4
$\begingroup$

In a $SU(2)$ gauge field theory with scalar field $\phi$ in the fundamental representation of the $SU(2)$ group with lagrangian $$\mathcal{L} = -\frac{1}{2}TrF_{\mu\nu}F^{\mu\nu} + (D_{\mu}\phi)^\dagger(D^{\mu}\phi) + \mu^2\phi^\dagger\phi - \frac{1}{2}\lambda(\phi^\dagger\phi)^2,$$ we can pick a vev $\phi_{0} = \frac{1}{\sqrt{2}}(0 \;\; v)^T$, with $v^2 = \frac{2\mu^2}{\lambda}$, and break the symmetry doing $\phi = \varphi + \phi_0$.

Then, the new lagrangian has a couple mixing terms $ig\partial_{\mu}\varphi^{\dagger}A^{\mu}_{a}t^a\phi_0$ + h.c., which can be set to zero by choosing the unitary gauge. This implies that the field $\varphi$ satisfies the conditions $\varphi^{\dagger}t^a\phi_0 = 0$, where $t^a$ are the generators of $SU(2)$ picked so that $t_a = \frac{1}{2}\sigma_a$, where $\sigma_a$ are the Pauli matrices.

The unitary gauge imposes constraints on the field $\varphi$. However, if we consider $\varphi^{\dagger}t^1\phi_0 = 0$ and $\varphi^{\dagger}t^3\phi_0 = 0$ and take $\varphi^{\dagger} = (\varphi^*_1 \;\; \varphi^*_2)$, I get that $$ \begin{pmatrix} \varphi^*_1 & \varphi^*_2 \end{pmatrix} \frac{1}{2}\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} \frac{1}{\sqrt2} \begin{pmatrix} 0 \\ v \end{pmatrix} = \frac{1}{2\sqrt2}\varphi^*_1v = 0 \Rightarrow \varphi^*_1 = 0. $$

Similarly, $$ \begin{pmatrix} \varphi^*_1 & \varphi^*_2 \end{pmatrix} \frac{1}{2}\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix} \frac{1}{\sqrt2} \begin{pmatrix} 0 \\ v \end{pmatrix} = -\frac{1}{2\sqrt2}\varphi^*_2v = 0 \Rightarrow \varphi^*_2 = 0. $$

I'm confused because this would imply that $\varphi = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$.

$\endgroup$
1
  • 2
    $\begingroup$ I have not kept track of your peculiar formal stunts, but the answer you should find in the unitary gauge is that $\varphi =(0,h)^T/\sqrt{2}$. $\endgroup$ Commented Nov 3, 2022 at 0:42

2 Answers 2

4
$\begingroup$

Your text should be illustrating the H-E-Brout phenomenon by parameterizing the Higgs doublet in the chiral parameterization, as opposed to the linear σ-model one you are using. They are, of course, equivalent, and it behooves you to catch your mistake in translating back to yours. (Observe $( \varphi_1^* \varphi_1+\varphi_2^*\varphi_2) \phi_0\neq 0$; they are not independent!)

The standard SM (Gürsey) parameterization is $$ \phi = UH\equiv e^{it^a \xi^a/v} \frac{1}{\sqrt2} \begin{pmatrix} 0 \\ v+h(x) \end{pmatrix}, $$ where the three ξs are the Goldstone bosons in the adjoint of SU(2), but the h, the dross higgs, is an SU(2) singlet. You must appreciate that $\phi$ has four components/d.o.f., and one of them, the "radial" one, does not transform under SU(2). Note if you think of the ξs as transformation parameters, no non-vanishing ξ can leave H and the SSB vacuum invariant!

It is then straightforward to see that, in any gauge, $$ D_\mu\phi= (\partial_\mu -igt^a A^a_\mu) UH= U(U^{-1}\partial_\mu U +\partial_\mu -igt^a ~U^{-1}A^a_\mu U) H. $$

The unitarity gauge is defined as the one in which all three goldstons are gauge-transformed to zero, i.e. U=1, and their gradients are absorbed by the gauge fields: in that gauge, $$ \leadsto \qquad (\partial_\mu -igt^a {\cal A}^a_\mu) \frac{1}{\sqrt2} \begin{pmatrix} 0 \\ v+h(x) \end{pmatrix}. $$

No goldstons, but the Higgs is still there! You may now read off the $\cal A_\mu^a$ mass terms from the lower diagonal entry of their $(D\phi)^\dagger \cdot D\phi$ matrix, etc...

(In the unitary gauge, $h/\sqrt{2}=\Re \varphi_2$. It turns out the h.c. term in your expression for mixing terms is not an innocent bystander at all! It is needed to cancel the Higgs component in your second expression which should not vanish individually, unlike what you guessed…)

$\endgroup$
2
$\begingroup$

Your results are right. That is exactly what it should be here. Quoted from Weinberg

$$0=\sum_{nm}\tilde{\phi}_m (t_\alpha)_{mn} v_n\qquad\qquad(21.1.2)$$ Eq. (21.1.2) shows that there are no Goldstone boson fields in unitarity gauge. Since the theory is gauge-invariant this means that there are no physical Goldstone bosons, whatever gauge we choose.

(21.1.2) is what you had enforced to let couple mixing term vanish. When you choose unitary gauge, you absorb the degree of freedom into the longitudinal mode of the gauge bosons. Notice that in the unbroken phase you have three gauge bosons which are vector fields each has only two transverse polarizations as they are massless. But if spontaneous symmetry breaking happens they obtain mass and the longitudinal polarized mode now is not unphysical as before, so the physical degree of freedoms are the same.

As a summary: in unitary gauge, 3 goldstone bosons vanish but 3 longitudinal polarized mode of the gauge bosons occur. The total physical degree of freedom is unchanged.

$\endgroup$
5
  • 1
    $\begingroup$ But... There is a non-Goldstone component, h, in the four components of 𝜑. This is what is confusing the OP, no? You are crucially misquoting Weinberg: His $\tilde \phi$ is not the raw, unrefined 𝜑 the OP is considering, any more than the earth is flat! $\endgroup$ Commented Nov 3, 2022 at 13:29
  • 1
    $\begingroup$ @CosmasZachos The non-goldstone component need not satisfy $\varphi^{\dagger}t^a\phi_0 = 0$ in unitary gauge and that's why it could exist. I suppose what you said is the true thing OP did not understand. Thanks. $\endgroup$ Commented Nov 3, 2022 at 14:06
  • $\begingroup$ I think so. Most good texts clarify this by dint of language. $\endgroup$ Commented Nov 3, 2022 at 14:29
  • $\begingroup$ Thank you both for your comments! I understand that after using the unitary gauge, both the real and imaginary parts of the first component and the complex part of the second component should vanish, leading to $\varphi = (0 \;\; h(x))^T/\sqrt2$ as @CosmasZachos commented above. What I don't understand is how the calculation works to get that result. I thought that it would be a consequence of the condition $\varphi^{\dagger}t^a\phi_0 = 0$, but clearly I'm not obtaining that result. Thanks! $\endgroup$
    – zequi
    Commented Nov 3, 2022 at 16:42
  • $\begingroup$ @zequi It is the sum of the mixing term you wrote + h.c. that vanishes, not each term individually! In the chiral language, you cannot make this mistake. $\endgroup$ Commented Nov 3, 2022 at 16:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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