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I have encountered a notation I am not familiar with, namely a tilde $\sim$ above a vector (i.e. a column vector), e.g. $\tilde{H}$. From the context, it is clear that it cannot mean transposition, complex conjugation or Hermitean conjugation. Is there any standard meaning of this notation? If it helps, I can mention that the notation occurs in the Lagrangian of two Higgs doublet models.

To give a specific example, consider e.g. equation 30 of Theory and phenomenology of two-Higgs-doublet models by Branco et al. $$\mathcal{L}_\mathrm{Yukawa}=\eta_{ij}^U\bar{Q}_{iL}\tilde{H}_1U_{jR}+\eta_{ij}^D\bar{Q}_{iL}H_1D_{jR}+...$$ Here $...$ denotes further terms which include e.g. $H_2$. Thus it is clear that the tilde is not used to distinguish the two Higgs doublets.

Another example of where the notation occurs is in eq.80 of Building and testing models with extended Higgs sectors by Ivanov. For a complex triplet $X=(\chi^{++},\chi^+,\chi^0)^T$, $\tilde{X}$ is defined as $\tilde{X}=(\chi^{0*},-\chi^{+*},\chi^{++*})^T$

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  • $\begingroup$ I've removed some comments that answered the question. Please use answers to answer the question. $\endgroup$ – rob Feb 13 at 20:27
  • $\begingroup$ @rob How am I supposed to do that if the question is closed? $\endgroup$ – knzhou Feb 13 at 22:43
  • $\begingroup$ @knzhou If you think a closed question deserves an answer, vote to reopen it. Please don't use a comment to answer a closed question. However, a comment explaining why the question is a good candidate for reopening will guide users who are reviewing the reopen queue towards your opinion. $\endgroup$ – rob Feb 14 at 1:49
  • $\begingroup$ Related. $\endgroup$ – Cosmas Zachos Mar 1 at 3:57
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This is the conjugate representation of SU(2). It is the backbone of the fermion masses in the SM and is detailed in standard SM texts.

That is to say, for the doublet, $$ \tilde H \equiv i\sigma_2 H^*, $$ so it transforms identically to H under SU(2)! (It, of course, reverses the hypercharge).

Behold: $$ \delta H = \frac{i}{2} \theta_a \sigma_a H \qquad \leadsto \\ \delta \tilde H = i\sigma_2 ( \frac{i}{2} \theta_a \sigma_a H )^* \qquad \\ = i\sigma_2 ( -\frac{i}{2} \theta_a \sigma_a^* H^* ) = \frac{i}{2} \theta_a \sigma_a ~ i\sigma_2 H ^* = \frac{i}{2} \theta_a \sigma_a \tilde H , $$ by virtue of $\sigma_2 \sigma_a^*+\sigma_a\sigma_2=0 $ for any a!

So, $\tilde H$ is a doublet just like H, except stood on its head, complex conjugated, with an extra - sign at the bottom.

As a result, the Yukawa coupling you exemplify, which is identical to the Yukawa of the SM for the standard Higgs doublet , provides masses for both the uplike quarks, the first term, and the downlike quarks.

That is so because $$ \langle H\rangle_0= \begin{pmatrix} 0\\ v \end{pmatrix} , \qquad \implies \qquad \langle \tilde H\rangle_0=\begin{pmatrix} v\\ 0 \end{pmatrix} , $$ so the v.e.v. of $\tilde H$ provides the uplike quark masses, just like that of the H for the downlike masses.


As for the Higgs triplet, I am not very experienced with its standard conventions. You dot your triplet (which is in the spherical basis) properly transformed to a Cartesian vector, now, to the Pauli vector, and re-express it in the spherical basis again to monitor the properties of the individual components, $$ \Delta=\begin{pmatrix} {\chi^{+}} &{\sqrt{2}} \chi^{++} \\ {\sqrt{2}}\chi ^0 & - \chi^{+} \end{pmatrix}= \chi^+ \sigma_3 +\chi^{++} \frac{\sigma_1+i\sigma_2}{\sqrt{2}} + \chi^0 \frac{\sigma_1-i\sigma_2}{\sqrt{2}} , $$ and read off the new (transformed) components in the adjoint object $$i\sigma_2 \Delta^* (-i\sigma_2) ,$$ noting the transposition effected. (Full disclosure: I seem to be getting a couple of errant - signs extra, contrasted to your expression. It may well be an artifact of the procedure. The v.e.v. is very much in the properly transposed position, as you should check!)

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    $\begingroup$ Superb answer! Just want to add one point: the $U$ term $\eta_{ij}^U\bar{Q}_{iL}\tilde{H}U_{jR}$ is symmetric under chiral rotation $H \rightarrow He^{\theta i}$ and $U_{jR} \rightarrow U_{jR}e^{\theta i}$ (thanks to $H^*$ embedded in $\tilde{H}$), while the $D$ term $\eta_{ij}^D\bar{Q}_{iL}HD_{jR}$ is NOT symmetric under chiral rotation $H \rightarrow He^{\theta i}$ and $D_{jR} \rightarrow D_{jR}e^{\theta i}$. This is where the two Higgs doublet model with Peccei-Quinn symmetry comes to rescue. $\endgroup$ – MadMax Mar 2 at 16:18
  • $\begingroup$ Thanks, I was dimly aware of it. I mentioned the reversal of weak hypercharge, but left it aside to prevent complication... $\endgroup$ – Cosmas Zachos Mar 2 at 16:24
  • $\begingroup$ @CosmasZachos Thank you very much for this! $\endgroup$ – Étienne Bézout Mar 2 at 18:35
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The meaning of the tilde is impossible to tell without further context. There is no standard meaning of the tilde and it is used all over (particle) physics to denote different things.

One possible guess is that the two different scalar fields are simply denoted $H$ and $\tilde H$. In that case, the tilde would not denote any operation but just distinguish the two different fields. There is no way to know if that is the correct interpretation without further context though.

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    $\begingroup$ Thank you for your comment. I have now updated the question to include a concrete example where the notation is used. $\endgroup$ – Étienne Bézout Feb 13 at 19:47

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