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
5,082 questions , 2,232 unanswered
5,353 answers , 22,786 comments
1,470 users with positive rep
820 active unimported users
More ...

  Some ambiguous points on Spontaneous Symmetry Breaking (SSB)?

+ 3 like - 0 dislike
1252 views

Almost in every textbook of condensed matter physics, the standard description of SSB could be formulated as follows:

Consider the lattice Heisenberg model in an external magnetic field $H=\sum_{ij}J_{ij}\mathbf{S}_i\cdot\mathbf{S}_j+hS_z$, where $h$ is the magnitude of magnetic field and $S_z=\sum_iS_i^z$. Now the average magnetization per site is a function of both magnetic field $h$ and number of lattice sites $N$, say $m\equiv \sum_i\left \langle S_i^z \right \rangle/N=m(N,h)$, where $\left \langle S_i^z \right \rangle\equiv tr(\hat{\rho }S_i^z)$ with $\hat{\rho }=e^{-\beta H}/tr(e^{-\beta H})$ the density operator. Then if $$\lim_{h\rightarrow 0}\lim_{N\rightarrow \infty }m(N,h)\neq 0$$, we say the system has SSB at temperature $T$. Now I get some questions:

(1)We know at finite $N$ and zero $h$, $m(N,h=0)=0$ due to spin-rotation symmetry. But there is no reason for that $$\lim_{h\rightarrow 0}m(N,h)=m(N,h=0)—[1]$$, right? Since the function $m(N,h)$ may not be continuous at $h=0$, from the math viewpoint.

(2)If Eq.[1] is correct, and hence $\lim_{h\rightarrow 0}m(N,h)=0$, then $\lim_{N\rightarrow \infty }\lim_{h\rightarrow 0}m(N,h)=0$, right?

(3)If Eq.[1] is wrong, say $\lim_{h\rightarrow 0}m(N,h)\neq m(N,h=0)$ and hence $\lim_{h\rightarrow 0}m(N,h)\neq0$, then what about $$\lim_{N\rightarrow \infty }\lim_{h\rightarrow 0}m(N,h)?$$ And why don't we use this identity to define SSB?

Thank you very much.

This post imported from StackExchange Physics at 2014-03-09 08:42 (UCT), posted by SE-user K-boy
asked Oct 7, 2013 in Theoretical Physics by Kai Li (980 points) [ no revision ]
What you write in point (2) is correct, since $m(N,h)$ is an analytic function of $h$ (and $\beta$) when $N$ is finite (this should be clear by looking at how $m(N,h)$ depends of $h$). This is actually the reason one has to take the thermodynamic limit first.

This post imported from StackExchange Physics at 2014-03-09 08:42 (UCT), posted by SE-user Heidar
@ Heidar, so you mean the function $m(N,h)$ is always continuous at $h=0$ for finite $N$? But is there any physical argument or mathematical proof for this point? Thanks.

This post imported from StackExchange Physics at 2014-03-09 08:42 (UCT), posted by SE-user K-boy
@K-boy: You can see that at finite $N$, the partition function (or the expectation value) is just the sum of a finite number of exponentials. This is therefore analytic, and the limit $h\to 0$ is continuous.

This post imported from StackExchange Physics at 2014-03-09 08:42 (UCT), posted by SE-user Adam
@ Adam, good point. But from the math viewpoint, even though the number of summation terms is finite, there is still possibility that some of the terms are singularity. For example, the partition function $$Z=\sum_{\alpha=1}^{d} e^{-\beta E_\alpha(h)}$$ with $d=2^N$ the dimension of the Hilbert space of $N$ spin-1/2 system and $E_\alpha(h)$ the eigenenergy of $H$, if there exist some $E_\alpha(h)$ which are not continuous functions of $h$, then the whole story may be not continuous on $h$. Is this possible?

This post imported from StackExchange Physics at 2014-03-09 08:42 (UCT), posted by SE-user K-boy
@K-boy: Usually, one expects that the eigenvalues are analytical functions of $h$ if the Hilbert space is finite (though I don't have a proof of that). Of course, you can always invent a model where that's not the case, but that's not typical. Where you are right is that even though the system is finite, and that the eigenvalues are all analytical, the partition function can be non-analytical in $h$ in the limit $\beta\to \infty$ (you can show that with only one quantum spin). This corresponds to a kind of quantum phase transition associated with a level-crossing.

This post imported from StackExchange Physics at 2014-03-09 08:42 (UCT), posted by SE-user Adam
@ Adam Thanks for your good comment.

This post imported from StackExchange Physics at 2014-03-09 08:42 (UCT), posted by SE-user K-boy

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
...