I feel like a ping pong ball on the question of whether or not total energy of a pair of electrons changes when they become entangled or disentangled.(adsbygoogle = window.adsbygoogle || []).push({});

So, I can argue myself into both positions on this question, entanglement/disentanglement do [or do not] imply energy changes in the system. Help please!

- I began with a mental model similar to electrons in orbitals. Emit a photon and drop into a lower energy state until reading the ground state. A pair of electrons unentangled, emit and drop to the triplet state, then emit again and fall to the singlet (ground) state.

- In an earlier thread, CGK elegantly showed me how to write the Hamiltonian which does not include spin states, and the wavefuncitions which do include spin.

cgk said: ↑This may be a stupid question, but does this really require QED (photons) for a faithful description? At least in the non-relativistic case I see little problem in writing down the Coulomb Hamiltonian [tex]H = \frac{p_1^2}{2m}+\frac{p_2^2}{2m}+\frac{e^2}{e\pi\epsilon_0\Vert \vec r_1 - \vec r_2\Vert}[/tex] and then just using it to propagate your initial wave function via the time-dependent Schrödinger equation [tex]i \hbar \partial_t \Psi(\vec x_1, \vec x_2,t) = H \Psi(\vec x_1, \vec x_2,t).[/tex] The initial conditions of the two electrons (momentum, position, spin, etc) would then be determined by how you set up the initial wave function at t=0 which you propagate.

Without external potentials this can probably even be solved exactly.

How is the spin included, then? It works like this: The spin of the system is not in the Hamiltonian, but in the wave function. An electronic wave function needs to be anti-symmetric: [itex]\Psi(\vec x_1, \vec x_2)=-\Psi(\vec x_2,\vec x_1)[/itex], where the x-vectors are the combined spin-space positions: [itex]\vec x_i = (\vec r_i, s_i)[/itex] and [itex]s_i[/itex] can only be "up" or "down" (or "alpha"/"beta").

Now, the anti-symmetry of the total wave function can result, for example, from a spatially symmetric wave function with an anti-symmetric spin-wave function: [tex]\Psi(\vec x_1,\vec x_2)=\Phi_+(\vec r_1,\vec r_2)S_-(s_1,s_2)[/tex] where [itex]\Phi_+(\vec r_1,\vec r_2)=\Phi_+(\vec r_2,\vec r_1)[/itex] and [itex]S_-(s_1,s_2)=-S_-(s_2,s_1)[/itex]. This would be a singlet wave function (there is only one way to realize an anti-symmetric spin wave function: [tex]S_-(s_1,s_2)=\frac{1}{\sqrt{2}}\big(A(s_1) B(s_2)-B(s_1)A(s_2)\big),[/tex]

where A(up)=1, A(down)=0, B(up)=0, and B(down)=1).

Or it can result from a anti-symmetric spatial wave function and a symmetric spin wave function (this would be a triplet: there are thee ways to make it: A(s1)*A(s2), B(s1)*B(s2), and (A(s1)*B(s2)+B(s1)*A(s2))).[1] While there are indeed no spin-terms in the Hamiltonian, the two kinds of wave functions also have different spatial parts. This is where the energies of the different spin states come from---it is an indirect effect (note, for example, that in an anti-symmetric spatial wave function the two electrons can never be at the same place, while in an symmetric spatial wave function they can!).

Since fermionic wave functions need to be anti-symmetric, this means, to some degree, that the two electrons are already entangled at the outset. However, if they are far separated, their spin state does not matter because the terms in the Hamiltonian which connect them (the "exchange interaction") depend on the overlap of the wave functions. All the spins functions will lead to the same energy if they are far apart. This is different when they come close together.

So, the exchange interaction and the anti-symmetry of the wave function is always there. It is just so that it can be considered negligible in certain circumstances.

- In Quantum Mechanics, lecture 8, Leonard Susskind said that although entangled, there is nothing Bob can do that changes Alice's density matrix. If there was energy associated with entanglement, then if Bob became disentangled, it would change Alice's density matrix.

(2) plus (3) convinced me that there can not be an energy change associated with enganglement/disentanglement.

- In Quantum Mechanics, Lecture 9, Susskind discussed an example. He started with an unentangled state |ud> and said that it could be expressed as a linear superposition

|ud>= |singlet>+|triplet>. Then the pair could interact with a radiation field with states |0>=no photon, |1>=photon. Start with [|singlet>+|triplet>] ⊗ |0> which then evolves to |singlet | 0> + |singlet | 1> in which the triplet emits a photon and becomes a singlet.

That brings me full circle again, in which the energy of the photon emited by the triplet is an energy delta.

But going back to cgk's Hamiltonian, if there is energy associated with entanglement, why is there not a term for it in the Hamiltonian?

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# Energy delta of entanglement

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