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I The Relation Between Wavefunctions in Dirac Equation

  1. Nov 27, 2017 #1
    Can the wave function in four dimensions be expressed as e^i(kx+ky+kz-wt)?
     
    Last edited: Nov 27, 2017
  2. jcsd
  3. Nov 27, 2017 #2

    vanhees71

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    This is (nearly) the plane-wave solution for the free Schrödinger equation. The general plane-wave solution is (using natural units ##\hbar=1##)
    $$u_{\vec{k}}(t,\vec{x})=\frac{1}{(2 \pi)^{3/2}} \exp[-\mathrm{i}(\omega t-\vec{k} \cdot \vec{x})], \quad \omega=\frac{\vec{k}^2}{2m}.$$
    It's not representing a pure state of the particle since it's not a square-integrable function, but you can build any true wave-function solution for a free particle via an integral
    $$\psi(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{k} A(\vec{k}) u_{\vec{p}}(t,\vec{x}).$$
    The function ##A(\vec{p})## is given by the initial condition ##\psi(t=0,\vec{x})=\psi_0(\vec{x})##, via the inverse Fourier transformation
    $$A(\vec{p})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{k} u_{\vec{k}}^*(t=0,\vec{x}) \psi_0(\vec{x}).$$
     
  4. Nov 28, 2017 #3
    Alright, thank you. That's helpful.
     
  5. Nov 28, 2017 #4

    dextercioby

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    The Dirac equation for the free particle is solved in many books, I would recommend Albert Messiah's classic from 1962 (English Translation), second volume.
     
  6. Nov 28, 2017 #5

    vanhees71

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    Note however that in relativistic QT, you cannot make sense of the wave function in "1st quantization" as is no problem in non-relativistic physics. The reason is that relativistic QT of interacting particles is automatically a many-body theory with a non-fixed particle number. Old-fashioned books which treat realtivistic wave mechanics as it's successful in the non-relativsitic case for the Schrödinger or Pauli equations, waste your time. The modern treatment of relativistic QT is in terms of QFT, which makes everything much simpler and clearer than the hand-waving 1st-quantization approach which leads to a very cumbersome reinterpretation in terms of Dirac's hole theory, which indeed can be formulated to the end for QED, but it's very cumbersome, and formulating QED (let alone the non-Abelian gauge theory of the rest of the standard model) as a QFT is much more transparent.
     
  7. Nov 29, 2017 #6
    I have been looking at some of the solutions to the Dirac equation and although I understand how the resulting spinors are obtained, i.e. via the matrix multiplication etc., I am however perplexed by what they represent physically. E.g. an up-spin 'particle' spinor can take the form ( 1, 0, p/(E+m), 0)T or something similar. However, what does the momentum - energy term p/(E+m) actually mean when contained in a spinor? Surely its not a probability coefficient of any kind?
     
  8. Nov 29, 2017 #7

    vanhees71

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    It's just the bispinor-valued coefficient of a plane-wave mode of the Dirac equation. When quantizing the field it occurs in the mode deomposition of the Dirac-field operator,
    $$\hat{\psi}(t,x)=\int_{\mathbb{R}^3} {\mathrm{d}^3 \vec{p}}{(2 \pi)^3} \sum_{\sigma= \pm 1/2} [\hat{a}(\vec{p},\sigma) u_{\vec{p},\sigma} \exp(-\mathrm{i} x \cdot p) + \hat{b}^{\dagger}(\vec{p},\sigma) v_{\vec{p},\sigma} \exp(+\mathrm{i} x \cdot p)]_{p^0=+\sqrt{\vec{p}^2+m^2}}.$$
    Here I assumed that the spinors are normalized as
    $$\sum_{\sigma=\pm 1/2} u_{\vec{p},\sigma} \bar{u}_{\vec{p},\sigma}=\gamma_{\mu} p^{\mu} +m,$$
    $$\sum_{\sigma=\pm 1/2} v_{\vec{p},\sigma} \bar{v}_{\vec{p},\sigma}=\gamma_{\mu} p^{\mu} -m,$$
    and then creation and annihilation operators obey the anticommutator relations
    $$\{\hat{a}_{\vec{p},\sigma},\hat{a}^{\dagger}_{\vec{p}',\sigma'} \} = \{\hat{b}_{\vec{p},\sigma},\hat{b}^{\dagger}_{\vec{p}',\sigma'} \}=(2 \pi)^3 \delta^{(3)}(\vec{p}-\vec{p}') \delta_{\sigma \sigma'}$$
    with all other combinations in the anticommutators vanishing.

    The ##\hat{a}_{\vec{p},\sigma}## (##\hat{b}_{\vec{p},\sigma}##) annihilate a particle (antiparticle) of momentum ##\vec{p}## with the spin-##z## component being ##\sigma## in its rest frame.
     
  9. Nov 30, 2017 #8
    Thank you Vanhees, I am familiar with the expressions above; the first being the Fourier expansion in terms of the annihilation and creation operators for particles and antiparticles. The second, normalization expressions giving the roots of the K.G Equation. However, your answer is purely mathematical. I still don't get what it means Physically for a bi-spinor to contain the p/(E+p) terms. For example, without evoking Q.F.T., how would Dirac have interpreted these terms?
     
  10. Nov 30, 2017 #9
    I don't know if $$p/(E+m)$$ has some particular physical meaning. Exactly, how your spinor look like depends on your chosen normalization and also on the adopted representation of the gamma matrices. Different representation can be more or less convenient for particular applications. The one you refer to is for example convenient then one wants to an expansion in $$p/m$$, i.e. a non-relativistic reduction.
     
  11. Nov 30, 2017 #10
    Okay, thank you very much for that.
     
  12. Dec 1, 2017 #11

    vanhees71

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    Well, there is some physics meaning in the ##u##'s and ##v##'s. The point is that spin is not as simple in relativistic as in non-relativistic physics. The reason is the structure of the Poincare group in comparison to the Galileo group. In relativistic physics, for massive particles, the usual meaning of the spin is defined in the restframe of the particle, i.e., it defines how the zero-momentum single-particle states transform under rotations. In the case of Dirac particles, that's of course the ##\mathrm{SU}(2)## fundamental representation, SU(2) being the covering group of the "classical rotation group", which mathematically is SO(3). So if you say, an electron has a certain spin-##z## component ##m \in \{1/2,-1/2\}##, it refers to the component of the total angular momentum in the restframe of the electron. Then you get the states for non-zero momentum by applying the boosts in the considered representation, which in the case of Dirac particles is ##(1/2,0) \oplus (0,1/2)##, which is an irreducible representation for the orthochronous Lorentz group but a reducible representation of the proper orthochronous Lorentz group. The reason is that you want a field with spin 1/2 which leads to a representation of the Poincare group that also admits spatial reflections. The two components referring to the irreducible reps. of the proper orthochronous group are Weyl or chiral fermions. This becomes important for the treatment of the weak interaction in the Standard Model, where the left-handed parts of fermions and quarks are isodoublets but the right-handed parts singulets, i.e., the weak interaction only couples to the left-handed parts of the Dirac particles in the Standard model (and for the quarks also to the flavor rather than the mass eigenstates).

    For details on the representation theory of the Poincare group, see Weinberg, Quantum Theory of Fields, Vol. 1 or Appendix B of my notes:

    https://th.physik.uni-frankfurt.de/~hees/publ/lect.pdf
     
  13. Dec 1, 2017 #12
    Thanks for taking the trouble to reply to me. I can understand some of that but I will read further as you suggest. Thank you.
     
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