Electron spin in magnetic field is magnetic dipole moment - PowerPoint PPT Presentation

EE201/MSE207 Lecture 11 Electron spin in magnetic field 𝜈 is magnetic dipole moment Interaction energy 𝐼 = − 𝜈𝐶 (as for compass needle) 𝐶 is magnetic field, dot-product 𝜈 = 𝛿 spin 𝜈 = 𝛿 orbital 𝑀 or 𝑇 𝛿 is called gyromagnetic ratio 𝛿 cl = 𝑟 In classical physics: charge ( −𝑓 for electron) 2𝑛 mass In quantum mechanics: For orbital motion 𝛿 orbital = 𝛿 𝑑𝑚 For spin 𝛿 spin = 𝑕 𝛿 𝑑𝑚 g -factor 𝑕 ≈ 2.0023 for free electron, different values in materials

In more detail 𝑛 Angular momentum 𝜕 𝑟 𝑀 = 𝑛𝑤𝑆 = 𝑛𝑆 2 𝜕 𝑆 Classical physics Magnetic moment 𝜈 = 𝐽𝐵 = 𝑟 𝑈 𝜌𝑆 2 = 𝑟 𝜕 2𝜌 𝜌𝑆 2 𝛿 cl = 𝜈 𝑀 = 𝑟 current area Therefore 2𝑛 Quantum, orbital motion 𝜈 𝑨 = −𝑜 𝑓ℏ The same 𝛿 , 𝑀 𝑨 = 𝑜ℏ  2𝑛 magnetic quantum number (usually 𝑛 ) The smallest value (quantum 𝜈 𝐶 = 𝑓ℏ (Bohr magneton) of magnetic moment) 2𝑛 Quantum, spin 𝑇 𝑨 = ±ℏ/2 , but still 𝜈 𝑨 ≈ ±𝜈 𝐶 (a little more, 𝜈 𝑨 ≈ ±1.001 𝜈 𝐶 ), so 𝛿 spin ≈ 2𝛿 cl (i.e., 𝑕 ≈ 2 ). (Different people use different notations; sometimes 𝑕 = −2 , sometimes 𝛿 is positive, sometimes 𝑕 is called gyromagnetic ratio, etc.)

Evolution of spin in magnetic field 𝐶 𝑒𝑀 𝜈𝐶 = −𝛿𝐶 𝐼 = − 𝑇 torque 𝜈 × 𝐶 𝑒𝑢 = 𝛿𝑀 × 𝐶 𝜈 𝑒𝑀 𝑦 𝑒𝑀 𝑧 Classically, precession 𝑒𝑢 = 𝛿𝑀 𝑧 𝐶 𝑨 𝑒𝑢 = −𝛿𝑀 𝑦 𝐶 𝑨 as in a gyroscope. (assume 𝐶 = 𝐶 𝑨 𝑙 ) Larmor precession, 𝜕 = 𝛿𝐶 . 𝑒 2 𝑀 𝑦 𝑒𝑢 2 = −𝛿 2 𝐶 𝑨 2 𝑀 𝑦 ⇒ |𝜕| = |𝛿𝐶 𝑨 | 𝐶 Quantum mechanics 𝑇 𝑨 = −𝛿𝐶 ℏ 1 0 𝐼 = −𝛿𝐶 Assume 𝐶 = 𝐶 𝑙 0 −1 2 𝑗ℏ 𝑒𝜓 𝑒𝜓 𝑒𝑢 = − 𝑗 𝛿 is negative, −𝛿 is positive 𝑒𝑢 = 𝐼𝜓 ⇒ 𝐼𝜓 ℏ 𝑒𝑏 𝑢 = 𝑗𝛿𝐶 𝑏 𝑢 𝑏 0 𝑓 𝑗𝛿𝐶𝑢/2 𝜓 𝑢 = 𝑏 𝑢 𝑒𝑢 2 ⇒ ⇒ 𝜓 𝑢 = 𝑐 𝑢 𝑒𝑐 𝑢 = − 𝑗𝛿𝐶 𝑐 0 𝑓 −𝑗𝛿𝐶𝑢/2 𝑐 𝑢 𝑒𝑢 2 𝑏 2 + 𝑐 2 = 1 Both 0 1 and 1 0 are eigenvectors

Average values 𝑏 𝑓 𝑗𝛿𝐶𝑢/2 𝑏 = 𝑏(0) 𝜓 𝑢 = 𝑐 𝑓 −𝑗𝛿𝐶𝑢/2 𝑐 = 𝑐(0) 𝑗𝛿𝐶𝑢 2 2 = 𝑏 2 and also precession about z-axis. Obviously 𝑇 𝑎 = const , since 𝑏𝑓 Nevertheless, let us check formally. 𝑐 ∗ 𝑓 𝑗𝛿𝐶𝑢/2 ℏ 𝑏 𝑓 𝑗𝛿𝐶𝑢/2 1 0 = 𝜓 𝑇 𝑨 𝜓 = 𝑏 ∗ 𝑓 −𝑗𝛿𝐶𝑢/2 , 𝑇 𝑨 𝑢 = 2 0 −1 𝑐 𝑓 −𝑗𝛿𝐶𝑢/2 = ℏ 2 ( 𝑏 2 − 𝑐 2 ) Yes, does not depend on 𝑢 . Similarly 𝑐 ∗ 𝑓 𝑗𝛿𝐶𝑢/2 ℏ 𝑏 𝑓 𝑗𝛿𝐶𝑢/2 0 1 = 𝜓 𝑇 𝑦 𝜓 = 𝑏 ∗ 𝑓 −𝑗𝛿𝐶𝑢/2 , 𝑇 𝑦 𝑢 = 2 1 0 𝑐 𝑓 −𝑗𝛿𝐶𝑢/2 = ℏ 2 𝑏 ∗ 𝑐 𝑓 −𝑗𝛿𝐶𝑢 + 𝑏𝑐 ∗ 𝑓 𝑗𝛿𝐶𝑢 = ℏ Re(𝑏 ∗ 𝑐 𝑓 −𝑗𝛿𝐶𝑢 ) oscillations with frequency 𝜕 = 𝛿𝐶 𝑐 ∗ 𝑓 𝑗𝛿𝐶𝑢/2 ℏ 𝑏 𝑓 𝑗𝛿𝐶𝑢/2 0 −𝑗 = ℏ Im(𝑏 ∗ 𝑐𝑓 −𝑗𝛿𝐶𝑢 ) 𝑇 𝑧 𝑢 = 𝑏 ∗ 𝑓 −𝑗𝛿𝐶𝑢/2 , 𝑗 0 𝑐 𝑓 −𝑗𝛿𝐶𝑢/2 2

Dynamics of average values = ℏ = ℏ Re 𝑏 ∗ 𝑐 𝑓 −𝑗𝛿𝐶𝑢 , 𝑏 2 − 𝑐 2 , = ℏ Im(𝑏 ∗ 𝑐𝑓 −𝑗𝛿𝐶𝑢 ) 𝑇 𝑧 𝑢 𝑇 𝑦 𝑢 𝑇 𝑨 𝑢 2 = ℏ 𝑏 = cos( 𝛽 2) If then 𝑇 𝑦 𝑢 2 sin 𝛽 cos(𝛿𝐶𝑢) 𝑐 = sin( 𝛽 2) = − ℏ 𝑇 𝑧 𝑢 2 sin 𝛽 sin(𝛿𝐶𝑢) = ℏ 𝑇 𝑨 𝑢 2 cos 𝛽 Dynamics of 〈 𝑇 𝑢 〉 is the same as in the classical case, 𝜕 = 𝛿𝐶 . A way to visualize (as if) However, remember that actually 𝐶 3 𝑇 2 = 4 ℏ ≈ 0.87 ℏ 𝛽 ℏ and if we measure 𝑇 𝑦 , 𝑇 𝑧 , or 𝑇 𝑨 , we always get ± 2 .

Measurement 𝑏 𝑓 𝑗𝛿𝐶𝑢/2 𝜓 𝑢 = 𝑐 𝑓 −𝑗𝛿𝐶𝑢/2 ℏ 2 with probability 𝑏 𝑓 𝑗𝛿𝐶𝑢/2 2 = 𝑏 2 . If we measure 𝑇 𝑨 , then we get + If we measure 𝑇 𝑦 , then we get + ℏ 2 with probability 2 𝑗𝛿𝐶𝑢 1 , 1 𝑏 𝑓 ℏ 2 = 〈𝜓 𝑦+ |𝜓〉 2 = 2 𝑄 𝑇 𝑦 = + = 𝑐 𝑓 −𝑗𝛿𝐶𝑢 2 2 2 = 1 2 𝑏𝑓 𝑗𝛿𝐶𝑢/2 + 𝑐 𝑓 −𝑗𝛿𝐶𝑢/2 2 = 1 2 𝑏 + 𝑐 𝑓 −𝑗𝛿𝐶𝑢 2 ℏ 2 = 1 2 𝑏 − 𝑐 𝑓 −𝑗𝛿𝐶𝑢 2 𝑄 𝑇 𝑦 = − frequency 𝜕 = 𝛿𝐶 Similarly, Special case: 𝑏 = 𝑐 = 1 2 , then 1 1 𝑄 𝑇 𝑦 = + ℏ 2 = cos 𝛿𝐶𝑢 2 = 2 + 2 cos(𝛿𝐶𝑢) (in general, amplitude can be smaller and also 1 1 𝑄 𝑇 𝑦 = − ℏ 2 = sin 𝛿𝐶𝑢 2 = 2 − 2 cos 𝛿𝐶𝑢 extra phase shift) Quantum coherent oscillations: oscillations of probability (if measured) If we measure 𝑇 𝑦 , the state will be collapsed onto 𝑦 -axis ( ± ℏ 2 ), 𝑇 𝑨 and 𝑇 𝑧 components will be lost.

Another way to consider evolution (in 𝑦 -basis) (not included into this course) 𝑨 -basis ⟶ 𝑦 -basis 𝐼 = −𝛿𝐶 ℏ −𝛿𝐶 ℏ 1 0 0 1 ⟶ 0 −1 1 0 (𝑦) 2 2 𝑨 (Rabi oscillations in a qubit) 𝑗ℏ 𝑒𝜓 𝑏 𝑦 𝑒𝑢 = 𝐼𝜓 𝜓 𝑦 = 𝑐 𝑦 𝑗 ℏ 𝑒𝑏 𝑦 = −𝛿𝐶 ℏ 2 𝑐 𝑦 𝑒𝑢  oscillations with frequency 𝛿𝐶/2 , 𝑗 ℏ 𝑒𝑐 𝑦 = −𝛿𝐶 ℏ probabilities will oscillate with freq. 𝛿𝐶 2 𝑏 𝑦 𝑒𝑢

(not included into this course) Experimental measurement of spin (Stern-Gerlach experiment, 1922) inhomogeneous magnetic field produces N force onto a magnetic moment 𝐼 = −𝛿𝐶 𝑇 neutral atoms angular momentum (or spin) S magnetic field gyromagnetic ratio 𝐺 = −𝛼𝐼 = 𝛿𝛼(𝐶 𝑇) If 𝐶 is inhomogeneous (not constant), then So, force depends on 𝑇 . 𝜖𝐶 𝑧 𝜖𝐶 𝑦 𝜖𝐶 𝑨 If 𝐶 𝑦, 𝑧, 𝑨 = 𝐶 0 + 𝛽𝑨 𝑙 − 𝛽𝑦 𝑗 𝜖𝑦 + 𝜖𝑧 + 𝜖𝑨 = 0 ), (for magnetic field not important then 𝐺 = 𝛿𝛽 𝑇 𝑨 𝑙 − 𝑇 𝑦 𝑗 not important (oscillates because of Larmor precession about z-axis, so zero on average) 𝐺 𝑨 = 𝛿𝛽𝑇 𝑨 spin-up is deflected down ( 𝛿 < 0 ), (particle should be neutral because otherwise spin-down is deflected up charge will circle in magnetic field)

Addition of spins (similar for angular momenta) (1) + 𝑇 𝑨 𝑇 (1) + (2) 𝑇 = 𝑇 (2) 𝑇 𝑨 = 𝑇 𝑨 Two particles (vectors added) (scalars added) 2 + 𝑇 2 2 + 2 𝑇 2 = 𝑇 1 𝑇 (1) 𝑇 (2) However, not simple for the total spin 𝛽 ↑↑ Two particles with spin 1/2 𝛽 ↑↓ 𝜓 = 𝛽 ↑↑ ↑↑ + 𝛽 ↑↓ ↑↓ + 𝛽 ↓↑ ↓↑ + 𝛽 ↓↓ ↓↓ = 𝛽 ↓↑ 𝛽 ↓↓ 1 2 + 1 2 = 1 (𝑇 𝑨 = 1 ∙ ℏ) , 𝑡 = 1 ↑↑ 𝑛 = ↑↓ 𝑛 = 0 these states are not eigenstates of the total spin 𝑇 2 ↓↑ 𝑛 = 0 ↓↓ 𝑛 = − 1 2 − 1 2 = −1 (𝑇 𝑨 = −1 ∙ ℏ) , s = 1 Eigenstates ( ↑↓ − ↓↑ )/ 2 𝑡 = 0 (called singlet), 𝑛 = 0 of total spin: ↑↑ 𝑛 = 1 𝑡 = 1 𝑛 = 0 ( ↑↓ + ↓↑ )/ 2 (easy to check) (triplet) 𝑛 = −1 ↓↓

Addition of arbitrary spins (or angular momenta) Addition of arbitrary spins 𝑡 1 and 𝑡 2 is quite complicated. Possible values range from 𝑡 1 + 𝑡 2 to |𝑡 1 − 𝑡 2 | (integer ladder). Eigenvectors are given by Clebsch-Gordan coefficients. Example quarks: 𝑡 = 1 2 two quarks: 1 2 + 1 2 = 1 or 0 (mesons: vector and pseudoscalar) three quarks: 1 or 0 + 1 2 = 1 2 or 3 2 Delta, proton, Omega, neutron, etc. etc.