Anderson’s article Saturday, May 6 2006
Blogroll and Condensed Matter Physics and Symmetry 7:49 pm
Background: I guess quantum mechanics would do. Since, as usual, high-energy physics is dominating the posts here, it’s time for some condensed matter stuff. Let us start with what is arguably the most famous article in CMP.
Link : P. W. Anderson, “More is different”, Science 177, 393 (1972)
(via the blog http://nanoscale.blogspot.com/ which is one of the very few blogs dealing with condensed matter physics.)
This article outlines what I call the “Anderson philosophy behind CMP” 
In case you didn’t know, P.W.Anderson(Physics Nobel Laureate(1977) - is arguably among the most famous physicists in condensed matter physics. You can find his website at Princeton here and here.
And people who are into condensed matter physics would find this link to Journal Club for Condensed Matter Physics from Bell-labs interesting ….
Posted by : Loganayagam.R.
May 19, 2006 at 11:12 pm
I have one question related to symmetry breaking in the context of stat-mech. Consider a classical magnet with short range exchange interaction above its transition temperature. Now in principle it is possible that you turn off all sources of external magnetic field exactly. Since symmetry breaking always require an infinitesimal external magnetic field (that’s what decides the direction of overall magnetization), what would happen if you take such a magnet below its T_c. It seems to me that symmetry would remain unbroken even in the thermodynamics limit for such a case. You may well argue that always tiny fields are present but I am considering a question of principle, not practice … but I may well be wrong.
Now what happens if you treat everything quantum field theoretically. For simplicity consider only photons and electrons as the only existing particles in the problem (enough for exchange interaction).
Is there some non-zero probability that symmetry might get broken now?
ps:these thoughts was generated after reading a section of Leggett’s ‘Problems in Physics’ which also raises this question in classical context. Incidentally, its fourth chapter is very much along the lines of Anderson’s ‘More is Different’.
June 17, 2006 at 6:13 pm
Tarun:
I think the configuration you are referring to would correspond to an unstable state. And yes, in principle just like in any other case of other unstable equilibirum the system would remain there if there is absoltely no external disturbance (here magnetic field).
This case may be of very little interest even from a theoretical viewpoint because I suppose the spins would align in a specific direction even in the presence of the minutest fields
June 17, 2006 at 6:27 pm
Maybe I dont quite comprehend how big a thing this concept of broken symmetry is in detrmining the flavour of physics as we go to higher length sales, but I must confess that I didnt find the article very impressive.
Most of it was a little too obvious. I dont know what Anderson had in mind but I am sure no theoretical particle physicist, however arrogant (and stupid) he might be, would have insisted that a theory of everything would imply that rich and diverse fields like chemistry ( leave alone molecular biology and the rest) would all become redundant or diminish in importance. It is well accepted and understood that problems at different scales cannot all be tackled using a reductionist approach. ALso, that theories at all length scales have unique and exciting features to them is not something that any respectable scientist would set out to deny.
June 17, 2006 at 9:08 pm
Venky,
As I understand it, the question Tarun has raised deserves a bit deeper answer.
If it is a finite number of spins and you treat them classically and exactly, it is clear that each and every spin will stay as it is left initially.(Since the Hamiltonian is trivially zero in the absence of the magnetic field.) For such a system, the very concept of temperature is ill-defined.
In fact, for the thermal description to be brought in, a thermodynamic limit is to be taken. Which is precisely what Anderson is insisting in the article - often, understanding large N systems require that we take the N → ∞ limit, take the concepts which arise only in that limit (e.g, broken symmetry) and apply it back to finite N systems (See the second para in page no.4 of the above article).
Now, coming back - from the way Tarun is posing the question, the question is what happens when a)thermodynamic limit is taken b) Quantum effects are taken into account c) if the system is treated quantum field theoretically - in which we combine the continuum description with a quantum description ? Let us anlyse these cases one by one.
a)Tarun’s opinion is that in the classical case, the symmetry would be unbroken even in the thermodynamic limit- i.e, the system will be in the “thermodynamic” state of zero “macroscopic” magnetisation. The problem I have with this idea is that this particular state is really a singularity in the configuration space - it has infinite (free)energy for one and hence one can argue that this state is to be excised out of the configuration space. And once this intermediate state is gone, we really have two configuration spaces and hence two separate systems - between which we have to choose thus leading to broken symmetry.
Let me clarify further - I see this as a case analogous to what one meets in study of limits in calculus. We all know of cases in calculus when the limit is not well-defined because the limit of a function depends on the way you approach the limit point. In such cases, you just cannot say by fiat that the limit is the average of the left-limit and the right-limit. In the same way, the thermodynamic limit at zero magnetic field is different depending on the way you approach that zero magnetic field. Thus, the limit is strictly speaking singular and ill-defined at zero field and the only we way we can proceed is to break the symmetry and choose one of the ground states. In that sense, the singularity forces us to break the symmetry and this is the source of broken symmetry as I understand it. I might well be wrong, and I would certainly be interested in what Legget has to say about this in his book, which unfortunately I’m yet to see.
(contd.)
June 19, 2006 at 6:12 am
(contd.)
Now, of course, we do have an experimental way of having symmetry breaking/phase transition - to take a classical system and cool it below its transition temperature. It is possible that there are such “supercooled” systems which are in a kind of unstable equilibrium (that venky mentions). But, my doubts revolve around a question of principle - in an infinite system at the thermodynamic limit, how much sense does it make to actually include the non-equilibrium state as a possible state of the system ? May be you can do some kind of (non-equilibrium) statistical mechanical analysis, but can you really do thermodynamics ?
And venky, I don’t agree with your assesment that
For example, I will be very much interested in trying to understand how such a ssytem returns to equilibrium, in formulating models which capture the essentials of such a system far from equilibrium.
b)& c) What tarun asks next is something I would like to understnd better too - His question is what new physics does a quantum mechanical/qunatum field theoretical treatment bring in ? It is not immediately clear that everything will be as before.
For example, for a finite system with many classical ground states(say think of a particle in a double well in 1+1d), quantum mechanics can actually lead to a unique ground state due to quantum tunneling. A similar thing is also possible in field theory - the instantons business in QFT reminds us that.
Hence, the question - how does the actual symmetry breaking take place in QFT ? Tarun asks,
Now, this question makes me uncomfortable for a similar reason as before. Anyway, I am seriously out of my depth here and that is the reason I was hesitant to answer tarun before. So, I will stop my comments here…
June 20, 2006 at 1:05 pm
It is not so clear to me.How can the Hamiltonian be zero when there is a nearest neighbour interaction. Secondly why cant we define a temperature for a finite system? A system which is in thermal equilibirum with a heat bath maintained at a certain temperature is said to have the same temperature, right?
Isnt the singularity that you are tallking about arise because the spins are infinite in number? Why cant an infinite system possess infinte energy?
Just wanted to clarify if you had meant unstable equilibirum there ?
June 20, 2006 at 11:40 pm
Oops ! Sorry, it was my mistake. I forgot that you need an Ising interaction for a ferromagnetic transition. Anyway, my point is still this - if you treat the finite system exactly and classically, thermodynamic description doesnot enter the picture - what is the temperature for a two-spin system for example ?
You still have infinite degrees of freedom sitting in the heat bath ! And you can’t treat it “exactly classically” after you put a heat bath. And then there are systems for which the canonical ensemble simply doesnot exist/is inequivalent to microcanonical description….
Of course, in practice we do approximate finite systems with a lot of particles by an infinite system and assign the temperature of that infinite system to that finite one, which is one example of Anderson’s statement about taking the conceps from N → ∞ limit and applying it back to finite N systems. But, the question is - as a matter of principle,
Or to take a more simple example that Anderson mentions in the article - a collection of finite number of nucleons doesnot in principle have a “shape” but in practice, nucleus in some cases does to a very good approximation behave like it has a shape.
Yes, right. You caught me on that one. What matters of course, is the free-energy density -
if extensivity holds,the total free energy of almost any infinite system diverges ! So, I’ve to throe out the baby with the bathwater if I want to excise that point for that reason…
Now that you point it out, I am a bit more unsure/confused than before. I should probably do a bit more of thinking/reading on it. Meanwhile, what do you think is the criterion we should use to define what enters the configuration-space of an infinite system ? We obviously can’t allow arbitrarily singular configurations - our ability to say anything about the system would be seriously impaired that way and more importantly, arbitrarily singular configurations are unphysical. Now, if as you’ve mentioned (and I’ve acquiesced) infinite free-energy is not a good criterion, what is a good criterion then ?
Yes, unstable equilibrium. And by the way, there does exist a field called Non-equilibrium thermodynamics.
June 21, 2006 at 9:21 pm
Some comments:
1) I too agree that the question lies outside the realm of equilibrium
thermodynamics. Thus it is probably a question of kinetics as well as
energetics. Also since a classical ising model does not have any dynamics
one would need to specify some kind of mechanism for spin-flip etc say
Glauber dynamics/kawasaki dynamics (don’t worry if you don’t know these
terms, it is just artificial way to give spins a dynamics ) . For quantum
mechanical spins, dynamics is always there but if one wants to treat
heat-bath also quantum mechanically then it may become quite difficult to
proceed (even for equilibrium stat-mech such models are difficult to
solve).
One may restate the original question more sharply : do one need an
EXTERNAL magnetic field to take the system from its non-equilibrium state
to the equilbrium one. Since we are taking thermodynamic field the
problems related to defining temperature do not arise anymore than in
other contexts. Also in this problem there is no question of order of
limits. The external magnetic field is identically zero. Symbolically (h=
magnetic field, N = system size) for Ising-like symmetry:
a) Lt h->0, Lt N-> infinity, T< T_c {spin system} —– Sym Breaking with
magnetisation determined by h-> 0+ or 0-
b) Lt N-> infinity, Lt h-> 0, T< T_c {spin system} —– NO Sym break
c) h=0 Lt N-> Infinity, T< T_c {spin system} ——- ??
It is the part c) that we are interested in here. It may well be that
there is no universal answer and the answer depends on initial
configuration as well as dynamics chosen to describe the system.
2) nayagam wrote:
(I don’t know command for quoting someone, help please) ( <blockquote> tag added - Tom. )
for equilibrium stat mech and in particular for Ising model on a lattice,
ofcourse one needs to include all configuartions in the partition
function. If one is below T_c, the m=0 configuration would not contribute
substantially as it would have much more energy than the true equilibrium
configuration. But we can see that we can’t answer the question of
approach to equilibrium by writing partition function alone.
ofcourse talking in general about which configuration to include, there is
some problem if one writes a field theory instant of a lattice theory. The
partition function is formally same as path integral. For ising sym, one
can write down the phi-4 theory. If one now allows all possible
configurations one may get ultraviolet divergences which you can remedy by
putting a lattice (`regularization’). Even then probably one can’t define
partition function in a very mathematical precise way and probably put
some conditions like differntiability etc.
But having said that, the non-equilibrium configuration we are concerned
about is not singular at all. In fact it just corresponds to all spins
being scrambled in some way to give net zero magnetization. Its only
`singular’ characteristic is that it is highly improbable from energetics
point of view.
3) Regd. the effect of quantum mechanics there is one nice
(unrelated) example that shows how QM can lead to ordering in systems
which are expected to be disordered classically.
probably people reading this are aware of frustration of classical spins
on triangular lattice with antiferromagnetic (AF) spin exchange. Basically
triangular lattice is non-bipartite so one can’t `satisfy’ all bonds (just
draw one triangle with spins at the vertices and this becomes eveident).
Now quantum mechanically, the disordering effect of such frustration could be
reduced as spins can be in superposition of `up’ and `down’
configurations. And it can be proved that at T=0 (when quantum
effects prevail) there is a phase transition separating disordered and
ordered phases.
Apart from this, quantum mechanically the order and the associated phase
transition may be very differnt from the `usual’ phase transitions e.g. in
Quantum hall effect the order is topological in nature (i.e. there is no
local order parameter, the ordering is defined in terms of loops/contours
and such. A spin-system analogy would be where < m > =0 on both sides of
phase transition , but the average of product of spins defined along
closed paths show some non-trivial behaviour unlike the disordered case)
It’s just my (wild) speculation that something like this could happen in
our problem and I may be completely wrong (in the frustration example in
the previous para the order is usual in the sense as local magnetization
becomes non-zero across the phase transition)
Note (24/6) : I’ve removed the broken comment and have uploaded the original comment that Tarun had mailed me - Tom .
June 21, 2006 at 9:56 pm
Hi tarun,
What I was worried about was not the order of limits at all but about the side from which the limit is taken , but rather that the limit h → 0+, N → ∞ and the limit h → 0- , N → ∞ were different. So, in what sense can we say that the limit h → 0 , N → ∞ exists ?
And since I can’t figure out what was the problem that you faced with posting the previous comment, just check out our FAQ “How do I type math ?” and see whether that is helpful. WordPress does behave sometimes weirdly and sorry for the inconvenience….
Update (24/6) : I’ve uploaded your comment above and have deleted the broken comments. The problem seems to be that whenever you type a less than sign (<
June 22, 2006 at 9:45 pm
I think you need to explain why this would prevent us from defining temperature for a system with finite number of spins. Dosent the statement that a system has a temperature T just imply that the probability that it has energy E is proportional to exp(-E/kT)?
And why is that?
July 29, 2006 at 3:17 pm
Sorry for replying so late. I still need quiet some time to think about the kind of issues that have been raised.
Anyway, here we go again.
Well, the answer is I guess it depends - which is indeed a frustrating answer. The point is that it depends on what you take as your starting point of equilibrium statistical mechanics .
You can probably take it to be the study of a deterministic system which is rendered probabilistic in a very specific way(given by boltzmann distribution) by some kind of an external system(a heat bath). If I understand you rightly, this is what you suggest above and your definition of temperature is that it is the parameter specifying the probability distribution. This in particular lands you directly on the canonical ensemble and you can probably proceed to do standard stat.mech. without much difficulty.
But, to me, this approach does raise some questions - a) What is the origin of probablistic dynamics in the bath of classical equilibrium statistical mechanics ? (Even if you perchance dismiss it off as a question outside the realm of equilibrium statistical mechanics, you cannot deny the fact that it is a question that ought to be addressed.)The dynamics of a particular classical system coupled to a very big classical system is by very definition deterministic.
My attempt at answering that question is as follows - The deterministic system(here, the very big system) completely mimics an ideal probablistic system(called the heat bath) at the N → ∞ limit. So, even though no real system is really a heat bath - they can be treated so by approximating them with their siblings at the N → ∞ limit.
The second question is this
b) Say you’re talking about a finite number of spins and you’ve defined equilibrium statistical mechanics using canonical description - Do you really have the means to establish the equivalence of this description with the rest of equilibrium statistical mechanics ? (say the microcanonical description ? )
Just to clarify what I’m worrying about - think of calculating the entropy of a finite number of non-interacting spins under an external magnetic field. A micro-canonical calculation involves counting the number of microstates for a fixed energy macrostate and then taking its logarithm. You end up with an expression with logarithms of factorials.
A canonical calculation proceeds by calculating the partition function , getting the free energy by taking it’s logarithm and then getting the entropy by differentiating it. My problem is this - the two expressions of the entropy are different ! The question is what is really the entropy of the system ? (In case you think entropy is not directly measurable, it is easy to establish that this ambiguity extends to all thermodynamic variables including say temperature.)
Usually, we’re not much worried about this issue - we take the factorials in the microcanonical expression, apply Stirling’s approximation and it’s easy to establish that both expressions are same under this approximation. But, then, to invoke Stirling’s approximation we really need the integers inside the factorials to be large, i.e, we need N → ∞ limit.
The point is that if the two descriptions contradict each other - which one would you choose ? The principle of apriori probablilty is more clearly stated in microcanonical description, but canonical description is the more convenient. Or should we talk of a microcanonical temperature as distinct from canonical temperature and which temperature do experimentalist dealing with finite systems measure in the lab ?
August 6, 2006 at 1:24 am
I cant see how that could be true. In the first place, the large system (acting as a heat bath)is energetically closed or rather, undergoes changes in energy(being in contact with the original system under consideration) of a scale that is negligible to its total energy. So it dosent make sense to talk of probabilistic distribtuion of energies.
The way I understand is that a large system would have to be in thermal equlibirum with some other equally sized or even bigger system and in such a configuration, its energy distribution would give the temperature. Once the temperature is obtained, the system can be isolated (and closed) and can be used as a reservoir for systems that are much smaller than itself (like the original system).
I also dont see the problem with this apparent disparity in canocial/microcanonical description. As a matter of fact, why should the two models give the same result when they describe different situations -one isolated and the other in contact with a heat bath. It is fortuitous that they give nearly the same answer for most systems we deal with (like classical gas) becuase we can easily approximate the degrees of freedom to be infinity.
April 10, 2008 at 7:52 am
AT FINITE TEMPERATURE THERE WILL BE FLUCTUATIONS WHICH WILL DRIVE THE SYSTEM TO A LOW ENERGY STATE IN COMPETITION WITH THE INTERACTION.