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Feynamn-Stuckelberg Interpretation of E<0

  1. Nov 12, 2015 #1
    Hi,
    I dont understand the F-S interpretation of the E<0 energys of the Dirac Equation. I´m mainly reading about this in Francis Halzen´s book: Quarks and Leptons sec. 3.5. The book states:
    And then says:
    [tex] e^{-i(-E)(-t)} = e^{-iEt} [/tex]
    Is he referring to the operator that evolves H eigenstates in time:
    [tex]U= I-i\frac{H}{ħ}\Delta t[/tex]
    So to evolve backwards in time am I asuming [tex]U= I+i\frac{H}{ħ}\Delta t[/tex]
    Is he adding a minus to time? Can you even do this?
    I dont understand!!
     
    Last edited: Nov 12, 2015
  2. jcsd
  3. Nov 12, 2015 #2

    DrClaude

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    For a time-independent Hamiltonian, you have ##\hat{U}(t,t') = e^{-i \hat{H} (t' -t) / \hbar}##, such that
    $$
    \hat{U}(t,t') \psi(t) = \psi(t')
    $$
    There is nothing said about the order of ##t## and ##t'##. So yes, changing the sign of ##t## in ##e^{-i E t}## is equivalent to evolving the system backwards in time.
     
  4. Nov 12, 2015 #3
    So, for the sake of physical coherence, he is forcing E<0 solutions to go back in time (t→-t) , which is the same as saying: Its a E>0 solution going foward in time??
     
  5. Nov 12, 2015 #4

    vanhees71

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    That's no surprise. I don't know of any other example in physics, where a pretty simple thing is made complicated by referring to esoterics rather than maths, than the Stueckelberg trick to interpret the negative-frequency modes in relativistic quantum theory.

    The trick consists in recognizing that an interpretation of relativistic wave equations in terms of single-particle wave functions is inconsistent with the physics described when it comes to collisions of particles at relativistic energies, because then you always may create new particles and/or destroy particles in the initial state. That's why it is most convenient to formulate relatistic quantum theory in terms of quantum field theory, i.e., a many-body theory from the very beginning.

    Then the Dirac equation for free particles becomes an equation for a fermionic field operator (fermionic, because only with canonical equal-time anticommutation relations you get an energy spectrum that bounded from below and thus only then a stable groundstate exists; this is an example for the famous spin-statistics theorem which says that for relativistic particles you must necessarily quantize the corresponding fields as bosons (fermions) if the fields have integer (half-integer) spin).

    Now you can expand the field operator in terms of energy-momentum eigenmodes of the Dirac equation. The only "tricky" thing is to write down annihilation operators in front of the positive-frequency modes and creation operators in front of the negative-frequency modes. Then you have for the field operator
    $$\hat{\psi}(t,\vec{x})=\sum_{\sigma \in \{-1/2,1/2\}} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \frac{1}{\sqrt{(2 \pi)^3 2 E_{\vec{p}}}} \left [\hat{a}(\vec{p},\sigma) \exp(-\mathrm{i} E_{\vec{p}} t+\mathrm{i}\vec{p} \cdot \vec{x}) + \hat{b}^{\dagger}(\vec{p},\sigma) \exp(+\mathrm{i} E_{\vec{p}}-\mathrm{i} \vec{p} \cdot \vec{x}) \right ].$$
    So you have two sorts of particles of the same mass, with annihilation operators ##\hat{a}## and ##\hat{b}##. All energy eigenvalues are positive ##E(\vec{p})=\sqrt{m^2+\vec{p}^2}##. As it turns out, the two sorts of particles have the same mass, positive energy (as it must be for a theory with stable ground state), and opposite charges. After normal ordering, energy, momentum, and charge are given by the operators
    $$\hat{H}=\sum_{\sigma} \mathrm{d}^3 \vec{p} E(\vec{p})[\hat{N}_a(\vec{p},\sigma)+\hat{N}_b(\vec{p},\sigma)],$$
    $$\hat{\vec{P}} = \sum_{\sigma} \mathrm{d}^3 \vec{p} \vec{p} [\hat{N}_a(\vec{p},\sigma)+\hat{N}_b(\vec{p},\sigma)],$$
    $$\hat{Q} = \sum_{\sigma} \mathrm{d}^3 \vec{p} [\hat{N}_a(\vec{p},\sigma)-\hat{N}_b(\vec{p},\sigma)],$$
    where the occupation-number operators are given by
    $$\hat{N}_a(\vec{p},\sigma)=\hat{a}^{\dagger}(\vec{p},\sigma)\hat{a}(\vec{p},\sigma), \quad \hat{N}_b(\vec{p},\sigma)=\hat{b}^{\dagger}(\vec{p},\sigma) \hat{b}(\vec{p},\sigma).$$
    The Dirac field thus always describes particles and their antiparticles.

    The Stueckelberg interpretation of the negative-frequency modes is thus precisely to avoid esoteric ideas about things going "backwards in time". This doesn't make sense. The only idea is to introduce an creation instead of an annihilation operator in front of the negative-frequency mode. So the other sign in the exponential is just right to describe a single-particle creation instead of an annihilation process when the field operator is applied to any state in Fock space.
     
  6. Nov 12, 2015 #5
    I know nothing about quantum field theory. For know I´ll accept that Dirac´s single particle descrption is 'wrong', at least to treat collisions, and that the'going backwards in time' interpretation its just a way of solving this inconsistency.

    The Lagrangian approach has more to do with quantum field theory? Cause I´m reading about it, but I´m only using it as a way to obtain Dirac´s equation. Yet when they (the class I´m in) introduce Feynamn´s Diagrams they use the Lagrangian formulation, this is, the interaction term of the Lagrangian instead Dirac, although they don´t fully explain us why. All of a sudden Dirac's solutions account for two different particles and I have all this kind of rules on how to write the interactions between particles.
    Any good book to recommend?

    Thanks for the replies, I now know I know nothing.
     
  7. Nov 12, 2015 #6

    bhobba

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  8. Nov 12, 2015 #7
    Is it not true that some authors regard normal ordering as mathematically suspect? IOW, a dodge to force physically realistic results from the equations? Sorry if this is considered thread drift.
     
  9. Nov 12, 2015 #8

    vanhees71

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    Be warned that sometimes textbooks with a lot of formulae seem to be awkward to read on an ebook format (except pdf's, which are of course good), because it seems as if many ebook formats can't handle the formulae correctly, although I don't own a kindle and read ebooks only on my Android tablet, and I don't have this specific book as an ebook.
     
  10. Nov 12, 2015 #9

    vanhees71

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    Normal ordering has troubles with gauge theories, if not done in a gauge covariant way. Of course, you can also just omit normal ordering and live with the trivial "tadpole" divergences in the Feynman rules. Then you have to renormalize them as any divergent loop diagram.

    On the other hand when working with free particles (which you also need in the interaction picture for interacting particles) there is not more math obscurity with normal ordering as with the entire QFT business, which is anyway up to now not mathematically rigorously definable (for physically relevant models in 4 spacetime dimensions). For the free-field observables, normal ordering is thus fine within the physicists' robust maths used in perturbative evaluations of QFTs. Here it has the advantage of giving the correct finite operators for energy, momentum, angular momentum, boost operators, and the charge operator, where you can read off the signs of the various contributions from their expressions in terms of the occupation-number operators (which are always positive semidefinite).
     
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