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I am working with pennyLane library in Python and I have a circuit that involves several operations to 10 qubits. I want to measure the occupation probability of a target state. The function qml.probs() returns the probabilities |βŸ¨π‘–|πœ“βŸ©|^2 of measuring the computational basis state |π‘–βŸ© given the current state |πœ“βŸ©.

If I use qml.probs(wires=range(10)) the measurement is with respect to |+π‘–βŸ©/|-π‘–βŸ©, and I need to get the result with respect to the basis |0⟩/|1⟩. I have tried [qml.probs(op=qml.PauliZ(i)) for i in range(10)] , qml.probs(op=[qml.expval(qml.PauliZ(i)) for i in range(10)]) but it gives the individual result of each qubit instead of the whole 10-qubit state, or it doesn't give back the expected result.

I think that there must be an option to get the result I want, and will be related to the "op" parameter that I am not using correctly, so I would be grateful if someone could help me.

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1 Answer 1

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If you have 10 qubits there are $2^{10}$ possible measurement outcomes. Hence, qml.prob(wires=10) will return an array with $2^{10}$ elements. Each element of an array is the probability of observing the state $|i\rangle$ where $i \in \{0,\ldots,2^{10}-1\}$. The measurement is done with respect to the computational basis. Given that you wrote something that doesn't quite make sense, I feel like it is worth pointing out that $|i\rangle$ is an integer representation of a binary string of length 10.

Perhaps, you want to take a look at a simpler example with 2 qubits:

dev = qml.device("default.qubit", wires=2)
@qml.qnode(dev)
def circuit():
   qml.Hadamard(wires=1)
   return qml.probs(wires=[0, 1])
circuit()

This code generates the state $$\tag{1} |0\rangle \left(\frac{|0\rangle + |1\rangle}{\sqrt{2}}\right) = \frac{|00\rangle}{\sqrt{2}} + \frac{|01\rangle}{\sqrt{2}}.$$ Given that the measurement is in the computational basis, your possible outcomes are: $|0\rangle \equiv |00\rangle, |1\rangle \equiv |01\rangle, |2\rangle \equiv |10\rangle, |3\rangle \equiv|11\rangle$. From (1) it is clear that qml.probs(wires=[0,1]) will yield an array [0.5, 0.5, 0, 0].

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  • $\begingroup$ Please don't write $|i\rangle / |-i\rangle$ because this notation doesn't make any sense. $\endgroup$
    – MonteNero
    Jul 11, 2022 at 7:53
  • $\begingroup$ True, thank you. $\endgroup$ Jul 12, 2022 at 8:09

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