adaptiveshadows

Locally-biased classical shadows with an adaptive update

How do I run this?

Install requirements, and navigate to folder adaptiveshadows/python. Then call python sim.py <name> where <name> is the molecule and encoding joined by an underscore. Molecules are: h2, lih, beh2, h2o, nh3 and encodings are jw, parity, bk. So one example of <name> is h2_jw.

Current results:

Take a Hamiltonian, and allow access to the ground state 1000 times. Record the energy difference between the estimated energy and the true ground energy.

Molecule	Encoding	Derand	Adaptive	Adaptive RSME	Diagonalize	Simulation
H2	JW	0.06	0.003, 0.04, 0.08, 0.09, 0.12, 0.15, 0.09, 0.004, 0.08, 0.03, 0.046	0.08	1sec	10sec
	Parity	0.03	0.02, 0.047, 0.054, 0.058, 0.036, 0.085, 0.029, 0.031, 0.01, 0.053	0.05
	BK	0.06	0.12, 0.03, 0.01, 0.03, 0.05, 0.04, 0.01, 0.02, 0.017, 0.2	0.08
LiH	JW	0.03	0.07, 0.027, 0.07, 0.01, 0.05, 0.02, 0.03, 0.04, 0.06, 0.02, 0.04	0.04	1sec	1min50sec
	Parity	0.03	0.01, 0.05, 0.07, 0.01, 0.05, 0.07, 0.03, 0.04, 0.03, 0.05	0.05		1min30sec
	BK	0.04	0.02, 0.009, 0.1, 0.09, 0.06, 0.1, 0.008, 0.1, 0.02, 0.02	0.07		1min40sec
BeH2	JW	0.06	0.02, 0.057, 0.05, 0.07, 0.06, 0.04, 0.055, 0.07, 0.07, 0.11	0.06	7sec	6min20sec
	Parity	0.09	0.09, 0.051, 0.06, 0.06, 0.05, 0.05, 0.07, 0.06, 0.1, 0.0009	0.06		5min40sec
	BK	0.06	0.07, 0.06, 0.05, 0.05, 0.08, 0.07, 0.06, 0.07, 0.07, 0.06	0.06		6min
H2O	JW	0.12	0.12, 0.12, 0.13, 0.1, 0.1, 0.12, 0.10, 0.09, 0.09, 0.1	0.11	15sec	6min45sec
	Parity	0.22	0.16, 0.056, 0.12, 0.1, 0.12, 0.1, 0.12, 0.09, 0.11, 0.09, 0.11	0.11		6min10sec
	BK	0.20	0.028, 0.11, 0.11, 0.1, 0.11, 0.09, 0.12, 0.11, 0.1, 0.11	0.10		6min10sec
NH3	JW	0.18	0.11, 0.076, 0.0507, 0.12, 0.14, 0.18, 0.16, 0.14, 0.14, 0.14	0.13	5min30	28min
	Parity	0.21	0.079, 0.086, 0.17, 0.14, 0.14, 0.14, 0.18, 0.15, 0.16, 0.16	0.14		26min
	BK	0.12	0.06, 0.09, 0.13, 0.13, 0.13, 0.14, 0.06, 0.15, 0.03, 0.10	0.11		27min

How do we generate a measurement basis?

One qubit at a time update idea

Consider $H = \sum_P \alpha_P P$ on $n$ qubits. Set $\mathcal{B}={X,Y,Z}$.

Let $i:[n]\to[n]$ be a random bijection. This gives an ordering of the qubits $i(1), \dots, i(n)$. Write $i(j)$ for $j\in[n]$.

We will chose $\beta_{i(j)}:\mathcal{B}\to\mathbb{R}^+$ in an adaptive style.

Let's do this slowly:

• $j=1$.
  • Set $A = { P \in H | P_{i(1)} \in \mathcal{B} }$.
  • Minimise $\Delta(\beta_{i(1)}) = \sum_{P\in A} \alpha_P^2 / \beta_{i(1)}(P_{i(1)})$.
  • Pick $B_{i(1)}$ from $\beta_{i(1)}$.
• $j>1$.
  • Set $A = { P \in H | P_{i(j)} \in \mathcal{B} \textrm{ and } P_{i(j')} \in {I, B_{i(j')}} \textrm{ for } j'<j }$.
  • Minimise $\Delta(\beta_{i(j)}) = \sum_{P\in A} \alpha_P^2 / \beta_{i(j)}(P_{i(j)})$.
  • Pick $B_{i(j)}$ from $\beta_{i(j)}$.

Now set $B = \otimes_{i\in[n]} B_i$

Actually, the minimisation procedure can be done analytically. Given $j$ and $A$ do the following:

• Set $A = A_X \cup A_Y \cup A_Z$ where $A_B = { P\in A | P_{i(j)}=B }$
• Set $c_B = \sum_{P \in A_B} \alpha_P^2$
• Set $c = \sum_B \sqrt{c_B}$
• If $c=0$ set $\beta_{i(j)}(B)=1/3$ else set $\beta_{i(j)}(B) = \sqrt{c_B} / c$