split_non_commuting transform accepts any type of measurement about pennylane

[rmoyard]

Hello @abhishekabhishek, thank you for double checking!

Your understanding of the problem is correct, you should remove the limitation concerning those measurements. And yes the main issue is that wires needs to be translated to operators and it seems that you are on a good track!

Let me know if anything is not clear

[rmoyard]

Hi @abhishekabhishek, with the old device it is indeed part of the pipeline and called under the hood. With the new device it is not used and split non commuting is not necessary because it can handle non commuting measurements.

I would suggest to mainly use the transform directly on a tape (qml.tape.QuantumTape) and for integration testing you can apply it directly on the QNode. Even if it is not necessary, the transform will be applied as part of the transform program.

     @qml.qnode(dev) 
     def circuit(params): 
         qml.RX(params[0], wires=0) 
         qml.RY(params[1], wires=1) 
         return qml.expval(qml.PauliZ(0) @ qml.PauliZ(1)), qml.expval(qml.PauliY(0)))

split_non_commuting(circuit)([0.1, 0.2])

In the test that you showed, I would suggest to explicitly apply the transform on the QNodes.

Let me know it that helps!

[abhishekabhishek]

Hi @Alex-Preciado,

I am currently working on this issue and just wanted to double check my understanding to make sure I'm on the right path. As I understand currently, the issue is the following:

Consider the following qnode:

dev = qml.device("default.qubit", wires=2)

@qml.qnode(dev)
def circuit(x):
    qml.RX(x,wires=0)
    return qml.probs(wires=[0, 1]), qml.expval(qml.PauliZ(0)), qml.expval(qml.PauliX(0))
    
>>> print(qml.draw(circuit)(0.5))

0: ──RX(0.50)─┤ ╭Probs  <Z>  <X>
1: ───────────┤ ╰Probs

Applying circuit = qml.transforms.split_non_commuting(circuit) should return a circuit s.t.

>>> print(qml.draw(circuit)(0.5))

0: ──RX(0.50)─┤ ╭Probs  <Z>
1: ───────────┤ ╰Probs 
\
0: ──RX(0.50)─┤ <X>
1: ───────────┤

But the transform currently doesn't support SampleMP, ProbabilityMP or CountsMP which can take either wires or obs. I think I have a solution which I'm currently implementing and testing i.e. using qml.PauliZ(wires[0])@qml.PauliZ(wires[1])@[email protected](wires[len(wires)-1]) as observables when either of these are used only with wires argument and I'm looking into the reorder_fn to determine if I need to make any changes to it.

Please let me know if you think I missed something 🙂

Cheers,
Abhi

[Alex-Preciado]

Hi @abhishekabhishek , thank you so much for looking into this 🚀 . @rmoyard will be providing some guidance here shortly.

[abhishekabhishek]

Thanks @rmoyard, I am currently writing tests for the modified transform but got a little confused by the existing integration and autodiff tests for e.g. the one below:

pennylane/tests/transforms/test_split_non_commuting.py

Lines 288 to 311 in f435f87

	@pytest.mark.autograd
	def test_split_with_autograd(self):
	"""Test that results after splitting are still differentiable with autograd"""
	dev = qml.device("default.qubit", wires=3)
	@qml.qnode(dev)
	def circuit(params):
	qml.RX(params[0], wires=0)
	qml.RY(params[1], wires=1)
	return (
	qml.expval(qml.PauliZ(0) @ qml.PauliZ(1)),
	qml.expval(qml.PauliY(0)),
	qml.expval(qml.PauliZ(0)),
	)
	def cost(params):
	res = circuit(params)
	return qml.math.stack(res)
	params = pnp.array([0.5, 0.5])
	res = cost(params)
	grad = qml.jacobian(cost)(params)
	assert all(np.isclose(res, exp_res))
	assert all(np.isclose(grad, exp_grad).flatten())

It seems that in this test, the split_non_commuting transform does not get called on the Qnode. Am I missing something here? Does the transform automatically gets applied to the Qnode when more than one expval are returned?

[albi3ro]

Closed by #4972