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Spinors and Parity

  1. Sep 25, 2012 #1
    I'm currently reading about parity and it's role in QFT and im trying to understand an argument of why parity exchanges right-handed and left-handed spinors. At page 94 in


    David Tong states that

    "Under parity, the left and right-handed spinors are exchanged. This follows from the
    transformation of the spinors under the Lorentz group. In the chiral representation, we
    saw that the rotation and boost transformations for the Weyl spinors u+- are
    [tex]u_{\pm} \to e^{i \vec \phi \cdot \vec \sigma/2} u_{\pm}, \ \ u_{\pm} \to e^{ \pm i \vec \chi\cdot \vec \sigma/2} u_{\pm}[/tex]
    Under parity, rotations don’t change sign. But boosts do flip sign. This confirms that
    parity exchanges right-handed and left-handed spinors
    [tex] P: u_{\pm} \to u_{\mp}."[/tex]

    I have a few questions to this statement. First of all what does he mean when he say that rotations does not change sign under parity, while boosts does? Any way to see the meaning of this formally (mathematically)?

    And how does this confirm that left-handed spinors are changed to right-handed spinors under parity?
  2. jcsd
  3. Sep 25, 2012 #2


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    The parity operation is a space inversion, x → -x, y → -y, z → -z. Polar vectors are those vectors who components change sign under a space inversion. Examples are position, velocity, momentum, electric field, etc. Axial vectors are those vectors whose components remain unchanged under a space inversion. Examples are angular momentum, magnetic field, etc. The cross product of two polar vectors is an axial vector. And a rotation is described by an axial vector. Thus rotations commute with the parity operation.
  4. Sep 26, 2012 #3
    Which two polar vectors describe the rotation? As far as I know, at least for infinitesimal rotations, a rotation about an axis θ is described by
    [tex] d\vec r = \vec \theta \times \vec r d\theta[/tex]
    and why should not this be a polar vector? Is there any reason that the vector θ should change sign? I mean it's not a function of the vector r like momentum etc.

    And what specifically does one mean when one says that a boost changes sign? That the boost occurs in another direction?

    I still don't see the connection to the conlusion that is drawn, namely that parity changes left-handed to right-handed spinors.
  5. Sep 26, 2012 #4


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    Because it's a cross product.
    It's a position vector.

    Write out the matrices. The parity operation P is spatial inversion:[tex]\left(\begin{array}{cccc}-1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&1\end{array}\right)[/tex]
    A typical spatial rotation R is [tex]\left(\begin{array}{cccc}cos θ&sin θ&0&0\\-sin θ&cos θ&0&0\\0&0&1&0\\0&0&0&1\end{array}\right)[/tex] Write out RP and PR and see that they are the same. Same θ.

    On the other hand, a boost B is [tex]\left(\begin{array}{cccc}cosh χ&0&0&sinh χ\\0&0&0&0\\0&0&0&0\\sinh χ&0&0&cosh χ\end{array}\right)[/tex] Write out BP and PB and see that they are not the same. The sign of χ must also change.
  6. Sep 26, 2012 #5


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    I suspect this is mostly convention. Only bilinear expressions in spinors are observable, like e.g. the expectation value of spin, which has to be even, so that a sign change of the spinor is not observable.
  7. Sep 26, 2012 #6


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    Not the sign of the spinor. The sign of the rotation. Under a spatial reflection, an element of the rotation group maps into itself.
  8. Sep 26, 2012 #7


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    I find the arguments in the link are pretty handwaving. First of all spinors are quantum fields, i.e., fermionic field operators. The parity transformation acts as a unitary operator on these field operators according to the rule
    [tex]\psi(t,\vec{x}) \rightarrow U(P) \psi(t,\vec{x}) U^{\dagger}(P) = \gamma^0 \psi(t,-\vec{x}).[/tex]
    Now by definition [itex]\psi_R[/itex] is the eigenvector of [itex]\gamma^5=\mathrm{i} \gamma^0 \gamma^1 \gamma^2 \gamma^3[/itex] with eigenvalue +1, i.e.,
    [itex]\psi_R=\frac{1+\gamma^5}{2} \psi[/itex]. Acting with the parity operator on this projection yields
    [tex]\psi_R(t,\vec{x}) \rightarrow \gamma^0 \psi_R(t,-\vec{x}).[/tex]
    Now we check for left/righthandedness by acting with [itex]\gamma^5[/itex] on this transformed spinor, leading to
    [tex]\gamma^5 \gamma^0 \psi_R(t,-\vec{x})=-\gamma^0 \gamma^5 \psi_R(t,-\vec{x}) = -\gamma^0 \psi_R(t,-\vec{x}).[/tex]
    This means that the space-reflected spinor field is a eigenvector of [itex]\gamma^5[/itex] to the eigenvalue -1, i.e., a lefthanded field.

    With this of course, you can check the behavior of the 16 bilinear covariant forms [itex]\overline{\psi} \Gamma \psi[/itex] under space reflections, identifying the "currents" with [itex]\Gamma=1[/itex] as scalar, [itex]\Gamma=\mathrm{i} \gamma^5[/itex] as pseudoscalar, [itex]\Gamma=\gamma^{\mu}[/itex] as vector [itex]\Gamma=\gamma^{\mu} \gamma^5[/itex] as axial vector, and [itex]\Gamma=\sigma^{\mu \nu}[/itex] as 2nd-rank tensor.
  9. Sep 26, 2012 #8


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    Again, the essential issue is group theory,not spinors. The original statement was,
    This is nothing more nor less than saying, if a rotation takes rr' then the same rotation takes -r → -r'.
    You don't need to master the behavior of Dirac bilinear covariants to understand that a reflection takes a right-handed object into a left-handed one.
  10. Sep 26, 2012 #9


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    Then you have to explain, why you call the eigenspinors of [itex]\gamma^5[/itex] left or right handed. This is justified by the very fact that parity maps the chirality eigenstates to chirality eigenstates of the opposite chirality.

    I don't see, what this should have to do with the transformation behavior of angular momentum. Maybe you mix up helicity and chirality. It's important to distinguish these two concepts, which are the same only for massless particles!
  11. Sep 26, 2012 #10


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    If you look at the OP's reference, and the quote as given by the OP, Tong is working with the transformation property of the two Weyl spinors, not the bilinears.
    All you have to do is accept the sentence I underlined, and that is what we have been discussing. That's all there is to it!
  12. Sep 26, 2012 #11


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    I also use the spinors, not the bilinears. I've only made this remark, because somebody has brought up the issue of the bilinears (more precisely these are sesquilinear forms of the Dirac spinor).

    To stress it again, and that's important to note, chirality ("handedness") has not a priori something to do with rotations or angular momentum or spin. The property of being an eigenspinor of [itex]\gamma^5[/itex] does not imply a certain helicity (except for massless particles, but in nature there are no known massless Dirac particles!).
  13. Sep 26, 2012 #12
    I have to say it's kind of hard to follow you guys. I'm new to the subject and i'm not learned in group theory and neither do I know what a 'bilinear is'. I'm just trying to understand Tongs argument on why the parity operator changes a right-handed weyl spinor into a left-handed one.

    Back to your answer Bill_K;

    But θ is just the orientation of the axis which you are rotating about, it's not the position of _something_. Isn't parity defined as the operation where you space invert every physical entity your system, an active transformation for which the coordinate system stays unchanged?

    Alright so parity and rotations does not commute, while rotations and boost does, but what does that have to do with the sign change?

    And how does the sign change account for the fact that

    [tex] u_{\pm} \to u_{\mp}[/tex]
    under parity?
  14. Sep 26, 2012 #13
    I found them handwavy as well, that's why I try to make sense of them. Do you have any other recommendations for reading up on this? You say that parity act on the fields according to

    [tex]\psi(t,\vec{x}) \rightarrow U(P) \psi(t,\vec{x}) U^{\dagger}(P) = \gamma^0 \psi(t,-\vec{x}).[/tex]

    but how does one justify this? Furthermore I thought


    was defined as the projected part of a general spinor, i.e.

    [tex] \psi_R = P \psi.[/tex]

    (BTW: How do you just quote one sentence at the time?)
    Last edited: Sep 26, 2012
  15. Oct 2, 2012 #14


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    In Srednicki's text,
    [tex]\psi(t,\vec{x}) \rightarrow U(P) \psi(t,\vec{x}) U^{\dagger}(P) = \gamma^0 \psi(t,-\vec{x})[/tex] is derived by requiring that an electron creation operator [itex]b^\dagger_s(\vec p)[/itex] transform into [itex]b^\dagger_s(-\vec p)[/itex] (that is, same spin, opposite momentum) up to an overall phase factor, and similarly for a positron creation operator.
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