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: [itex] V = 1/2 \lambda^2 (|\phi|^2 -\eta^2)^2 [/itex].
Scalar field is small and we can expand its modulus around [itex] \eta [/itex]:

[tex]

\phi(x) = |\phi(x)| e^{i \alpha(x)} = (\eta + \frac{1}{\sqrt{2}} \phi(x)) e^{i \alpha(x)}
[/tex]

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: [itex] V = 1/2 \lambda^2 (|\phi|^2 -\eta^2)^2 [/itex].
Scalar field is small and we can expand its modulus around [itex] \eta [/itex]:

[tex]

\phi(x) = |\phi(x)| e^{i \alpha(x)} = (\eta + \frac{1}{\sqrt{2}} \phi(x)) e^{i \alpha(x)}
[/tex]

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


[tex]

\phi(x) = |\phi(x)| e^{i \alpha(x)} = (\eta + \frac{1}{\sqrt{2}} \rho(x)) e^{i \alpha(x)}.
[/tex]

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.