# Tag Info

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Yes, the measurements on the ancillary qubits are unnecessary. You can just discard those qubits instead of measuring them.

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TL;DR No. Non-local gates and measurements between causally disconnected observers violate the no-signalling theorem and are therefore impossible. No-signalling theorem The following protocol allows Bob to send one classical bit $b$ to Alice faster than light. Alice prepares two sets of qubits $x_i$ and $y_i$ in the $|0\rangle$ state. Bob prepares one set of ...

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The IBM histogram is correct. You seem to be under the impression that a reset is a postselection. Applying a reset to a qubit cannot change the expected measurement statistics on other qubits. That would violate the no-communication theorem. So the distribution before the reset is equal to the distribution after. The same is not true for a postselection, ...

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$|\text{scattered}\rangle$ refers to the state where the photon is no longer present (because it was absorbed/scattered by the detector). More generally, you can think of it as any state where the state of the photon has become entangled with things outside the system. The simplest way to model this state is to introduce an ancilla qubit to represent "...

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Asymmetric readout error While its hard to make strong claims about the noise characteristics of any given quantum device, one explanation for what you're observing is readout error. For superconducting qubits readout error tends to be asymmetric: the probability $p(0|1)$ of observing a "0" after performing measurement on a computational basis ...

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Not in general, no. Consider the state $$|\psi\rangle=\frac{1}{\sqrt{2}}\left(|0x\rangle+|1\bar x\rangle\right)|+\rangle$$ for $x\in\{0,1\}$. We have $\langle Z_i\rangle=0$, and $\langle Z_1Z_2Z_3\rangle=0$ (to see this most trivially, look at the third qubit). However, the first two qubits are an eigenstate of $Z_1Z_2$ of eigenvalue $(-1)^x$. From the ...

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The claim does not specify what protocols for distinguishing quantum states are acceptable. In particular, it does not state whether we are allowed to err or reserve judgment. Below, we note success probability for protocols allowed to err and compute success probability for an error-free protocol. The success probability of the former is not bounded by $4\... 2 Applying a reset to a qubit is equivalent to measuring it, and then applying a bit flip to it conditioned on the measurement result. def reset(qubit): if measure(qubit) == ON: X(qubit) For example, in this Quirk circuit, you can see that the post-reset state matches the state you'd get when conditioning on a measurement-via-ancilla+bit-flip of ... 2 Bob does not know the outcome of Alice's measurement. All he knows is that Alice would have obtained$|\psi\rangle$with probability$\frac12$and$|\psi^\perp\rangle$with probability$\frac12$for some orthonormal basis$|\psi\rangle$,$|\psi^\perp\rangle$. Therefore, his state is a mixture $$\rho_B = \frac12|\overline\psi\rangle\langle\overline\psi|+\... 2 \vec{a}\cdot\vec{\sigma} defines an operator$$ a_xX+a_YY+a_ZZ $$where X,Y,Z are the Pauli matrices. So, for example of \vec{a}=(0,a_Y,a_Z) then it has no component in the X plane, and we say it defines a measurement in the Y-Z plane. 2 Without loss of generality your question could also be phrased as "what would I measure in the bottom, second register if I waited to measure the second register until after performing the QFT on the upper first register?" In that case it might be seen that measuring the second register only affects the (unmeasurable) global phase of the system. ... 1 According to Equation 6 of the paper, the state one is considering is:$$\frac{\mathrm{i}\sin\left(\theta_d\right)}{2}|1\rangle\left(|\zeta\rangle|\psi\rangle-|\psi\rangle|\zeta\rangle\right)+\frac{\mathrm{e}^{-\mathrm{i}\theta_d}+\cos\left(\theta_d\right)}{2}|0\rangle|\psi\rangle|\zeta\rangle-\frac{\mathrm{i}\sin\left(\theta_d\right)}{2}|0\rangle|\zeta\... 1 The correction of errors during computations is required for large-scale quantum information processing. A qubit is encoded in a subspace of many physical qubits in quantum error correction so that faults can be actively rectified without altering the stored data. We must account for the fact that errors occur not only when something like a gate or ... 1 Yes, probabilities shown in a histogram are squares of absolute values of coefficients in a state vector (i.e. probability amplitudes). Note that the amplitudes are generally complex numbers, hence we square the absolute values and not the amplitudes themselves. 1 Yes, this is exactly what happens. Instead of saying "I want to measure this specific value," you specify a set of POVM elements asking "which of these values will I get?". When you are measuring the value of the first qubit alone to see if it is in state$|0\rangle$or$|1\rangle$, the two POVM elements can be specified by$\Pi^{(0)}=|0\...

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Yes, the reasoning is correct. In fact, it can be generalized beyond pure states. By definition, every mixed quantum state $\rho$ is a positive semidefinite operator with unit trace. Since every positive semidefinite operator is Hermitian, we may interpret $\rho$ as an observable. In this case, the expectation of observable $\rho$ in state $\sigma$  \...

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This looks like expected behavior by optimizer at level 3. Whatever backend you chose, level 3 is noise-adaptive, and most probably qubits chosen as 1 and 2 are the least noisy to perform the swap on, and later reinterpret the result during measurement. The qubit identities are not wrong in itself - the optimizer/transpiler will remember its mapping of ...

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I think there are two nested questions there: How can you simulate partial measurement? There are a few approaches, but one basic one is described here. In short, zero the amplitudes of the states which didn't happen (i.e. ones for which the qubit you measured has the opposite value than the one you measured) and then normalize remaining amplitudes so that \$...

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