Renormalisation in 1D plaquette like ising model.

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  • #1
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Hi guys,

I'm working through past papers and I have a problem with deriving the renormalised scaling of the following:

[PLAIN]http://dl.dropbox.com/u/16658950/helpme.JPG [Broken]

I'm doing the rescaling as I would for a 1D ising model decimated with l = 2 (so every other spin, but N=4 in this case (number of spins). I think the fact that it's N=4 means I'm missing something in my derivation :-S. Does anybody have any ideas where I could be going wrong or what I'm missing?

Cheers!
 
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  • #2
Mute
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So, you have a four spin system and the problem as stated asks you to compute the partition function of the system, then to calculate the rescaled coupling after decimating two of the spins. The first part is hopefully straightforward:

[tex]Z = \mbox{Tr}(\exp(-H))[/tex]

where Tr means trace, and is basically just a sum over the possible values of all the spins. To decimate the spins, we almost do the same thing, except we only sum over the spins to be decimated. So, the coarse grained hamiltonian is

[tex]\exp(-H'(S,S')) = \sum_{S_1,S_2} \exp(-H(S,S',S_1,S_2))[/tex]

Once you've performed the sum, your goal is to hammer on H' until it looks like H, with new coupling constants K' and [itex]K'_0[/itex] (plus possibly new interaction terms, in general).

So, what have you tried so far? Have you done the above steps, or have you been trying to do this some other way? Show us some work and we can help pinpoint where you're going wrong.

Edit: okay, I just worked through it myself. Some useful hints: you'll get [itex]exp(-H') = \exp(\mbox{something})[/itex], having written the right hand side to look like an exponential, and it won't be obvious that the "something" can be written to look like the original H. So, it's useful to remember that for a function f(S,S'), where S and S' are [itex]\pm 1[/itex], you may write

[tex]f(S,S') = \frac{1}{4}[f(1,1)(S-1)(S'-1) - f(1,-1)(S-1)(S'+1) - f(-1,1)(S+1)(S'-1) + f(S,S')(S+1)(S'+1)][/tex]

If you do that, expanding out those brackets, you can collect on [itex]SS'[/itex] terms, and the resulting coefficient will be your new K'. You also made need to use some hyperbolic trig identities to get K' into the form the problem asks for.
 
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  • #3
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So, you have a four spin system and the problem as stated asks you to compute the partition function of the system, then to calculate the rescaled coupling after decimating two of the spins. The first part is hopefully straightforward:

[tex]Z = \mbox{Tr}(\exp(-H))[/tex]

where Tr means trace, and is basically just a sum over the possible values of all the spins. To decimate the spins, we almost do the same thing, except we only sum over the spins to be decimated. So, the coarse grained hamiltonian is

[tex]\exp(-H'(S,S')) = \sum_{S_1,S_2} \exp(-H(S,S',S_1,S_2)[/tex]

Once you've performed the sum, your goal is to hammer on H' until it looks like H, with new coupling constants K' and [itex]K'_0[/itex] (plus possibly new interaction terms, in general).

So, what have you tried so far? Have you done the above steps, or have you been trying to do this some other way? Show us some work and we can help pinpoint where you're going wrong.
Hi there,
thanks for the reply. I've just been trying to latex up what I've done:

[tex]
Z_{N} & = \sum_{\{S\}} \exp(\sum_{<ij>}KS_{i}S_{j} + \sum_{i}K_{0}) \\
& = \sum_{S} \sum_{S'} \left( e^{2K_{0}} \sum_{S_{1}} \exp(K S S_{1}) \exp(K S_{1} S') \right) \left( e^{2K_{0}}\sum_{S_{2}} \exp(K S' S_{2}) \exp(K S_{2} S) \right) \\
= \sum_{S} \sum_{S'} e^{w'(S, S')} e^{w'(S', S)}[/tex]

is what I've written down thus far. The primed label the new rescaled action. I'm looking at it now getting confused with what I've done, most likely because of all the caffiene :p

The next step would be to write that

[tex]e^{w'(S,S')} = e^{2K_{0}} \sum_{\sigma=\pm} \exp(KS \sigma) \exp(K \sigma S') = e^{K'_{0}} \exp(K' S S') [/tex]

I can expand this, and its just equivalent to the 1D Ising model (I use the identity exp(K s1 s2) = cosh (K) + s1 s2 tanh (K) ). The solution they have is clearly not the case for the a 1D ising model with N spins. I also note that we seem to get a sort of proliferation maybe? that the bonding is doubled between S and S'? :\
 
  • #4
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Hurrah! Thank you!

I'm afraid there is more though :P I carried on (skipped that part earlier).

It then asks me to set y = tanh(x) and show that there are 3 fixed points.

There is clearly one at K=0 and one at K=infinity. I'm not sure about the 3rd though. It mentions that I don't have to explicitly calculate it.

EDIT: After some discussion with some folk we've come to the conclusion that you could say there exists a fixed point by simply taking the difference between a few different values of K and if we get a change from a + to a minus sign (which we do, see http://www.wolframalpha.com/input/?i=plot+%282*%28tanh%28x%29%29^2+%2F+%281%2B+%28tanh%28x%29%29^4%29+and+tanh%28x%29 ) then there must be a fixed point. Can't think of a better way!

EDIT2: Now I've just found there is an intermediate value theorem which can solve the associated cubic and says that y is between 0 and 1.
 
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  • #5
Mute
Homework Helper
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EDIT2: Now I've just found there is an intermediate value theorem which can solve the associated cubic and says that y is between 0 and 1.
What cubic? Setting y = tanh(K) in the expression the problem quotes will give you a quartic (after canceling out a y on both sides). The resulting quartic has 4 roots, but only two are real. It's easy to notice that y = 1 is a root of the quartic. You can then factor this out using long division, leaving you with a cubic. Is this the cubic you're talking about? This cubic will only have one real solution, so that's the third root you're looking for (but I guess you may have to show that the other two roots are complex).

By the way, the last hint in my post above (f(S,S') = ...) was incorrect the first time I wrote it. I've corrected it now. I worked through the problem myself, a slightly different way, and got an expression for K which is equivalent to the one in the problem. The nice thing about the one in the problem though, is that by writing in terms of y = tanh(K), when finding the fixed points you don't have to worry about missing any K = infinity fixed points. The way I did it the natural variable seems to be y = exp(K), but then you have to be careful not to forget the K = infinity fixed point.
 
  • #6
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What cubic? Setting y = tanh(K) in the expression the problem quotes will give you a quartic (after canceling out a y on both sides). The resulting quartic has 4 roots, but only two are real. It's easy to notice that y = 1 is a root of the quartic. You can then factor this out using long division, leaving you with a cubic. Is this the cubic you're talking about? This cubic will only have one real solution, so that's the third root you're looking for (but I guess you may have to show that the other two roots are complex).

By the way, the last hint in my post above (f(S,S') = ...) was incorrect the first time I wrote it. I've corrected it now. I worked through the problem myself, a slightly different way, and got an expression for K which is equivalent to the one in the problem. The nice thing about the one in the problem though, is that by writing in terms of y = tanh(K), when finding the fixed points you don't have to worry about missing any K = infinity fixed points. The way I did it the natural variable seems to be y = exp(K), but then you have to be careful not to forget the K = infinity fixed point.
You shouldn't be getting a quartic :) You should be getting a power y^5, and the roots are 0 and 1. Factor these out and you're left with a cubic. Your y = 0 fixed point corresponds to K=0, and y=1 corresponds to K=infinity. And that random theorem implies there is a route a real route between 0 and 1 for cubic that remains after factoring.

You most likely got a quartic because you cancelled through by a factor of y rather than factoring it out. Your imaginary roots lead me to believe that you rearranged incorrectly! :)

P.S. I had no idea what your hint was trying to tell me with the f's! I've never seen that notation. Thanks for the help non the less, the stuff you wrote motivated me to find the answer in the end - even if it was along another route!
 
  • #7
Mute
Homework Helper
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You shouldn't be getting a quartic :) You should be getting a power y^5, and the roots are 0 and 1. Factor these out and you're left with a cubic. Your y = 0 fixed point corresponds to K=0, and y=1 corresponds to K=infinity. And that random theorem implies there is a route a real route between 0 and 1 for cubic that remains after factoring.

You most likely got a quartic because you cancelled through by a factor of y rather than factoring it out. Your imaginary roots lead me to believe that you rearranged incorrectly! :)
Yes, I said in my post I cancelled the y first, as y = 0 was obviously a solution and we were looking for the other non-zero solutions. It is easy to see that y = 1 is another root, so you can factor that out, leaving a cubic, which is the cubic I wanted to make sure you were talking about - I wanted to make sure you hadn't missed a power of y.

The fact that you seem to think there are no imaginary roots suggests that you have rearranged incorrectly (or perhaps misinterpreted your results, or perhaps you somehow factored the quintic into a cubic with roots 0, 1 and a non-trivial one, plus a quadratic with two complex roots)! 0 and 1 are not roots of the cubic that results from factoring (y-1) out of y^4 - 2y + 1, which is the cubic I have been talking about. There are five roots to the original quintic equation: y = 0, 1, a very non-trivial real root, and two complex roots. You can verify this by typing in y = 2y^2/(1+y^4) into wolfram alpha: http://www.wolframalpha.com/input/?i=y+=+2y^2/(1+y^4) So, there are definitely complex roots to that quintic - if you don't get any complex roots, you've made a mistake somewhere.

(Note that in general an n degree polynomial has n roots (some may be double roots). Some of those roots may be complex.)

P.S. I had no idea what your hint was trying to tell me with the f's! I've never seen that notation. Thanks for the help non the less, the stuff you wrote motivated me to find the answer in the end - even if it was along another route!
Basically, the way I did the problem (which doesn't use the hint the problem gives you), you get

[tex]-\mathcal H' = \mbox{stuff} + 2\ln(\cosh(K(S+S')))[/tex]
which doesn't look like it has an SS' term hiding in it anywhere. However, since S and S' can only take on values [itex]\pm 1[/itex], the [itex]\ln(\cosh(K(S+S')))[/itex] can only take on up to four possible values. The f = ... hint was just saying you can write this function in terms of these four possible values times [itex](1/4)(S\pm1)(S'\pm 1)[/itex]. Expanding out the brackets let's you generate SS' terms without having to do anything crazy like Taylor expanding the log.
 
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  • #8
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Yes, I said in my post I cancelled the y first, as y = 0 was obviously a solution and we were looking for the other non-zero solutions. It is easy to see that y = 1 is another root, so you can factor that out, leaving a cubic, which is the cubic I wanted to make sure you were talking about - I wanted to make sure you hadn't missed a power of y.

The fact that you seem to think there are no imaginary roots suggests that you have rearranged incorrectly (or perhaps misinterpreted your results, or perhaps you somehow factored the quintic into a cubic with roots 0, 1 and a non-trivial one, plus a quadratic with two complex roots)! 0 and 1 are not roots of the cubic that results from factoring (y-1) out of y^4 - 2y + 1, which is the cubic I have been talking about. There are five roots to the original quintic equation: y = 0, 1, a very non-trivial real root, and two complex roots. You can verify this by typing in y = 2y^2/(1+y^4) into wolfram alpha: http://www.wolframalpha.com/input/?i=y+=+2y^2/(1+y^4) So, there are definitely complex roots to that quintic - if you don't get any complex roots, you've made a mistake somewhere.

(Note that in general an n degree polynomial has n roots (some may be double roots). Some of those roots may be complex.)



Basically, the way I did the problem (which doesn't use the hint the problem gives you), you get

[tex]-\mathcal H' = \mbox{stuff} + 2\ln(\cosh(K(S+S')))[/tex]
which doesn't look like it has an SS' term hiding in it anywhere. However, since S and S' can only take on values [itex]\pm 1[/itex], the [itex]\ln(\cosh(K(S+S')))[/itex] can only take on up to four possible values. The f = ... hint was just saying you can write this function in terms of these four possible values times [itex](1/4)(S\pm1)(S'\pm 1)[/itex]. Expanding out the brackets let's you generate SS' terms without having to do anything crazy like Taylor expanding the log.
You're right. Sorry. I don't know what I did with that quintic, I was very tired at the time and appear to have lost my scrap work!

Apparently, there are 3 fixed points. I suspect these are real fixed points. We have found 2 real ones, and 3 imaginary non physical ones. I'm also supposed to deduce which one is the non trivial root. I figured a non trivial root corresponds to the critical point, which would be repulsive in both directions. This is not the case for either K=0 (attractive) or K=inf (repulsive). Slightly confusing!

EDIT: Sorry, didn't read the wolframalpha link properly! There is a third real root in there!

Thanks again!
 

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