# Taylor expansion for SSB

1. Mar 27, 2013

### 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 $\eta$:

$$\phi(x) = |\phi(x)| e^{i \alpha(x)} = (\eta + \frac{1}{\sqrt{2}} \phi(x)) e^{i \alpha(x)}$$

How did he do that expansion???

2. Mar 27, 2013

### Fredrik

Staff Emeritus
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.

3. Mar 27, 2013

### fzero

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

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

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.

4. Mar 27, 2013

### 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.

5. Mar 28, 2013

### LayMuon

here is the attachment with that page from the book.

#### Attached Files:

• ###### mmm1.pdf
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6. Mar 28, 2013

### 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.

7. Mar 30, 2013

### LayMuon

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