There is a well known connection between statistical mechanics in D spatial dimensions and quantum field theory in D-1 spatial dimensions. Changing the temperature in statistical mechanics corresponds to changing the coupling constants in the QFT. Changing the temperature in QFT corresponds to changing the system size of the classical system in the Euclidean time direction. I'm wondering about the relation between these two distinct notions of temperature (using the Ising model as a concrete example).

Let me sketch the argument taking the classical Ising model Hamiltonian in 2D to a quantum system. The classical Hamiltonian which appears in the partition function is $$H_\text{cl} = -\sum_{i,j} \beta_x \,\sigma_3(i,j)\sigma_3(i+1,j) + \beta_y \,\sigma_3(i,j)\sigma_3(i,j+1),$$ where the $\beta_x,\beta_y$ are coupling constants in the x and y directions, each which contains a factor of the inverse temperature $\beta$ since this appears in the exponent of the partition function.

We can take the y direction to be Euclidean time, and think of the transfer matrix between rows as being a time translation operator $e^{-H\tau}$ where $\tau$ is the y lattice spacing.

To figure out $H$ we can take the limit where $\tau\rightarrow 0$, but to keep large scale properties the same we have to also take $\beta_y\rightarrow \infty, \beta_x\rightarrow 0$ in a way that involves a new parameter $\lambda$. This parameter can be thought of as containing information on the original temperature.

Doing this procedure (the Hamiltonian limit) we get the 1+1D quantum Hamiltonian $$H = -\sum_i \sigma_1(i)+\lambda \sigma_3(i)\sigma_3(i+1).$$

Now my question is where does the size of the original lattice $L$ in the time (y) direction come to play? To get back to the original partition function we look at $\text{Tr}\, e^{-LH}$. But thinking of it as a quantum system this $L$ plays the role of inverse temperature. But all of the temperature information is supposed to be encoded in $\lambda$, with $\lambda=1$ marking the position of the critical point.

Did we secretly take $L\rightarrow\infty$ in the Hamiltonian limit, in which case we should use vacuum expectation values to talk about original system (and this is why the critical point only depends on $\lambda$)? In talking about the statistical mechanics on a finite lattice is it fair to use $\text{Tr}\, e^{-LH}$, which seems like it has effects due to two "temperatures"?

  • $\begingroup$ I ended up finding a discussion of this in Cardy's Scaling and Renormalization book. In the treatments I have seen they are indeed taking $L\rightarrow\infty$, and the ordinary phase transition appears by varying $\lambda$. The quantum temperature is indeed distinct from the stat mech temperature, and both appear as axes in a renormalization flow diagram. There can be additional critical points and cross over behavior. $\endgroup$
    – octonion
    Mar 18, 2016 at 21:03
  • $\begingroup$ Thanks for the follow-up. So you are calling the 'quantum temperature' L, if I'm not mistaken. Then does lambda contain what you call the 'stat mech temperature?' Or are there parameters in addition to these two that I am missing? $\endgroup$
    – Rococo
    Mar 18, 2016 at 21:29
  • 1
    $\begingroup$ Yes those are the same parameters I established in the premise of the question. But I was looking for some insight into the effects of these two 'temperatures.' The 2D Ising model does not even have an ordered phase if there is finite $L$, even if the other direction is infinite. And in D greater than 2 there is a cross over between two critical points (one for small L that acts like the D-1 Ising model). These things weren't clear to me when I wrote the question. $\endgroup$
    – octonion
    Mar 19, 2016 at 4:44
  • $\begingroup$ There is some confusion with dimensions. To clarify forget about the continuum limit and simply consider the transfer matrix. Imagine the classical lattice is $L\times M$, what should be the dimension of the, say row to row transfer matrix? $\endgroup$
    – lcv
    Dec 15, 2018 at 11:22
  • $\begingroup$ Ops.. I didn’t notice how old the question was $\endgroup$
    – lcv
    Dec 15, 2018 at 11:28

1 Answer 1


I am not completely sure I am understanding your question, but it looks like the basic confusion is in what limits exactly one takes in this mapping, and how the variables are rescaled. I am nearly following the treatment of Sachdev in the beginning of his Quantum Phase Transitions book, but adapted to a 2D Ising chain and with your variables.

Okay, so your original variables are:

$\tau$: lattice spacing (in x and y)

$N$: number of lattice sites (in x and y)

$\beta_x$: coupling in x divided by T

$\beta_y$ coupling in y divided by T

Now, you are going to take the limit where $\tau \rightarrow 0$, as you said, and do so in a particular way. Specifically:

-Take $\tau \rightarrow 0$ and $N \rightarrow \infty$, such that $N\tau=L$ (the system length) remains constant.

-Take $\tau\rightarrow 0$ and $\beta_y \rightarrow 0$, such that $\beta_y/\tau =E_{0,y}$ remains constant, where $E_{0,y}$ is the ground state energy per unit length along the y-direction. Do the same with $\beta_x$ and $E_{0,x}$.

-Take $\xi$ as fixed, where $\xi$ is a macroscopic property of interest. For example, one can take $\xi=(a/2)e^{2\beta_y}$, where $\xi$ is the correllation length (Sachdev derives this).

This amounts to keeping the large length scales (L and $\xi$) constant while the microscopic one $\tau$ goes to zero.

So the game one plays is to write the transfer matrix expression for the partition function purely in terms of $L$, the $E_{0}$s, and $\lambda$, at which point it is not changed by taking the thermodynamic limit. Once you do this, you will find that the transfer matrix T has a form like:

$T=e^{\tau H'(E_{0,x},E_{0,y},\xi)}$

where $H'$ is some function of the scaled parameters (which will become the new Hamiltonian). This means that the expression for the partition function becomes:

$Z=tr(T^N)=tr(e^{N\tau H'})=tr(e^{L H'})$

at which point the expression for the partition function is completely using the scaled variables, and you can take the thermodynamic limit without changing its form.

At this point, $H'=E_{0,x}\sum_i \sigma_1(i) \sigma_1(i+1)+(1/2\xi)\sigma_3(i)+E_{0,y}$. Then you can drop the constant term $E_{0,y}$ and combine the other two terms into a parameter $\lambda$ to get your expression.

So it looks to me like you were maybe confused about $N$ and $L$, which I hope I've named according to the same convention as you. $N$, the number of lattice sites, does indeed diverge in this limit. However, the partition function expression uses $L$, the system length, which stays constant in this limit. $L$ does play the role of inverse temperature of the system in question, while $\lambda$ can contain something like a characteristic temperature $T_c$ for a phase transition. This means that the phase of the system is determined by $L\lambda$, the only combination of parameters available.

  • $\begingroup$ Thank you for the response. Your $L$ is what I was calling $N$ in the post, I will just call it $L$ now for clarity. I fully understand if $L$ is constant and $\tau$ goes to zero the number lattice points must diverge. But I do not see where $L$ entered into your derivation of the form of $H'$. That seems to be determined by fixing $\xi$. Your statement at the end is essentially my question, how is $L$ playing the role of inverse temperature? And are you sure about your statement that the combination $L\lambda$ determines the phase? $\endgroup$
    – octonion
    Mar 13, 2016 at 19:36
  • $\begingroup$ Hi octonion, I have added a little bit in the body to show explicitly how $L$ comes in, I hope this helps. Regarding your last question, $L$ and $\lambda$ are the only parameters that are around, so I don't see any possible alternative. $\endgroup$
    – Rococo
    Mar 13, 2016 at 19:58
  • $\begingroup$ Hi Rococo, I appreciate the time you spent on this answer and I think it might be helpful to some readers. I'm aware of the procedure to take the Hamiltonian limit. For reference I read Mussardo's book on statistical field theory and a paper by Fradkin and Susskind Phys Rev D 17 (1978) 2637. There is an argument for $\lambda=1$ being the critical point, which I may edit into my question when I have time. It is possible I am being stupid, so I will think about your answer, but right now it seems you aren't hitting on my question. $\endgroup$
    – octonion
    Mar 13, 2016 at 20:24
  • $\begingroup$ Yes, I agree that it is possible that I am misunderstanding your question. If this is the case, you might think about how to reword it so that I or someone else is more sure exactly what you are looking for. $\endgroup$
    – Rococo
    Mar 13, 2016 at 20:59
  • $\begingroup$ That said, for my current reading of your two questions, my response would be, in one phrase each: "no, we don't necessarily take L to infinity," and "yes, in this mapping the original temperature is mapped into the parameter in the Hamiltonian, while the system length in mapped onto the temperature." If that seems to be missing the point entirely, than I clearly don't understand what you are looking for and you may have to look elsewhere. $\endgroup$
    – Rococo
    Mar 13, 2016 at 21:03

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