6.17

The Wave Function and Size

• The wave function for the one zero zero case is proportional to e^{-r/a0}. Decreasing a0 shortens the wave function.

Limitations of the Model

• The model of hydrogen-like atoms with a single electron does not apply when the number of electrons is large due to electron-electron interactions.

• Analyzing a neutral helium atom (two electrons) requires considering the Coulomb repulsion between the electrons, making the potential more complex.

Effect of a Magnetic Field on a Hydrogen Atom

Classical Perspective

• In a magnetic field, an electron experiences torque, given by \tau = \mu \times B, where \&mu is the magnetic dipole moment and B is the magnetic field.

• For an electron moving in a circular loop, the angular momentum L is m(r \times v), where m is mass, r is the position vector, and v is the velocity.

Defining \&mu

• \&mu is defined as the product of current I and area A, with its direction given by the right-hand rule for the current density vector.

• For a circular orbit, the area A is \pi r^2.

• The magnetic field B is oriented in the z-direction in the discussed diagram.

Calculating \&mu

• The current I is the charge q over the period T, where q is the charge of the electron.

• The period T is 2\pi r / v, where v is the speed of the electron.

Magnitude of \&mu

• The magnitude of \&mu is the charge on the electron times the speed of the electron times r divided by 2.

• \mu = (e * v * r)/2

Expressing \&mu in terms of Angular Momentum

• Multiplying the numerator and denominator by m gives.

• \mu = (e * m * v * r) / (2m) = (e * L) / (2m).

• Where L is the angular momentum.

• The magnetic dipole moment is related to the angular momentum vector of the electron.

Direction of \&mu and L

• The \&mu vector is opposite to the L vector.

• \mu = - (e / 2m) * L

Torque and Precession

• The net torque on the rotating electron causes its orbit to precess.

• Precession is the rotation of the axis around which the electron is rotating.

• The torque vector is perpendicular to both \&mu and B.

• The torque causes precession, which is a rotation of the axis that the electron is rotating around.

Bird's Eye View Analysis

• Analyzing the situation from a bird's eye view shows the component of L perpendicular to B, denoted as L_{\perp}.

• L_{\perp} = |L| * \sin(\theta), where \theta is the angle between \&mu and B.

• There will be a change in the L vector parallel to the torque that Newton's law for momentum.

• The change in L with respect to time: \tau = dL/dt

• The delta L is always perpendicular to L, causing the L vector to move in a circle

Frequency of Precession

• The angular displacement of the precession is \,delta \theta.

• Delta L is equal to perpendicular times delta theta.

• Delta L = L perpendicular * delta theta

• If I do (\delta L)/(\delta t), then I see that I have the angular velocity of the second, right, which will be the radius per second.

• Magnitude of omega vector: w_p.

• On the left hand side, we have (\delta L)/(\delta t). So that is the the torque, which is just \&mu \times B.

• So the magnitude of the torque |\tau| = |\mu| * |B| * \sin(\theta).

• Known equations

  • L_{\perp} = |L| * \sin(\theta)

  • |w_p| = (\delta \theta)/(\delta t)

Solving for the Precession of Velocity

• The magnitude of w_p is the charge of the electron times the magnetic field magnitude divided by two times the mass.

• w_p = (q * B) / 2m

• This is called the Larmor frequency, where the angular momentum vector precesses about the magnetic field.

Applying Force and Resonance

• Applying a force at the same frequency as the natural frequency of the electron in the presence of the field results in resonance.

Mass Spring System

• The mass spring system has a certain natural frequency. Resonance means there's a a large transfer energy per per cycle.

• If I apply a driving force, right, then I'm going to get a response in the amplitude.

• f = f_0 * cos(\sqrt(k/m) * t). That would be the resonant frequency.

• The natural frequency is at the resonant frequency, that's when you're transferring the most amount of energy.

Natural Frequency

• With the right frequency, get the maximum energy transfer.

Equilibrium

• The amplitude doesn't grow without bound because the electron is also radiating away energy at the same time that this work is being done.

• The large energy transfer is recycled, but but eventually, we reach an equilibrium where the work in per cycle equals the the radiation out of the electron.

ESR Experiment

• This is what's going on in the ESR experiment. The momentum we're talking about is an intrinsic spin angle momentum of the electron.

• Electrons have certain intrinsic properties: mass, charge, and spin angle momentum (vector S).

• Effective spin of an electron can also cause precession.

Magnetic Field

• The magnetic field can't do any work on the electron because the force is perpendicular to the motion.

• Magnetic fields can never do work on charged particles.

Quantum Mechanical Analysis

• Quantum mechanics does not involve forces, torques, positions, or velocities of the electron.

• Instead, consider the Hamiltonian, which includes the potential energy of the electron-magnetic field system.

• The magnetic potential energy is U = -\mu \cdot B.

Work to Rotate \&mu

• W = \int{\thetai}^{\thetaf} \tau{\text{ext}} \cdot d\theta.

• This work is equal to the change of potential energy

• Potential energy is the work required to change the configuration of the system

Potential Energy with a dot Notation

• \Delta U = -\mu \cdot B{\text{final}} - (-\mu \cdot B{\text{initial}}).

• The external torque is the opposite of the internal torque which is the the torque that the field is exerting on the magnetic particle

Hamiltonian

• The new term in the Hamiltonian is \&mu \cdot B.

• Expressing \&mu in terms of angular momentum L leads to simplifications.

• Choosing the magnetic field to be in the z-direction simplifies the dot product.

• The new Hamiltonian commutes with L_z, meaning they share eigenfunctions.

• The solutions for the old problem involve a phi part e to that of im five, where m had to be an integer.

EigenValue

• The new energy is the old energy plus the angular momentum

• The old energy, E_n = -13.6 eV / n^2, is the energy of the hydrogen atom without any magnetic field present.

Energy level splitting

• Energy levels are split when a magnetic field is applied.

• For n = 2, l can be 0 or 1. If l = 1, ml can be -1, 0, or +1. The first energy level has L = 0. So ml has to equal zero

• The quantum number raises or low the energy.

• This is called the normal Zeeman splitting.

• Energy levels of the electron are divided.

• When we're looking at photons that are emitted, we have one photon that's emitted with that energy, another photon emitted with that energy, and another photon emitted with that energy.

• The war quantum model split into three since the energy can raise or lower.

• The colors of light, you see, emitted from the hydrogen atom. They will be split if you turn on a magnetic field.

• Without the magnetic field, this time, we're not here.

• So then we just have a transition from five to two.

• This one is the color that is you turn on the magnetic field and magically, this line splits. So now that you have three lines.