# Symbolic circuits with pytket-qujax#

In this notebook we will show how to manipulate symbolic circuits with the pytket-qujax extension. In particular, we will consider a QAOA and an Ising Hamiltonian.

See the docs for qujax and pytket-qujax.

from pytket import Circuit
from pytket.circuit.display import render_circuit_jupyter
from jax import numpy as jnp, random, value_and_grad, jit
from sympy import Symbol
import matplotlib.pyplot as plt
import qujax
from pytket.extensions.qujax import tk_to_qujax

# QAOA#

The Quantum Approximate Optimization Algorithm (QAOA), first introduced by Farhi et al., is a quantum variational algorithm used to solve optimization problems. It consists of a unitary $$U(\beta, \gamma)$$ formed by alternate repetitions of $$U(\beta)=e^{-i\beta H_B}$$ and $$U(\gamma)=e^{-i\gamma H_P}$$, where $$H_B$$ is the mixing Hamiltonian and $$H_P$$ the problem Hamiltonian. The goal is to find the optimal parameters that minimize $$H_P$$. Given a depth $$d$$, the expression of the final unitary is $$U(\beta, \gamma) = U(\beta_d)U(\gamma_d)\cdots U(\beta_1)U(\gamma_1)$$. Notice that for each repetition the parameters are different.

## Problem Hamiltonian#

QAOA uses a problem dependent ansatz. Therefore, we first need to know the problem that we want to solve. In this case we will consider an Ising Hamiltonian with only $$Z$$ interactions. Given a set of pairs (or qubit indices) $$E$$, the problem Hamiltonian will be:

$$$H_P = \sum_{(i, j) \in E}\alpha_{ij}Z_iZ_j,$$$

where $$\alpha_{ij}$$ are the coefficients. Let’s build our problem Hamiltonian with random coefficients and a set of pairs for a given number of qubits:

n_qubits = 4
hamiltonian_qubit_inds = [(0, 1), (1, 2), (0, 2), (1, 3)]
hamiltonian_gates = [["Z", "Z"]] * (len(hamiltonian_qubit_inds))

Notice that in order to use the random package from jax we first need to define a seeded key

seed = 13
key = random.PRNGKey(seed)
coefficients = random.uniform(key, shape=(len(hamiltonian_qubit_inds),))
print("Gates:\t", hamiltonian_gates)
print("Qubits:\t", hamiltonian_qubit_inds)
print("Coefficients:\t", coefficients)
Gates:	 [['Z', 'Z'], ['Z', 'Z'], ['Z', 'Z'], ['Z', 'Z']]
Qubits:	 [(0, 1), (1, 2), (0, 2), (1, 3)]
Coefficients:	 [0.6794174  0.2963785  0.2863201  0.31746793]

## Variational Circuit#

Before constructing the circuit, we still need to select the mixing Hamiltonian. In our case, we will be using $$X$$ gates in each qubit, so $$H_B = \sum_{i=1}^{n}X_i$$, where $$n$$ is the number of qubits. Notice that the unitary $$U(\beta)$$, given this mixing Hamiltonian, is an $$X$$ rotation in each qubit with angle $$\beta$$. As for the unitary corresponding to the problem Hamiltonian, $$U(\gamma)$$, it has the following form:

$$$U(\gamma)=\prod_{(i, j) \in E}e^{-i\gamma\alpha_{ij}Z_i Z_j}$$$

The operation $$e^{-i\gamma\alpha_{ij}Z_iZ_j}$$ can be performed using two CNOT gates with qubit $$i$$ as control and qubit $$j$$ as target and a $$Z$$ rotation in qubit $$j$$ in between them, with angle $$\gamma\alpha_{ij}$$. Finally, the initial state used, in general, with the QAOA is an equal superposition of all the basis states. This can be achieved adding a first layer of Hadamard gates in each qubit at the beginning of the circuit.

With all the building blocks, let’s construct the symbolic circuit using tket. Notice that in order to define the parameters, we use the Symbol object from the sympy package. More info can be found in this documentation. In order to later convert the circuit to qujax, we need to return the list of symbolic parameters as well.

def qaoa_circuit(n_qubits, depth):
circuit = Circuit(n_qubits)
p_keys = []

# Initial State
for i in range(n_qubits):
circuit.H(i)
for d in range(depth):
# Hamiltonian unitary
gamma_d = Symbol(f"γ_{d}")
for index in range(len(hamiltonian_qubit_inds)):
pair = hamiltonian_qubit_inds[index]
coef = coefficients[index]
circuit.CX(pair[0], pair[1])
circuit.Rz(gamma_d * coef, pair[1])
circuit.CX(pair[0], pair[1])
p_keys.append(gamma_d)

# Mixing unitary
beta_d = Symbol(f"β_{d}")
for i in range(n_qubits):
circuit.Rx(beta_d, i)
p_keys.append(beta_d)
return circuit, p_keys
depth = 3
circuit, keys = qaoa_circuit(n_qubits, depth)
keys
[γ_0, β_0, γ_1, β_1, γ_2, β_2]

Let’s check the circuit:

render_circuit_jupyter(circuit)

# Now for qujax#

The pytket.extensions.qujax.tk_to_qujax function will generate a parameters -> statetensor function for us. However, in order to convert a symbolic circuit we first need to define the symbol_map. This object maps each symbol key to their corresponding index. In our case, since the object keys contains the symbols in the correct order, we can simply construct the dictionary as follows:

symbol_map = {keys[i]: i for i in range(len(keys))}
symbol_map
{γ_0: 0, β_0: 1, γ_1: 2, β_1: 3, γ_2: 4, β_2: 5}

Then, we invoke the tk_to_qujax with both the circuit and the symbolic map.

param_to_st = tk_to_qujax(circuit, symbol_map=symbol_map)

And we also construct the expectation map using the problem Hamiltonian via qujax:

st_to_expectation = qujax.get_statetensor_to_expectation_func(
hamiltonian_gates, hamiltonian_qubit_inds, coefficients
)
param_to_expectation = lambda param: st_to_expectation(param_to_st(param))

# Training process#

We construct a function that, given a parameter vector, returns the value of the cost function and the gradient. We also jit to avoid recompilation, this means that the expensive cost_and_grad function is compiled once into a very fast XLA (C++) function which is then executed at each iteration. Alternatively, we could get the same speedup by replacing our for loop with jax.lax.scan. You can read more about JIT compilation in the JAX documentation.

For the training process we’ll use vanilla gradient descent with a constant stepsize:

seed = 123
key = random.PRNGKey(seed)
init_param = random.uniform(key, shape=(len(symbol_map),))
n_steps = 150
stepsize = 0.01
param = init_param
cost_vals = jnp.zeros(n_steps)
cost_vals = cost_vals.at[0].set(param_to_expectation(init_param))
for step in range(1, n_steps):
cost_vals = cost_vals.at[step].set(cost_val)
param = param - stepsize * cost_grad
print("Iteration:", step, "\tCost:", cost_val, end="\r")
Iteration: 1 	Cost: 0.24282376
Iteration: 2 	Cost: -0.05965677
Iteration: 3 	Cost: -0.20374677
Iteration: 4 	Cost: -0.29387552
Iteration: 5 	Cost: -0.35805708
Iteration: 6 	Cost: -0.40720624
Iteration: 7 	Cost: -0.445078
Iteration: 8 	Cost: -0.47320744
Iteration: 9 	Cost: -0.49311572
Iteration: 10 	Cost: -0.5066574
Iteration: 11 	Cost: -0.51566684
Iteration: 12 	Cost: -0.52163947
Iteration: 13 	Cost: -0.52564937
Iteration: 14 	Cost: -0.52840817
Iteration: 15 	Cost: -0.53036916
Iteration: 16 	Cost: -0.53181255
Iteration: 17 	Cost: -0.5329127
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Iteration: 31 	Cost: -0.5381187
Iteration: 32 	Cost: -0.53827465
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Iteration: 48 	Cost: -0.5402081
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Iteration: 91 	Cost: -0.54446423
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Iteration: 95 	Cost: -0.5448335
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