6.19

Splitting in Magnetic Fields

Energy Level Splitting

• When a magnetic field is applied to a hydrogen atom (or other atoms), energy levels split.
• Example: For l = 2, there are five states.
  • Splitting leads to seven energy levels but ten states due to degeneracy.
  • Degeneracy: 1, 1, 2, 2, 2, 1, 1, totaling 10 states.
• Spin up and spin down states contribute to this; three energy levels are degenerate.

Quantum Number Representation

• States can be listed by quantum numbers Lz and Sz.
• L_z ranges from +2 to -2.
• For each L_z, there's a spin-up (+\frac{1}{2}) and spin-down (-\frac{1}{2}) state.
• This alternative view confirms the existence of 10 states (5 x 2).

Formula for Number of States

• The number of states is given by: 2(2l + 1).
• For l = 2 and spin s = \frac{1}{2}, this is 2(2(2) + 1)(\frac{1}{2} + 1) = 2 * 5 * 2 = 10.
• General formula: Number of states = 2l + 1.
• Example: If l = 1, number of states = 3 (positive, negative, and zero).

Degeneracy and Measurement

• If a measurement yields a degenerate energy level, the collapse isn't complete.
• The system could be in any linear combination of degenerate states.
• Complete state specification requires measuring Lz and Sz separately.
• Full state description requires five quantum numbers: n, l, ml, s, ms.

Quantum Number Relationships

• Maximum value of l is n - 1.
• Example: A valid state could be n = 5, l = 3, ml = -2, s = \frac{1}{2}, ms = \pm \frac{1}{2}.
• Turning on a magnetic field in this state results in 2(3) + 1 = 7 states.

Energy Levels and Splitting

• The splitting is due to orbital angular momentum.
• With n = 5, energy E5 = -\frac{E1}{5^2}.
• Spin is an inherent property; it cannot be turned off, just like mass or charge cannot.

Energy Shift in a Magnetic Field

• New energy in field: E = E0 \pm \muz B_z
• Can be written as: E = E0 + \frac{q Bz}{2m} \hbar m_l.
  • Where \muz = \frac{q}{2m} Lz with a negative sign.
  • Lz = ml \hbar.

Bohr Magneton

• Bohr magneton \mu_B = \frac{q \hbar}{2m}.
  • q is the magnitude of the electron charge.
• Energy shift becomes: E = E0 \pm \muB Bz ml.

Lande g-factor

• Splitting due to spin includes a Lande g-factor.
• Complete energy shift: \Delta E = g \muB Bz m_s.
• Total splitting (including orbital and spin): \Delta E = \muB Bz (ml + g ms).
• The g-factor accounts for the difference between spin and orbital angular momentum contributions.

g-factor values

• For orbital angular momentum: \mu = \frac{q}{2m}L. For spin: \mu = -2 \frac{q}{2m}S (S is spin).
• The g-factor is not exactly 2; it's approximately 2.001 for electrons.
• G-factor values: Electrons (-2.001), Protons (+5.58), Neutrons (-3.3).

Implications of g-factor

• Even though neutrons are neutral, they exhibit a relationship between spin and magnetic dipole moment.
• Spin is a separate physical quantity; its connection to charge varies.
• The sign indicates vector direction; negative means antiparallel moments.

Simplified Energy Equation

• \Delta E = \muB B (ml + g m_s).
• Since g is not exactly 2, states are not perfectly degenerate.
• The splitting is slightly different, requiring precise measurements to observe.

Experimental Evidence for Spin

• Force on a magnetic dipole is proportional to the gradient of the magnetic field.

Force and Torque on Magnetic Dipole

• Torque exists due to opposing forces on the dipole.
• Net force is zero only if the magnetic field is uniform.
• If B depends on position, net force is non-zero. Example:
  • If B = B_0 x^2, the force is greater where the field is stronger.

Force Equation

• F_x = -\frac{\partial U}{\partial x}, where U is potential energy.
• General relationship: F = -\nabla U, where \nabla is the gradient operator.

Potential Energy

• Potential energy: U = -\mu \cdot B = -(\mux Bx + \muy By + \muz Bz).
• Gradient operator: \nabla = \hat{x} \frac{\partial}{\partial x} + \hat{y} \frac{\partial}{\partial y} + \hat{z} \frac{\partial}{\partial z}.
• If Bx and By are zero, Fz = -\frac{\partial}{\partial z} (\muz B_z).

Stern-Gerlach Experiment

• First experiment to demonstrate quantized spin angular momentum.
• Non-uniform magnetic field setup: North pole above a South pole.
• Magnetic field is stronger near sharp edges.

Experiment Side View

• Magnetic field lines are denser near the top of the magnet.
• Since magnetic field is a function of z,B_z = f(z).

Experiment Setup

• A beam of electrons is sent through the magnetic field.
• Detect the electrons on the other side.
• Angular momentum is equal to zero, since r \times p = 0.
• \muz = \muB g m_s \frac{dB}{dz}.
• Since m_s has two possible values: \pm \frac{1}{2}, leading to two forces.

Quantum Mechanical Results

• Measuring z-component of the magnetic dipole moment collapses quantum state.
• Electrons curve up or down, creating two points on the screen.

Classical Result (contrast)

• Classically, mu_z could have any value, creating a continuum.
• Experimental results differ significantly from classical predictions.

Total Angular Momentum

• J = L1 + L2, where L represents any kind of angular momentum (spin or orbit).
• In the case of an electron, J = L + S.

Two-Particle System

• L1 and L2 are the momentum angular of two different particles.

Free particle states

• \psi = j{kl}(r) Y{lm}(\theta, \phi)
• Where the indices k and l specified the quantized angular momentum

Eigenfunction for Two Noninteracting Particles

• A two particle system might have the following trial wave function:
  • \psi1 = j{k1 l1}(r) Y{l1 m}(\theta, \phi)
  • \psi2 = j{k2 l2}(r) Y{l2 m}(\theta, \phi)
• If the two particles are non-interacting, then then the Hamiltonian is separable
  • H = H1 + H2 where the two Hamiltonians commute, i.e. [H1, H2] = 0
  • because H1 is a function of r1, \theta1, \phi1 and H_2 uses completely different variables.
• Because of this, separation of variables can be used and we expect solutions of the following form
  • \Psi = f(r1, \theta1, \phi1) \times f(r2, \theta2, \phi2)
• The state of the system the can have six quantum numbers specified
  • k1, l1, m{l1}, k2, l2, m{l2}
  • \Psi = j{k1 l1}(r1) \times j{k2 l2}(r2) \times Y{l1 m{l1}}(\theta1, \phi1) \times Y{l2 m{l2}}(\theta2, \phi2) \times const

Measuring Total Angular Momentum

• If we make a measurement of the total angular momentum of the system we can ask if the following values hold true for j given l1 and l2
• Since angular momentum vectors we can specify its z component, resulting in a cone (distribution in the values of the x and y values)

Adding Angular Momentum for Two Independent degrees of freedom

• The cone of the first state can be added to the cone of the second

Calculating J Values

• J = l1 + l2
• The total angular momentum of any system is quantized
• We expect a j quantum number and a m_j quantum number that can be specified. The magnitude of J will be quantized by the following formula:
  • |J| = \hbar^2 j (j+1)
  • And Jz = mj \hbar
• The maximum j quantum number is when the two vectors align in the same direction
  • j{max} = l1 + l_2
• The minimum j is when they point in opposite directions
  • j{min} = |l2 - l_1|

Example Calculation

• l1 = 3 and l2 = 2
• then j{max} = 5 and j{min} = 1
• The j states must be between zero and 5, i.e. you cannot obtain a j=6 and j=0 etc.
• j = 5,4,3,2,1 are the only possibilities.

Measuring M

• Then number of states are described as: (2j + 1) states. And has a total of 3, 5, 7, 9, 11 which is 35 states.
• If we are to make a measurement of the total angular momentum and the jz

Representation Discussion

• As long as the particles are not interacting, the individual angular momentum quantum numbers is valid.
  • But if they are then individual measurements cannot be made
• We can write eigenstates in terms of the individual or the collective values.
  • <l1, m1, l2, m2|
  • <j, jz, l1, l_2|
• J does not commute with l1 and l2.

The Case of 2 NonInteracting Particles

• The free wavefn of the 2 partiles can be broken down to:
  • Wave fn 1 = 2 * l1 + 1 = 7 states
  • Wave fn 2 = 2 * l2 + 1 = 5 states
  • The total # of eigenstates = 7 * 5 - 35 which can be specified with four quantum numbers
• But j does not commutate with l1 and l2 (and vise versa).
  • [j^2, l_1] != 0
  • [j^2, l_2] != 0
• Meaning, once in this lock state, if one tries to measure total momentum, we get a distribution (can't have both individual angular momentums and total be set).

Given Example

• l1 = 3, l2 = 2
• ml1 = -2, ml2 = + 1
• 5 3, 2 -2, +1. Then jz = ml1 + ml2 = -1
• If a measurement is made on J, the magnitude of the distribution with these of these five stats. That distribution in the uncup representation is:
  • Linear combination w/ phis the coupled quantum numbers
  • Phi = a * (5 3, + b * (4 - 1) +….
• The z component can be added

Additions Properties

• When j^2 is together (ie commutes) with l1^2 and l2^2, it signifies that those quantum numbers can be specified simultaneously. This is, however, it does not with l1z, and l2z. So knowing both total angular momentum and l1 and l2 means having a distribution for what j^2 is (vis versa).
• However, since the jz can be specified at the same time as mlz, and ml2z, the jz has to follow these rules: If l1z = -2 while l2z = 1, then jz MUST be - 3. It's because these components all simply add (even though the angles of the vectors dont).
• jz does commute with l1 z and l_2 z