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Parity Operations and CPT Theorem

  1. Nov 6, 2013 #1
    Hey all,

    I have a four part question:

    1. The problem statement, all variables and given/known data

    Let ψ represent a wave function where x,y,z are spatial coordinates and t is time. The particles [itex]\pi^{-}[/itex], [itex]\pi^{0}[/itex], [itex]\pi^{+}[/itex] are pions ([itex]\pi[/itex] mesons). The parity inversion operation is represented by

    3. The attempt at a solution

    Parities involve a simple change in sign with regards to the components.

    ∴Pψ(x,y,z,t)=ψ(-x.-y,-z,-t)

    Part B

    1. The problem statement, all variables and given/known data

    The time reversal of above qs is represent by?

    3. The attempt at a solution

    I am not sure if this is inversion of the components (xyzt) or inverting the sign of the pi mesons. My answer is but not sure:

    Tψ(x,y,z,t)=ψ(-x.-y,-z,t)

    Part C

    1. The problem statement, all variables and given/known data

    For first question, charge conjugation is what?

    3. The attempt at a solution

    C[itex]\pi^{-}[/itex] = [itex]\pi^{+}[/itex], C[itex]\pi^{+}[/itex]=[itex]\pi^{-}[/itex], C[itex]\pi^{0}[/itex] = [itex]\pi^{0}[/itex]

    This seems straight forward but maybe too straight forward?

    Part D

    1. The problem statement, all variables and given/known data

    According to the CPT theorem, if P is violated in an experiment and T is not, then we know what?

    3. The attempt at a solution

    Since CP are always grouped, the answer would be:

    C is also violated?

    Help anyone,

    Thanks
     
  2. jcsd
  3. Nov 6, 2013 #2

    haruspex

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    I really am not qualified to write on this topic, but here are some thoughts.
    Pψ(x,y,z,t)=ψ(-x,-y,-z,-t) looks like PT to me. Why isn't it just Pψ(x,y,z,t)=ψ(-x,-y,-z,t)?
    Similarly, why isn't Tψ(x,y,z,t)=ψ(x,y,z,-t). Maybe I'm being naive.
    As I understand it, the CPT theorem says that if a certain interaction occurs then the interaction derived from reversing all of C, P and T also occurs. I.e. you cannot violate all of C, P and T at once. That leads me to a different conclusion from yours.
     
  4. Nov 6, 2013 #3
    Hi haruspex,

    Thanks for the insight in parity and time inversion operators. Parity inverts spatial co-ordinates and time inverts time co-ordinate so you are correct.

    In relation to CPT, can you please elaborate? I know that Cobalt (60) is a proof of weak interactions being able to violate parity. Therefore it leads me to believe my original conclusion stands. What do you think?
     
  5. Nov 6, 2013 #4

    vanhees71

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    If this is a physics question, then your idea about time-reversal is not correct. In quantum theory the time-reversal operation is necessarily realized as an antiunitary transformation, i.e., if formulated in terms of the wave function, it involves complex conjugation, i.e., for a scalar particle
    [tex]\psi(t,\vec{x}) \stackrel{\hat{P}}{\rightarrow} \psi^*(-t,\vec{x}).[/tex]
     
  6. Nov 6, 2013 #5
    Hi vanhees71,

    Yes this is a physics physics problem.

    Are you inverting the time component by applying Parity operator? If so then doesn't Parity on a wavefunction invert spatial co-ordinates and leave time same? Similarly the time reversal operator T, inverts time but not spatial co-ordinates. Correct?
     
  7. Nov 6, 2013 #6

    haruspex

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    Thanks vanhees for jumping in with the antiunitary bit. I said I wasn't qualified.
    My understanding of the symmetry violations is this:
    It used to be thought all was symmetric. E.g. if a certain interaction could occur, then it could occur equally well with all the charges flipped, but all else the same. This was found to be false, but it still seemed to be the case that if you flipped both C and P then the corresponding interaction happened as easily. Then that was found to be false. The theory today is that the only combination of reversals that is guaranteed to produce a symmetry is flipping all three.
    But I struggle to apply this here. What does it mean to say that P is violated and T is not in an experiment? Does it mean that if I flip P only I get a different result, but if I flip T only I get the same result? Does it say anything about flipping P and T together?
    My gut feel is that it is a trick question and the answer will be that it does not tell you anything about C violation.
     
  8. Nov 6, 2013 #7
    Hi Vanhees,

    The answer I believe will be one of the two:

    either C is also violated (due to P violation),
    CPT is all together violated.

    The question may be a trick question. However, I don't see the answer being the latter option as in the question it says T is not violated. If experiment was in fact run and the process was observed then how can all of CPT violate? Isn't that impossible for an observed process? Additionally, I believe T can be regarded separately but CP cannot (ie. you cannot isolate P from C vice versa) therefore if P violates then so does C. Correct?
     
  9. Nov 6, 2013 #8

    haruspex

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    The issue is this: can P be violated but neither C nor T?
     
  10. Nov 6, 2013 #9
    Hi haruspex,

    I think the Cobalt (60) experiment's sucess was for this reason; it was possible to violate P for an occurring process. This resulted in left and right handed night of certain particles such as neutrinos (known as helicity). The experiment proved weak interactions to occur whilst violating P (Parity).

    The next question, can P be violated resulting in consequent C violation? Well a brief search has led me to the following file:

    https://www.kvi.nl/~scholten/TV-lectures/OsakaTV2_RGET.pdf

    See page (slide) 19 in which is makes mention of the Parity of Cobalt (60) being violated and then an arrow sign (which I take as thus/therefore/consequently/which leads to) saying charge is also violated fully.

    From this we can establish that violating P also violates C which together is known as CP violation.

    The next, next question, can CP violate without T violating? I think it can but I shall leave that up for discussion.
     
  11. Nov 6, 2013 #10

    haruspex

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    No, that's wrong. P violation occurs if you only invert P and things change. Likewise C violation. If you invert both, symmetry may be observed (no CP violation), or it might not be (CP violation). The CPT theorem says that if you invert all three then symmetry will be observed; and that is all it says.
    So violation of any one is equivalent to violation of the other two simultaneously.
    That said, I have come to the conclusion your original answer is correct, though maybe not for the reasons you offered. Consider this table:
    Code (Text):

    C  P  T  Result
    0  0  0  X
    1  1  1  X (CPT)
    0  1  0  Y (P violation)
    1  0  1  Y (CT violation)
    1  1  0  X (no T violation, so no CP violation)
    0  0  1  X etc.
    1  0  0  Y
    0  1  1  Y
     
    So you can see that flipping C only flips the result.
     
  12. Nov 6, 2013 #11
    Hi haruspex,

    Thanks for the table and explanation. I am still unsure of the conclusion. Violating P violates C (when T is not violated) is the answer i gave but in the table it says

    1 1 0 X for C, P, and no T violations, respectively.

    Since the system is left invariant, C is not violated when P is violated so where does that leave me? CPT doesn't completely violate and nor does C need to necessarily violate if P violates when T does not. Is momentum not conserved (I thought it was) or is the conclusion non-conclusive?
     
  13. Nov 6, 2013 #12

    haruspex

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    No, that's not how to read my table. Maybe using 0's and 1's is confusing. + and - might have been better. The line
    0 0 0 X
    means that with some particular experiment we get result X. I arbitrarily label the charge, parity and time settings in the experiment as the 0 values. So the line
    0 0 1 X
    says we flipped time only and still got result X. No T violation.
    1 1 0 X
    says we flipped C and P but not T, and still got result X: no CP violation.
    0 1 0 Y
    says we flipped P only and got a different result: P violation.
    Etc.

    There I'm out of my depth. What does 'left invariant' mean in this context, how do you know it's left invariant, and how does it follow that C is not violated when P is?
     
  14. Nov 7, 2013 #13
    Invariant means it has not changed state (ie. X->X and not X->Y). If P is 1 (inverting=violating) the result changes X->Y without C changing to 1 (C=0, P=1, X->Y) then how can we say C will always violate (invert) when P does? Isn't that what my original conclusion says? I am probably missing something real obvious here so would like it to be clarified without bothering you too much. You've helped me tons and I appreciate it.

    Thank you
     
  15. Nov 7, 2013 #14

    haruspex

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    Let me put it this way. It seems to me that in any given experiment there are four possibilities:
    1. Complete symmetry, no violations anywhere.
    2. C violation and PT violation.
    3. P violation and CT violation.
    4. T violation and CP violation.
    Is that clearer?
     
  16. Nov 7, 2013 #15
    Right so in 3, P violates which in turn violates C and that's my original conclusion?
     
  17. Nov 7, 2013 #16

    haruspex

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    Well, I'm not sure what you mean by 'in turn' here. Since T is not violated, CP is not violated. For P to be violated but not CP we need C violated (so that the two violations of P and C cancel in CP). Is that what you meant?
     
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