Spin

Question 1

Magnetic Dipole Moment

Description

Answer 1

-a rotating charge produces a magnetic dipole moment
-an electron rotates around a path of radius r
-the current through the loop is:
I = e/T = e / (2πr/v)
-magnetic moment:
μ = IA = evr/2
= e/2m * mvr
= e/2m * L
=>
|μL = - e/2m |L

Question 2

Dipole in a Magnetic Field

Answer 2

|μL = - e/2m |L
-if we place the diploe in a magnetic field, it starts to precess;
|d|L/dt| = |-e/2m|Lx|B|
=>
dL/dt =e/2me*LBsinθ
-for a small change in angular momentum:
d|L = Lsinθdφ
-sub in:
dφ/dt = eB/2m
-where m is the rest mass of an electron

Question 3

Potential Energy of a Dipole in a Magnetic Field

Formula

Answer 3

U = - |μ . |B

Question 4

Force Acting on a Dipole in a Magnetic Field

Formula

Answer 4

|F = -∇(-|μ . |B)

-force is non-zero only if the field is non-uniform

Question 5

Stern-Gerlach Experiment

Description

Answer 5

• sent a beam of atoms through an inhomogeneous magnetic field and measured the trace on a screen
• they were hoping to detect a splitting of the beam, which would be evidence for the quantisation of angular momentum
• but experiment shows a clear two-fold splitting of the beam

Question 6

Stern-Gerlach Experiment

Prediction

Answer 6

-the ground state of the hydrogen atom is described by the following quantum numbers:
n=1, l=0, ml=0
-let |B=B(z)^z
|F = -e/2m ∂B/∂z Lz ^z
-in quantum experiments:
|Lz = ℏ*ml
-therefore a possible results would be:
|F| = e/2m ∂B/∂z ℏ*ml
-where ml=l, ..., -l
-this is 2l+1 possible values of force so in general we would expect the beam to split 2l+1 times
-so for the ground state of hydrogen where l=0 we predict
2(0)+1 = 1 
i.e. a single beam, no splitting

Question 7

Stern-Gerlach Experiment

Spin

Answer 7

• the observed results are consistent with some sort of angular momentum with quantum number 1/2
• but this can’t be the usual angular momentum that is produced by an electron orbiting around the nucleus
• in order to explain the observations of the Stern-Gerlach experiment we postulate an additional type of angular momentum, spin

Question 8

What is spin?

Answer 8

• described the same way as angular momentum but physically spin has nothing to do with the electron’s orbital motion around the nucleus
• spin is an intrinsic property of the electron and exists even if the electron is stationary

Question 9

Mathematical Description of Spin

Answer 9

-spin quantum numbers: s, ms
-spin operators:
^Sz, |^S² ≡ ^Sx²+^Sy²+^Sz²
-eigenvectors:
{|s,ms⟩}

|^S² |s,ms⟩ = ℏ²s(s+1) |s,ms⟩
^Sz|s,ms⟩ = ℏms |s,ms⟩
-s can take integer (0,1,2,…) or half integer(1/2,3/2,…) values
ms = -s, -s+1, …. , s-1 , s

Question 10

General Formula for Spin Magnetic Moment

Answer 10

|μs = -g * q/2m * |S

-where q is the charge on the particle and m is the mass of the particle

Question 11

Complete Wave Function of the Electron

Answer 11

-the full wave function is:
|nlmlsms⟩
-for an electron, s=1/2 always so we only need to know the projection ms

Question 12

Full Degeneracy of the Hydrogen Energy Levels

Answer 12

gn = 2n²

Question 13

Spin of Fermions

Answer 13

-half integral spin

Question 14

Spin of Bosons

Answer 14

-integer spin

Question 15

Spins of Object Made up of Multiple Particles

Answer 15

-just add the spins of the constituent particles

Question 16

Why don’t we account for proton spin when calculating the spin of a hydrogen atom?

Answer 16

-because:
μs = - g * q/2m * |S
where g is the gyromagnetic ratio, q is the charge, m is the particle mass and |S is the spin
-since the mass of a proton is&raquo_space;> mass of an electron we don’t need to consider the proton in the Stern-Gerlach experiment

Question 17

Values of ms for an Electron

Answer 17

-an electron has spin +1/2 so ms is either -1/2 or +1/2

Question 18

Hilbert Space of Electron Spins

Answer 18

-the Hilbert space of electron spins is equal to;
span{|s,ms⟩} = { |1/2,+1/2 ⟩, |1/2, -1/2⟩}
-so the two vectors |1/2,+1/2⟩=|up⟩ and |1/2,-1/2⟩=|down⟩ are basis vectors for the Hilbert space

Question 19

Spin Operators in the Electron Spin Basis

^|S² and ^Sz

Answer 19

-^|S² and ^Sz are easily calculated since:
^|S² |s,ms ⟩ = ℏ²s(s+1)|s,ms ⟩
^Sz |s,ms ⟩ = ℏms |s,ms ⟩
-just substitute in the values of s and ms for the up an down vectors to produce 2x2 matrix representations of the operators in terms of the basis {|up⟩ , |down⟩}

Question 20

Spin Operators in the Electron Spin Basis

^S+ and ^S-

Answer 20

^S+ = ^Sx + ^Sy
^S- = ^Sx - ^Sy
-the general formula for applying the raising and lowering operators (^S+ and ^S-) is:
^S± |s,ms⟩ =ℏ² √[s(s+1) - ms(ms±1)] |s,ms±1⟩
-applying this to the basis vectors of the electron spin Hilbert space:
^S+ |up⟩ = ^S- |down⟩= 0 , since they both give vectors that are not in the Hilbert space
^S+ |down⟩= ℏ |down⟩
^S- |up⟩ = ℏ |up⟩
-it is easy to construct the 2x2 operator matrices for ^S+ and ^S- respectively from these results

Question 21

Spin Operators in the Electron Spin Basis

^Sx and ^Sy

Answer 21

-using the raising and lowering operators:
^S+ = ^Sx + ^Sy
^S- = ^Sx - ^Sy
-we can write formula for ^Sx and ^Sy:
^Sx = 1/2 (^S+ + ^S-) and ^Sy = 1/2(^S+ - ^S-)
-just substitute in the matrix form of ^S+ and ^S- for the matrix forms of ^Sx and ^Sy

Question 22

Spin Operators in the Electron Spin Basis

Pauli Matrices

Answer 22

^Sx = ℏ/2 * σx
^Sy = ℏ/2 * σy
^Sz = ℏ/2 * σz
-where σx, σy and σz are the Pauli matrices, 2x2 matrices with entries a, b, c, d:
σx: 0, 1, 1, 0
σy: 0, -i, i, 0
σz: 1,0,0,-1