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P and T reversal symmetry

  1. Aug 3, 2010 #1
    I wanted to understand what symmetries the standard model lagrangian has, and what "effective" (don't know what to call this) symmetries our universe has due to the vacuum state breaking the symmetry.

    Unfortunately, I'm having trouble extracting that information from the Lagrangian of the different interactions.

    What term shows that Parity is violated, and Time reversal is violated? And are these violated directly in the lagrangian, or only due to the non-symmetric vacuum state?
  2. jcsd
  3. Aug 4, 2010 #2
    C, P, and T violation are all in the theory before symmetry breaking. The easiest to recognize is the P violation. The SU(2) part of the electroweak force involves a left chiral projector, [itex]P_L = \frac{1-\gamma_5}{2}[/itex]. Under P, [itex]\gamma_5[/itex] changes sign, switching the projector to a right chiral one. So, this force violates P maximally, since the chiral states are a compete orthonormal basis for spin.

    If we're ignoring symmetry breaking, we can work in the gauge basis for all fermions. In this basis, it becomes clear that charge conjugating the P-conjugated SU(2) interaction returns the original expression. This means that this force also violates C maximally.

    The harder thing to see is the T violation. In fact, you rarely hear it spoken of this way. Usually physicists speak about CP violation. However, in a Lorentz invariant theory (as the standard model is with or without symmetry breaking) the two are the same, as CPT cannot be broken.

    In the standard model, the only source of CP violation is the CKM matrix, which describes the relationship between the gauge and mass eigenstates and usually appears in the SU(2) gauge sector. However, the ultimate source of that matrix is the simultaneous flavor diagonalization of the Yukawa couplings of the quarks to the Higgs. Since this diagonalization may be performed whether or not the Higgs develops a vev, the CP violation must be inherent in the Yukawa couplings themselves. Thus, even if we keep electroweak symmetry unbroken and work in the gauge basis, there will still be CP violating effects. We will simply see them as being in the Higgs sector rather than the gauge one.
  4. Aug 4, 2010 #3
    I thought the gamma matrices were originally chosen to satisfy a certain algebra, which allowed one to write the Dirac equation. Why would a change of handedness of the coordinate system affect, or even be related to, our choice of the gamma matrices at all? (Sorry, I haven't done any particle physics since undergrad and even then I didn't understand it terribly well.)

    To make sure I understand this, let me try playing with some of it.

    from wikipedia
    The charged current part of the Lagrangian is given by
    [tex]\mathcal{L}_C=-\frac g{\sqrt2}\left[\overline u_i\gamma^\mu\frac{1-\gamma^5}2M^{CKM}_{ij}d_j+\overline\nu_i\gamma^\mu\frac{1-\gamma^5}2e_i\right]W_\mu^++h.c.[/tex]

    Charge conjugation will not affect a contracted term like
    [tex]\overline u_i M^{CKM}_{ij} d_j[/tex]
    (or is that a "pseudo-scalar"? how can I tell?), and since
    [tex]W_\mu^+ \rightarrow -W_\mu^+[/tex]
    means the entire term changes sign.

    Am I understanding that correctly?
    What about the gamma matrix here, and what about g?

    I don't really follow this part. If someone just gave me a matrix and said: what does this look like after CP operations? I don't know how to do that. This is probably related to my parity question above. Can someone explain the mechanics of doing such operations in a bit more depth?
  5. Aug 4, 2010 #4
    The Dirac matrices are chosen to satisfy the Clifford algebra, which is necessary to correctly form the Dirac equation. But, what is actually relevant here is that they act like a 4-vector under Lorentz transformations and parity flips. Under parity, the spacial components of any 4-vector change sign and the time component remains the same. If we contract two 4-vectors that transform in this way under parity, we find that the result is parity invariant. Thus, a term like
    [tex]A_\mu \overline{e}\gamma^\mu e,[/tex]
    which, of course, is the standard QED interaction, will be invariant under parity.

    [itex]\gamma_5[/itex] changes things a bit, as it is defined to be [itex]\gamma_5 = i\gamma^0\gamma^1\gamma^2\gamma^3[/itex]. Given the properties of [itex]\gamma^\mu[/itex], we can see that [itex]\gamma_5[/itex] will change sign under parity. Since, in the weak interaction, it multiplies a [itex]\gamma^\mu[/itex], that part of the interaction will differ by an overall sign after a parity flip. The total effect of that is to change the left chiral projector, [itex]P_L = \frac{1-\gamma_5}{2}[/itex], into the right chiral one, [itex]P_R = \frac{1+\gamma_5}{2}[/itex], meaning that the interaction violates P.

    I don't have time right now to go into this much detail on the other two. I'll try to get back to them later. But, I'll point out that the expression you have for the weak force there is in the mass basis, not the gauge basis. If you work in the gauge basis, there is no [itex]M^{CKM}[/itex] in the gauge interaction term. And, charge conjugation does affect the fermions.
  6. Aug 4, 2010 #5
    I think I'm misunderstanding something fairly basic here. So let me try to write my thoughts down and maybe that will point out where my misunderstanding lays.

    Okay, we need the dirac matrices to satisfy the anti-commuting relations:
    [tex]\{ \gamma^\mu , \gamma^\nu \} = \gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu = 2g^{\mu\nu} I[/tex]
    where [itex]I[/itex] is the identify matrix, and [itex]g^{\mu\nu}[/itex] is the minkowski metric.

    There are infinitely many sets of matrices that satisfy these relations. Given a basis, a new basis can be obtained by a similarity transformation:
    [tex]\gamma^{\bar{\mu}} = S^{-1} \gamma^\mu S[/tex]

    Now considering linear combinations:
    [tex]\gamma^{\bar{\mu}} = \Lambda^{\bar{\mu}}{}_{\mu} \gamma^\mu[/tex]
    then enforcing the commutation relation (and remembering that while for a given [itex]\mu[/itex] a [itex]\gamma^\mu[/itex] is a matrix, with a given [itex]\mu,\bar{\mu}[/itex] a [itex]\Lambda^{\bar{\mu}}{}_{\mu}[/itex] is just a number.
    [tex]2g^{\bar{\mu}\bar{\nu}} I =
    \{ \gamma^{\bar{\mu}}, \gamma^{\bar{\nu}} \} =
    \Lambda^{\bar{\mu}}{}_{\mu} \gamma^\mu
    \Lambda^{\bar{\nu}}{}_{\nu} \gamma^\nu
    \Lambda^{\bar{\nu}}{}_{\nu} \gamma^\nu
    \Lambda^{\bar{\mu}}{}_{\mu} \gamma^\mu
    \Lambda^{\bar{\mu}}{}_{\mu} \Lambda^{\bar{\nu}}{}_{\nu}
    \{\gamma^\mu, \gamma^\nu \} = \Lambda^{\bar{\mu}}{}_{\mu} \Lambda^{\bar{\nu}}{}_{\nu} 2g^{\mu\nu} I [/tex]
    So the only linear transformations allowed are those that preserve the minkowski metric. Such as rotations, parity transformations, time reversal, Lorentz boosts.

    Okay, hopefully I understood correctly up to that point. And I learned a little bit about the basis by working that out. But I'm still confused on the next step. We can choose any gamma basis we want, right? Now if I change my coordinate system basis, why should this affect my choice in gamma basis? Maybe I'm fundementally misunderstanding how to think about what these gamma matrices "are".

    Once I understand this requirement between the two basis choices, the rest of your comments on the gamma matrices make sense to me now that I see we can make a new basis with not just unitary transformations, but also linear combinations.

    I'd love to hear more if you have time. Thank you for helping so far.

    Yes, but after contraction, I was hoping it would cancel.
    So the
    [tex]\overline u_i M^{CKM}_{ij} d_j[/tex]
    term is a pseudo-scalar and not a true scalar?

    EDIT: Oh, I see. I can't really pull the gamma matrices out like that. Hmm.
    Last edited: Aug 4, 2010
  7. Aug 5, 2010 #6
    To see that the gamma matrices need to transform under parity as a vector, consider the Dirac equation in momentum space
    [tex]\left(\gamma^\mu p_\mu-m\right)\psi = 0.[/tex]
    [itex]p^\mu[/itex] is a vector and [itex]m[/itex] is a scalar. Thus, for the Dirac equation to remain true under a parity inversion, [itex]\gamma^\mu[/itex] must also transform as a vector. Then, everything else I've said follows.
  8. Aug 9, 2010 #7
    Despite playing with this for awhile now, I am still confused on this point.

    In your argument above, you are implicitly calling [itex]\gamma^\mu p_\mu[/itex] invariant, and a scalar. But it is a matrix. The only matrix that has the same value in all inertial frames is the metric (or matrices proportional to, or formed from the metric such as: [itex]\delta^{\mu}{}_{\nu} = g^{\mu\alpha}g_{\alpha\nu}[/itex]).

    So in what sense can we consider [itex]\gamma^\mu p_\mu[/itex] invariant?


    And still lost on a previous question as well: How does our choice of coordinate system restrict our choice of gamma basis? Isn't the dirac equation just as valid with any choice of gamma basis; so shouldn't it not matter if we don't transform the basis when we change coordinate systems?

    It really looks to me that the gamma basis is a separate choice from the coordinate basis. If not, and they are linked somehow, this seems to be a profound statement on geometry somehow. I'd like to understand this in more depth.
  9. Aug 9, 2010 #8
    I think you may be confusing space-time tensors with spin-space matrices. [itex]\gamma^\mu p_\mu[/itex] is a spin matrix, but must be a space-time scalar for the Dirac equation to not be frame-dependent.

    Consider a Lorentz transformation, [itex]\Lambda[/itex]. The spinor representation is [itex]S(\Lambda)[/itex], and [itex]S(\Lambda)\psi(p^\mu) = \psi'(\Lambda^\mu_{\phantom{\mu}\nu}p^\nu)[/itex]. Then, the Lorentz transformation of the Dirac equation gives
    [tex]\begin{align}S(\Lambda)\left(\gamma^\mu p_\mu - m\right)\psi(p^\mu) &= \left(S(\Lambda)\gamma^\mu S^{-1}(\Lambda)p_\mu - m\right)S(\Lambda)\psi(p^\mu)\\ &= \left(\gamma^\mu\Lambda^\nu_{\phantom{\nu}\mu}p_\nu-m\right)\psi'(\Lambda^\mu_{\phantom{\mu}\nu}p^\nu).\end{align}[/tex]

    As for bases, the choice of coordinate basis in no way restricts the choice of gamma basis. However, under a coordinate change, a given gamma basis will be transformed into a generically different gamma basis in the new frame, as defined by the spinorial Lorentz transformation. Then, the vectorial Lorentz transformation takes you back to your original basis. But, none of this in any way restricts your choice of gamma basis.

    I'm still intending to get back to the other parts of your original question; but, I need a chance to sit down and find a relatively concise way to do it.
  10. Aug 16, 2010 #9
    I'm sorry, I've reread through your message once a day and tried to think it through, but I still don't understand. I must be misunderstanding something spectacularly obvious and simple, but I just can't find it.

    I still don't understand what you mean by scalar in reference to a matrix here. Do you mean the components of that matrix have to be the same in all coordinate systems? It seems like only a lesser constraint of the matrix must result in physically equivalent solutions is necessary.

    To help us find some common terminology so my error can be pointed out easier, let's write out all the indices. I'll use (a,b) to refer to these implied indices. So now we have:
    [tex]\gamma(a,b)^\mu p_\mu[/tex]
    And we have a choice of basis for the gamma's, and I'd like to include that explicitly somehow to aid discussion. So let's define each of [itex]\gamma^0,\gamma^1,\gamma^2,\gamma^3[/itex] to be some constants that satisfy the appropriate algebra: literally constants, the components of the matrices are the same in all coordinate systems. Then we can explicitly include factors for the choice of gamma basis. So the choice of basis comes down to a choice of a unitary matrix I'll call U.
    [tex]U(a,b)\gamma(b,c)^\mu U^{-1}(c,d) p_\mu[/tex]

    Now, from my choice of defining those symbols, the gamma's can't change when we change coordinate system. What you seem to be saying is that for some reason, when I change coordinate system, I must change my choice in U... that somehow my choice of basis is somehow tied to my coordinate choice. But at the same time you say the opposite as well.

    For example:
    What!? I don't understand how those two statements don't conflict.
    Yes, I understand I must be missing something very basic here. I'm hoping someone can help me see, and it will all click.

    From my viewpoint, solving the equation
    [tex](\gamma^\mu p_\mu - m) \phi(p_\mu) = 0[/tex]
    gives the correct physical answers as long as the choice of [itex]\gamma^0,\gamma^1,\gamma^2,\gamma^3[/itex] satisfy the necessary algebra. So if we change coordinate system and don't change our choice of [itex]\gamma^0,\gamma^1,\gamma^2,\gamma^3[/itex] the equation will still give the correct physics. No? What am I missing?

    If I do a parity transformation in my coordinate system, why am I forced to change my basis of gamma? Either I'm free to choose the basis, or I'm not. I really truly don't understand.

    I look forward to that very much. I'm sorry I'm struggling so much, but your help is very much appreciated.
  11. Aug 16, 2010 #10
    I think, perhaps, the piece that you're missing is that the choice of representation for [itex]\gamma^\mu[/itex] also fixes the choice of representation for the spinor. When you change basis, the components of the spinor must rearrange as well. What I've tried to convey above is what happens under a Lorentz transformation with no other change of basis.

    The point that I'm trying to draw out is that a [itex]\gamma[/itex] matrix basis is not frame invariant. Under a boost along the z-axis, for instance, [itex]\gamma^0[/itex] in the boosted frame will be a linear combination of [itex]\gamma^0[/itex] and [itex]\gamma^3[/itex] in the stationary frame. I'm not sure what more I can say to justify this beyond appealing to the need for the Lagrangian of the theory to be Lorentz invariant. At some point you might simply need to work out a simple example (pick a basis, work out the transformation matrix for a boost along one axis, and apply it) to see that this is true.
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