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
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

  Question about Landau's phase transition theory

+ 0 like - 0 dislike
723 views

I have a question about Landau's theory of quantum phase transition. In his model, the free energy is assumed to be

\begin{equation} F = f_0 + \alpha (T-T_c) \Delta^2 + \beta \Delta^4 \end{equation}

The ground state of the system depends strongly on the sign of $T-T_c$. In this way, we find that the scaling exponent near the critical point is $1/2$, which may be somewhat different from that in experiments -- as a results, we need renormalization group method to understand the discrepancy. This is a theory that has been accepted by this community.

OK, now my question is why in the second term the coefficient is $\alpha (T-T_c)$, instead of $\alpha (T-T_c)^\gamma$, where $\gamma$ is a constant, e.g., $\gamma = 3/5$ or $1/3$. This is perhaps a trivial problem, but has never been discussed explicitly in standard textbooks. The answer to this problem is not so straightforward for most of us.

A related question maybe like that: how to prove this point in experiments. Thanks very in advance.

This post imported from StackExchange Physics at 2014-05-04 11:12 (UCT), posted by SE-user Ming Gong
asked May 3, 2014 in Theoretical Physics by Ming Gong (0 points) [ no revision ]

2 Answers

+ 0 like - 0 dislike

As I remember, one of the most important problems of Landau theory that it can't describe critical points in a right manner, because simply it's not taking in account fluctuations, this answers your both questions. Moreover, when Landau proposed his theory, there was not suitable technologies to make accurately experiments to support or falsify landau theory very close to critical points, only letter it was understood that new theories as Izing model is required for right explanation of near crirical pount behaviour.

This post imported from StackExchange Physics at 2014-05-04 11:12 (UCT), posted by SE-user TMS
answered May 3, 2014 by TMS (40 points) [ no revision ]
+ 0 like - 0 dislike

In principle, since Landau theory is phenomenological, one could imagine that the quadratic term could have a non trivial dependence on the microscopic parameter (for example $(T-T_c)^{0.3}$). However, there are strong arguments that show that it is not the case when one is trying to justify Landau theory microscopically.

The main argument is that this kind of singular behavior is possible only if fluctuations at all length scales are taken into account. Indeed, if only some fluctuations are included, then the partition is just the sum of a finite number of exponential, and it is thus analytical. Furthermore, picking by hand the correct exponent (using the OP's exponent $\gamma$) renders the theory useless, since one is putting by hand the correct behavior, which is then no better than the scaling hypothesis.

Also, all calculations made for deriving microscopically the Landau free energy show that the coefficient are analytical function of the microscopic parameters (though the dependence on $T-T_c$ might be quadratic and not linear depending on the symmetries of the problem). See for example the case of the Ising model or Gorkov's derivation of the Ginzburg-Landau free energy for a superconductor.

This post imported from StackExchange Physics at 2014-05-04 11:12 (UCT), posted by SE-user Adam
answered May 3, 2014 by Adam (115 points) [ 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:
p$\hbar\varnothing$sicsOverflow
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
...