It's generally stated in the textbooks that when the Higgs field acquires a certain VEV the corresponding symmetry is spontaneously broken. For example in QFT in a Nutshell by A. Zee:

But none of the $SU(2)$ transformations leaves $\binom{0}{v}$ invariant: The vacuum expectation value of φ spontaneously breaks the entire $SU(2)$ symmetry

I was wondering why this should be the case. We have a Lagrangian of the form

$$ \bar \Psi_L \Phi \psi_R ,$$

where $\Psi_L$ is the lepton doublet and $\Phi$ the Higgs doublet. This term is obviously invariant under $\Psi_L \rightarrow U \Psi_L$ and $\Phi \rightarrow U \Phi $:

$$ \bar \Psi_L \Phi \psi_R \rightarrow \bar \Psi_L U^\dagger U \Phi \psi_R = \bar \Psi_L \Phi \psi_R$$

Regarding Zee's explanation: A general $\Phi= \binom{\Phi_1}{\Phi_2}$ isn't invariant under $SU(2)$ transformations either, but this is not what is important here. Only the complete term in the Lagrangian must be invariant and it is regardless of if we put in some vev or a general $\Phi$.

Now, if we put in a certain $\Phi= \binom{0}{v}$, we still have

$$ \bar \Psi_L \Phi \psi_R= \bar \Psi_L \binom{0}{v} \psi_R \rightarrow \bar \Psi_L U^\dagger U \binom{0}{v} \psi_R = \bar \Psi_L \binom{0}{v} \psi_R .$$

The same holds true if we write this term a little different, as it is usually done, using $\Psi_L = \binom{\nu}{e}$:

$$ \bar \Psi_L \Phi \psi_R = \binom{\bar \nu}{\bar e}^T \binom{0}{v} \psi_R = 0 \cdot \bar \nu \psi_R + v \cdot \bar e \psi_R $$

The symmetry is still there only "hidden" a little bit.

What exactly breaks here the symmetry and how can it be shown explicitly?

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    $\begingroup$ The whole point of spontaneous symmetry breaking is that the symmetry is not broken on the level of the Lagrangian. $\endgroup$
    – ACuriousMind
    Commented Apr 19, 2015 at 12:17

2 Answers 2


I will amplify the above point that numbers do not transform under symmetries--- only multiplets do. Your QFT text should explain this clearly in introducing SSB. If not, try the excellent S Coleman, Aspects of Symmetry, 1985.

Of course the symmetry is still there, so just hidden: a change of variables can alter neither the physics nor the math! The currents and charges involved are conserved, albeit now realized in the nonlinear Nambu-Goldstone mode. The question you are essentially asking is how to "see" the apparent breaking. Absorbing the relevant dimensions into prefactors, the v.e.v. v is just a number, like 5 or 37. So while <Φ>= (0,v) does transform under the symmetry, by itself, in isolation, so in the electron mass term, it does not, any more than 5 or 37 transform under a symmetry.

Let me dramatize the point sparing you the evidently confusing physics scaffolding, just considering two real 2-vectors (A,B) and (a,b). Their dot product (A,B).(a,b)=Aa+Bb is a scalar, invariant under a rotation by θ, take it to be infinitesimal, for simplicity, so a → a+θb, A → A+θB, b → b -θa, B → B -θA. So the O(θ) variation of it under the rotation vanishes.

Changing variables to b= 37+b' would not change the vector, nor their scalar product, Aa+Bb'+37B, which sure looks asymmetric, now (although you now where it came from, so it is symmetric). A fortiori, if you chose the vector <(a,b)>=(0,37), their scalar product would also be invariant under rotations, even though it sure looks like 37 B, now, if you knew not its past: that it is a specific value of a 2-vector (a,b).

In the shifted variables, δa= θ(37+b'), and δb'= -θa, and of course, you do not transform the number 37. Still, the above asymmetric dot product is invariant. It looks asymmetric, so "broken", what with this freaky extra term of 37B, but you may work backwards and modify your linear transformation laws to make it invariant, under the one we just used, and then work backwards to "see" the symmetry.

Now what Tony Zee is suggesting, a bit elliptically, is that it would be conceivable, under different circumstances, to break/hide some symmetries, but not others. For example, if one had a 3 vector v.e.v. (0,0,v), there would be a symmetry rotating first and second components to each other, and still realized linearly, leaving this v.e.v. invariant. If you repeated all this for 3-vectors, you'd still see a manifest, residual (Wigner-Weyl mode) symmetry in your scalar product between the first and second components. In the standard model, however, Zee points out, this does not happen: the entire SU(2) breaks/hides this way.


First, in order to show the invariance of the term $\bar{\psi}_L \phi \psi_R$, you have to transform respectively:

$\psi_L \to e^{-i \alpha^a(x) \frac{\sigma_a}{2} -i \alpha(x) \frac{Y}{2}} \psi_L$

$\psi_R \to e^{-i \alpha(x) \frac{Y}{2}} \psi_R$

$\phi \to e^{-i \alpha^a(x) \frac{\sigma_a}{2} -i \alpha(x) \frac{Y}{2}} \phi$

(a=1,2,3) so that the invariance requires the term $e^{i \frac{\alpha(x)}{2} (y_{L} - y_\phi - y_R)}$ to be equal to 1, i.e. $y_{L} - y_\phi - y_R = 0$ where $y_L,y_R$ and $y_\phi$ are respectively the weak-hypercharge of the left-handed doublet $\psi_L$, right-handed singlet $\psi_R$ and the Higgs doublet. (Your $U$ matrix was just $e^{-i \alpha^a(x) \frac{\sigma_a}{2}}$). In passing, the hypercharge of the Higgs ($y_\phi=1)$ is chosen accordingly explaining why the positively charged complex field is the upper component of the Higgs field ($Q = T_3 + \frac{Y}{2}$ with $T_3 = \frac{\sigma_3}{2}$).

Second, after the spontaneous symmetry breaking, the term is reduced to $v \bar{e}_R e_L$ (+the Higgs coupling) using $\psi_L = \begin{pmatrix}\nu_L\\ e_L\end{pmatrix}$ as you wrote. Now if you transform it under the symmetry, since the rule of transformation for the left-handed component of the electron differs from the one of the right-handed component, the term is no more invariant ($v$ being a simple number not affected by the transformation).


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