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$$V(\phi)=\frac{\lambda}{4}(\phi^\dagger \phi-\frac{1}{2}\nu^2)^2 $$

and states that because of this, $$\langle 0 | \phi(x) |0 \rangle=\frac{1}{\sqrt{2}}\begin{pmatrix}v \\ 0 \end{pmatrix}$$

However, I thought the proper procedure would be to write $$\phi=q+\tilde{\phi}$$, and substitute this

$$\tilde{V}(\tilde{\phi})=\frac{\lambda}{4}((q+\tilde{\phi})^\dagger (q+\tilde{\phi})-\frac{1}{2}\nu^2)^2 $$

Then you would calculate the minimum of the quantum effective potential $$\tilde{V}_{eff}(\tilde{\phi})$$, which will give you an equation that gives you 'q' in terms of 'ν' and your other couplings. You then plug this value into 'q' to get

$$\tilde{V}(\tilde{\phi})=\frac{\lambda}{4}((q(\nu,\lambda)+\tilde{\phi})^\dagger (q(\nu,\lambda)+\tilde{\phi})-\frac{1}{2}\nu^2)^2 $$

Moreover, any field M that multiplies the Higgs in an interaction now has a component

$$q(\nu,\lambda)*M$$

Did Srednicki skip all these steps? How can you just say that the vacuum expectation value of a field is the minimum of the classical potential? Is he renormalizing all his couplings so that the minimum of the classical potential is the minimum of the quantum effective potential?