# The mathematics of entanglement

I've finally managed to get a grasp on the Bell test experiments and all that they imply about our reality. Now I'm curious about the mathematical derivation which allowed Schrodinger to predict the existence of entangled entities.

Can someone please provide an explanation of these equations or direct me towards good sources of information on the subject? Thanks.

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This follows immediately from Schrodinger's demonstration of the equivalence of wave mechanics to Heisenberg's matrix mechanics, although I did not read the original paper, and I couldn't find it in a quick search online. There are many ways to argue this, and I will just give a few of the more obvious ones.

### Theoretical justifications

The property of entanglement is the statement that a general wavefunction for 2 particles is a function of their 6 coordinates, not two functions of their 3 coordinates. Schrodinger toyed with the idea that the wavefunction is a physical wave, but abandoned it once he came in contact with the Bohr/Heisenberg matrix mechanics.

In matrix mechanics, you treat many particles using independent X operators and P operators which have the property that $[X_j,P_k]= i\delta_{jk}$. In wave mechanics, you interpret $P$ as a differential operator for functions of the $X$ coordinates, and $X_j$ as multiplying by the appropriate coordinate, and this gives a representation of the algebraic canonical commutation relations. The number of dimensions is determined by the number of commuting X coordinates, and is the same as the classical configuration space, half of phase space, ignoring the momenta.

Adding more coordinates is adding more commuting X operators, and more independent P operators, so that the wavefunction grows in dimension without bound, and there is no natural restriction you can choose. A measurement of the $X_i+X_j$ operator will prepare a state which is not entangled in the $X_i+X_j$ $X_i-X_j$ basis, but this state is entangled in the $X_i$ $X_j$ basis.

The sum of product wavefunctions of the form $\psi(x,y)=f(x)g(y)$ does not respect the product form, so if you restrict yourself to unentangled wavefunctions, the evolution operator cannot be a general linear map. The constraint which produces entangled wavefunctions is that

$$\partial_x \partial_y \log(\psi(x,y)) = 0$$

Which is the differential constraint that asserts that the logarithm is a function of x plus a function of y only. These product wavefunctions are only eigenfunctions of Hamiltonians of the form $H_1(x)+H_2(y)$, where one H is a function of x alone, and the other is a function of $y,p_y$ alone, so that it is impossible to couple such systems to each other.

Adding more particles adds more canonical coordinates, but so does adding more dimensions to space, so that the particle has more independent directions to move in. The two operations are indistinguishable in matrix mechanics, because you don't specify what the dimensions mean until you specify the Hamiltonian and the observable interpretation.

### Helium atom

This is a theoretical argument with a startling conclusion, that particles have entangled states, so it is important to verify it experimentally. The ground state of Helium provides a particularly clean test. Two independent electrons will both have to sit in an S-wave cloud around the nucleus, and repel each other heavily by their electrostatic interaction. By admitting non-product wavefunctions, you can make the amplitude for the two electrons to be both on the same side of the nucleus be smaller, and the amplitude for the two electrons to be on opposite sides bigger, without spoiling the overall spherical symmetry of the two electron system altogether. The result is that the Helium atom is 30% lower in energy in the entangled true ground state than it would be in the artificial unentangled ground state. These calculations are best done using the variational method, and the entangled ground state will predict the ionization energy for Helium as precisely as you like (I believe you can get to 1% accuracy with paper and pencil calculations--- the once-ionized comparison state is easy because it is the exactly solvable He+ ion, essentially the same as the H atom).

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Thanks for the detailed response, where can I more info on these operators you refer to? – Gearoid Murphy Nov 27 '11 at 14:31
@Gearoid Murphy: If you mean the use of the word "operator" in my answer, it is described in Dirac's quantum mechanics book, in Feynman's lectures, in Sakurai's quantum mechanics book, on Wikipedia's page on Matrix Mechanics, and in many other places (the ones I listed are accurate and well presented, in my opinion). – Ron Maimon Nov 27 '11 at 22:15

try the book of sakurai. i am adding link, to the specific book, you can get it in your school lab for sure

http://www.amazon.com/Modern-Quantum-Mechanics-Revised-Sakurai/dp/0201539292

pages 226 - 229

pay attention it is the first book (as showed in the link) and not the advanced one.

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