• 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
5,026 questions , 2,180 unanswered
5,344 answers , 22,686 comments
1,470 users with positive rep
815 active unimported users
More ...

  normal degeneracy and the “span” of a symmetry group

+ 3 like - 0 dislike

In Tinkham's "Group Theory and Quantum Mechanics", Tinkham defines normal degeneracy so that the span of the action of the Hamiltonian's symmetry group on any energy eigenstate yields all possible eigenstates with that same energy.

He then goes on to say that under the condition of normal degeneracy, the degenerate subspaces form irreducible representations of the symmetry group.

I'm having trouble connecting the dots. By the definition of the symmetry group it is clear that the span of the action of the symmetry group on some energy eigenstate $v$, i.e.:
W_v \equiv span(\{\rho(g)v : \forall g \in G \}) 
(where $\rho(g)$ is the operator associated with the group element $g$ of the symmetry group $G$) does contain only eigenstates of the same energy eigenvalue as $v$, and indeed does form a representation, since, for any $w \in W_v$ and $g_j \in G$ ($G$ being finite):
\rho(g_j) w = \rho(g_j) \sum_i w_i \rho(g_i)v = \sum_i w_i \rho(g_j) \rho(g_i) v = \sum_i w_i \rho(g_j g_i) v \in W_v
It is only left to show that this representation is irreducible. I'm stuck here.

asked Feb 19, 2015 in Theoretical Physics by chris [ no revision ]

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).
To avoid this verification in future, please log in or register.

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

Your rights