Entanglement, Quantum Spin, and the Density Matrix

Jim Mahoney, Sep 2011

Consider two classic quantum experiments. 

The first is the double slit experiment. A stream of particles is sent through two slits, and the resulting pattern as they arrive at a screen beyond is examined. If the particle is going through one slit or the other with 50/50 probability, no interference is seen. However, if the particles are in a superposition of the top and bottom slit states, then the wavefunction goes through both, interferes with itself, and an interference pattern is observed. 

The second experiment is Bell’s Inequality, using two beams of particles sent in opposite directions, with corresponding particles in an entangled state. Measurement on one side collapses the wavefunction on the other side in a non-local way.

Can we combine these two experiments to send information via the non-local effect? Special relativity forbids this, and so we will be left with something analogous to a proposed perpetual motion which we know won’t work due to the second law of thermo. But finding the flaw can be instructive.

The idea is to do different sorts of measurements on one side to produce different wavefunction distributions on the other said, in particular, a 50/50 mixed state probability vs a pure superposition state, and then use a double slit apparatus to detect the difference between these two.

The specific proposal is as follows :
	1. Create a stream of pairs of spin 1/2 particles in a total spin 0 entangled state. Send pair members towards +Y and -Y.
	2. Measure the -Y one in two different ways, in an attempt to get two measurable different states at +Y.
		case 1: measure the Z component of spin of the -Y particle
		case 2: measure the X component of spin of the -Y particle
	3. On the right, we split the stream with a magnet along the Z axis to get two beams. 
	    Those two beams are then sent towards the top and bottom slit of a 
	    double slit experiment to see if they interfere or not.

[[ put a stylized picture of the setup here ]]

The verbal argument for why this can send information through the non-local effect runs like this:
	In case 1, each particle at +Y is in a definite eigenstate along Z. 
		Therefore after the Z split it either goes up or down, with 50/50 probability.
		Therefore there is no interference between the beams when the go through the two slits.
	In case 2, each particle at +y is in a definite eigenstate along Y.
		Therefore after the Z split each one goes both up and down, since it has both up and down components.
		Therefore each beam goes through both slits, each interferes with itself, 
		and there is an observable interference pattern.

It turns out (no surprise) that this argument is wrong, and the results of the measurement at -Y cannot be observed at +Y. 

The first problem is that since the particles going through the slits remain in distinguishable states (spin up through the top slit, spin down through the bottom slit), they will never interfere. But that problem can be fixed: simply pass one of the beams slowly (so as not to split it) through a rotating magnetic field so that it’s spin is that same as the other. “Slowly” means (dt)(dE) is small compared to Plank’s constant, where dt is time to rotate and beam and dE is the energy difference between the spin states in the magnetic field. In that case, the beam won’t split, and the particles going through the top and bottom slit are in the same spin state and are indistinguishable.

But even with that problem fixed, the scheme won’t work. It turns out that the math formalism describing this situation isn’t quite what the hand waving suggests. 

Your mission, should you describe to accept it: do the math.

We start with the math formalism for a spin 1/2 particle. The typical basis along the Z axis is

\[VerticalSeparator]↑⟩ =\[VerticalSeparator](Subscript[s, z])=+(()/2)⟩ 
\[VerticalSeparator]↓⟩ =\[VerticalSeparator](Subscript[s, z])=-(()/2)⟩

The wavefunction |ψ⟩ of a spin 1/2 particle is made up of these two , so

\[VerticalSeparator]ψ⟩ = α \[VerticalSeparator]↑⟩ + β \[VerticalSeparator]↓⟩

which as a typical column vector of two complex numbers can be written

(({{⟨↑\[VerticalSeparator]ψ⟩}, {⟨↓\[VerticalSeparator]ψ⟩}})) = (({{α}, {β}}))

where |α((|)^2) is the probability of being in the up state and |β((|)^2)is the probability of being down.

The Z component of spin for these spin 1/2 particles is  ±/2, so in this basis the matrix for the Z spin operator (Subscript[S, z])is

(Subscript[S, z])=(({{⟨↑\[VerticalSeparator](Subscript[S, z])\[VerticalSeparator]↑⟩, ⟨↑\[VerticalSeparator](Subscript[S, z])\[VerticalSeparator]↓⟩}, {⟨↓\[VerticalSeparator](Subscript[S, z])\[VerticalSeparator]↑⟩, ⟨↓\[VerticalSeparator](Subscript[S, z])\[VerticalSeparator]↓⟩}}))=(()/2)(({{1, 0}, {0, -1}}))= (()/2)(Subscript[σ, z])

where (Subscript[σ, z]) is one of the three Pauli matrices.

Just as energy and time are connected in quantum mechanics, the spin and rotation angle are of a quantum particle are connected. And just as (e^(-i E/)) moves a time independent wavefunction through time, the rotation operator for a particle is

(Overscript[R, ->]) = (E^(-(I /)(Overscript[S, ->])·(Overscript[θ, ->])))

For rotation around the Y axis for example this gives

(Subscript[R, y])=(E^(-(I /)(Subscript[S, y])(Subscript[θ, y])))= (E^(-I/2 (Subscript[θ, y])(Subscript[σ, y])))

where the Pauli spin matrix for the Y direction is

(Subscript[σ, y])=(({{0, -I}, {I, 0}}))

Exponentiating a matrix may look a bit weird, but since the exponential can be written as a power series

(E^a)=1 + a + (1/(2!))(a^2)+(1/(3!))(a^3)+…

and since powers of the Pauli matrices are particularly simple (((Subscript[σ, x])^2)=1), the (-i (Subscript[σ, x])) terms can be lumped together to give the typical complex expansion of the exponential into sines and cosines.

(Subscript[R, y])= (E^(-I/2 (Subscript[θ, y])(Subscript[σ, y]))) = (E^((-(Subscript[Iσ, y])) ((Subscript[θ, y])/2)))
    =cos(θ/2)-I (Subscript[σ, y])sin(θ/2)=(({{cos(((Subscript[θ, y])/2)), -sin(((Subscript[θ, y])/2))}, {sin(((Subscript[θ, y])/2)), cos(((Subscript[θ, y])/2))}}))

We can use this last result to rotate our Z basis by 90°to get the states with (Subscript[S, x])=±(()/2). Plugging in (Subscript[θ, y])=π/4 gives

R((Subscript[θ, y])=90°)=(1/(√2))(({{1, -1}, {1, 1}}))

so

\[VerticalSeparator](Subscript[s, x])=+(()/2)⟩ = R((Subscript[θ, y])=90°)\[VerticalSeparator]↑⟩=(1/(√2))(({{1, -1}, {1, 1}}))(({{1}, {0}}))=(1/(√2))(({{1}, {1}}))
\[VerticalSeparator](Subscript[s, x])=-(()/2)⟩ = R((Subscript[θ, y])=90°)\[VerticalSeparator]↓⟩=(1/(√2))(({{1, -1}, {1, 1}}))(({{0}, {1}}))=(1/(√2))(({{-1}, {1}}))

You can check this by confirming that these two states are eigens of (Subscript[S, x]). (Or for that matter get these states by solving for the eigens of (Subscript[S, x]).)

Two spin 1/2 quantum particles can be in spin eigenstates with S= or S=0. If the particles are labeled 1 and 2, we define the total spin as

(Subscript[S, total])=(Subscript[S, 1])+(Subscript[S, 2])

which when acting on a state with both particles in the up state gives

(Subscript[S, total])\[VerticalSeparator](Subscript[↑, 1])⟩ \[VerticalSeparator](Subscript[↑, 2])⟩=((Subscript[S, 1])+(Subscript[S, 2]))\[VerticalSeparator](Subscript[↑, 1])⟩ \[VerticalSeparator](Subscript[↑, 2])⟩
=(Subscript[S, 1])\[VerticalSeparator](Subscript[↑, 1])⟩ \[VerticalSeparator](Subscript[↑, 2])⟩+(Subscript[S, 1])\[VerticalSeparator](Subscript[↑, 1])⟩ \[VerticalSeparator](Subscript[↑, 2])⟩=(()/2)\[VerticalSeparator](Subscript[↑, 1])⟩ \[VerticalSeparator](Subscript[↑, 2])⟩+(()/2)\[VerticalSeparator](Subscript[↑, 1])⟩ \[VerticalSeparator](Subscript[↑, 2])⟩=\[VerticalSeparator](Subscript[↑, 1])⟩ \[VerticalSeparator](Subscript[↑, 2])⟩

so that state is equivalent to the up state of a spin 1 particle. Likwise, |(Subscript[↓, 1])⟩ |(Subscript[↓, 2])⟩= |(Subscript[↓, 1]),(Subscript[↓, 2])⟩=|↓↓⟩=|--⟩ (choose your notation; they’re all equivalent) is the down state of a spin 1 particle.

The other two states, |↓↑⟩ and |↑↓⟩, are not eigens of (Subscript[S, total]). (Check this for yourself). However, there are two linear combinations that are: |↑↓⟩+|↑↓⟩, the symmetric combination, has total spin 1, while the antisymetric superposition,  |↑↓⟩-|↑↓⟩, has total spin 0. (Again, you should check this. Note that written this way these states aren’t normalized.)

... coming ; see for example http://en.wikipedia.org/wiki/Density_matrix

Basically this is a way to combine the 50/50 statisticle ensembles of the result of the measurement into a single “thing” that we can compare in both cases.

... and not finding ‘em.