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A Why must a polar vector parameterized by spinor be timelike?

  1. May 4, 2016 #1
    I heard (somebody told me and I also read from some paper) that a polar vector whose components are parameterized by the Dirac spinor [itex]\bar\psi\gamma^\mu\psi[/itex] must be a timelike vector. Why is so? I think a general polar vector can either be timelike or spacelike, isn't it? Is that because a spinor parameterized polar vector is not the most general polar vector, that is, [itex]\bar\psi\gamma^\mu\psi[/itex] can only parameterize a timelike polar vector or what?

    So, in contrast, must a spinor parameterized axial vector (pseudovector) [itex]\bar\psi\gamma^\mu\gamma\psi[/itex] be spacelike? Why is so?

    P.S. [itex]\psi[/itex] denotes the components of a Dirac spinor expressed in terms of a column matrix, [itex]\bar{\psi}=\psi^\dagger\gamma^0[/itex] is the adjoint spinor to [itex]\psi[/itex], [itex]\gamma^\mu[/itex] denote the Dirac matrices, and [itex]\gamma=\gamma^0\gamma^1\gamma^2\gamma^3[/itex].

    In addition, please help me confirm the following. Does the notion that under spatial inversion the components of a polar vector change a sign while the components of an axial vector (a pseudovector) keep invariant only apply to Riemannian geometry? Because it's like in pseudo-Riemannian geometry, such as Minkowski space, under spatial inversion the spatial components of a vector change a sign while the temporal component of it keeps invariant.
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  3. May 9, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
  4. May 9, 2016 #3


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    No idea about the topic, but maybe the quantum mechanics forum works better.
  5. May 23, 2016 #4
    I recently got a clue to my question. Is that because the Noether current of the Dirac Lagrangian [itex]\mathcal{L}=\frac{\mathrm{i}}{2}[\bar{\psi}\gamma^\nu\partial_\nu\psi-\partial_\nu\bar{\psi}\gamma^\nu\psi]-\mathrm{m}\bar{\psi}\psi[/itex] under a U(1) gauge transformation is [itex]J^\mu=-\bar{\psi}\gamma^u\psi[/itex] and thus identifies [itex]\bar{\psi}\gamma^u\psi[/itex] as being ``proportional" to the quantum probability current 4-vector of a charged spin 1/2 field? Because the classical electric current 4-vector [itex]j[/itex] is equal to [itex]\rho u[/itex] ( [itex]\rho[/itex] is the charge density per unit volume) and [itex]u=\frac{dx}{d\tau}[/itex], (with [itex]x[/itex] being the worldline of the charged fermions and [itex]\tau[/itex] being the proper time of the fermions) the velocity 4-vector of the fermions, must be timelike, [itex]j[/itex] must be timelike. And the quantum probability current 4-vector should be consistent with the classical electric current 4-vector, so [itex]\bar{\psi}\gamma^u\psi[/itex] must be timelike. As a result, though a general polar vector can be timelike or spacelike or even lightlike, [itex]\bar{\psi}\gamma^u\psi[/itex] only parameterize a timelike polar vector and thus are not the most general parameterization of a polar vector. Am I right?

    As for a spinor parameterized axial vector [itex]\bar{\psi}\gamma^\mu\gamma\psi[/itex], I still have no clue about whether it's spacelike or timelike or generic (can be spacelike or timelike or lightlike).

    Is there anybody knowing about this who can answer me? I guess people researching on Quantum Field Theory should know this.
    Last edited: May 23, 2016
  6. May 23, 2016 #5


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    The vector you are talking about is the charge current and therefore it must be timelike. Indeed, it can be proved that it is timelike by using the Dirac equation and the fact that the mass ##m## is a real (not imaginary) number.
  7. May 25, 2016 #6
    Really? That [itex]\bar{\psi}\gamma^\mu\psi[/itex] is timelike can be proved directly by using the Dirac equation? I can't see this after some trials. I can only see the Dirac equation [itex]\mathrm{i}\gamma^\nu\frac{\partial\bar{\psi}}{\partial x^\nu}+\mathrm{m}\bar{\psi}=0, -\mathrm{i}\gamma^\nu\frac{\partial\psi}{\partial x^\nu}+\mathrm{m}\psi=0[/itex] results in [itex]\frac{\partial(\bar{\psi}\gamma^\mu\psi)}{\partial x^\mu}=0[/itex], which I don't think entails [itex]\bar{\psi}\gamma^\mu\psi[/itex] to be timelike. I can't directly prove [itex]\bar{\psi}\gamma^0\psi>\sum_{k=0}^3\bar{\psi}\gamma^k\psi[/itex].

    I can only make sense that [itex]\bar{\psi}\gamma^\mu\psi[/itex] is timelike. I just think the Dirac equation describes the wave of a particle with mass m and the worldline trajectory of any particle with nonvanishing mass must be timelike. [itex]\bar{\psi}\gamma^\mu\psi[/itex] gives the probability current of the wave and should reflect the real particle trajectory so should be timelike. Further, I think a wave satisfying the Dirac equation must satisfy
    the Klein-Gordon equation [itex](\frac{\partial^2}{c^2\partial t^2}-\nabla^2)\psi+\frac{\mathrm{m}^2c^2}{\hbar}\psi=0[/itex], which is in accord with [itex]E^2=\mathrm{m}^2c^4+c^2\textbf{p}^2[/itex], which is only satisfied by the energy-momentum of a relativistic particle with nonvanishing mass, so the worldline trajectory of the described particle must be timelike.

    On the other hand, I think by the same argument the probability current [itex]j^\mu=\frac{\mathrm{i}\hbar}{2\mathrm{m}}(\phi^*\partial^\mu\phi-\partial^\mu\phi^*\phi)[/itex] of the wave satisfying the Klein-Gordon equation should also be timelike, right? But I do not know how to prove it directly, either.

    So can you give me further clue on how to prove it directly? I can't see where in the proof that m is real is invoked. Thank you.
    Last edited: May 25, 2016
  8. May 26, 2016 #7
    I think the question you are asking is, how/why the Noether current is timelike vector. You have already answered this question by looking at the classical picture and drawing an analogy. Although I, too, would love to know why [itex]j^\mu j_\mu > 0[/itex] (in appropriate signature) mathematically.
    Last edited: May 26, 2016
  9. May 27, 2016 #8


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    Well, the proof is quite difficult:

    You cannot prove it, because in the Klein-Gordon case it is not true.
  10. May 27, 2016 #9


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    It's not too hard to prove for free-particle plane-wave solutions of the Dirac equation. Once you include the electromagnetic field, I'm not sure how much more complicated it gets.

    A plane-wave solution to the Dirac equation [itex](-i \alpha \cdot \nabla + \beta m) \Psi = i \frac{\partial}{\partial t} \Psi[/itex] is (using the most common representation of the Dirac matrices):

    [itex]\Psi = e^{i \vec{p} \cdot \vec{x} - i E t} \left(\begin{array}\\ \chi \\ \frac{\vec{\sigma} \cdot \vec{p}}{E+m} \chi \end{array} \right)[/itex]

    where [itex]\chi[/itex] is a two-component spinor obeying the Klein-Gordon equation and where [itex]\vec{\sigma}[/itex] is the Pauli spin matrices. Then

    [itex]\bar{\Psi} \gamma^0 \Psi = \Psi^\dagger \Psi = \chi^\dagger (1 + \frac{(\vec{\sigma}\cdot p)^2}{(E+m)^2)} \chi = \chi^\dagger \chi \frac{2E}{E+m}[/itex] (which uses facts about spin matrices, and the fact that [itex]p^2 + m^2 = E^2[/itex])

    [itex]\bar{\Psi} \gamma^j \Psi = \Psi^\dagger \alpha_j \Psi = \chi^\dagger (\sigma_j \frac{\vec{\sigma} \cdot \vec{p}}{E+m} + \frac{\vec{\sigma} \cdot \vec{p}}{E+m} \sigma_j) \chi = \frac{2 p_j}{E+m} \chi^\dagger \chi[/itex]

    So [itex]\bar{\Psi}\gamma^\mu \Psi = \frac{2 P^\mu}{E+m} \chi^\dagger \chi[/itex]

    which is clearly a time-like 4-vector. It's square is [itex] \frac{4 m^2}{(E+m)^2} (\chi^\dagger \chi)^2 > 0[/itex]
  11. May 30, 2016 #10


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    The hard problem, of course, is to prove it for arbitrary superpositions of plane wave solutions.
  12. May 30, 2016 #11


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    Yes, that looks messy, unless there is a clever way to do it.
  13. May 30, 2016 #12
    Not true. My original question is to ask why [itex]\bar{\psi}\gamma^\mu\psi[/itex] is timelike. When I inquired this, I didn't think of invoking that [itex]\bar{\psi}\gamma^\mu\psi[/itex] is involved with a Noether current though I have ever read about the latter in several different places.

    Some time ago I asked my graduate advisor why he needn't constrain [itex]\bar{\psi}\gamma^\mu\psi[/itex] to be timelike in one of his proofs which requires the vector to be timelike to get the result in his paper, then he replied me [itex]\bar{\psi}\gamma^\mu\psi[/itex] has already been timelike. I kept keeping his answer in doubt until afterwards I read that [itex]\bar{\psi}\gamma^\mu\psi[/itex] is timelike in a paper which uses spinor to work on something in physics. Recently I also read in a literature about spinor which says [itex]\bar{\psi}\gamma^\mu\psi[/itex] is timelike. Then I started to dispel away my doubt and believe that [itex]\bar{\psi}\gamma^\mu\psi[/itex] is timelike, but I kept wondering why because I think a generic polar vector can be either timelike or spacelike or lightlike. This question has haunted my mind several times but I kept not being able to figure out the answer.

    On the other hand, I doubt all Noether currents are timelike. I guess the Noether current which is interpreted as the energy-momentum (which is not a tensor in the case of gravity, meaning it's coordinate or gauge dependent) of the gravitational field in General Relativity is not necessarily timelike because that the speed of light is the biggest reachable speed doesn't apply to spacetime, the background where physical events take place.
    Last edited: May 30, 2016
  14. May 30, 2016 #13
    Apparently that is true (especially with a counter example mentioned in this very thread :wink:). It seems that the equation of motion has some role to play in deciding the nature of the 4-vector. Currently, I am trying to prove it in a general sense i.e given a Lagrangian and thus the on-shell constraint, how can we decide the nature of the current.

    The reason I am stressing on the current formalism is because we also need to find the nature of [itex]\bar{\psi}\gamma^\mu\gamma\psi[/itex] (although stevendaryl's approach should answer that too).
  15. May 31, 2016 #14


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    They are certainly not. For example, the Dirac current for imaginary ##m## is spacelike. Or to take a more physical example, the Klein-Gordon current (with positive ##m^2##) is not timelike at some points when you take a superposition of different frequencies, even if all frequencies are positive.
    Last edited: May 31, 2016
  16. Jun 16, 2016 #15
    This paper, at first glance, looks very intriguing, but as I read it in detail, I found it is so hard to understand its derivations because it uses the technique of differential topology, of which I don't have systematical knowledge. I'm not sure why the paper keeps stressing that set transversal to S is open and dense with respect to some smooth weak topology. Is this what explains the probability that a Bohmian electron will always spend time in that transversal set so that the probability it touches S (when its speed is the speed of light) is zero? It's like the paper mainly uses differential topology to prove a Bohmian electron can never reach the speed of light. It doesn't invoke that the mass m is real at all.

    I've heard, in a string theory book, tachyon is a hypothetical particle with imaginary mass. So is the current associated with tachyons an example of the spacelike Dirac current you talk about? If so, then the spin of tachyon should be 1/2. Tachyon is a quantum particle so its property can't be envisaged by the classical view--there is no corresponding imaginary mass in classical mechanics to provide intuition. But, in the latter example, is it possible that a frequency is negative? Is it also a whim in quantum field theory?
  17. Jun 16, 2016 #16
    I wonder, given a manifold which has spin structures so that we can construct a spinor bundle on it, whether any spinor field on the bundle satisfies the Dirac equation. I don't think the answer is yes, because if that's the case, the Dirac equation is an identity, instead of an equation which needs to be solved for the spinor ##\psi(x)##. Then for any spinor field which doesn't satisfy the Dirac equation, [itex]\bar{\psi}\gamma^\mu\psi[/itex] isn't necessarily timelike?

    Does [itex]\bar{\psi}\gamma^\mu\gamma\psi[/itex] present a Noether current? If so, what is the Lagrangian giving this Noether current?
    Maybe [itex]\mathcal{L}=\frac{\mathrm{i}}{2}[\bar{\psi}\gamma^\nu\gamma\partial_\nu\psi-\partial_\nu\bar{\psi}\gamma^\nu\gamma\psi]-\mathrm{m}\bar{\psi}\psi[/itex], which gives the Euler-Lagrangian equations [itex]\mathrm{i}\gamma^\nu\gamma\frac{\partial\bar{\psi}}{\partial x^\nu}+\mathrm{m}\bar{\psi}=0, -\mathrm{i}\gamma^\nu\gamma\frac{\partial\psi}{\partial x^\nu}+\mathrm{m}\psi=0[/itex], which result in [itex]\frac{\partial(\bar{\psi}\gamma^\mu\gamma\psi)}{\partial x^\mu}=0[/itex]. But is there such a Lagrangian [itex]\mathcal{L}=\frac{\mathrm{i}}{2}[\bar{\psi}\gamma^\nu\gamma\partial_\nu\psi-\partial_\nu\bar{\psi}\gamma^\nu\gamma\psi]-\mathrm{m}\bar{\psi}\psi[/itex] in literature and is there actually such particles in nature whose corresponding field is described by it?
  18. Jun 16, 2016 #17


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    Yes it does. Page 13, the beginning of third paragraph:
    "Another remark concerns the mass m that appears in the Dirac equation. It played no role in our reasoning whether m>0 or m=0."

    The string-theory tachyon has spin zero and is not described by the Dirac equation.
  19. Jun 16, 2016 #18
    I am sorry I don't understand that completely. I leave it to the experts here.

    Yes. In literature it corresponds to the axial symmetry. See equation 4.101 in http://www.damtp.cam.ac.uk/user/tong/qft/four.pdf. The current is conserved for zero mass.

    That is a very interesting Lagrangian and the equations of motion (there is typo in equation corresponding to [itex]\bar{\psi}[/itex]) certainly imply [itex]\frac{\partial(\bar{\psi}\gamma^\mu\gamma\psi)}{\partial x^\mu}=0[/itex]. It is certainly a Noether current corresponding to the (global) symmetry [itex]\psi\to e^{-i\alpha}\psi[/itex] which is conserved irrespective of the mass. I have not seen such Lagrangian in literature (but that doesn't mean that it do not exist).

    Just for the information, it is true in Bosonic string theory. Once you include supersymmetry, the tachyons disappear from the spectrum of the theory.
    Last edited: Jun 16, 2016
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