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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.

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  • $\begingroup$ If you want to look at entanglement Hamiltonians, systems whose ground state is unentangled might not be the best starting point ... $\endgroup$ – Norbert Schuch Apr 8 at 19:30
  • $\begingroup$ 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. $\endgroup$ – Ricky Pang Apr 9 at 15:26
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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 $\mu\rightarrow +\infty$.

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  • $\begingroup$ 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? $\endgroup$ – Ricky Pang Apr 7 at 16:07
  • $\begingroup$ 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). $\endgroup$ – Christophe Apr 8 at 7:40
  • $\begingroup$ 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. $\endgroup$ – Ricky Pang Apr 8 at 10:58
  • $\begingroup$ 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... $\endgroup$ – Christophe Apr 8 at 13:14
  • $\begingroup$ 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. $\endgroup$ – Ricky Pang Apr 8 at 13:33

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