Mathematical explanation of why Higgs has a vev If Higgs ($\phi$) is a complex doublet:
$ \phi_{1}+i\phi_{2}$
$\phi_{3}+i\phi_{4}$
how do I show that $\phi_3$ has a vev but the others do not?
$V(\phi)=\mu^{2}(\phi^{\dagger}\phi)+\lambda(\phi^{\dagger}\phi)^{2}$ where $\phi^{\dagger}=1/\sqrt{2}(\phi_{1}-i\phi_{2})$
How do I determine these following values:
$\dfrac{dV}{d\phi_{1}}=0$, $\dfrac{dV}{d\phi_{2}}=0$, $\dfrac{dV}{d\phi_{3}}=v$, $\dfrac{dV}{d\phi_{4}}=0$
This is what I have done but I'm not sure it's correct, because I'm getting zero for all four:
$\dfrac{dV}{d\phi_{1}}=\mu^{2}\phi_{1}+\lambda\phi_{1}^{3}+\lambda\phi_{1}\phi_{2}^{2}=0=>\phi_{1}(\mu^{2}+\lambda\phi_{1}^{2}+\lambda\phi_{2}^{2})=0=>\phi_{1}=0$
 A: You can't show that the Higgs vev affects one particular component or the real and not imaginary part - all the components and phases are equally good. In fact, the potential only depends on 
$$ R = \phi^\dagger \phi = (\phi_1)^2+(\phi_2)^2+(\phi_3)^2+(\phi_4)^2 $$
so only the value of $R$ is determined by minimizing the potential. Its derivative is
$$ 0 = \frac{\partial V}{\partial R} = \mu^2+2\lambda R $$ 
which implies that 
$$R = -\mu^2/2\lambda$$
Note that $\mu^2$ should be negative – or you wrote a wrong sign in front of it – if you want to have solutions at a positive $R$. This wrong sign is the reason why you concluded that there are no nonzero solutions: indeed, with the positive coefficients in all terms of the potential, zero is the only stationary point. 
The "length" of $\phi$ may be calculated as the square root of $R$.
But once you know the magnitude of $R=\phi^\dagger \phi$, all the directions in the 4-dimensional space are equally good, so there is a whole 3-real-dimensional sphere of solutions. All of them are related by the $SU(2)$ or $SU(2)\times U(1)$ gauge transformations to any other solution from the 3-sphere worth of solutions. So by choosing an appropriate gauge transformation, we can make $\phi_3$ nonzero while others may be chosen to vanish.
