• 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.


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


(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,054 questions , 2,207 unanswered
5,347 answers , 22,726 comments
1,470 users with positive rep
818 active unimported users
More ...

  Higgs VEV on the language of coherent states

+ 2 like - 0 dislike

Is it correct to represent Higgs VEV as the coherent state?

I mean, suppose translational invariant coherent state

|\alpha\rangle = Ne^{\alpha \hat{a}^{\dagger}_{\mathbf p =0}}|\alpha =0\rangle, \quad N: \quad \langle \alpha|\alpha\rangle = 1

of massless particles with zero dispersion relation. In principle, it is possible to state that the non-shifted Higgs doublet operator has VEV on $|\alpha \rangle$ state. But is this correct?

asked Apr 28, 2016 in Theoretical Physics by NAME_XXX (1,060 points) [ revision history ]

With a cutoff, something like this holds with an appropriate Bogoliubov transformation. But a coherent state cannot be translation invariant. In the renormalized version the Bogoliubov transformation makes no longer sense.

@ArnoldNeumaier : when writing about absence of translational invariance of the coherent state, did You mean nonzero mass case? If no, could You explain why the coherent state isn't translation invariant?

With a cutoff, the translational symmetry is not present. Without a cutoff, the B-transformation that causes the VEV to shift makes no longer sense as an operator (it is divergend, not unitarily implementable).

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:
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