Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

  Goldstone phase initial value

+ 3 like - 0 dislike
711 views

Suppose we have relativistic system with spontaneously broken symmetry. For simplicity, let choose broken $U(1)$ symmetry:
$$
L = \frac{1}{2}(\partial_{\mu}\varphi )^{2} - V(|\varphi |) , V(|\varphi | = \varphi_{0}) = 0 \qquad (1)
$$

Let us parametrize Goldstone degree of freedom, $\varphi = \varphi_{0}e^{i\theta}$. Then instead of $(1)$ we will get

$$
L = \frac{\varphi_{0}^{2}}{2}(\partial_{\mu}\theta )^{2} \qquad (2)
$$

In fact, this leaves us with undetermined VEV of Goldstone degree of freedom. This VEV corresponds to the coherent state of $\theta$-particles with zero momentum. 

Suppose also that instead of $(1)$ we have the lagrangian with small explicit symmetry breaking term, which after substitution of anzats $\varphi = \varphi_{0}e^{i\theta}$ leads to appearance $-\frac{\varphi_{0}^{2}m^{2}}{2}\theta^2$ extra term in $(2)$:

$$
L' = \frac{\varphi_{0}^{2}}{2}(\partial_{\mu}\theta )^{2} - \frac{\varphi_{0}^{2}m^{2} \theta^{2}}{2} \qquad (2')
$$

The first question: in case of absense of explicit symmetry breaking, can we fix this VEV by some properties of underlying theory? Or it is completely undertermined and may take arbitrary values from $0$ to $2 \pi$?

The second question: in case of presence of explicit symmetry breaking term, can we immediately state that VEV of $\theta$ field is zero due to presence of mass term in $(2')$ which makes zero contribution in classical energy of $\theta$ field only if $\rangle \theta \langle = 0$?

The third extra question. Suppose that we know only effective theory with $\theta$ field and its interactions with other fields, and the lagrangian takes the form

$$
L{''} = \frac{1}{2}(\partial_{\mu}\theta )^{2} - \sum_{i}\partial_{\mu}\theta J^{\mu}_{i}c_{i}, \qquad (3)
$$

where $J_{\mu}^{i}$ denotes some vector or pseudovector currents, in dependence of what $\theta$ is. May we strictly state that due to scale invariant form of the lagrangian $(3)$ $\theta$ field arises in corresponding effective theory in a way nothing but as goldstone degree of freedom?

asked Jan 5, 2016 in Phenomenology by NAME_XXX (1,060 points) [ revision history ]
edited Jan 5, 2016 by NAME_XXX

Or it is completely undertermined and may take arbitrary values from 00 to 2π2π?

It is completely arbitrary.

n case of presence of explicit symmetry breaking term, can we immediately state that VEV of θθ field is zero due to presence of mass term in (2′)(2′) 

Seems so. This sometimes is called vacuum alignment.

But why it is completely arbitrary? $U(1)$ group is compact group.

But why it is completely arbitrary? U(1)U(1) group is compact group.

Why not? All vacuua are completely congruent to each other if you look at the potential. And what does it have to do with compactness here?

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysics$\varnothing$verflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...