Does Electron Entanglement Affect Total Energy?

anorlunda
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Messages
11,326
Reaction score
8,750
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.
  1. 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.
  2. 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 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} and then just using it to propagate your initial wave function via the time-dependent Schrödinger equation i \hbar \partial_t \Psi(\vec x_1, \vec x_2,t) = H \Psi(\vec x_1, \vec x_2,t). 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: \Psi(\vec x_1, \vec x_2)=-\Psi(\vec x_2,\vec x_1), where the x-vectors are the combined spin-space positions: \vec x_i = (\vec r_i, s_i) and s_i 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: \Psi(\vec x_1,\vec x_2)=\Phi_+(\vec r_1,\vec r_2)S_-(s_1,s_2) where \Phi_+(\vec r_1,\vec r_2)=\Phi_+(\vec r_2,\vec r_1) and S_-(s_1,s_2)=-S_-(s_2,s_1). This would be a singlet wave function (there is only one way to realize an anti-symmetric spin wave function: S_-(s_1,s_2)=\frac{1}{\sqrt{2}}\big(A(s_1) B(s_2)-B(s_1)A(s_2)\big),
    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.
  3. 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.
  4. 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.
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!

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?
 
Physics news on Phys.org
So you have a pair of particles in some triplet state.

After a time, the two particle state decays to the lower energy singlet state.

Whether or not the particles are entangled is incidental to the change of energy. You can entangle a pair of particles with no change in energy.

One example of this would be two atoms (elastically) bouncing off of each other (in a closed system) Because the atoms interact with each other, they become entangled.
 
  • Like
Likes bhobba
jfizzix said:
So you have a pair of particles in some triplet state.

After a time, the two particle state decays to the lower energy singlet state.

Whether or not the particles are entangled is incidental to the change of energy. You can entangle a pair of particles with no change in energy.

One example of this would be two atoms (elastically) bouncing off of each other (in a closed system) Because the atoms interact with each other, they become entangled.

Thanks for the reply.

You say that the singlet state is a lower energy than the triplet, but that neither the singlet nor triplet are necessarily lower energy than the product (unentangled) state. That sounds contradictory.

Why would the pair initially in a product state (unentangled) fall into either triplet or singlet unless those were lower energy states?
 
I don't think it's so much falling as it is evolving. The triplet states all have angular momentum 1 and the same energy. Two of those states are separable, and one is entangled. Depending on what sort of forces are driving this system, it can evolve from a separable state to an entangled state and back again without changing energy.
 
  • Like
Likes anorlunda
anorlunda said:
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.
  1. 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.
  2. 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.
  3. 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.
  4. 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.
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!

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?
You have to distinguish between the entangled singlet case(2-dimensional HS) that has no energy associated to the entanglement and the case involving a triplet in which entanglement may imply energy changes/interactions.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!
Back
Top