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Confusion about duality transformation in 1+1D Ising model in a transverse field

+ 5 like - 0 dislike
18778 views

In 1+1D Ising model with a transverse field defined by the Hamiltonian
\begin{equation}
H(J,h)=-J\sum_i\sigma^z_i\sigma_{i+1}^z-h\sum_i\sigma_i^x
\end{equation}
There is a duality transformation which defines new Pauli operators $\mu^x_i$ and $\mu^z_i$ in a dual lattice
\begin{equation}
\mu_i^z=\prod_{j\leq i}\sigma^x_j
\qquad
\mu_i^x=\sigma^z_{i+1}\sigma^z_{i}
\end{equation}
then these $\mu_i^x$ and $\mu_i^z$ satisfy the same commutation and anti-commutation relations of $\sigma^x_i$ and $\sigma^z_i$, and the original Hamiltonian can be written in terms of $\mu_i^x$ and $\mu_i^z$ as
\begin{equation}
H(J,h)=-J\sum_i\mu_i^x-h\sum_i\mu_i^z\mu_{i+1}^z
\end{equation}

At this stage, many textbooks will tell us since $\sigma$'s and $\mu$'s have the same algebra relations, the right hand side of the last equation is nothing but $H(h,J)$. My confusions are

1) Does that the operators having the same algebra really imply that $H(J,h)$ and $H(h,J)$ have the same spectrum? We know for a given algebra we can have different representations and these different representations may give different results. For example, the angular momentum algebra is always the same, but we can have different eigenvalues of spin operators.

2) This is related to the first confusion. Instead of looking at the algebra of the new operators, we can also look at how the states transform under this duality transformation. In the eigenbasis of $\mu_i^x$, if I really consider it as a simple Pauli matrix, the state $|\rightarrow\rangle$ corresponds to two states in the original picture, i.e. $|\uparrow\uparrow\rangle$ and $|\downarrow\downarrow\rangle$. The same for state $|\leftarrow\rangle$. In the $\mu_i^z$ basis, the correspondence is more complicated. A state corresponds to many states in the original picture, and the number  of the corresponding states depend on the position of this state. Therefore, this duality transformation is not unitary, which makes me doubt whether $H(J,h)$ and $H(h,J)$ should have the same spectrum. Further, what other implication may this observation lead to? For example, doing one duality transformation is a many-to-one correspondence, then doing it back should still be a many-to-one correspondence, then can we recover the original spectrum? 

3) Another observation is we in the above $\mu_i^z$ involves a string of operators on the left side, we can equally define it in terms of a string of operators on the right side, so it seems there is an unobservable string. What implication can this observation lead to? Is this unobservable string related to the unobservable strings in Levin-Wen model?

asked Sep 12, 2014 in Theoretical Physics by Mr. Gentleman (255 points) [ revision history ]
recategorized Sep 12, 2014 by dimension10

1 Answer

+ 5 like - 0 dislike

1) In general, an algebra can have many representations. In this case, however, if you assume that there is a unique joint +1 eigenstate of the \(\sigma_i\)'s, that determines the representation uniquely. [All the other states can be found this state by applying products of \(\sigma_i^x\)to it. And from the anti-commtation of \(\sigma_i^x\) and \(\sigma_i^z\) you know that applying \(\sigma_i^x\) to a state must flip the sign of the eigenvalue of \(\sigma_i^z\)]. And as you can easily check, both the \(\sigma_i^z\) and the \(\mu_i^z\) do, indeed, have a unique +1 eigenstate, so the Hilbert space has the same structure and there is a unitary transformation between the two representations.

Wherefore, then, the paradoxical many-to-one nature of the transformation which you highlight in question 2).? The problem is that you are considering an infinite system. The Hilbert space of an infinite system is not really well-defined (it would have uncountable dimension!) Much better to consider a finite system. Then you have to specify the boundary conditions. The most natural choice is periodic boundary conditions, i.e. identify spins 1 and spin N+1. But that leads to some awkwardness regarding how to define the duality transformation -- how do you deal with the strings of \(\sigma^x\)'s that are supposed to go off to infinity in the definition of \(\mu^z\)? There are ways to make it work but you end up having to consider two different sectors of Hilbert space separately and it's a mess.

So instead, let's do open boundary conditions -- just keep the ends of the chain separate and leave off the interaction terms in the Hamiltonian \(\sigma_0^z \sigma_1^z \)and \(\sigma_N^z \sigma_{N+1}^z\)that would couple the ends of the chain to something that isn't there. Then we can leave the duality transformation more or less as is, with the exception of \(\mu_N^x = \sigma_N^z \sigma_{N+1}^z\) . That doesn't work because there's no (N+1)-th spin. Instead, let's just define \(\mu^x_N= \sigma_N^z\). You can check this doesn't affect the algebra. It does, however, resolve the problem you were raising about many-to-one mappings. Consider, as in the question, the state \(|\rightarrow \rangle_{\mu}\) . Then the only state it corresponds to in the \(\sigma\) representation is \(|\uparrow\uparrow\rangle_{\sigma}\). The state \(|\downarrow \downarrow \rangle_{\sigma}\), on the other hand, by contrast to the naive infinite-system analysis) now satisfies \(\mu^x_N = -1\), hence it is not the same as \(|\rightarrow \rangle_{\mu}\). (To answer your third question: the string is actually not unobservable

So in conclusion, the non-unitarity you were seeing was simply a manifestation of the fact that the infinite system is not well-defined. Looking at a finite system instead resolves the difficulties.

There's another interesting way of looking at this problem: H(J,h) has two degenerate ground states when \(J \gg h\) (spontaneous symmetry breaking) and only one ground state when \(h\gg J\). So it was inevitable that an attempt to create an exact mapping that swaps J and h was going to run into difficulties. The reason why everything works out once we go to a finite system with boundary conditions as described is left as an exercise to the reader.

answered Sep 12, 2014 by Dominic Else [ no revision ]

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