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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 angular momentum 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! Electrons, 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 wave function! 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 electron 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: