# Entanglement Hamiltonian of a subsystem and the reduced density matrix

I encounter a problem in finding the entanglement Hamiltonian of a subsystem. Suppose my system consists of two sites and two fermions, and the Hamiltonian of the full system is the following: $$\begin{equation} \hat{H} = -\mu \sum_{i} c^{\dagger}_{i} c_{i} ~~,~~ \mu = const ~~,~~ \mu \geq 0 \end{equation}$$ Where $$c^{\dagger}_{i}$$ is the fermionic creation operator of site $$i, i = A,B$$. Then, let me partition my system as two pieces. The first site is subsystem A and the second site is subsystem B. Besides, I denote $$| 0 \rangle$$ as no fermion at the site and $$|1 \rangle$$ as 1 fermion at the site. Therefore, the ground state ( lowest energy configuration) is the following state: $$\begin{equation} |G.S. \rangle = |1\rangle_{A} \otimes |1\rangle_{B} = |11\rangle \end{equation}$$ It means that our full system is filled with fermions and the energy is $$E = -2 \mu$$, which is the lowest energy state that we can achieve. The density matrix of the full system at the ground state is: $$\begin{equation} \rho = |G.S. \rangle \langle G.S.| = |11\rangle \langle 11| \end{equation}$$ Having the density matrix of the system with respect to its ground state, we compute the reduced density matrix of subsystem A: $$\begin{equation} \begin{split} \rho_{A} &= \text{tr}_{B}(\rho) \\ &= \langle 0_{B} | 11\rangle \langle 11|0 \rangle_{B} + \langle 1_{B} | 11\rangle \langle 11|1 \rangle_{B} \\ &= |1 \rangle_{A} \langle 1|_{A} \end{split} \end{equation}$$ Apart from using the definition of reduced density matrix, we can use the partition function to calculate the density operator of subsystem A: $$\begin{equation} \rho_{A} = \frac{e^{-H_{A}}}{Z_{A}} \end{equation}$$ where $$H^{A}$$ is the entanglement Hamiltonian of subsystem A and the partition $$Z_{A} = \text{tr}(e^{-H_{A}})$$. I try to use $$H_{A} = -\mu c^{\dagger}_{A}c_{A}$$ but it gives us the correct result once the $$\mu \rightarrow \infty$$:

$$\begin{equation} \begin{split} \rho_{A} = \frac{e^{-H_{A}}}{Z_{A}} &= \frac{1}{Z_{A}} |0\rangle_{A} \langle 0|_{A} + \frac{e^{\mu}}{Z_{A}} |1 \rangle_{A} \langle 1|_{A} ~~,~~ Z_{A} = \text{tr}(e^{\mu c^{\dagger}_{A} c_{A}}) = 1 + e^{\mu} \\ \lim_{\mu \rightarrow \infty} \rho_{A} &= |1 \rangle_{A} \langle 1|_{A} \end{split} \end{equation}$$

Therefore, I want to ask that what is the correct entanglement Hamiltonian of subsystem A in this case? Why I cannot directly use $$H_{A} = -\mu c^{\dagger}_{A} c_{A}$$ as the entanglement Hamiltonian of subsystem A? Why we need to care about the density matrix with respect to the ground state only? I would appreciate if someone could explain more on entanglement hamiltonian and the density operator of ground state.

• If you want to look at entanglement Hamiltonians, systems whose ground state is unentangled might not be the best starting point ... – Norbert Schuch Apr 8 at 19:30
• Yes, you are right, @NorbertSchuch. I found that my example is not that good since the the ground state is a product state rather than entangled state. – Ricky Pang Apr 9 at 15:26

## 1 Answer

In the canonical ensemble, the density matrix of the system is given by $$\rho={1\over Z}e^{-H/k_BT}$$ By choosing to define the density matrix as $$\rho=|G.S.\rangle\langle G.S.|$$, you implicitly considered the case of a temperature $$T=0$$. As you have shown, the reduced density matrix is then $$\rho_A=|1\rangle_A\langle 1|_A$$ In your example, the two subsystems do not interact, i.e. $$H=H_A+H_B$$, so $$Z=Z_AZ_B$$ and $$\rho={1\over Z_AZ_A}e^{-(H_A+H_B)/k_BT}$$ It follows that $$\rho_A={1\over Z_A}e^{-H_A/k_BT}$$ What you defined as the entanglement Hamiltonian is therefore $$H_A/k_BT$$. Your chemical potential is actually $$\mu/k_BT$$. Since you have set the temperature to $$T=0$$, the entanglement Hamiltonian is $$\lim_{T\rightarrow 0} H_A/k_BT$$ which is equivalent to take the limit $$\mu\rightarrow +\infty$$.

• Thank you for your comment, @Christophe. So, can I say that I implicitly choose $T \rightarrow 0$ when I compute the density matrix using the ground state only? Besides, if I use the first excited state to compute the density matrix, how can I modify the density operator in canonical ensemble? – Ricky Pang Apr 7 at 16:07
• Using the ground state only for the density matrix is indeed equivalent to thermal equilibrium at zero temperature. It helps to understand why your entanglement Hamiltonian corresponds to the limit $\mu\rightarrow +\infty$. A density matrix computed using only the excited state cannot be interpreted as a thermal equilibrium ($T\rightarrow +\infty$ implies equal probability of the ground state and excited state). – Christophe Apr 8 at 7:40
• May I ask why thermal equilibrium is important in our case? The reason of asking this question is that in the above derivation, I do not assume my system is in equilibrium. But I agree that there should have some relation between this quantum system with statistical mechanics. – Ricky Pang Apr 8 at 10:58
• Probably because my field is Statistical Physics :-) so the argument is quite natural for me. If you are interested in Quantum Information, the argument is probably less convincing for you... – Christophe Apr 8 at 13:14
• I think your argument is natural for me if we think in statistical physics way. It also solve why the chemical potential going to infinity. – Ricky Pang Apr 8 at 13:33