Temperature in the Hamiltonian limit

Question

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"?

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.