EDIT: Additional question at the end

I am trying to illuminate how the "unphysical" gauge bosons $W^{1}_{\mu},W^{2}_{\mu},W^{3}_{\mu},B_{\mu}$ will be the "physical" $W^{\pm},Z_{\mu},A_{\mu}$ when diagonalizing the mass matrix. Notice that it is in Euclidean time, so we do not have to care about the Lorentz indices. Furthermore $\sigma(x)$ is the Higgs field and $v$ is the vacuum expectation value.

After the symmetry breaking $$ SU(2)_L\times U(1)_Y \rightarrow U(1), $$

and inserting the vacuum expecation value, I got the following Lagrangian (just the dynamical part):

$$ \frac{1}{2}D_{\mu}\phi^{\dagger}D_{\mu}\phi = \frac{1}{2}\partial_{\mu}\sigma \partial_{\mu} \sigma + \frac{(v+\sigma)^2}{8}(g^2W^{1}_{\mu}W^{1}_{\mu} + g^2W^{2}_{\mu}W^{2}_{\mu} + (gW^{3}_{\mu} - g'B_{\mu})(gW^{3}_{\mu} - g'B_{\mu})) . $$ $W^{\pm}=W^{1}_{\mu}\pm W^{2}_{\mu}$ is clear, but retrieving $Z_{\mu}$ and $A_{\mu}$ not. I tryied the following, since the last part of the Lagrangian can be written like:
$$ (W^{3}_{\mu},B_{\mu}) \begin{pmatrix}g^2 & -gg'\\-gg'& g'^{2} \end{pmatrix} \begin{pmatrix}W^{3}_{\mu}\\B_{\mu} \end{pmatrix} $$ The diagonlized matrix reads $$ M_D=\begin{pmatrix}0 & 0\\0& g^2 +g'^{2} \end{pmatrix} $$ and does not give the right linear combinations of $Z_{\mu}$ and $A_{\mu}$, which are given in my literature as $$ A_{\mu} = \frac{g'W^{3}_{\mu} + g B_{\mu}}{\sqrt{g^2+g'^2}},\qquad Z_{\mu} = \frac{gW^{3}_{\mu} - g' B_{\mu}}{\sqrt{g^2+g'^2}} $$ My question is now, how to get these combinations, it looks like I am close, but only close. And the other question where comes the normalization conditions for the field from?



I finally found the linear combinations, mass eigenstates, like they are in the literature, by inserting not only the diagonlized mass matrix $M_D$, but by inserting $M = PM_DP^{-1}$ As I was looking at the covariant derivative to find out how the fields couple to the Higgs doublet I was wondering how I could possibly turn the following matrix into mass eigenstates of the gauge fields:

$$ \frac{i}{2}\begin{pmatrix}gW^{3}_{\mu} + g'B_{\mu} & 0\\ 0& g W^{3}_{\mu} + g'B_{\mu}\end{pmatrix} $$

again, cheers!

  • $\begingroup$ In the last two formulae, you meant the squared couplings under the square roots and not the couplings themselves, didn't you? When you make this fix, the eigenstate combinations clearly work out. $\endgroup$ – Luboš Motl Jul 11 '14 at 9:45
  • $\begingroup$ A couple more mistakes... The Higgs is a complex scalar field, so you don't need the leading factor of $\frac12$ and $W^\pm = \frac{1}{\sqrt{2}} (W^1 \pm i W^2)$ $\endgroup$ – innisfree Jul 11 '14 at 9:46
  • $\begingroup$ You have already noticed in the second equation that the neutral combination that get's mass is proportional to $gW_\mu^3-g^\prime B_\mu$, then you just need to be sure that is canonically normalized by dividing by $\sqrt{g^2+g^{\prime\,2}} so that you get $\cos\theta=g/\sqrt{g^2+g^{\prime\,2}}$ and it's all done. $\endgroup$ – TwoBs Jul 11 '14 at 9:54
  • $\begingroup$ inserting the diagonlized matrix between the to vectors one just gets, $(g^2 + g'^2)B_{\mu}B_{\mu}$. It would be nice if somebody wrote one line down, so I finally see it. thanks! $\endgroup$ – nerdizzle Jul 11 '14 at 11:06
  • $\begingroup$ ah, i think i see your problem. When you diagonalise the matrix, you cannot simply replace the old matrix with the new diagonalised one in your formulas. You must use the new basis in which the matrix is diagonal! i.e. you can make the subsitutation $M=U^{-1} M^{diag} U$, the $U$'s rotate the $B, W$ to the Z-boson and the photon. $\endgroup$ – innisfree Jul 11 '14 at 14:29

You have noticed already that $$ \mathcal{L}_{mass}\propto \left[g^2(W^1_\mu W_\mu^{1}+W^2_\mu W_\mu^{2})+(gW_\mu^3-g^\prime B_\mu)^2\right] $$ with the kinetic terms for $W^{i}_\mu$ and $B_{\mu}$ canonically normalized. Therefore the neutral linear combination of $W^3_\mu$ and $B_\mu$ that gets mass is proportional to $(gW_\mu^3-g^\prime B_\mu)$ and the proportionality constant $1/\sqrt{g^2+g^{\prime\,2}}$ fixed by the fact that you want keep the fields canonically normalized (that is, you are doing a rotation from $(W_\mu,B_\mu)^T$ to $(Z_\mu,A_\mu)^T$) $$ \mathcal{L}_{mass}\propto \left[g^2(W^1_\mu W_\mu^{1}+W^2_\mu W_\mu^{2})+(g^2+g^{\prime\,2})\frac{(gW_\mu^3-g^\prime B_\mu)^2}{g^2+g^{\prime\,2}}\right]= \left[g^2(W^1_\mu W_\mu^{1}+W^2_\mu W_\mu^{2})+(g^2+g^{\prime\,2})Z_\mu^2\right] $$ where you see that $$ \frac{g}{\sqrt{g^2+g^{\prime\,2}}}=\cos\theta_W $$ and $$ \frac{m_W^2}{m_{Z}^2\cos^2\theta_W}=1\,. $$ in agreement with the literature. Indeed, only the $Z$-boson and $W^{\pm}$ get masses, as the photon is massless: the neutrual combination that gets mass must be identified with $Z$-boson

| cite | improve this answer | |
  • $\begingroup$ Hello @TwoBs. This is very interesting, but then how do we recognize that there is a $W^+$ and $W^-$ in nature from the equations. Of course, $W_1^2 + W_2^2\sim W^+W^-$, but why would we decide to make this in the form of $W^+W^-$ instead of saying that there are two "neutral" W bosons with a mass : the "W1" and the $W2$ bosons ? $\endgroup$ – Mathieu Krisztian May 4 at 20:16
  • $\begingroup$ One can use whatever basis to perform the calculations, even $W^{1,2}$ basis would be ok. It would corresponds to the real bi-dimensional representation of SO(2). Instead the $W^{+-}$ basis corresponds to the irreducible one-dimensional complex representation of $U(1)$. I'd say the latter is nicer because it trivializes the constraints from the $U(1)$ charge conservation, in the sense that in this basis the $Q$ operator is diagonal so that the states have definite charges. In real doublet $W^{1,2}$ instead the $Q$ is a 2x2 rotation with frequency given by the charge, less useful in general imo $\endgroup$ – TwoBs May 5 at 21:26
  • $\begingroup$ thank you for your comment $\endgroup$ – Mathieu Krisztian May 7 at 22:42
  • $\begingroup$ It looks like there is a problem the determinant of the matrix of @nerdizzle is null... So it is not possible to diagonalize it. Indeed the terms of his matrix are : $g^2$, $-gg'$, $-gg'$, $g'^2$. This makes Determinant=0. $\endgroup$ – Mathieu Krisztian Jul 5 at 11:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.