before the electroweak phase transition of the Universe, all weak bosons are massless just like the photons, and with the temperatures well above TeV, the Higgs-particle hasn't given any mass to the charged leptons yet. would there be a way of distinguishing electrons, muons and taus?

physics bonus question: in quantum mechanics, what property of the Hamiltonian operator is responsible for the conservation of probability?

maths bonus question: can you show that $\exp(A)$ is unitary if $A$ is anti-Hermitean, $A^+=-A$, where $A^+$ is the complex-conjugated, transposed matrix $A$?

maths bonus question: if $U=\exp(A)$ should be unitary, it's enough to show that $UU^+=\mathrm{id}$. let's substitute the exponential series for matrices,

ReplyDelete\begin{equation}

UU^+ = \exp(A)\exp(A)^+ =

\left(\sum_i\frac{A^i}{i!}\right)\left(\sum_j\frac{A^j}{j!}\right)^+

\end{equation}

which becomes by applying the Cauchy product,

\begin{equation}

=\sum_i\sum_j^i\frac{A^{i-j}(A^+)^j}{(i-j)!j!}

\end{equation}

because $(A^j)^+ = (A^+)^j$, and by noticing that

\begin{equation}

\frac{1}{(i-j)!j!} = \frac{1}{i!}{i\choose j}

\end{equation}

one can apply the binomial series,

\begin{equation}

=\sum_i\frac{1}{i!}\sum_j^i{i\choose j}A^{i-j}(-A)^j = \exp(A-A) = \mathrm{id}

\end{equation}

of course you can make any Hermitean operator $H$ anti-Hermitean by multiplication with $\mathrm{i}$, $(iH)^+ = -iH$, such that $\exp(iH)$ is unitary, which is widely used in quantum mechanics.

physics bonus question: it's the Hermiticity of $H$ which causes the time evolution $U=\exp(\mathrm{i}Ht)$ to be unitary (see above!). Then, the integral over the square of the wave-function is always conserved:

ReplyDelete\begin{equation}

\psi\rightarrow\exp(\mathrm{i}Ht)\psi

\end{equation}

such that

\begin{equation}

\int\mathrm{d}^3x\:\psi^*\psi

\end{equation}

is constant because the time evolution with $\exp(\mathrm{i}Ht)$ cancels. non-Hermitean Hamiltonian operators can, for instance, describe particle decay or creation, $H+\mathrm{i}\Gamma$ at the rate $\Gamma$, because then the time-evolution $U=\exp((\mathrm{i}H-\Gamma)t)$ is not unitary and therefore not probability conserving.

main question: the leptons are massless, but still they have different coupling constants for doing weak interactions. by doing scattering experiments one could distinguish electrons, muons and taus.

ReplyDeleteYou have not convinced me on the main physics question. The leptons all have the same electric charge, -1, (anti-leptons +1). Further, they all have the same weak hypercharge (up to left-right asymmetry): Y=2(Q-T_3). Where T_3 is the third component of its weak isospin: Y=-1 for left handed, -2 for right handed. So you could certainly tell a left handed from a right handed lepton, (which are decoupled in this phase of the universe since they have no mass).

DeleteIn terms of different coupling constants, I think what you may be referring to is the fact that the muon and electron have different anomalous couplings, (g-2)? But the loop corrections that make these different are in fact mass terms. So shouldn't be present..

In terms of actually answering this question, my response is, well, it depends. In the strictest sense of the standard model, the answer is no, you can not tell the difference between electrons, muons and taus. However, we know the standard model to be incomplete.. neutrinos have mass. How do the neutrinos get those masses? If your model involves neutrinos picking up simple Dirac masses generated through the higgs mechanism, then the answer would still be no, they would be massless at this point in the Universe. If your mass is generated through some other mechanism, then maybe. Depends on your model.