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Consider a particle bound to a fixed center by a spherically symmetric potential \(V(r)\). a. Prove $$ |\psi(0)|^{2}=\left(\frac{m}{2 \pi \hbar^{2}}\right)\left\langle\frac{d V}{d r}\right\rangle $$ for all \(s\) states, ground and excited. b. Check this relation for the ground state of a three-dimensional isotropic oscillator, the hydrogen atom, and so on. (Note: This relation has actually been found to be useful in guessing the form of the potential between a quark and an antiquark. See Moxhay and Rosner, J. Math. Phys., 21 (1980) 1688.)

Short Answer

Expert verified
The equation holds for both 3D harmonic oscillator and hydrogen atom.

Step by step solution

01

Understanding the Formula

We need to prove the relation \(|\psi(0)|^2=\left(\frac{m}{2 \pi \hbar^{2}}\right)\left\langle\frac{d V}{d r}\right\rangle\) for any state with angular momentum quantum number \(l = 0\). This is typically referred to as an \(s\)-state.
02

Express the States in Spherical Coordinates

For an \(s\)-state, the wave function has no angular dependence, i.e., \(\psi(r, \theta, \phi) = \psi(r)\). The radial wave function satisfies a spherical symmetric Schrödinger equation with the form\[-\frac{\hbar^2}{2m}\frac{d^2 u(r)}{dr^2} + V(r)u(r) = Eu(r),\]where \(u(r) = r\psi(r)\) and \(E\) is the energy.
03

Relation for Expectation Values

For the given wave function, the expectation value of the radial derivative of \(V(r)\) can be expressed as:\[ \left\langle\frac{d V}{d r}\right\rangle = \int_0^\infty \psi^*(r) \frac{d V}{d r} \psi(r) r^2 dr. \]
04

Orthogonality and Normalization

For \(s\)-states, the integral of the wave function over all space gives\[ \int_0^\infty |\psi(r)|^2 r^2 dr = 1. \]The property involves evaluating the potential derivative specifically at the origin, where \(\psi(0)\) applies.
05

Verification for 3D Harmonic Oscillator

The wave function for the ground state of a 3D isotropic oscillator is\[ \psi(r) = \left( \frac{m \omega}{\pi \hbar} \right)^{3/4} e^{-\frac{m \omega r^2}{2 \hbar}}. \]The potential is \(\frac{1}{2} m \omega^2 r^2\), so \(\frac{dV}{dr} = m \omega^2 r\).Calculate \(|\psi(0)|^{2}\) and check against the given formula.
06

Verification for Hydrogen Atom Ground State

The ground state wave function for the hydrogen atom is\[ \psi(r) = \frac{1}{\sqrt{\pi a_0^3}} e^{-r/a_0} \]and the potential is \(V(r) = -\frac{e^2}{4 \pi \epsilon_0 r}\), with \(\frac{dV}{dr} = \frac{e^2}{4 \pi \epsilon_0 r^2}\).Substitute \(|\psi(0)|^{2}\) into the original equation and check the relation.

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Key Concepts

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

Quantum States
Quantum states are the fundamental building blocks in quantum mechanics. They describe the state of a quantum system, which can be anything from an electron in an atom to a photon of light. These states are represented mathematically by wave functions, which encapsulate all the possible information about a system's physical properties.

For any quantum state, wave functions can be used to calculate probabilities. For example, the square of the absolute value of the wave function \( |\psi(\mathbf{r})|^2 \) gives the probability density of finding a particle at a given position in space.

In the context of spherically symmetric potentials, quantum states are particularly interesting due to the unique properties these potentials impose on wave functions. For instance, in an s-state, which has zero angular momentum (\(l = 0\)), the wave function depends only on the radial distance from a fixed center, simplifying calculations significantly.
Schrödinger Equation
The Schrödinger Equation is the cornerstone of quantum mechanics, providing a way to predict the behavior of quantum systems over time. This equation looks similar to wave equations in classical physics but operates in the realm of quantum probabilities.

In three dimensions, the time-independent Schrödinger Equation for a particle in a potential \( V(r) \) is given by:\[ -\frac{\hbar^2}{2m}\frac{d^2 u(r)}{dr^2} + V(r)u(r) = Eu(r) \]Here, \( u(r) \) is the radial wave function, \( \hbar \) is the reduced Planck's constant, and \(m\) is the mass of the particle.

This form of the equation is particularly useful for spherically symmetric potentials, where the complexity of angular variables can be eliminated, focusing solely on the radial component. The solutions to this differential equation give wave functions that describe possible quantum states and their associated energies.
Spherically Symmetric Potential
A spherically symmetric potential is a type of potential energy field where the force acting on a particle depends only on the distance from a fixed center and not on the direction. This feature makes it highly appealing for modeling systems like atoms and certain quantum mechanical scenarios.

Some common examples include:
  • Gravitational potentials.
  • Coulomb potentials (e.g., in hydrogen atoms).
  • Harmonic oscillator potentials in three dimensions.
In these scenarios, the potential \( V(r) \) is only a function of \( r \), thus simplifying the mathematics involved. This symmetry reduces the three-dimensional problem to a one-dimensional one, dealing only with radial distances.

Spherically symmetric potentials often result in wave functions with no angular dependence, leading to the special case of s-states \( (l=0) \) where we focus exclusively on the radial distributions of positions and energies.
Wave Function
The wave function, denoted as \( \psi(\mathbf{r}) \), is a mathematical description of a quantum system. It is a crucial concept in quantum mechanics because it contains all the accessible information about a quantum particle.
  • The magnitude of the wave function gives the probability amplitude for a system's state.
  • The square of its absolute value \( |\psi(r)|^2 \) gives the probability density for finding a particle at a particular location.
For many quantum systems under spherically symmetric potentials, the wave function simplifies to depend solely on the radial distance, especially in s-states.

Determining the wave function involves solving the Schrödinger Equation, which, when solved, provides insights into the likely positions and energies of particles within a potential field. This equation's solution, the wave function, is foundational, guiding understanding and predictions about quantum system behavior.

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Most popular questions from this chapter

Estimate the ground-state energy of a one-dimensional simple harmonic oscillator using $$ \langle x \mid \tilde{0}\rangle=e^{-\beta|x|} $$ as a trial function with \(\beta\) to be varied.

Consider an atom made up of an electron and a singly charged \((Z=1)\) triton \(\left({ }^{3} \mathrm{H}\right)\). Initially the system is in its ground state \((n=1, l=0)\). Suppose the system undergoes beta decay, in which the nuclear charge suddenly increases by one unit (realistically by emitting an electron and an antineutrino). This means that the tritium nucleus (called a "triton") turns into a helium \((Z=2)\) nucleus of mass \(3\left({ }^{3} \mathrm{He}\right)\). a. Obtain the probability for the system to be found in the ground state of the resulting helium ion. b. The available energy in tritium beta decay is about \(18 \mathrm{keV}\) and the size of the \({ }^{3} \mathrm{He}\) atom is about \(1 \AA\). Check that the time scale \(T\) for the transformation satisfies the criterion of validity for the sudden approximation.

A one-electron atom whose ground state is nondegenerate is placed in a uniform electric field in the \(z\)-direction. Obtain an approximate expression for the induced electric dipole moment of the ground state by considering the expectation value of ez with respect to the perturbed state vector computed to first order. Show that the same expression can also be obtained from the energy shift \(\Delta=-\alpha|\mathbf{E}|^{2} / 2\) of the ground state computed to second order. (Note: \(\alpha\) stands for the polarizability.) Ignore spin.

Compute the Stark effect for the \(2 s_{1 / 2}\) and \(2 p_{1 / 2}\) levels of hydrogen for a field \(\mathscr{E}\) sufficiently weak so that \(e \mathscr{E} a_{0}\) is small compared to the fine structure, but take the Lamb \(\operatorname{shift} \delta(\delta=1057 \mathrm{MHz})\) into account (that is, ignore \(2 p_{3 / 2}\) in this calculation). Show that for \(e \mathscr{E} a_{0} \ll \delta\), the energy shifts are quadratic in \(\mathscr{E}\), whereas for \(e \mathscr{E} a_{0} \gg \delta\) they are linear in \(\mathscr{E}\). Briefly discuss the consequences (if any) of time reversal for this problem. This problem is from Gottfried (1966), Problem 7-3.

A one-dimensional potential well has infinite walls at \(x=0\) and \(x=L\). The bottom of the well is not flat, but rather increases linearly from 0 at \(x=0\) to \(V\) at \(x=L\). Find the first-order shift in the energy levels as a function of principal quantum number \(n\).

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