Entanglement Hamiltonian for two 1/2 spin system

Consider a two 1/2-spin state $$|\phi\rangle=\frac{1}{\sqrt{2}}(|1\rangle_A\otimes|1\rangle_B-|0\rangle_A\otimes|0\rangle_B),$$ cut in half (so we get A- and B-subsystem), the reduced density matrix for A-subsystem is simply $$\rho_A=\frac{1}{2}(|1\rangle_A\langle1|_A+|0\rangle_A\langle0|_A).$$ It is assumed (call it the first assumption) that $\rho_A$ can be written in terms of the entanglement Hamiltonian $$\rho_A=\frac{e^{-\mathcal{H}_A}}{\text{Tr}_A(e^{-\mathcal{H}_A)}}$$ and it is further assumed (call it the second assumption) that the single spin entanglement Hamiltonian has the form $$\mathcal{H}_A=-hS^z_A,$$ where $h$ is a constant characterizing the effective external magnetic field, and $S^z_A=\frac{1}{2}\left( {\begin{array}{*{20}{c}} 1&0\\ 0&-1 \end{array}} \right)$ is the spin operator for A in z direction. However, here with my form of $\rho_A=\frac{1}{2}\left( {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right)$, it seems that I get $h=0$.

Here are my questions:

1. Are my first and second assumptions correct? (currently I only know this kind of assumptions is true for free fermion systems, see e.g. P5-7 of Peschel's paper: arXiv:0906.1663v3)

2. If the assumptions are correct, it seems inevitably $h=0$, and the Hamiltonian vanishes. How shall I interpret this?

3. If I change my $|\phi\rangle$ to $|\phi\rangle=|1\rangle_A\otimes|1\rangle_B$, then $\rho_A=\left( {\begin{array}{*{20}{c}} 1&0\\ 0&0 \end{array}} \right)$ and $h\rightarrow\infty$. How should I interpret this?

Thanks so much!

You can always write any density matrix in the form $\rho = e^{-H_A}$ in a limiting sense, as long as you allow some of the parameters entering $H_A$ to become infinite. This follows from the identity $$\rho = e^{\ln \rho},$$ but notice that if $\rho$ has some eigenvalues equal to zero, then the eigenvalues of the exponent become infinite (or are not defined, to a mathematician). Strictly speaking you have to consider the limiting procedure in which the eigenvalues of $\rho$ are finite but tend to $0$. This corresponds to including in $H_A$ a term $\lambda P_0$, where $P_0$ is the orthogonal projection onto the eigenspace of $\rho$ with eigenvalue $0$, and then letting $\lambda\to \infty$.
Your second assumption that $H_A = -h S^z_A$ is only correct if the reduced density matrix is diagonal in the $\{|0\rangle,|1\rangle\}$ eigenbasis of $S_A^z$. For a maximally entangled state this is always true, since the reduced density operator is diagonal in all bases (it is identity). However this is not so for the majority of entangled and non-entangled states, e.g. a simple (unnormalised) example is $$|\psi\rangle = |00\rangle + |01\rangle + |11\rangle.$$
About the interpretation, it is easier to think of $h$ as analogous to an inverse temperature. Or even better, think of it as a magnetic dipole energy measured in units of temperature, so we could also write $h = \mu/k_BT$, where $\mu$ is the magnetic energy of a spin in an external magnetic field, and $k_B T$ is the thermal energy. Therefore, $h\to 0$ means that the temperature $T\to\infty$. This creates a correspondence between the intuition that 1) high temperature leads to more randomness and 2) stronger entanglement leads to more (local) randomness.
On the other hand, a non-entangled state corresponds to a pure reduced density matrix and therefore $h\to\infty$, or $T\to 0$. In the thermal picture, as $T\to 0$ the system is in its ground state and there are no more thermal fluctuations, only quantum fluctuations. In the entanglement picture, if there is no entanglement there is no local classical uncertainty due correlations with another sub-system, but there is still quantum uncertainty for measurements of observables that that do not commute with $\rho_A$ (i.e. quantum fluctuations).
• Thanks for your elaborate answer. Previously I thought in the way that we have let $T=1$ in $\rho_A$ and there is no more sense of temperature in $h$. Oct 17 '14 at 16:38
• @Antonio_phy Setting $k_B T=1$ just means that you use units of energy equal to $k_B T$, it doesn't necessarily mean that the temperature cannot change. However, if you like you can think of $T$ being constant (but finite, importantly), and imagine varying the external magnetic field instead of the temperature. The results are the same, because the only thing that matters is the dimensionless ratio $\mu/k_BT$. Oct 17 '14 at 16:43