I'm trying to understand the way that the Higgs Mechanism is applied in the context of a $U(1)$ symmetry breaking scenario, meaning that I have a Higgs complex field $\phi=e^{i\xi}\frac{\left(\rho+v\right)}{\sqrt{2}} $ and want to gauge out the $\phi_2$ field that is causing my off-diagonal term, in normal symmetry breaking. I present the following transformation rules that hold in order to preserve local gauge invariance in Spontaneous Symmetry breaking non 0 vev for $\phi_1$ :

$$ \begin{cases} \phi\rightarrow e^{i\theta}\phi\\ A_{\mu}\rightarrow A_{\mu}-\frac{1}{q}\partial_{\mu}\theta\\ D_{\mu}=\partial_{\mu}+iqA_{\mu} \end{cases} $$

As I understand, the Higgs gauge fixing mechanism is used to specify the transformations that gauge $\phi_2$ away. The idea is that we want to look for the angle that gives us a Higgs field with one real degree of freedom, as in

$$ \phi\rightarrow e^{i\theta}\phi=e^{i\theta}e^{i\xi}\frac{\left(\rho+v\right)}{\sqrt{2}}=e^{i\theta^{'}}\frac{\left(\rho+v\right)}{\sqrt{2}} $$

so

$$\begin{cases} \phi\rightarrow e^{i\theta}\phi=e^{i\theta}e^{i\xi}\frac{\left(\rho+v\right)}{\sqrt{2}}=e^{i\theta^{'}}\frac{\left(\rho+v\right)}{\sqrt{2}}\\ A_{\mu}\rightarrow A_{\mu}-\frac{1}{q}\partial_{\mu}\theta^{'} \end{cases} $$

where $\theta^{'}=\theta+\xi$. I omit some factors of $v$ on the exponential for the time being. That's what I see my books doing. For $\theta^{'}=0$ this becomes

$$ \begin{cases} \phi\rightarrow\frac{\left(\rho+v\right)}{\sqrt{2}}\\ A_{\mu}\rightarrow A_{\mu} \end{cases} $$ and the rest is the derived desired interactions and terms in general. I note that this does not preserve local gauge invariance because

$$\begin{cases} \phi\rightarrow e^{i\theta^{'}}\frac{\left(\rho+v\right)}{\sqrt{2}}\neq e^{i\theta^{'}}\phi\\ A_{\mu}\rightarrow A_{\mu}-\frac{1}{q}\partial_{\mu}\theta^{'} \end{cases} $$ So is this transformation the one that we do or have I wronged somewhere and it can be done correctly via a legit gauge transformation?

I'm trying to understand the way that the Higgs Mechanism is applied in the context of a $U(1)$ symmetry breaking scenario, meaning that I have a Higgs complex field $\phi=e^{i\xi}\frac{\left(\rho+v\right)}{\sqrt{2}} $ and want to gauge out the $\xi$ field that is causing my off-diagonal term, in normal symmetry breaking. I present the following transformation rules that hold in order to preserve local gauge invariance in Spontaneous Symmetry breaking non 0 vev for $\rho$ :

$$ \begin{cases} \phi\rightarrow e^{i\theta}\phi\\ A_{\mu}\rightarrow A_{\mu}-\frac{1}{q}\partial_{\mu}\theta\\ D_{\mu}=\partial_{\mu}+iqA_{\mu} \end{cases} $$

As I understand, the Higgs gauge fixing mechanism is used to specify the transformations that gauge $\xi$ away. The idea is that we want to look for the angle that gives us a Higgs field with one real degree of freedom, as in

$$ \phi\rightarrow e^{i\theta}\phi=e^{i\theta}e^{i\xi}\frac{\left(\rho+v\right)}{\sqrt{2}}=e^{i\theta^{'}}\frac{\left(\rho+v\right)}{\sqrt{2}} $$

so

$$\begin{cases} \phi\rightarrow e^{i\theta}\phi=e^{i\theta}e^{i\xi}\frac{\left(\rho+v\right)}{\sqrt{2}}=e^{i\theta^{'}}\frac{\left(\rho+v\right)}{\sqrt{2}}\\ A_{\mu}\rightarrow A_{\mu}-\frac{1}{q}\partial_{\mu}\theta^{'} \end{cases} $$

where $\theta^{'}=\theta+\xi$. I omit some factors of $v$ on the exponential for the time being. That's what I see my books doing. For $\theta^{'}=0$ this becomes

$$ \begin{cases} \phi\rightarrow\frac{\left(\rho+v\right)}{\sqrt{2}}\\ A_{\mu}\rightarrow A_{\mu} \end{cases} $$ and the rest is the derived desired interactions and terms in general. I note that this does not preserve local gauge invariance because

$$\begin{cases} \phi\rightarrow e^{i\theta^{'}}\frac{\left(\rho+v\right)}{\sqrt{2}}\neq e^{i\theta^{'}}\phi\\ A_{\mu}\rightarrow A_{\mu}-\frac{1}{q}\partial_{\mu}\theta^{'} \end{cases} $$ So is this transformation the one that we do or have I wronged somewhere and it can be done correctly via a legit gauge transformation?

I'm trying to understand the way that the Higgs Mechanism is applied in the context of a U(1) symmetry breaking scenario, meaning that I have a Higgs complex field $\phi=e^{i\xi}\frac{\left(\rho+v\right)}{\sqrt{2}} $ and want to gauge out the $\phi_2$ field that is causing my off-diagonal term, in normal symmetry breaking. I present the following transformation rules that hold in order to preserve local gauge invariance in Spontaneous Symmetry breaking non 0 vev for $\phi_1$ :

$$ \begin{cases} \phi\rightarrow e^{i\theta}\phi\\ A_{\mu}\rightarrow A_{\mu}-\frac{1}{q}\partial_{\mu}\theta\\ D_{\mu}=\partial_{\mu}+iqA_{\mu} \end{cases} $$

As I understand, the Higgs gauge fixing mechanism is used to specify the transformations that gauge $\phi_2$ away. The idea is that we want to look for the angle that gives us a Higgs field with one real degree of freedom, as in

$$ \phi\rightarrow e^{i\theta}\phi=e^{i\theta}e^{i\xi}\frac{\left(\rho+v\right)}{\sqrt{2}}=e^{i\theta^{'}}\frac{\left(\rho+v\right)}{\sqrt{2}} $$

so

$$\begin{cases} \phi\rightarrow e^{i\theta}\phi=e^{i\theta}e^{i\xi}\frac{\left(\rho+v\right)}{\sqrt{2}}=e^{i\theta^{'}}\frac{\left(\rho+v\right)}{\sqrt{2}}\\ A_{\mu}\rightarrow A_{\mu}-\frac{1}{q}\partial_{\mu}\theta^{'} \end{cases} $$

where $\theta^{'}=\theta+\xi$. I omit some factors of $v$ on the exponential for the time being. That's what I see my books doing. For $\theta^{'}=0$ this becomes

$$ \begin{cases} \phi\rightarrow\frac{\left(\rho+v\right)}{\sqrt{2}}\\ A_{\mu}\rightarrow A_{\mu} \end{cases} $$ and the rest is the derived desired interactions and terms in general. I note that this does not preserve local gauge invariance because

$$\begin{cases} \phi\rightarrow e^{i\theta^{'}}\frac{\left(\rho+v\right)}{\sqrt{2}}\neq e^{i\theta^{'}}\phi\\ A_{\mu}\rightarrow A_{\mu}-\frac{1}{q}\partial_{\mu}\theta^{'} \end{cases} $$ So is this transformation the one that we do or have I wronged somewhere and it can be done correctly via a legit gauge transformation?

I'm trying to understand the way that the Higgs Mechanism is applied in the context of a $U(1)$ symmetry breaking scenario, meaning that I have a Higgs complex field $\phi=e^{i\xi}\frac{\left(\rho+v\right)}{\sqrt{2}} $ and want to gauge out the $\phi_2$ field that is causing my off-diagonal term, in normal symmetry breaking. I present the following transformation rules that hold in order to preserve local gauge invariance in Spontaneous Symmetry breaking non 0 vev for $\phi_1$ :

$$ \begin{cases} \phi\rightarrow e^{i\theta}\phi\\ A_{\mu}\rightarrow A_{\mu}-\frac{1}{q}\partial_{\mu}\theta\\ D_{\mu}=\partial_{\mu}+iqA_{\mu} \end{cases} $$

As I understand, the Higgs gauge fixing mechanism is used to specify the transformations that gauge $\phi_2$ away. The idea is that we want to look for the angle that gives us a Higgs field with one real degree of freedom, as in

$$ \phi\rightarrow e^{i\theta}\phi=e^{i\theta}e^{i\xi}\frac{\left(\rho+v\right)}{\sqrt{2}}=e^{i\theta^{'}}\frac{\left(\rho+v\right)}{\sqrt{2}} $$

so

$$\begin{cases} \phi\rightarrow e^{i\theta}\phi=e^{i\theta}e^{i\xi}\frac{\left(\rho+v\right)}{\sqrt{2}}=e^{i\theta^{'}}\frac{\left(\rho+v\right)}{\sqrt{2}}\\ A_{\mu}\rightarrow A_{\mu}-\frac{1}{q}\partial_{\mu}\theta^{'} \end{cases} $$

where $\theta^{'}=\theta+\xi$. I omit some factors of $v$ on the exponential for the time being. That's what I see my books doing. For $\theta^{'}=0$ this becomes

$$ \begin{cases} \phi\rightarrow\frac{\left(\rho+v\right)}{\sqrt{2}}\\ A_{\mu}\rightarrow A_{\mu} \end{cases} $$ and the rest is the derived desired interactions and terms in general. I note that this does not preserve local gauge invariance because

$$\begin{cases} \phi\rightarrow e^{i\theta^{'}}\frac{\left(\rho+v\right)}{\sqrt{2}}\neq e^{i\theta^{'}}\phi\\ A_{\mu}\rightarrow A_{\mu}-\frac{1}{q}\partial_{\mu}\theta^{'} \end{cases} $$ So is this transformation the one that we do or have I wronged somewhere and it can be done correctly via a legit gauge transformation?

I'm trying to understand the way that the Higgs Mechanism is applied in the context of a U(1) symmetry breaking scenario, meaning that I have a Higgs complex field $\phi=e^{i\xi}\frac{\left(\rho+v\right)}{\sqrt{2}} $ and want to gauge out the $\phi_2$ field that is causing my off-diagonal term, in normal symmetry breaking. I present the following transformation rules that hold in order to preserve local gauge invariance in Spontaneous Symmetry breaking non 0 vev for $\phi_1$ :

$$ \begin{cases} \phi\rightarrow e^{i\theta}\phi\\ A_{\mu}\rightarrow A_{\mu}-\frac{1}{q}\partial_{\mu}\theta\\ D_{\mu}=\partial_{\mu}+iqA_{\mu} \end{cases} $$

As I understand, the Higgs gauge fixing mechanism is used to specify the transformations that gauge $\phi_2$ away. The idea is that we want to look for the angle that gives us a Higgs field with one real degree of freedom, as in

$$ \phi\rightarrow e^{i\theta}\phi=e^{i\theta}e^{i\xi}\frac{\left(\rho+v\right)}{\sqrt{2}}=e^{i\theta^{'}}\frac{\left(\rho+v\right)}{\sqrt{2}} $$

so

$$\begin{cases} \phi\rightarrow e^{i\theta}\phi=e^{i\theta}e^{i\xi}\frac{\left(\rho+v\right)}{\sqrt{2}}=e^{i\theta^{'}}\frac{\left(\rho+v\right)}{\sqrt{2}}\\ A_{\mu}\rightarrow A_{\mu}-\frac{1}{q}\partial_{\mu}\theta^{'} \end{cases} $$

where $\theta^{'}=\theta+\xi$. I omit some factors of $v$ on the exponential for the time being. That's what I see my books doing. For $\theta^{'}=0$ this becomes

$$ \begin{cases} \phi\rightarrow\frac{\left(\rho+v\right)}{\sqrt{2}}\\ A_{\mu}\rightarrow A_{\mu} \end{cases} $$ and the rest is the derived desired interactions and terms in general. I note that this does not preserve local gauge invariance because

$$\begin{cases} \phi\rightarrow e^{i\theta^{'}}\frac{\left(\rho+v\right)}{\sqrt{2}}\neq e^{i\theta^{'}}\phi\\ A_{\mu}\rightarrow A_{\mu}-\frac{1}{q}\partial_{\mu}\theta^{'} \end{cases} $$

Am I looking at this the right way?

I'm trying to understand the way that the Higgs Mechanism is applied in the context of a U(1) symmetry breaking scenario, meaning that I have a Higgs complex field $\phi=e^{i\xi}\frac{\left(\rho+v\right)}{\sqrt{2}} $ and want to gauge out the $\phi_2$ field that is causing my off-diagonal term, in normal symmetry breaking. I present the following transformation rules that hold in order to preserve local gauge invariance in Spontaneous Symmetry breaking non 0 vev for $\phi_1$ :

$$ \begin{cases} \phi\rightarrow e^{i\theta}\phi\\ A_{\mu}\rightarrow A_{\mu}-\frac{1}{q}\partial_{\mu}\theta\\ D_{\mu}=\partial_{\mu}+iqA_{\mu} \end{cases} $$

As I understand, the Higgs gauge fixing mechanism is used to specify the transformations that gauge $\phi_2$ away. The idea is that we want to look for the angle that gives us a Higgs field with one real degree of freedom, as in

$$ \phi\rightarrow e^{i\theta}\phi=e^{i\theta}e^{i\xi}\frac{\left(\rho+v\right)}{\sqrt{2}}=e^{i\theta^{'}}\frac{\left(\rho+v\right)}{\sqrt{2}} $$

so

$$\begin{cases} \phi\rightarrow e^{i\theta}\phi=e^{i\theta}e^{i\xi}\frac{\left(\rho+v\right)}{\sqrt{2}}=e^{i\theta^{'}}\frac{\left(\rho+v\right)}{\sqrt{2}}\\ A_{\mu}\rightarrow A_{\mu}-\frac{1}{q}\partial_{\mu}\theta^{'} \end{cases} $$

where $\theta^{'}=\theta+\xi$. I omit some factors of $v$ on the exponential for the time being. That's what I see my books doing. For $\theta^{'}=0$ this becomes

$$ \begin{cases} \phi\rightarrow\frac{\left(\rho+v\right)}{\sqrt{2}}\\ A_{\mu}\rightarrow A_{\mu} \end{cases} $$ and the rest is the derived desired interactions and terms in general. I note that this does not preserve local gauge invariance because

$$\begin{cases} \phi\rightarrow e^{i\theta^{'}}\frac{\left(\rho+v\right)}{\sqrt{2}}\neq e^{i\theta^{'}}\phi\\ A_{\mu}\rightarrow A_{\mu}-\frac{1}{q}\partial_{\mu}\theta^{'} \end{cases} $$ So is this transformation the one that we do or have I wronged somewhere and it can be done correctly via a legit gauge transformation?