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Dirac Sea of antiparticles

  1. Mar 2, 2012 #1
    I'm not sure if my interpretation is correct, but this Dirac Sea interpretaton does as far as I understand this, tell us that every energy level from -infinity to a certain energy level E<0 is filled with anti-particles. And this should be true for every single location in the universe.

    If this would be true, how is it possible that we can even travel through a vaccuum? I mean, not only the outer space vaccuum but even the vaccuum between two molecules here on Earth. If every single spot containts more than an infinity amount of -for instance- positrons, and these particles have a mass greater than zero, we would have an infinity mass density in every single location, which should make it impossible for any motion.

    So how is motion possible if the dirac sea is true?
     
  2. jcsd
  3. Mar 2, 2012 #2

    bhobba

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    The same way an electric current can pass through a wire when all the electron shells are filled - you simply occupy the lowest energy the sea will allow you to.

    Its not an issue - that theory is long dead being replaced by QFT.

    Thanks
    Bill
     
  4. Mar 3, 2012 #3

    Bill_K

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    Yes, to add emphasis, there is no Dirac Sea. This was an early idea that turned out to be incorrect. They were not antiparticles, they were supposed to be normal particles in negative energy states, and antiparticles were the vacancies ("holes"). But Quantum Field Theory replaced these negative energy states with positive energy antiparticles. (And no, they do not travel backwards in time!)

    The idea of having states fully occupied up to a certain energy, the Fermi level, survives in condensed matter scenarios, e.g. conductors and neutron stars, but negative energy states don't occur in Quantum Field Theory.
     
  5. Mar 3, 2012 #4
    Frankly speaking I never understood how QFT cures the "disease" of Dirac sea completely, I think the infinite ground state energy is the remnant of Dirac sea concept, which doesn't seem to be a much less serious problem.
     
  6. Mar 4, 2012 #5

    bhobba

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    Actually I think its the first indication 'naive' field theory is sick and you really need an EFT with a cut off.

    Thanks
    Bill
     
  7. Mar 5, 2012 #6

    Bill_K

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    An interesting question to ponder is why these issues still come up in the 21st century, since they were resolved like 75 years ago. I think the reason is that most textbooks feel the need to start off with a historical introduction, and so in the first chapter the student is led through all the ancient simple ideas. They still carry appeal just because they are so simple, and the fact that they turned out to be wrong is somehow overlooked.

    I have yet to meet a graduate student who wasn't secretly in love with the old Bohr model of the atom, in which electrons travel in little circular orbits, quantized orbits because an integral number of wavelengths must fit around the circumference. And they become delighted when I suggest they go home and generalize the idea to include elliptical orbits! :smile:

    The "infinite energy of the vacuum" is an aspect of the factor ordering problem that arises in the process of quantization, when you take a classical Hamiltonian and replace all the dynamical variables with operators. Ambiguity arises because the order of the operators is not specified, and this makes a genuine difference. A field is resolved by Fourier transform into an infinite number of harmonic oscillators. Each oscillator has a classical Hamiltonian H = ½(k2 + x2). A straightforward substitution leads to the quantum Hamiltonian H = ½ω(akak* + ak*ak) with eigenvalues (n + ½)ħω. The ½ħω term is the ground state energy, and an infinite number of oscillators leads to an infinite energy of the vacuum.

    (A common claim at this point is that the zero point of energy doesn't matter and we can simply discard the ½ħω. This is also wrong - the zero point of energy does matter!) The correct way is to reorder the operators to H = ωak*ak which has eigenvalues nħω and no vacuum energy.
     
  8. Mar 5, 2012 #7
    How does the zero point of energy matter? In QM, only differences in energy matter. In GM, we have a cosmological constant which can cancel out any zero point energy. In QFT, we renormalize away the infinite energies anyways.
     
  9. Mar 6, 2012 #8

    Bill_K

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    Khashishi, It's true in classical mechanics that the zero point of energy doesn't matter. You say a particle has total energy E = T + V where T is its kinetic energy and V is the potential energy it acquires from being in an external potential V(x), and you can always add an arbitrary constant to V and get the same equations of motion, and it doesn't matter whether the total E is positive or negative.

    This is no longer true in other areas like quantum mechanics for example, where the frequency of the wavefunction is proportional to the energy. Suppose you have a photon with frequency ω. What is its energy? Is it E = ħω? Or is it E = ħω + arb const. It's ħω of course. And the energy of a particle at rest is E = mc2 and the energy density of the electromagnetic field is ½(E2 + B2) and none of these expressions have arbitrary constants in them.

    Especially when you come to general relativity, energy is a quantity that must be properly dealt with. It's the source of the gravitational field (actually the energy density is T00, a component of the stress energy tensor), and it must be positive.

    Similarly the electric charge and current are the source of electromagnetism. Suppose the Dirac sea was real and the vacuum state had a nonzero charge density from all those particles in the Dirac sea. An infinite charge density is a real problem to handle, but here's a further question: are those charges sitting still? Or are they moving. A charge density defines a rest frame, namely the frame in which there is no current.

    If the vacuum state determined a unique rest frame, your Lorentz invariant theory has just lost its Lorentz invariance. Likewise if the vacuum state had a nonzero energy density, is it sitting still? There would have to be a rest frame in which it's sitting still. The vacuum state would not be Lorentz invariant for that reason either. And it must be!
     
  10. Mar 6, 2012 #9
    I may be wrong (I'm trying to play devil's advocate here), but I think in QM and QFT energy is still defined up to a constant.

    If we add a constant in the Lagrangian in QFT, for example, all physical predictions remains the same as all expectation values are unchanged (same in QM).

    Casimir effect too doesn’t depend on the zero point energy, but only to the difference of the vacuum energy inside and outside the conductors plates.

    Even the Dirac sea isn't really a problem: for Lorentz invariance, if all the negative energy states are filled, then the vacuum is invariant (a Lorentz transformation only permutes the negative energy states among themselves). Concerning infinite charge, a photon can't interact with a negative energy state (it can't change this state into another negative energy one as they are all filled) unless it has enough energy to convert a negative energy state into a positive energy one creating then a pair of particle anti-particle.

    In conclusion I think in QFT Dirac sea could be seen as a useless and overcomplicated description of vacuum, and as such ignored even though it could work as well as the usual standard model vacuum.

    In general relativity though things are different as energy is a source of gravity. I think this only is another reason for which QFT can't describe gravity.

    Please show me how I'm wrong :smile:

    Ilm
     
    Last edited: Mar 6, 2012
  11. Mar 7, 2012 #10

    Bill_K

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    Ok! The vacuum state must be Lorentz invariant, do you agree with that? Well the point is, the only vector which is Lorentz invariant is the zero vector. And that fact by itself rules out the Dirac sea.

    You want to include all the negative energy electron states. I'm not sure how to define 'all' but certainly that includes states that are moving. And you say this vacuum state is going to be Lorentz invariant. First, in order for it to be rotationally invariant all the moving electron states must be the same in each spatial direction, so the total current density at any point must add up to zero. The 4-current in the vacuum state can only be something like (Q, 0, 0, 0) where Q is an enormous charge density.

    Now do a Lorentz boost. When you do this the components of the 4-current mix, and therefore in the new rest frame the current density will no longer be zero. Thus in the new frame the vacuum is no longer rotationally invariant. The only way out of the contradiction is to suppose that Q is zero. The 4-current in the vacuum state must be (0, 0, 0, 0), i.e. the zero vector. That rules out the sea entirely.

    Historically the idea of the Dirac sea was discarded when it was realized a) how unnecessarily complicated it was and b) yet how limited it was. It doesn't work at all for bosons, of course. If a bose particle had negative energy states there would be an immediate and catastrophic collapse to the state of lowest energy. Also there's a problem even for fermions, namely that it treats particles and antiparticles in totally different ways. But we now understand they must be treated on an equal footing. Positrons are supposed to be holes in a sea of electrons. Why not the other way around, and explain electrons as holes in the sea of positrons? Can't have it both ways!
     
  12. Mar 8, 2012 #11
    Again, for the sake of discussion I'll try to contradict you ^^

    I agree Dirac sea should be Lorentz invariant.

    Let [itex]b^{\dagger}(p)[/itex] be the creation operator (from the vacuum defined by [itex]b(p)|0\rangle=0[/itex]) of a fermion with momentum [itex]p[/itex] and negative energy [itex]\epsilon=-\sqrt{m^2 + p^2}[/itex]. Then the Dirac sea [itex]|\Omega\rangle[/itex] would be

    [itex]
    |\Omega \rangle = \prod_{p\in ℝ^3}\,b^{\dagger}(p) |0 \rangle\, ,
    [/itex]

    where the product is taken with some arbitrary but fixed ordering to chose the phase.

    If we now perform a (proper orthochronous) Lorentz transformation [itex]\Lambda[/itex]

    [itex]
    |\Omega\rangle \rightarrow \prod_{p\in ℝ^3}\,b^{\dagger}(\Lambda p) |0 \rangle\,,
    [/itex]

    which is a reordering of the product and so differs from [itex]|\Omega\rangle[/itex] for at most a sign*.

    Well, I'm actually really not sure of what i wrote, your argument seems to me pretty convincing xD

    I'm not so sure about this.

    To collapse in the minimum energy state** it's necessary a dynamical process allowing the transition. For example a photon with some energy have an infinity of states with lesser energy, but it can't decay into them unless it's scattering with charged particles.

    Here It seems to me that in some way it's just a matter of convention.

    Choosing the electrons as particles is like choosing a "forward" direction of time to propagate particle. The opposite chose would be the one corresponding to the time reversal of the previous chose.

    Unless we introduce the time reversal violation it seems to me really arbitrary what we chose as the forward direction of time.

    Ilm

    * If we add the spin I think [itex]|\Omega\rangle[/itex] would transform with the direct product of spin [itex]1/2[/itex] representation so we would need to add to his definition a projection opertor on his scalar component?

    ** I think this would be the one with minimum [itex]\epsilon^2[/itex], just as the minimum momentum along say the [itex]z[/itex] axe is [itex]p_z=0[/itex] and not [itex]p_z=-\infty[/itex]. But let's suppose it's different for the energy so that the minimum is [itex]\epsilon=-\infty[/itex].
     
    Last edited: Mar 8, 2012
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