jcYoon's Physics

[jcyoon]

Thurs 2007-11-01 4:53 AM

Dear Professor Zhi-qiang Shi,

Thanks for your email. I think I was not clear with the normalization factor. Please let me try with explicit mathematical expressions.

The Lorentz invariance of Lagrangian can be substantiated either by proving the Lagrangian is Lorentz scalar or by comparing the values of Lagrangian evaluated for the reference frames under Lorentz transformations such as the rest frame and the boosted frame of particle.

For the weak interaction Lagrangian $\overline{\psi}\gamma^{\mu}(1 + \gamma^{5}\psi$, when the Lagrangian is evaluated in the rest frame with
u_{rest} = \sqrt{m} \left( \begin{array}{c} \xi & \xi \end{array} \right),
it gives
2m \xi^{\dagger} \sigma^{\mu} \xi,
while in the boosted frame with
u_{boosted} = \left( \begin{array}{c} \sqrt{p \cdot \sigma}\xi &
\sqrt{p \cdot \overline{\sigma}}\xi \end{array} \right),
it gives
2 (p \cdot \overline{\sigma}) \xi^{\dagger} \sigma^{\mu} \xi.

Therefore, the normalization factor for the rest frame and boosted frames are different as the normalization factor (p \cdot \overline{\sigma}) is not Lorentz invariant. We may overcome this issue by ignoring its Lorentz variance, but we are using this Lorentz-variant factor that vanishes when we make the relativistic approximation to neglect the left-handed helicity
u_{boosted} = \left( \begin{array}{c} \sqrt{p \cdot \sigma}\xi &
\sqrt{p \cdot \overline{\sigma}}\xi \end{array} \right)
\rightarrow
\sqrt{E} \left( \begin{array}{c} \xi & 0 \end{array} \right).
Note that if the normalization factor had been Lorentz-invariant, then there would not have been such a relativistic approximation.

Sincerely yours,
J.C. Yoon

[Zhi-qiang Shi]

Thurs 2007-11-01 7:14 PM

Dear Dr. J. C. Yoon,
Thanks you for your elaborate reply.

1. As for my statement in mail dated October 11, the plane wave solutions of Dirac equation should reduce to the eigenfunction of \Sigma_z in its rest frame, like Eq. (3.2) in Relativistic Quantum Mechanics by Bjorken and Drell. However, the solution given by you does not satisfy this condition. A correct solution should be Eq. (2) or (3) in my paper (calculation on the lifetime of polarized muons in flight).

2. Both vector and axial vector (pseudovector) are not Lorentz scalar, see Eq. (2.38) in Relativistic Quantum Mechanics by Bjorken and Drell. However, the Lagrangian is the product of vector (or axial vector) and axial vector (or vector), and so the Lagrangian is Lorentz invariant. For example, the Lagrangian of purely leptonic process is given by
L(x)~[\overline{\psi}_a\gamma_\mu(1+\gamma_5)\psi_b][\overline{\psi}_c\gamma_\mu(1+\gamma_5) \psi_d].
Under proper Lorentz transformation, we have
L’(x’)~ a_\mu^\nu a_\sigma^\mu [\overline{\psi}_a\gamma_\nu(1+\gamma_5)\psi_b] [\overline{\psi}_c\gamma_\sigma(1+\gamma_5) \psi_d].
Because
a_\mu^\nu a_\sigma^\mu=\delta^\nu_\sigma ,
(see Eq. (2.3) in Relativistic Quantum Mechanics by Bjorken and Drell ), we obtain
L’(x’)=L(x)
that is to say that the Lagrangian is Lorentz invariant.

Sincerely yours,
Zhi-qiang Shi

[jcyoon]

Fri 2007-11-02 6:16 AM

Dear Professor Zhi-qiang Shi,

Thanks for your patient and elaborate reply, which was quite helpful for me to understand your point.

1. The Dirac solution in the rest frame that I have used is from the standard textbook of Peskin and Schroeder, Eq. (3.47) on page 45 in terms of chiral representation. The $\Sigma_{z}$ for the chiral representation can be given by
\Sigma_{z}= \left( \begin{array}{cc} \sigma_{z} 0 & 0 \sigma_{z} \end{array} \right)
And for the Dirac solution in the rest frame with spin-up along the z direction
\xi = \left( \begin{array}{c} 1 & 0 \end{array} \right)
We have the eigenvalue of +1 for $\Sigma_{z}$
\Sigma_{z}u_{rest} = +1 u_{rest}

2. I agree that the Lorentz invariance of the weak interactions can be shown from your argument. What I have pointed out is that the practical Dirac solution is not properly normalized to retain the Lorentz invariance that is supposed to be guaranteed from the proof of infinite Lorentz transformations.

Let us make my mistake clear.
My claim of Lorentz violation of the weak interactions Lagrangian (not the Standard Model) is incorrect, in the sense that its Lorentz invariance can be proved from using infinite Lorentz transformations. The correct statement should be that the weak interactions Lagrangian is proven to be Lorentz-invariant, but the Dirac solutions we use fail to retain the Lorentz invariance as its normalization factor is Lorentz variant, which is later exploited by the relativistic approximation.

I will be thinking about it more, but I am glad that you brought up a critical point to me, which I dearly appreciate.

Sincerely yours,
J.C. Yoon

[Zhi-qiang Shi]

Fri 2007-11-02 6:45 PM

Dear Dr. J. C. Yoon,

Thank you for your email.

1. According to the Dirac solution that you have used, in the boosted frame the solution is given by
u_{boosted}=\sqrt{E} \left( \begin{array}{c}\sqrt{p\cdot \sigma}\xi&\sqrt{p\cdot\overline{\sigma}} \xi \end {array} \right),
In the rest frame, p=0, we should obtain
u_{rest} =\sqrt{m} \left( \begin{array}{c}0 & 0 \end{array} \right),
It is not the way you given by using
u_{rest} =\sqrt{m} \left( \begin{array}{c} \xi & \xi \end{array} \right).
Please you consider again.

2. You said: “the practical Dirac solution is not properly normalized to retain the Lorentz invariance that is supposed to be guaranteed from the proof of infinite Lorentz transformations.” “the Dirac solutions we use fail to retain the Lorentz invariance as its normalization factor is Lorentz variant.” What does it mean? I think that the Dirac solutions are assuredly not Lorentz invariant. The symmetry theory only requires that Dirac equation and the Lagrangian are Lorentz invariant. The normalization factor is decided by using normalizing condition, not Lorentz invariance.

Sincerely yours,
Zhi-qiang Shi

[jcyoon]

Sat 2007-11-03 5:26 AM

Dear Professor Zhi-qiang Shi,

Thanks for your prompt and kind reply.

1. In the notation from my statement following Peskin and Schroeder, p is p^{\mu} and thus in the rest frame we have p^{\mu} = (m,0) which give the same solution in the rest frame
u_{rest} =\sqrt{m} \left( \begin{array}{c} \xi & \xi \end{array} \right).
Note that \sqrt{E} should be dropped in the solution in the boosted frame.

2. In my statement 2 I was equivocally talking about two different Lorentz invariances: Lagrangian and matrix element. What I meant to say was that though the Lagrangian is Lorentz invariant under an infinitesimal Lorentz transformation, the matrix element is not necessarily guaranteed to be Lorentz invariant, since the matrix elements for the rest frame and boosted frame are different due to normalization factors.

Sincerely yours,
J.C. Yoon

[Zhi-qiang Shi]

Mon 2007-11-05 6:35 PM

Dear Dr. J. C. Yoon,
Thanks for your patient and elaborate reply.

I agree that though the Lagrangian is Lorentz invariant under an infinitesimal Lorentz transformation, the matrix element is not necessarily guaranteed to be Lorentz invariant.

Our discussion is successful. Now, your perspectives have completely accorded with mine. It is a nice opportunity to discuss with you about the Lorentz invariant. Both of us can learn many something in this constructive discussion. I am very gratified.

Sincerely yours,
Zhi-qiang Shi

[jcyoon]

Wed 2007-11-07 5:39 AM

Dear Zhi-qiang Shi,

I am glad to receive your reply and I sincerely appreciate the great opportunity for me to learn from you through our discussion.

If you don't mind, I would like to post our helpful discussion on my web site, for which I will be very grateful.

But if it is not a good idea, please feel free to let me know so.

Sincerely yours,
J.C. Yoon

[Zhi-qiang Shi]

Wed 2007-11-07 6:04 PM

Dear Dr. J. C. Yoon,
I am pleased to post our discussion on your web site. I hope that more physicists can share our productive discussion.
Sincerely yours,
Zhi-qiang Shi