A quantum description of EPR

The starting point for a quantum description of the EPR setup is to return to the idea of the quantum state, $\ket \psi$, which contains all the information possible about a quantum system.

Quantum theory asserts that the anti-correlations of spins in Bohm's EPR scheme arises because the positron, electron pair are part of one quantum state. The two particles are "entangled" in single state.

Entanglement is not particularly rare, at least at the microscopic scale. We typically treat an atom, with its many electrons as a single quantum state. The Pauli exclusion principle, responsible for the filling rules of atomic shells, is a consequence of the characteristics of a many-particle quantum state of identical fermions -- half-integer spin particles like the electron.

Prelude: Quantum State for 2 particles?

In Bohm's spin version of the EPR experiment $$\pi^0 \to e^- + e^+$$ the two particles fly off in opposite directions. Let's assume$\color{red}{}^*$ we can bring one of the particles from each decay to a detecting station on the right, call this particle 1, and one to a detecting station on the left, call it particle 2.



At the right detector station we can make measurements of the $S_z$ component of spin of particle 1 with a S-G magnet. (Or we could measure $S_x$, or $S_y$, or even $S_{\hat n}$). Associated with this measurement we have some sort of operator, let's call it $\hat S_{1z}$ because we are measuring particle 1, and not particle 2. We would expect to measure one of the eigenvalues, either $S_z\pm \hbar/2$ of particle 1.

  • [$\color{red}{}^*$] Experimentally, we could observe millions of $\pi^0$ decays, in which the particles rush off in arbitrary directions in space, but only measure the small fraction of these in which the particles happen to be emitted exactly in the directions of our two S-G magnets at left and right.
    -OR-
    Maybe we could deploy many electron waveguides (they work like optical fibers for light) and bring a larger fraction of particles emitted to the left into a detector for the left hemisphere, and a similar fraction to a detector in the right hemisphere...

Similarly, there is an analyzer (at B) with an associated operator $\hat S_{2z}$. (Though we could also measure other spin components.)

Problem 4.1

...invites us to consider how to write the state ket $\ket \tau$ for a two-particle system, in terms of the single particle wave functions of the particle arriving at 1, $\ket{\psi}_1$ and $\ket\phi_2$. Consider two possibilities (not necessarily normalized): $$\ket\tau = \ket\psi_1\ket\phi_2$$ or $$\ket\tau = \ket\psi_1+\ket\phi_2$$

Let's assume there is an operator $\hat S_{z1}$ that operates on $\ket \tau$, but only "cares about" the state of particle 1....

I tried to show that when $\ket \tau=\ket+_1\ket\phi_2$: $$\hat S_{1z}\ket\tau= \hat S_{1z}\ket+_1\ket\phi_2 =+\hbar/2\ket+_1\ket\phi_2=\hbar/2\ket \tau.$$ This means that such a product state is an eigenstate of the $\hat S_z$ operator.

But when the $\hat S_z$ goes to work on the additive state:$\ket \tau=\ket+_1+\ket\phi_2$: $$\begineq\hat S_{1z}\ket\tau &= \hat S_{1z}\big(\ket+_1 +\ket\phi_2\big) =+\hbar/2\ket+_1 + \ket\phi_2\\ &\neq \hbar/2\big(\ket+_1 +\ket\phi_2\big)\\ \hat S_{1z}\ket\tau &\neq \hbar/2\ket\tau. \endeq $$ The additive state does not behave like an eigenstate of $\hat S_{1z}$.

Digression on angular momentum in QM

We have not proved the following results. But a detailed treatment of angular momentum, and addition of angular momenta finds these results, which apply to both single particles as well as quantum systems of more than one particle:

  • The total angular momentum of any quantum system (of a single particle or a system of many particles) is quantized: There is a quantum number, $s$, that may take on values of ${0, \frac 12, 1, \frac32, 2, \frac52,....}$ characterizing any such system.
  • Individual kinds of particles have their own unique value of $s$. A photon has $s=1$. Both electrons and protons are $s=\frac12$ particles. A neutral $\pi^0$ pion is a $s=0$ particle.

    But for systems of more than one particle, there might be different configurations with different values of $s$, depending on the relative orientations of the spins of the two or more particles...

  • The eigenvalue of total angular momentum^2 for an quantum system $\ket \psi$ is $$\hat S^2\ket \psi = (s)(s+1)\hbar^2\ket\psi$$
  • All particles with integer values of $s$ are bosons and do not obey the Pauli exclusion principle. All particles with half-integer spins are fermions and obey the Pauli exclusion principle.
  • The projection of angular momentum in the $z$ direction of any quantum system is quantized: There is a quantum number, $m$, that may take on values between $-s \leq m \leq s$ in integer steps, such that an eigenstate of the $\hat S_z$ operator satisfies $$\hat S_z\ket \psi=m\hbar \ket.$$
  • Since $[\hat S^2,\hat S_z]=0$, the two operators are compatible, and their eigenfunctions are the same. Thus, we can label a state which is a simultaneous eigenstate of both $\hat S^2$ and $\hat S_z$ with its quantum numbers $s$ and $m$, as $\ket{sm}$.

We are most familiar with the electron which is a $s=1/2$, spin 1/2 particle (a fermion). Its possible $z$ projection quantum numbers are $-s \leq m \leq s$, that is: $$-1/2 \leq m \leq +1/2$$ And since the possible values of $m$ must be spaced by 1, this means $m=\pm 1/2$. And therefore the possible eigenvalues of $\hat S_z$ are $\pm \hbar/2$.

According to this new way of writing things, the $\hat S_z$ eigenstates of the electron could be labelled in terms of quantum numbers $s$ and $m$ as: $$\text{spin up: }\ket+\equiv \ket{s,m}=\ket{\frac12,+\frac12}\ \ \text{ and } \ \ \text{spin down: }\ket-\equiv\ket{\frac12,-\frac12}.$$

Possible angular momentum states of 2 electrons

We have seen that a two particle state can be written as a product of two single particle states, $\ket\psi_1\ket\phi_2$. We can easily see that with two possible states (spin up or down) for each particle, there must be four possible combinations of these states (I'm dropping the subscripts 1 and 2, but think of the order as containing that information...): $$\ket +\ket +,\ \ \ket +\ket -,\ \ \ket -\ket +,\ \ \ket -\ket -.$$ It can be shown that the following 3 states of a two particle system are the 3 eigenstates of the $\hat S^2$ operator, with quantum number $s=1$--the triplet states: $$\nonumber\ket{1,+1}=\ket+\ket+,$$ $$\nonumber\ket{1,0}=\frac{1}{\sqrt 2}\Big(\ket+\ket- {\color{red}+} \ket-\ket+\Big),$$ $$\nonumber\ket{1,-1}=\ket-\ket-.$$ There is one more state which can be constructed to be orthogonal to all 3 of the states above, which clearly has $S_z=0$, and it is the "singlet", which corresponds to the only $s=0$ state of a two particle system: $$\ket{0,0}=\frac{1}{\sqrt 2}\Big(\ket+\ket- {\color{red}-} \ket-\ket+\Big),$$ And this is the state that the electron and positron are in after the decay of the spin zero $\pi^0$ pion.

QM Explanation of anti-correlation

Returning to our simultaneous measurement of $z$ projection on particles 1 and 2:

The quantum mechanical explanation of the anti-correlation of the spins is:

  • The original state of the system after pion decay is the singlet state: $$\frac{1}{\sqrt 2}\Big(\ket+\ket- {\color{red}-} \ket-\ket+\Big).$$
  • Let's say that particle 1 is measured first, and that it is found to be spin up $\ket+_1$. Then as soon as the measurement occurs, the state collapses into: $$\Rightarrow \ket+_1\ket-_2,$$ and the measurement of particle 2 which follows must find, with 100% probability that particle 2 is in the spin down state $\ket -_2$.