• 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.


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

204 submissions , 162 unreviewed
5,030 questions , 2,184 unanswered
5,344 answers , 22,704 comments
1,470 users with positive rep
816 active unimported users
More ...

  Why does a spontaneously broken charge operator create exactly a one-Goldstone state?

+ 3 like - 0 dislike

It is commonly claimed [2] that, if $j^0$ is a charge (density) that generates a spontaneously broken symmetry transformation, then


It can be shown (cf. Weinberg Vol2, chap 19, equation (19.2.34)), that $j^0|\text{VAC}\rangle$ has nonzero overlaps with 1-Goldstone states [1], but how can I see it's not, e.g., a supeposition 1-Goldstone and 2-Goldstone states? Or is equation (1) meant to be taken as a defining formula for 1-Goldstone state? In any case, there seems to be another paradox: it can also be shown (cf. Weinberg Vol2, chap 19, equation (19.2.35)), that  $\phi|VAC\rangle$ has nonzero overlaps with 1-Goldstone states, where $\phi$ is the scalar field that acquires VEV in the model[1]. However, by Noether's theorem, $j^0$ is a quadratic function of $\phi$ and $\pi$ ($\pi$ being the canonical conjugate of $\phi$),  and if $\phi$ has a mode expansion at all, wouldn't $j^0$ have to contain a term that creates two particles?

Furthermore, in demonstrating the non-existence of parity doubling of hadronic spectrum, it is claimed $$j^0|h\rangle=|h, \text{1-Goldstone}\rangle\cdots(2),$$

where $|h\rangle$ is a 1-hadron state. Even if I take for granted that $j^0$ creats a 1-Goldstone when acting on vacuum, it's still not clear to me why it does not do anything to the hadron at all.

[1] But note in Weinberg's proof, nowhere did he explicitly defined what a 1-Goldstone state is. He only loosely defined 1-particle state as a state of which momentum is the only continuous index, and 1-Goldstone as a massless 1-particle state with nonzero overlap with both $j^\mu|\text{VAC}\rangle$ and $\phi|\text{VAC}\rangle$.

[2] For example, M. Schwartz, Quantum field theory and standard model

Also, Weinberg Vol 2, when talking about nonexistence of hadronic parity doubling:

asked Nov 3, 2015 in Theoretical Physics by Jia Yiyang (2,640 points) [ revision history ]
edited Nov 5, 2015 by Jia Yiyang

This is difficult to answer as the fields make sense only as interactive fields, about very little is known when one asks too precise questions. Where is (1) claimed?

By the way, it should be "cf." and "e.g.,", not "c.f." and "e.g,"

To your note [1]: Weinberg defines after (19.2.8) what a Goldstone boson is - a pole in the effective action corresponding to a massless mode of the second derivative of the potential. Corresponding to each such pole is an asymptotic particle (Peskin-Schroeder, end of Section 11.5) that participates in the S-matrix and defines a Hilbert space of asymptotic 1-boson states. Being massless, these states cannot decay, hence this 1-particle space is preserved by translations, hence agrees with the space of nonasymptotic 1-boson states.

@ArnoldNeumaier, I just added two references for equation (1) (and corrected c.f and e.g, of course :D)

about footnot [1], in his second proof of Goldstone theorem (the non-effective-potential, current-algebra-like approach), he writes down single Goldstone states such as $|B\rangle$, but without writing its full relation with field operators, except some partial information such as (19.2.34) and (19.2.35).

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