Wick rotation of the Schrödinger equation

Question

Studying the following paper: https://www.nature.com/articles/s41534-019-0187-2.pdf

Trying to figure out how $ E_T$ shows up from (1) and (2).

Any suggestion or guidance would be appreciated.

We focus on many-body systems that are described by Hamiltonians $H=\sum_{i} \lambda_{j} h_{i}$, with real coefficients, $\lambda_{i}$, and observables, $h_{i}$, that are tensor products of Pauli matrices. We assume that the number of terms in this Hamiltonian scales polynomially with the system size, which is true for many physical systems, such as molecules or the Fermi-Hubbard model. Given an initial state $|\psi\rangle$, the normalised imaginary time evolution is defined by $$ |\psi(\tau)\rangle=A(\tau) e^{-H \tau}|\psi(0)\rangle\tag1 $$ where $A(\tau)=1 / \sqrt{\left\langle\psi(0)\left|e^{-2 H \tau}\right| \psi(0)\right\rangle}$ is a normalisation factor. In the instance that the initial state is a maximally mixed state, the state at time $\tau$ is a thermal or Gibbs state $\rho_{T=1 / t}=e^{-H \tau} / \operatorname{Tr}\left[e^{-H t}\right]$, with temperature $T=1 / \tau$. When the initial state has a non-zero overlap with the ground state, the state at $\tau \rightarrow \infty$ is the ground state of $H$. Equivalently, the Wick rotated Schrödinger equation is, $$\frac{\partial|\psi(\tau)\rangle}{\partial \tau}=-\left(H-E_{\tau}\right)|\psi(\tau)\rangle,\tag2$$ where the term $E_{\tau}=\langle\psi(\tau)|H| \psi(\tau)\rangle$ results from enforcing normalisation. Even if $|\psi(\tau)\rangle$ can be represented by a quantum computer, the non-unitary imaginary time evolution cannot be naively mapped to a quantum circuit.

Answer 1

As hinted in the highlighted text, $E_T$ arises from the normalization factor $A(\tau)$. Specifically, differentiating $(1)$, we get two terms

$$ \begin{align} \frac{\partial|\psi(\tau)\rangle}{\partial\tau} &= \frac{\partial}{\partial\tau}\left(A(\tau)e^{-H\tau}|\psi(0)\rangle\right) \\ &= \frac{\partial A(\tau)}{\partial\tau}e^{-H\tau}|\psi(0)\rangle + A(\tau)\frac{\partial e^{-H\tau}}{\partial\tau}|\psi(0)\rangle.\tag3 \end{align} $$

Computing the derivative in the first term, we obtain

$$ \begin{align} \frac{\partial A(\tau)}{\partial\tau} &= \frac{\partial}{\partial\tau}\left(\frac{1}{\sqrt{\langle\psi(0)|e^{-2H\tau}|\psi(0)\rangle}}\right) \\ &= -\frac{1}{2\sqrt{\langle\psi(0)|e^{-2H\tau}|\psi(0)\rangle^3}} \langle\psi(0)|\frac{\partial}{\partial\tau}e^{-2H\tau}|\psi(0)\rangle \\ &= A(\tau)^3 \langle\psi(0)|e^{-H\tau}He^{-H\tau}|\psi(0)\rangle \\ &= A(\tau) \langle\psi(0)|e^{-H\tau}A(\tau)HA(\tau)e^{-H\tau}|\psi(0)\rangle \\ &= A(\tau) \langle\psi(\tau)|H|\psi(\tau)\rangle \\ &= A(\tau) E_T\tag4 \\ \end{align} $$

where we took advantage of the facts that $A(\tau)\in\mathbb{R}$, that $H$ is Hermitian, and that $e^{-H\tau}$ commutes with $H$. Finally, substituting $(4)$ into $(3)$, we find

$$ \begin{align} \frac{\partial|\psi(\tau)\rangle}{\partial\tau} &= A(\tau)E_T e^{-H\tau}|\psi(0)\rangle - A(\tau)He^{-H\tau}|\psi(0)\rangle \\ &= (E_T-H)|\psi(\tau)\rangle \end{align} $$

as advertised.