the energy eigenfunctions. (Do not bother with normalization.) Show that the energy eigenvalues are given by $$ E_{l m n}=\left(\frac{\hbar^{2}}{2 m_{e}}\right)\left[k_{m n}^{2}+\left(\frac{l \pi}{L}\right)^{2}\right] \quad(l=1,2,3, \ldots, m=0,1,2, \ldots) $$ where \(k_{m n}\) is the \(n\)th root of the transcendental equation $$ J_{m}\left(k_{m n} \rho_{b}\right) N_{m}\left(k_{m n} \rho_{a}\right)-N_{m}\left(k_{m n} \rho_{b}\right) J_{m}\left(k_{m n} \rho_{a}\right)=0 . $$ b. Repeat the same problem when there is a uniform magnetic field \(\mathbf{B}=B \hat{\mathbf{z}}\) for \(0<\) \(\rho<\rho_{a} .\) Note that the energy eigenvalues are influenced by the magnetic field even though the electron never "touches" the magnetic field. c. Compare, in particular, the ground state of the \(B=0\) problem with that of the \(B \neq 0\) problem. Show that if we require the ground-state energy to be unchanged in the presence of \(B\), we obtain "flux quantization" $$ \pi \rho_{a}^{2} B=\frac{2 \pi N \hbar c}{e} \quad(N=0, \pm 1, \pm 2, \ldots) $$

Consider the neutron interferometer. Prove that the difference in the magnetic fields that produce two successive maxima in the counting rates is given by $$ B=\frac{4 \pi \hbar c}{|e| g_{n} \hbar l} $$ where \(g_{n}(=-1.91)\) is the neutron magnetic moment in units of \(-e \hbar / 2 m_{n} c\), and \(\lambda \equiv\) \(\lambda / 2 \pi\). This problem was in fact analyzed in the paper by Bernstein, Phys. Rev. Lett., 18 (1967) 1102.

Consider a one-dimensional simple harmonic oscillator. a. Using $$ \left.\left.\begin{array}{l} a \\ a^{\dagger} \end{array}\right\\}=\sqrt{\frac{m \omega}{2 \hbar}}\left(x \pm \frac{i p}{m \omega}\right), \quad \begin{array}{l} a|n\rangle \\ a^{\dagger}|n\rangle \end{array}\right\\}=\left\\{\begin{array}{l} \sqrt{n}|n-1\rangle \\ \sqrt{n+1}|n+1\rangle \end{array}\right. $$ evaluate \(\langle m|x| n\rangle,\langle m|p| n\rangle,\langle m|\\{x, p\\}| n\rangle,\left\langle m\left|x^{2}\right| n\right\rangle\), and \(\left\langle m\left|p^{2}\right| n\right\rangle .\) b. Translated from classical physics, the virial theorem states that $$ \left\langle\frac{\mathbf{p}^{2}}{m}\right\rangle=\langle\mathbf{x} \cdot \nabla V\rangle \text { (3D) } \quad \text { or } \quad\left\langle\frac{p^{2}}{m}\right\rangle=\left\langle x \frac{d V}{d x}\right\rangle $$ Check that the virial theorem holds for the expectation values of the kinetic and the potential energy taken with respect to an energy eigenstate.

A one-dimensional simple harmonic oscillator with natural frequency \(\omega\) is in initial state $$ |\alpha\rangle=\frac{1}{\sqrt{2}}|0\rangle+\frac{e^{i \delta}}{\sqrt{2}}|1\rangle $$ where \(\delta\) is a real number. a. Find the time-dependent wave function \(\left\langle x^{\prime} \mid \alpha ; t\right\rangle\) and evaluate the (time-dependent) expectation values \(\langle x\rangle\) and \(\langle p\rangle\) in the state \(|\alpha ; t\rangle\), i.e. in the Schrödinger picture. b. Now calculate \(\langle x\rangle\) and \(\langle p\rangle\) in the Heisenberg picture and compare the results.

A particle of mass \(m\) moves in one dimension \(x\) under a potential energy \(V(x)\). a. For \(V(x)=-V_{0} b \delta(x), V_{0}>0, b>0\), find the bound-state energy eigenvalue \(E\). b. Generalize this to the "double delta function" potential $$ V(x)=-V_{0} \frac{b}{2}\left[\delta\left(x+\frac{a}{2}\right)+\delta\left(x-\frac{a}{2}\right)\right] $$ and find the bound-state energy eigenvalues and plot the corresponding eigenfunctions. Also show that you get the expected results as \(a \rightarrow 0\).

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