The Spin (continued). February 8, 2012

Transcription

1 The Spin continued. Magnetic moment of an electron Particle wave functions including spin Stern-Gerlach experiment February 8,

2 Magnetic moment of an electron. The coordinates of a particle include the spin variables S, m s and the position variables x, y, z. The Hamiltonian of a free particle Ĥ = p 2 2m does not depend on spin. It means that the spin will manifest itself only in the presence of a magnetic field. There are two linearly independent eigenstates: ψ + = ψ k r α = A e ikr 1 0 ψ = ψ k r β = A e ikr 0 1 2

3 S z ψ + = 2 ψ + S z ψ = 2 ψ Free - particle wave functions, including spin. four - fold degenerate. E k = 2 k 2 2m : ψ k+ r ψ k+ r ψ k r ψ k r

4 Bohr magneton. We remember from electrodynamics that a circular loop of a wire carrying a current I, which is of cross-sectional area A, produces a magnetic moment: µ = 1/cIA in CGS. For an election one finds that the magnetic moment is directly proportional to its angular momentum: µ = e mc S = e 2mc σ = µ Bσ = 2µ B S. The quantity e µ B = 2mc is called Bohr magneton. = erg gauss Let us compare this expression with that for the orbital motion. For the sake of simplicity let us consider a circular loop of radius 3

5 r, around which the electron is moving with the speed v. The corresponding current is: and the magnetic moment I = e v 2πr µ = e v 2πrc e vr πr2 ẑ = 2c ẑ where ẑ is the unit vector along ẑ axis. On the other hand L = mvr ẑ Thus µ = e 2mc L = µ B L

6 We see that the coefficient which is called gyromagnetic ratio is two times larger for the spin. This fact can be explained only in a relativistic theory. Consider now a magnetic dipole in a magnetic field. For the sake of simplicity let s represent it as a current loop tilted at the angle θ with respect to B. We also assume that the current in the loop is due to positively charged carriers q = e. We know that the magnetic field exerts the Lorenz force upon moving carriers F = q c [v B] 4

7 F B θ µ F F F

8 Looking at the figure we can see two facts: 1. If the field is uniform the net force exerted on the loop is The net torque exerted on the loop is net zero. Indeed, the forces on the two sloping sides do not exert the torque they are trying to stretch the loop but not rotate it! On the other hand the forces exerted on the horizontal sides are trying to rotate the loop in such a way that it becomes horizontal, i.e. the magnetic moment is aligned with µ. The torque N = µ B 5

9 One can calculate the potential energy of the magnetic dipole: V = θ 0 Ndθ = µ B = µb cos θ θ θ V = Ndθ = µb sin θdθ = µb cos θ + const. 0 0 If we put the dipole in a non-uniform field, the net force on it is no longer zero. Indeed, consider a loop in a non-uniform field. 6

10 B F I B F F The net downward force upon this loop is not zero! This force can be calculated as F = U = µb = µ B.

11 Stern-Gerlach Experiment. The Stern-Gerlach S-G experiment was originally performed in 1921 using a beam of silver atoms. The spin of a silver atom is due to its outer 5s electron. In the following we simply refer to electron spins in the beam. The magnet used in the experiment looks like this: 7

12 The predominant component of B is B z. Furthermore B z varies most strongly in z-direction so that B ê z B z z The force acting upon an electron and ultimately upon the

13 atom is F = µb ê z µ z B z z The predominant force on the atom is in the z direction. In addition the sign of this force is solely determined by the sign z. The z component of force causes atoms in the beam to be deflected. The strike the detection screen off-axis. If we know where the atoms strike the screen we can calculate F z. Indeed when atom passes through the magnet it gains z-component of the momentum. The expectation value of the momentum can be calculated according to the Ehrenfest theorem dp z dt = F z

14 Or x p z = F z t = F z v x Therefore = µ z B z x v x p z = 2µ B S z B z x v x = 2µ B S z B z x v x Since µ is directly proportional S the Stern-Gerlach apparatus is an instrument that measures the z component of a spin. We have to emphasize that incident beam carries an isotropic distribution of direction of magnetic moments. This means that from a classical point of view we should expect a continuous distribution of deflections corresponding to the values µ z continuously distributed between µ and µ in the original ion-spinpolarized beam. The pattern expected from the classical stand point should be continuous. However, experiment finds a beam

15 dividing into two discrete components. Thus experiment indicates the µ z or, equivalently, S z can assume only two values: S z = ± 2

16 Problem Consider a polarized beam containing electrons in the α z state sent through an S-G analyzer, which measures spin projection S ξ, such that axis ξ makes angle θ with z. Questions: a. What values will be found? b. With what probabilities will these values occur? c. What is the expectation value S ξ? 8

17 a. As in any spin measurements the only values that can be found for S z are 2 and 2, to prove it we have to find the eigenvalue of the operator: Ŝ ξ = 2 σ ξ = 2 σ ξ, where ξ = cos θ, sin θ is a unit vector. 1 0 Ŝ ξ = Ŝz cos θ + Ŝx sin θ = cos θ sin θ Or cos θ sin θ Ŝ ξ = sin θ cos θ 9

18 Let s solve the eigenvalue problem for this matrix: cos θ sin θ µ µ = λ sin θ cos θ ν ν To find the eigenvalues we solve: or cos θ λ sin θ sin θ cos θ λ = 0 λ 2 cos 2 θ sin 2 θ = 0, or λ 2 1 = 0 = λ = ±1 Thus the eigenvalues of Ŝz are: ± 2, Q.E.D. 10

19 b. To find the probabilities we have to find eigenstates of σ ξ and then represent α z as a superposition of these eigenstates. To find the eigenstate corresponding to S ξ = + 2 we solve: cos θ sin θ µ µ = +1 sin θ sin θ ν ν and require This yields: or Or µ 2 + ν 2 = 1 cos θµ + sin θν = µ ν sin θ = µ1 cos θ = 2ν sin θ 2 cos θ 2 = 2µ sin2 θ 2 ν = µ tan θ 2 11

20 . Using the normalization condition we get Or µ tan 2 θ 2 = 1, µ = cos θ 2 ν = sin θ 2

21 Therefore the eigenstate corresponding to S ξ = + 2 is: α ξ = cos θ 2 sin θ 2 Similarly the eigenstate corresponding to S ξ = 2 is: β ξ = sin θ 2 cos θ 2 We note that the choice of α ξ and β ξ is not unique. They both can be multiplied by an arbitrary phase factor. We now represent α z as: α z = aα ξ + bβ ξ, 12

22 or 1 0 = a cos θ 2 sin θ 2 + b sin θ 2 cos θ 2 or a cos θ 2 b sin θ 2 = 1 a sin θ 2 + b cos θ 2 = 0 This system of equation can be solved straightforwardly: a = cos θ 2 b = sin θ 2

23 Therefore α z = cos θ 2 α ξ sin θ 2 β ξ, and the probabilities to find S ξ equal to 2 or - 2 are: P + = cos 2 θ 2 P = sin 2 θ 2

24 c. The expectation value of S ξ can be found in two ways: 1. We can use operator Ŝξ and the eigenstate α: S ξ = α 2 σ ξα = 2 α σ z α cos θ + α σ x α sin θ Here we used: = 2 1 cos θ + 0 sin θ = 2 cos θ α σ z α = α σ x α = = 1 0 = = 1 = 0 13

25 2. We can use the probabilities P + and P and the eigenvalues of Ŝξ: S ξ = cos 2 θ/ sin 2 θ/2 2 = 2 cos θ Thus in both cases: S ξ = 2 cos θ