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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 $\beta_{i(1)}$.
  • $j&gt;1$.
    • Set $A = { P \in H | P_{i(j)} \in \mathcal{B} \textrm{ and } P_{i(j')} \in {I, B_{i(j')}} \textrm{ for } j'&lt;j }$.
    • 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 $\beta_{i(j)}(B) = \sqrt{c_B} / c$