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XVI. SPIN 1/2

A. revisiting the ladder operators

We have the ladder operators, so we know that eigenvalues of Lz must be of the form:

Lz = m¯h + m0¯h (444) with m being integer, and this way the algebra connects all states to each other. But what could m0 be? Could it be any number? Let’s think for a bit about the possible answer. First, in the midterm, we had a particle in a ring. The solutions for the particle in the ring were:

ψ(x)= eikx (445) such that 2πrk = 2πm. This implies:

rp = m¯h (446) so clearly m0 = 0 is allowed, and seems to make sense here. Could it be other things? When we turn our system upside down, since the is a vector, it seems that this makes:

Lz →−Lz (447) and m0 →−m0. If we want to stay within the same Hilbert subspace, what must m0 be? Zero works, since −0=0. So Lz = m¯h seems to be okay, with m being an integer. We would mape integers to integers. Any other values? m0 = 1/2 is a tempting possibility.

B. Constructing the spin-1/2 representation

But could there be other states? Other representations of the angular momentum operators? Let’s see, above we deduced that m0 must be an integer since m0 and −m0 must belong to the same representation and connected to each other with the ladder operators. But then, what about: m0 = 1/2. Since m0 − (−m0) = 1 which is an integer. This is allowed since a ladder operator can connect these states to each other. This is the half-integer spin! Half integer spins are the stuff of legends. There’s this vague statement about rotating them by 2π and them not returning to themselves. Let’s see what that’s all about. Starting with:

2 L = L−L+ + Lz(¯h + Lz) (448)

If we are considering a state with the maximum m = 1/2, then L+ |1/2 = 0, and:

ℓ = 1/2. (449)

Applying L− to this state would produce a state with Lz |ℓ = 1/2,m = −1/2 = −¯h/2 |m = −1/2. What would happen if we apply L− to this again? we get zero. Now, a particle with angular momentum ℓ = 1/2 can be written as a superposition of two states:

|ψ = ψ↑ |↑ + ψ↓ |↓ (450) you may wonder, why did I shift to the use of up and down arrows rather than the +1/2 and −1/2? Because the spin-1/2 is so common, that we have special notation for it. Why is it so common? All fermionic elementary particles have spin-1/2! , protons, quarks, neutrons, neutrinos. you name it! What more - there is no spatial iφ/2 representation that works. Look up at the Lz eigen states. e is simply not periodic! It is double valued, since it is essentially a square root of eiφ. Not legit for a ! All we can do for spin-1/2 particles is find their angular momentum matrices. These also have special names: the Pauli matrices. The constuction is really easy. Consider L±, these are 2X2 matrices:

0 c 0 0 L+ = , L− = (451)  0 0   c∗ 0  56

To find L, we return to the identity (448).

0 0 L−L+ = 2 (452)  0 |c|  The RHS of the identity is:

2 2 0 0 L − Lz(¯h + Lz) =h ¯ 1 3/2 1 1 = (453)  0 2 − − 2 ) 2   where 1/2 3/2= ℓ(ℓ + 1), and we obtain:

c = 1. (454)

Now it is easy to see what the Lx and Ly matrices are:

1 ¯h 0 1 1 ¯h 0 −i L = (L+ + L−)= , L = (L+ − L−)= . (455) x 2 2  1 0  y 2i 2  i 0 

The spin matrices are denoted as Lα = σα usually:

1 0 0 1 0 −i σ = , σ = , σ = (456) z  0 −1  x  1 0  y  0 i  Notice that the sigma matrices all square to 1:

1 0 σ2 = σ2 = σ2 = (457) x z y  0 1  What about the “two rounds to get back to itself” business? Let’s take an with spin up:

|ℓ = m = 1/2 = |↑ (458)

To rotate it around the z-axis by an angle φ, we can employ the exponent of Lz:

1 |↑ → ei 2 σz φ |↑ (459)

Now, the |uparrow is an exact eigenstate of σz with eigenvalue 1. Therefore it is an exact eigenstate of all powers of σz, and of exponents thereof. The upshot is that we can simply replace σz in the expression above with the eigenvalue, 1:

1 = ei 2 φ |↑ (460)

If we rotate the electron by 2π we get the surprising result:

= eiπ |uparrow = −|uparrow! (461)

Only by turning the electron again by 2π do we retrieve the original wave function. You can eaily notice that it doesn’t really matter whether we use the |↑ or |↓ states, both will get a −1 upon a 2π rotation. Indeed, by expanding the exponent, and using the squaring to 1 characteistic, we get:

1 φ φ ei 2 σz φ = cos + iσ sin . (462)  2  z  2  which at φ = 2π, is just −1.

C. Spin-1/2 rotation and spin-1/2 in a magnetic field

I have a lofty goal - get to the really intriguing topic of Bell inequalities. But on the way there, you are going to get a bit of Custer oil. We need to know something about how to diagonalize the spin-1/2 rotation matrices in an arbitrary direction. 57

The problem of dialgonalizing the spin-matrices in a particular direction has a huge physical significance. The electron is a charged body, and it is rotating (indeed in a mysterious way with half integer spin, but nontheless!). Therefore it has a dipole moment. To remind you a bit, a current loop has a dipole moment proportional to the current time the area covered by the loop:

= IA (463)

If an electron is moving in a circle, we can write: eω I = , A = πr2 (464) 2π and together this gives:

eωr2 = (465) 2 This may not look so familiar yet, but when we recall that L = mωr2 we obtain: eL = (466) 2m and if we are tempted to vectorize this a bit, then clearly:

eL = (467) 2m This is the magnetic moment of a charged particle in an orbit with a given angular momentum. This means that the magnetic moment itself is quantized. Consider, to be specific a magnetic field in the z direction B =zB ˆ . The hamiltonian corresponding to this is: e Hˆ = −B = −B Lˆ (468) 2m z

This hamiltonian is very easy to solve: the eigenstates are just the Lz momentum states. And the are just: e¯h Emz = −B mz (469) 2me

For an electron, the quantity multiplying the mz and B is known as the ’Bohr magneton’. Its numerical value is:

e¯h −23 B = = 0.9 10 J/T (470) 2me

For electrons, there is another source of a magnetic moment - their spin. Using the notation S for the electron spin, and the Pauli matrices as the representation of the angular-momentum spin operators, we first must concede that we don’t know what the radius of charge motion in the case of an electron is. Therefore, when we try to write the magnetic moment of the electron, we need a fudge factor:

eS s = g (471) 2me With g being the fudge factor. For an electron, the fudge factor happens to be very close to

g = 2. (472)

This is called the gyromagnetic ratio. Actually, it has quantum electrodynamics corrections to it which makes it slightly bigger than 2: g = 2(1+ α/2π + ...) with α = Ke2/c¯h ≈ (137)−1. Now we can write down the magnetic field hamiltonian on the spin itself: ge¯h 1 Hˆ = −B s = B nˆ σ (473) 2me 2 58 with the extra 1/2 coming from the conversion of the spin-operators to Pauli matrices. If the field is only in the z-direction we simply have:

ge ge¯h ge¯h 1 0 Hˆ = −B Sˆz = −B σz = −B (474) 2me me 4me  0 −1 

This is rather remarkable. If the spin is up along the field direction, then the is −Bz(r)B. If the spin is pointing in the opposite direction it is Bz(r)B. This is wierd - only two directions possible for the spin. And if the magnetic field is changing in space (hence the extra r) we would have a force that depends on the direction of the spin and will separate the spin-up from the spin-down electrons:

F = −∇E = mzB∇Bz(r). (475)

Are there indeed just two components for the electron spin? This was tested in the Stern Gerlach experiment. In this experiment a beam of electrons was passed under a pole of a magnet where the magnetic field has a strong gradient, and then hit a screen. Indeed what they so was that the beam split into two!

D. The not so nicely oriented magnetic field

Now some technicalities. What happens if our field is not pointing in the z-direction? Say it is pointing in the θ,φ direction. What is the Hamiltonian? g Hˆ = −B (sin θ cos φσx + sin θ sin φσy + cos θσz) (476) 2 B The matrix in the brackets is basically the unit vector in the direction of the field times the set of Pauli matrices. It looks pretty neat:

cos θ sin θe−iφ nˆ σ = (477)  sin θeiφ cos θ  First of all, what are the eigenvalues of this matrix? Just 1 and −1. How do I know? it is just a rotated Pauli matrix - we could have chosen to redefine the directionn ˆ as our z-direction, and then we would immediately see this. But we haven’t. So we are stuck with a rather unpleasant 2X2 diagonalization problem.

E. Solution by unitary rotations (skipped in class)

But the remedy is at hand - the same way thatn ˆ σ is just a rotation of the σz matrix, the eigenvectors are just a rotation of the |↑ , |↓ eigenkets. How can we rotate the z-direction to then ˆ direction? Not so difficult. We need to rotate about the direction pˆ = − sin φxˆ + cos φyˆ by an amount θ. The directionp ˆ is chosen since it is perpendicular to the initial z-direction i and finaln ˆ direction. Now, to rotate, we already know the trick. exp h¯ φSz rotates about the z-axis by an amount φ. Therefore, we can use:

i Rˆ = exp − θpˆ S (478)  ¯h  to rotate about thep ˆ direction. And this we simply do by:

θ Rˆ = exp −i pˆ σ (479)  2  And we expect that:

1 Rˆ |↑ = Rˆ = |↑ˆ (480) z  0  n and similarly for |↓. Note the minus sign in the definition of Rˆ. The reason for this minus sign (except for the fact that this is necessary to make the trick work!) is that the rotation essentially rotates the space representation in the 59 negative value of the argument. We discussed this in the second round of talking about angular momentum operators using the ’rotated cross’ diagram. But what is this funny exponent? Can we calculate it? Here there is a trick. Just expand the exponent in powers, and what do we get?

2 3 θ 1 θ 1 θ Rˆ = 1 − i pˆ σ + i − i pˆ σ. . . (481) 2 2!  2 3!  2

This looks a bit strange: where did all the higher powers ofp ˆ σ go to? You can verify that for any unit vector (ˆp σ)2 = 1, so it only appears linearly in the odd powers. But now, we have something that looks exactly like the Euler formula, but with an invading Pauli matrix. No problem, pretend that the Pauli matrix is part of the i: θ Rˆ = cos − ipˆ σ sin θ (482) 2 Which we can write:

θ θ −iφ ˆ cos 2 − sin 2 e R = θ iφ θ (483)  sin 2 e cos 2  And we can immediately read off the solution given below

F. Solution of the rotated field problem

In class we tested these solutions:

θ cos 2 |↑nˆ = θ iφ (484)  sin 2 e  and:

θ −iφ − sin 2 e |↓nˆ = θ (485)  cos 2  Here we see that indeed θ = π doesn’t give us the original state, but rather, an extra −1.