Update README.md

README.md

@@ -103,7 +103,7 @@ is the pushforward of the tangent vector $\frac{d \alpha}{d\tau}(\tau) \in T_\al
 
 By the time-dependent Schrödinger equation, the state $\ket{\psi(\alpha(t))} \in \mathcal{H}$ at some time $t$ will evolve according to $i \frac{d}{dt} \ket{\psi(\alpha(t))} = H \ket{\psi(\alpha(t))}$, which is satisfied (up to a constant) by the propagator $U(t_2 - t_1) = \exp[-i(t_2-t_1)H]$ given the Hamiltonian $H$. Here, the Hamiltonian $H$ is the infinitesimal generator of the one-parameter unitary group of time translations whose elements are the unitary transformations $U(t_2 - t_1):\mathcal{H} \to \mathcal{H}$ on the state space $\mathcal{H}$ for any $t_1, t_2 \in \mathbb{R}$. By performing a Wick rotation $\tau = it$, the state $\ket{\psi(\alpha(\tau))} \in \mathcal{H}$ at some imaginary time $\tau$ will evolve according to the imaginary-time Schrödinger equation $-\frac{d}{d\tau} \ket{\psi(\alpha(\tau))} = H \ket{\psi(\alpha(\tau))}$, which is satisfied (up to a constant) by the non-unitary propagator $U(\tau_2 - \tau_1) = \exp[-(\tau_2-\tau_1)H]$. Taking $\tau_1 = \tau$ and $\tau_2 = \tau + \delta\tau$, we propagate the state $\ket{\psi(\alpha(\tau))} \in \mathcal{H}$ by
  $$\ket{\psi(\alpha(\tau + \delta \tau))} = U(\delta\tau) \ket{\psi(\alpha(\tau))} = \ket{\psi(\alpha(\tau))} - \delta \tau H \ket{\psi(\alpha(\tau))} + \frac{(-\delta\tau)^2}{2}H^2 \ket{\psi(\alpha(\tau))} + \cdots \approx \ket{\psi(\alpha(\tau))} - \delta \tau H \ket{\psi(\alpha(\tau))} \in \mathcal{H}$$
-approximated to first order over the interval $[\tau, \tau + \delta \tau]$, which becomes exact in the limit $\delta\tau \to 0$. Enforcing a different normalization, we may also evolve the state $\ket{\psi(\alpha(\tau))} \in \mathcal{H}$ according to the imaginary-time Schrödinger equation $- \Delta\frac{d}{d\tau} \ket{\psi(\alpha(\tau))} = \Delta H \ket{\psi(\alpha(\tau))}$ with the deviations $\Delta \frac{d}{d\tau} = \frac{d}{d\tau} - \left\langle \frac{d}{d\tau} \right\rangle_{\psi(\alpha)}$ and $\Delta H = H - \langle H \rangle_{\psi(\alpha)}$, which is often more advantageous in a stochastic framework.
+approximated to first order over the interval $[\tau, \tau + \delta \tau]$, which becomes exact in the limit $\delta\tau \to 0$. Enforcing a different normalization, we may also evolve the state $\ket{\psi(\alpha(\tau))} \in \mathcal{H}$ according to the imaginary-time Schrödinger equation $- \Delta\frac{d}{d\tau} \ket{\psi(\alpha(\tau))} = \Delta H \ket{\psi(\alpha(\tau))}$ involving the deviations $\Delta \frac{d}{d\tau} = \frac{d}{d\tau} - \left\langle \frac{d}{d\tau} \right\rangle_{\psi(\alpha)}$ and $\Delta H = H - \langle H \rangle_{\psi(\alpha)}$, which is often more advantageous in a stochastic framework.
 
 To determine the actual form of the tangent vector $\frac{d \alpha}{d\tau}(\tau) \in T_\alpha \mathcal{M}$ at time $\tau$, we impose the constraint that the projection
  $$\left\langle \frac{d}{d\tau} \psi(\alpha(\tau)), \bigg[ \Delta \frac{d}{d\tau} + \Delta H \bigg] \psi(\alpha(\tau)) \right\rangle = 0$$
@@ -121,4 +121,4 @@ is the change in the parameters $\alpha(\tau) \in \mathcal{M}$ due to the non-un
 
 It must be noted that the initialization of the parameters can have a dramatic effect on the performance of the algorithm. The initial state $\ket{\psi(\alpha(0))}$ must be chosen such that $\langle \psi_0, \psi(\alpha(0)) \rangle \neq 0$, or else learning is not possible. The more overlap there is with the ground state, the more efficient the algorithm will be. With at least some overlap, we will expect that $\ket{\psi(\alpha(\tau))} \to \ket{\psi_0}$ as $\tau \to \infty$ for a sufficiently small time step $\delta\tau$. This can be seen by noting the change in the energy functional over the interval $[\tau, \tau + \delta \tau]$, by taking the expectation of $H$ in the state $\ket{\psi(\alpha(\tau + \delta\tau))} \approx \ket{\psi(\alpha(\tau))} - \delta \tau \Delta H \ket{\psi(\alpha(\tau))} = \ket{\psi(\alpha(\tau))} + \delta \tau \Delta \frac{d}{d\tau} \ket{\psi(\alpha(\tau))}$, i.e.
  $$E[\psi(\alpha(\tau + \delta\tau))] = E[\psi(\alpha(\tau))] - 2\delta\tau F^\dagger(\alpha) S^{-1}(\alpha) F(\alpha) + \mathcal{O}(\delta\tau^2)$$
-where $\mathcal{O}(\delta\tau^2)$ denotes the term involving $\delta\tau^2$. Since $\delta\tau > 0$ and $S(\alpha)$ is positive-definite, we have that the change in energy $E[\psi(\alpha(\tau + \delta\tau))] - E[\psi(\alpha(\tau))] < 0$ for a sufficiently small time step $\delta\tau$.
+where $\mathcal{O}(\delta\tau^2)$ denotes the term involving $\delta\tau^2$. Since $\delta\tau > 0$ and $S(\alpha)$ is positive-definite, the second term will be strictly negative, such that the change in energy $E[\psi(\alpha(\tau + \delta\tau))] - E[\psi(\alpha(\tau))] < 0$ for a sufficiently small time step $\delta\tau$.