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Superposition of two electrons

  1. Feb 26, 2006 #1
    I have a simple question. In quantum mechanics the superposition principle is given:

    psi=c1*psi1+c2*psi2+...

    Now, is it possible that psi1 is an electron A and psi2 is an electron B? I mean can we bring several electrons in superposition? Couldn't this violate the probability conservation? And can we say psi is a new particle created out of the two other electrons?
     
  2. jcsd
  3. Feb 26, 2006 #2

    ZapperZ

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    First of all, it would be very strange if your psi's represents different electrons IF the system contains only one. Where would the 2nd electron come from if you have, let's say, a hydrogen atom?

    Secondly, in multi-electron system, you can have the indistinguishibility phenomena kicking in, giving you the Fermi-Dirac statistics. In that case, a whole set of new rules comes into play and your total wavefunction must be asymmetric. Look further in your QM text for quantum statistics. You will see that the wavefunction can be in a superpostion, but it isn't the simple linear series the way you have written. In fact, you will see the beginning of 'entangled' states there.

    Zz.
     
  4. Feb 26, 2006 #3
    Superposition principle can occur when there is not interaction between particles. I mean, if in the hamiltonian there are not mixed variables, so you can make the state space as the tensor product of the spaces of individual particles, then you can apply the superposition principle. But if you have some like [tex]\hat S_1 \cdot \hat x_2^2[/tex] (something very strange) then you can´t.
     
  5. Feb 26, 2006 #4
    As I recognised we can't really take a linear combination, as you said ZapperZ. But what is then meant by the different psi's of:

    psi = c1*psi1 + c2*psi2 + ...

    And another question:

    I saw in my textbook that we can describe a many body system like the helium atom with a wave function psi(electron1, electron2). But isn't there a way to make a wave function like psi(WholeState). I mean, my psi(e1, e2)=psi(x1,y1,z1;x2,y2,z2) isn't really the same thing as psi(WS)=(x,y,z).
     
  6. Feb 26, 2006 #5

    ZapperZ

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    It means that a system has a probability to be in state psi1, another probability to be in a state psi2, etc... This is true even for a single particle. This is what Schrodinger was trying to illustrate with his cat.

    Eh?

    Zz.
     
  7. Feb 26, 2006 #6
    Ok, the state psi can be represented as a linear combination of several possible states psi1, psi2. If I understand this correctly.

    With the second question I mean this: let's say we have two free electrons described by the wavefunctions psiE1 and psiE2. Now what happens in nonrelativistic quantum mechanics (not quantum field theory) if the two electron wave functions "come" together? Do we then have an effetive state described by a wave function of the form psi(E1, E2)=psi(x1,y1,z1; x2,y2,z2) or can we build a wave function of the form psi(WholeSystem)=psi(x,y,z). Do you see the difference? I ask, if we can represent the system of the two electrons with a single wave function containing only the position variables x,y,z or can't we do that and have to describe the effective wave function with six position variables (three for each electron).
     
  8. Feb 26, 2006 #7

    ZapperZ

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    You construct something resembling the Slater determinant, and in the case of a 2-particle system, you end with either a single state, or a triplet state, i.e. the system can have two possible configuration depending on how the spins are alligned.

    Zz.
     
  9. Feb 26, 2006 #8
    Ah, thanks, I saw something in my text book called the Slater determinant. Didn't know, that is has something to do with all that. Thanks
     
  10. Feb 26, 2006 #9

    nrqed

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    You *must* have a wavefunction of the form Psi(x1,y1,z1;x2,y2,z2). The meaning of this wavefunction is the following: the square of its magnitude, Psi Psi* represents the probability of finding one of the two particles in a small volume element located at x1, y1,z1 and the other particle located at x2,y2,z2. Since there are two particles, there *must* be be 6 cooordinates. That's the answer to your question.

    Now, going a bit further....The question obviously becomes whether it's possible to write this wavefunction in terms of products of wavefunctions of one particle. In general, no. But if the two particles are not interacting, it is possible, although not in a trivial way. The answer depends on whether the two particles are distinguishable or undistinguishable, and if they are undistinguishable, it depends on whether they are fermions or bosons.

    Patrick
     
  11. Feb 27, 2006 #10
    If system's hamiltonian only contains spatial coordinates (such as {x,y,z} or else) you only need one wavefunction to describe the system. But, if you consider spin, since the total state space is built as [tex]\mathcal E = \mathcal{E}_{spatial} \otimes \mathcal{E}_{spin}[/tex] then, a base may be the tensor product of the basis, so the dimension of the whole space state will be higher.

    In fact, the spin degeneracy is '2s + 1' (for example, the degeneracy for electrons is 2 so we use +/- to distinguish that two different states with the same energy).

    For electrons, (s = 1/2) including the spin makes that we need two wavefunctions to describe the whole system. [tex]\Psi_+[/tex] and [tex]\Psi_-[/tex] or as an spinor [tex]\binom{\Psi_+}{\Psi_-}[/tex].

    This is only for a one-particle system and I didn't say yet nothing about Pauli, but its important to say that not all of the mathematical states are allowed, for fermions we must have antisymmetrical states and for bosons symmetrical states (linear symmetric/antisymmetric combinations of the basis vectors since the eigenfunctions of the whole hamiltonian don't have to have symmetry well defined).

    Being more general, the number of wavefunctions needed is the dimension of the space state, but considering the symmetry we may need less functions
     
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