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

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

145 submissions , 122 unreviewed
3,930 questions , 1,398 unanswered
4,873 answers , 20,701 comments
1,470 users with positive rep
502 active unimported users
More ...

Electronic configuration for singlet and triplet states

+ 0 like - 0 dislike
114 views

Is there a difference in the electronic configuration for singlet and triplet states? For example, He atom has 1s2 configuration in its ground state (singlet state)

But what about when the He atom is excited. It can come in both states (Signlet and Triplet)

Can I tell from the excited electronic configuration whether the atom is in a singlet or a triplet state?

Thanks!

This post imported from StackExchange Physics at 2014-04-13 12:28 (UCT), posted by SE-user Haneen Su
asked Apr 12, 2014 in Theoretical Physics by Haneen Su (0 points) [ no revision ]

1 Answer

+ 1 like - 0 dislike

I am not sure that I get your question right, but let me try to answer according to my understanding.

The spin part of the two electron wave function of the singlet state $|0,0\rangle=|l=0,m=0\rangle$ is

$|0,0\rangle=(|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle)/\sqrt{2}$

The three triplet states look like this:

$|1,0\rangle=(|\uparrow\downarrow\rangle+|\downarrow\uparrow\rangle)/\sqrt{2}$

$|1,1\rangle=|\uparrow\uparrow\rangle$

$|1,-1\rangle=|\downarrow\downarrow\rangle$

If you are unfamiliar with the notation, $\uparrow$ denotes spin $+1/2$ and $\downarrow$ denotes spin $-1/2$; the numbers on the left denote: the total angular momentum $l$ of both electrons and its $z$ component $m$.

If the electrons are in the same energy state, they have to have different spins, that is, the state can be either $|0,0\rangle$ or $|1,0\rangle$, so in principle singlet and triplet are possible. If one of the electrons is excited, any of the four states is possible, since the spins don't have to be different any more. Therefore, if you know that both electrons have spin up or both have spin down, you can be sure, that it is a triplet state. If one spin is up and the other down, you cannot tell from the spin configuration whether it is a singlet or a triplet, since again both $|0,0\rangle$ or $|1,0\rangle$ are possible.

This post imported from StackExchange Physics at 2014-04-13 12:28 (UCT), posted by SE-user Photon
answered Apr 12, 2014 by Photon (70 points) [ no revision ]
I'd suggest using $\uparrow$ and $\downarrow$ instead of $+$ and $-$ to denote spins, otherwise they look too mixed up when adding and subtracting kets.

This post imported from StackExchange Physics at 2014-04-13 12:28 (UCT), posted by SE-user Ruslan
I was too lazy to look up the TeX commands for them and didn't know them by heart, to be honest.

This post imported from StackExchange Physics at 2014-04-13 12:28 (UCT), posted by SE-user Photon
\uparrow and \downarrow

This post imported from StackExchange Physics at 2014-04-13 12:28 (UCT), posted by SE-user Ruslan
Thanks, edited.

This post imported from StackExchange Physics at 2014-04-13 12:28 (UCT), posted by SE-user Photon

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$\hbar$ysicsO$\varnothing$erflow
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
...