How was the Taylor expansion for SSB in superconductors done?

[LayMuon]

I am reading about spontaneous symmtry breaking for superconductors and came a cross to this simple statement:

Here is the potential for complex scalar field: V = 1/2 \lambda^2 (|\phi|^2 -\eta^2)^2.
Scalar field is small and we can expand its modulus around \eta:

<br /> <br /> \phi(x) = |\phi(x)| e^{i \alpha(x)} = (\eta + \frac{1}{\sqrt{2}} \phi(x)) e^{i \alpha(x)}<br />

How did he do that expansion?

 

[Fredrik]

If that's just an expansion, then where did that $\eta$ come from? What is the relationship between $\eta$ and $\phi$?

I might help to provide an exact reference. Link directly to the relevant page at google books if that's possible.

 

[fzero]

I am reading about spontaneous symmtry breaking for superconductors and came a cross to this simple statement:

Here is the potential for complex scalar field: V = 1/2 \lambda^2 (|\phi|^2 -\eta^2)^2.
Scalar field is small and we can expand its modulus around \eta:

<br /> <br /> \phi(x) = |\phi(x)| e^{i \alpha(x)} = (\eta + \frac{1}{\sqrt{2}} \phi(x)) e^{i \alpha(x)}<br />

How did he do that expansion?

This expression is using $\phi$ for two related, but different quantities. It would be better to rewrite this as


<br /> <br /> \phi(x) = |\phi(x)| e^{i \alpha(x)} = (\eta + \frac{1}{\sqrt{2}} \rho(x)) e^{i \alpha(x)}.<br />

This formula defines a real scalar field $\rho$. You might try to rewrite the potential in terms of $\rho$ to get an idea of why one might want to do this field redefinition.

 

[LayMuon]

You are right, I think it was wrong in the text, there was no mentioning of this newly defined real field, no notation change, so I got confused.

 

[LayMuon]

If that's just an expansion, then where did that $\eta$ come from? What is the relationship between $\eta$ and $\phi$?

I might help to provide an exact reference. Link directly to the relevant page at google books if that's possible.

here is the attachment with that page from the book.

 

[fzero]

The book uses $\varphi$ (\varphi in TeX) to distinguish the newly defined field from $\phi$. I almost used it too, but figured the redefinition would be clearer with a completely different symbol.

 

[LayMuon]

The book uses $\varphi$ (\varphi in TeX) to distinguish the newly defined field from $\phi$. I almost used it too, but figured the redefinition would be clearer with a completely different symbol.

Yes, I was confused. For me phi is phi. it's interested how brain doesn't notice the difference even with close inspection.