In this case, we choose an autoregressive wavefunction ansatz, called sampling part \(\phi(n) := \langle n | \phi \rangle, \sum_n |\phi(n)|^2 = 1\),
and a extra correction factor, called extra part \(f_n\). Then total wavefunction \(\psi(n) = \langle n|\psi\rangle\) can be represent as
In general
The expectation of energy \(E\) is
\[\begin{aligned}
\langle E\rangle :=& \dfrac{\langle\psi|H|\psi\rangle}{\langle\psi|\psi\rangle}
=\dfrac{\sum_n|\phi(n)|^2\dfrac{\langle \psi | n\rangle \langle n|H|\psi\rangle}{|\phi(n)|^2}}{\sum_n |\phi(n)|^2 |f_n|^2}
= \dfrac{\bigg\langle f_n^*\dfrac{\langle n|H|\psi\rangle}{\langle n|\phi \rangle} \bigg\rangle_{n\sim |\phi(n)|^2}}{\big\langle |f_n|^2\big\rangle_{n\sim |\phi(n)|^2}}
\end{aligned}\]
The denominator is actually a constant, so remark it as \(B = \big\langle |\phi(n)|^2\big\rangle_{n\sim |\phi(n)|^2}, \widetilde{f}_n := f_n/\sqrt B\). Define local energy as
\[\begin{split}\begin{aligned}
E_{\rm loc} &= \dfrac{f_n^*}{B}\dfrac{\langle n|H|\psi\rangle}{\langle n | \phi \rangle}
= \dfrac{f_n^*}{B} \dfrac{\sum_m\langle n | H | m\rangle \langle m|\psi\rangle}{\phi(n)}\\
&=\sum_m \dfrac{f_n^*}{\sqrt{B}}\dfrac{f_m}{\sqrt{B}}\dfrac{H_{nm}\phi(m)}{\phi(n)}
=\sum_m \widetilde{f}_n^*H_{nm}\widetilde{f}_m\dfrac{\phi(m)}{\phi(n)}
\end{aligned}\end{split}\]
Formally, the correction factor provides a (nonlinear) transformation of the Hamiltonian.
The gradient can be calculated as
\[\begin{split}\begin{aligned}
\partial_\theta \langle E\rangle &= \dfrac{\partial}{\partial\theta}\dfrac{\langle\psi|H|\psi\rangle}{\langle\psi|\psi\rangle}
= \bigg[\dfrac{\langle\partial_\theta\psi|H|\psi\rangle}{\langle\psi|\psi\rangle} - \dfrac{\langle \psi | H| \psi\rangle}{\langle \psi|\psi\rangle^2}\langle\partial_\theta\psi|\psi\rangle\bigg]+\mathrm{c.c.}\\
&=2\Re\bigg[\dfrac{\sum_n\partial_\theta\langle \psi|n\rangle\langle n|H|\psi\rangle}{\sum_n|\psi(n)|^2} - \langle E\rangle \dfrac{\sum_n\partial_\theta\langle\psi|n\rangle\langle n|\psi\rangle}{\sum_n|\psi(n)|^2}\bigg] \\
&=2\Re\bigg[ \dfrac{\sum_n|\phi(n)|^2\dfrac{\partial_\theta\langle \psi|n\rangle}{f_n^*\langle \phi|n\rangle}\dfrac{f_n^*\langle n|H|\psi\rangle}{\langle n|\phi\rangle}}{\sum_n|\psi(n)|^2} - \langle E\rangle\dfrac{\sum_n|\phi(n)|^2\dfrac{f^*_nf_n\partial_\theta\langle\psi | n\rangle}{f^*_n\langle \phi|n\rangle}}{\sum_n|\psi(n)|^2} \bigg] \\
&=2\Re\bigg[ \bigg\langle\dfrac{\partial_\theta\langle \psi|n\rangle}{\langle \psi|n\rangle} \bigg(\dfrac{1}{B}\dfrac{f_n^*\langle n|H|\psi\rangle}{\langle n|\phi\rangle}-|f_n|^2\bigg)\langle E\rangle\bigg\rangle_n \bigg] \\
&=2\Re\big[ \big\langle\partial_\theta\ln\langle\psi|n\rangle (E_{\rm loc}-\dfrac{|f_n|^2}{B}\langle E\rangle) \big\rangle_n \big]\\
&=2\Re\big[ \big\langle(\partial_\theta\ln(f_n\phi(n))^*\rangle) (E_{\rm loc}-|\widetilde{f}_n|^2\langle E\rangle) \big\rangle_n \big]\\
\end{aligned}\end{split}\]
With some symmetry
We consider state \(\ket{N,S,M}\) which is eigenvector of operators \(\{N,S,S_z\}\)
, Let Spin Flip operator \(U_{\rm SF}:=\mathrm{e}^{\mathrm{i}\mathrm{\pi}(S_x-N/2)}\), can flip spins, such as
\[U_{\rm SF} \ket{N,S,M} = (-1)^{N/2-S}\ket{N,S,-M}\]
For states with \(M=0\), then \(N_\alpha = N_\beta = N/2\), it leads to
\[U_{\rm SF}\ket{N,S,0} = (-1)^{N_\alpha-S}\ket{N,S,0}\]
For example, with the basis set \(\{ \ket{n_\alpha m_\beta} := \ket{n}\otimes \ket{m}:\ket{n},\ket{m}\in\{\ket{0},\ket{1}\},\ket{0} = \begin{bmatrix}1\\0\end{bmatrix},\ket{1} = \begin{bmatrix}0\\1\end{bmatrix} \}\),
the matrix elements like
\[\begin{split}[U_{\rm SF}] = \begin{bmatrix} 1&0&0&0\\ 0&0&1&0\\ 0&1&0&0\\ 0&0&0&-1\end{bmatrix}\end{split}\]
then \(U_{\rm SF} \ket{1_\alpha 1_\beta} = -\ket{1_\alpha 1_\beta}\) can be verified. In conclusion
\[U_{\rm SF}\ket{n} = \eta_n \ket{n_{\rm SF}} =: |\bar{n}\rangle\]
Where \(|n_{\rm SF}\rangle\) is the state whose spins be flipped in state \(\ket{n}\). If target state \(\ket{\psi}\) with \(N\) electrons has determinate eigenvalue \(\eta\) of operator \(U_{\rm SF}\)
(\(\eta\) is defined by yourself. such as H-chain(\(n=50\)), \(N_\alpha\) is 25, if the target state is siglet, then \(\eta = (-1)^{25-0}=-1\))
\[U_{\rm SF}\ket{\psi} = \eta \ket{\psi}, \ U_{\rm SF} = \bigotimes_{i=1}^{N/2}U_{\rm SF}\]
Define projector \(P_\eta = \dfrac{1}{2}(I+\eta U_{\rm SF})\), which \(I\) is unit operator, it is easy to show that \(P_{\eta}^2 = I, [P_{\eta} , H]=0\),
for our symmetry-projected NQS
\[\ket{\psi_\eta} = \dfrac{P_{\eta}\ket{\psi}}{\sqrt{\langle \psi | P_\eta | \psi\rangle}}, \ \langle n|\psi\rangle = f_n \langle n|\phi\rangle ,\ \langle\phi|\phi\rangle =1\]
the expectation of energy is
\[\begin{split}\begin{aligned}
\langle E\rangle = \dfrac{\langle{\psi}|{H{P_\eta}}|{\psi}\rangle}{\langle{\psi}|{{P_\eta}}|{\psi}\rangle}&= \dfrac{\sum_{n}\langle{\psi}|{n}\rangle\langle{n}|{HP_\eta}|{\psi}\rangle}
{\sum_{n}\langle{n}|{\psi}\rangle\langle{n}|{P_\eta}|{\psi}\rangle} \\
&= \dfrac{\sum_n|\langle{n}|{\phi}\rangle|^2\dfrac{\langle{\psi}|{n}\rangle\langle{n}|{HP_\eta}|{\psi}\rangle}{|\langle{n}|{\phi}\rangle|^2}}
{\sum_n|\langle{n}|{\phi}\rangle|^2\dfrac{\langle{\psi}|{n}\rangle\langle{n}|{P_\eta}|{\psi}\rangle}{|\langle{n}|{\phi}\rangle|^2}}\\
&=\dfrac{\bigg\langle \dfrac{\langle{\psi}|{n}\rangle\langle{n}|{HP_\eta}|{\psi}\rangle}{|\langle{n}|{\phi}\rangle|^2}\bigg\rangle_n}{\bigg\langle\dfrac{\langle{\psi}|{n}\rangle\langle{n}|{P_\eta}|{\psi}\rangle}{|\langle{n}|{\phi}\rangle|^2}\bigg\rangle_n}\\
&=\dfrac{\bigg\langle \dfrac{f_n^*\langle{\phi}|{n}\rangle\langle{n}|{HP_\eta}|{\psi}\rangle}{|\langle{n}|{\phi}\rangle|^2}\bigg\rangle_n}
{\bigg\langle\dfrac{f_n^*\langle{\phi}|{n}\rangle\langle{n}|{P_\eta}|{\psi}\rangle}{|\langle{n}|{\phi}\rangle|^2}\bigg\rangle_n}\\
&=\dfrac{\bigg\langle \dfrac{f_n^*\langle{n}|{HP_\eta}|{\psi}\rangle}{\langle{n}|{\phi}\rangle}\bigg\rangle_n}{\bigg\langle\dfrac{f_n^*\langle{n}|{P_\eta}|{\psi}\rangle}{\langle{n}|{\phi}\rangle}\bigg\rangle_n}=\langle E_{\rm loc}(n)\rangle_n
\end{aligned}\end{split}\]
Define \(B = 2\bigg\langle \dfrac{f_n^*\langle n|P_\eta|\psi\rangle}{\langle n|\phi\rangle} \bigg\rangle_n, \ \widetilde{f}_{n} = f_n/\sqrt{B}\), Then
\[\begin{aligned}
P_{\rm loc}(n) = \dfrac{1}{B} f_n^*\dfrac{\langle{n}|{P_\eta}|{\psi}\rangle}{\langle{n}|{\phi}\rangle}
= \dfrac{1}{B} f_n^*\dfrac{\langle{n}|{\psi}\rangle+\eta\langle n|\bar{\psi}\rangle}{\langle{n}|{\phi}\rangle}
= \dfrac{1}{2B}(|f_n|^2+\eta f_n^*f_{\bar{n}}\dfrac{\langle{\bar{n}}|{\phi}\rangle}{\langle{n}|{\phi}\rangle})
\end{aligned}\]
local-energy is
\[\begin{split}\begin{aligned}
E_{\rm loc}(n) &= \dfrac{2f_n^*}{B}\dfrac{\langle{n}|{HP_\eta}|{\psi}\rangle}{\langle{n}|{\phi}\rangle} = \dfrac{f_n^*}{\langle P_{\rm loc}\rangle_n}\dfrac{\sum_m\langle{n}|{H}|{m}\rangle\langle{m}|{P_\eta}|{\psi}\rangle}
{\langle{n}|{\phi}\rangle}\\
&=\dfrac{1}{2}\dfrac{2f_n^*}{B}\dfrac{\sum_m H_{nm}(\langle{m}|{\psi}\rangle+\eta\langle m|\bar{\psi}\rangle )}{\langle{n}|{\phi}\rangle}\\
&=\dfrac{f_n^*}{\sqrt{B}}\dfrac{\sum_m H_{nm}(\frac{f_m}{\sqrt{B}}\langle{m}|{\phi}\rangle+\eta \frac{f_{\bar{m}}}{\sqrt{B}}\langle{\bar{m}}|{\phi}\rangle)}{\phi(n)}\\
&=\dfrac{\sum_m \widetilde{f}_n^* H_{nm}(\widetilde{f}_m\langle m|\phi\rangle + \eta \widetilde{f}_{\bar{m}}\langle \bar{m}|\phi\rangle)}{\phi(n)}
\end{aligned}
gradient of :math:`\langle E \rangle` is\end{split}\]
\[\begin{split}\begin{aligned}
\partial_\theta\langle E\rangle =& \dfrac{\partial}{\partial \theta}\dfrac{\langle{\psi}|{H\textcolor{purple}{P_\eta}}|{\psi}\rangle}{\langle{\psi}|{\textcolor{purple}{P_\eta}|}{\psi}\rangle} \\
=& 2\Re \Bigg[ \dfrac{\langle{\partial_\theta\psi}|{HP}|{\psi}\rangle}{\langle{\psi}|{P}|{\psi}\rangle}-\dfrac{\langle{\psi}|{HP}|{\psi}\rangle}{|\langle{\psi}|{P}|{\psi}\rangle|^2}\times \langle{\partial_\theta \psi}|{P}|{\psi}\rangle \Bigg]\\
=&2\Re \Bigg[ \dfrac{\sum_n\langle{\partial_\theta\psi}|{n}\rangle\langle{n}|{HP}|{\psi}\rangle}{B} - \dfrac{\sum_n\langle{\psi}|{n}\rangle\langle{n}|{HP}|{\psi}\rangle}{B} \big\langle (\partial_\theta\ln (f_n\phi(n))^*) P_{\rm loc}(n)\big\rangle_n\Bigg]\\
=&2\Re \Bigg[ \big\langle (\partial_\theta\ln (f_n\phi(n))^*) E_{\rm loc}\big\rangle_n-\langle E\rangle \big\langle (\partial_\theta\ln (f_n\phi(n))^*) P_{\rm loc}\big\rangle_n\Bigg]\\
=&2\Re \big[ \big\langle (\partial_\theta\ln (f_n\phi(n))^*) (E_{\rm loc}-\langle E\rangle P_{\rm loc})\big\rangle_n\big]
\end{aligned}\end{split}\]