Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Free Electrons In Proximity

  1. Feb 4, 2015 #1

    anorlunda

    User Avatar
    Science Advisor
    Gold Member

    Consider two free electrons passing close to each other. They start in a product state, independent and not entangled. Then they might have an event which emits a photon and fall into a entangled triplet state or a single state, each of which have lower energy than the product state.

    Is the probability of such an event something that we can write analytical QM expressions to calculate, or is it part of the standard model's coupling coefficients that are determined experimentally?

    If analytical, what is a good source for me to study?

    Am I correct in saying that the probability of such an event is a function of the distance between the electrons? Might the probability be a function of the relative velocities, or other factors too?
     
  2. jcsd
  3. Feb 4, 2015 #2

    cgk

    User Avatar
    Science Advisor

    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.
     
  4. Feb 4, 2015 #3

    BruceW

    User Avatar
    Homework Helper

    Interesting question. the electrons repel each other... so they can't form a bound state in the typical sense. But then, it seems like there should be some effect due to the electron degeneracy pressure, even if there are only two electrons... maybe this page will help? http://en.wikipedia.org/wiki/Exchange_interaction
     
  5. Feb 4, 2015 #4

    anorlunda

    User Avatar
    Science Advisor
    Gold Member

    Not dumb cgk, but your Hamiltonian doesn't include the energies in the spins. Also, I think you would need at least two Hamiltonians, one for two electrons in a product state initially, then two entangled electrons plus one photon post-event.

    BruceW, thank you. That Wikipedia article speaks nearly exactly to the point, and it gives Hamiltonians including spin states. (It also refers to Dirac's book as a source.) However, the article considers only the case of two electrons in orbitals around a proton; not free electrons. That suggests two things.

    1) Drop the terms in the Hamiltonian for the proton, giving the free electron case as a simplification. Valid???

    2) A premise in my OP was that these (exchange interaction) events do occur between free electrons. Perhaps that's not true at all. Perhaps it happens only to electrons in the same valence shell.

    I believe that the question remains partially unanswered.

    Hmmm, when one prepares a population of electron pairs in the triplet or singlet state for experimental purpose, how is that preparation accomplished?
     
  6. Feb 5, 2015 #5

    cgk

    User Avatar
    Science Advisor

    anorlunda, the Hamiltonian I wrote down includes both the main component of the energies arising for different spin states and the "exchange interaction" (the exchange interaction is not real---it is a particular way of looking at the Coulomb interaction for systems with anti-symmetric wave functions. See below.).

    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.


    [1] If you make a super-position of both kinds of wave functions, you may get something which is not in an spin-eigenstate (but of course can still be propagated and would be a valid wave function, as long as it is anti-symmetric in total).
     
  7. Feb 5, 2015 #6

    anorlunda

    User Avatar
    Science Advisor
    Gold Member

    CGK,

    I apologize for misreading your first reply. Thanks for the very thorough second reply. It does indeed seem to answer my question, but it will take me some time to digest it.
     
  8. Feb 5, 2015 #7

    BruceW

    User Avatar
    Homework Helper

    I think I understand now... We are constrained in our initial choice of wavefunction (to satisfy spin-statistics theorem). And when the electrons get closer, their potential energy increases more rapidly than the classical Coulomb interaction because of the form of the two-particle wavefunction that we were constrained to. is that about right?
     
  9. Feb 16, 2015 #8

    anorlunda

    User Avatar
    Science Advisor
    Gold Member

    A follow up question please CGK.


    Now I am thinking of a pair of electrons approaching each other, but still separated by distance r. They pass, become entangled, and are now receeding, and once again at distance r. The spin terms should be equally negligable at the approaching r as the receeding r, yet the system was unentangled on approach and entangled on recession.

    I am thinking that the energy of the system as expressed by the Hamiltonian should be different when receeding than approaching. On approach, we have a superposition of symmetric and antisymmetric cases, but when receeding we definitely have either symmetric or antisymmetric.

    I guess I'm having trouble where you said, "The spin of the system is not in the Hamiltonian, but in the wave function." because I want to see the change in energy between the approach and receeding states reflected in the Hamiltonian.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Free Electrons In Proximity
Loading...