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"Postselection" refers to the process of conditioning on the outcome of a measurement on some other qubit. (This is something that you can think of for classical probability distributions and statistical analysis as well: it is not a concept special to quantum computation.) Postselection has featured quite often (up to this point) in quantum mechanics ...


10

As the other answer conveyed (and to which I am just trying to provide some clarification), post-selection is about just looking at a subset of possible measurement outcomes. To my mind, this falls into two different cases, as below. Yes, they are different aspects of the same thing, but they are used very differently by two different communities. ...


4

Your intuition is correct for a single qubit, in that if I measure $$\alpha\vert 0 \rangle + \beta\vert 1 \rangle$$ I would get either $\vert 0 \rangle$ or $\vert 1 \rangle$. But since the qubits are in a large entangled state, the relevant information stored in the ratios of different probabilities is still held fixed, and the $\frac{C}{\lambda_j}$ factors ...


3

You are half right, in that the $C$ factor is only kept there for (what I assume being) explanatory purposes. However, the $1/\lambda_j$ factors definitely stays there after postselection. One way to see this is that you can think of those factors as attached to the other registers, so that the state is equivalently written as $$ \left(\sum_j\beta_j\sqrt{1-...


1

You can use save_statevector with conditional = True, like that: from qiskit import QuantumCircuit from qiskit.providers.aer import AerSimulator import numpy as np circ = QuantumCircuit(2, 2) 𝜓 = [np.sqrt(0.8), np.sqrt(0.2)] circ.initialize(𝜓, 0) circ.initialize(𝜓, 1) circ.measure(1, 1) circ.save_statevector(conditional = True) simulator = AerSimulator(...


1

As I understand it, the key caveat to postselection, at least in regards to retrocausal effects, is that Bob already knows what state Alice will post-select. Since he already knows what she will do, and she always does so with 100% success rate, then he can just treat his qubit as already having underwent the measurement and received a pure state, so that he ...


1

CW from self-answer, and also because this is more of an extended comment than an answer. Let $A$ be the adjacency matrix of the Cayley graph of our group $\mathcal{H}$ of order $N$. Notice that $A$ is square-hermitian. Further let $\mathbb{I}_N$ be the $N\times N$ identity matrix. It occurs to me that I am, in a sense, asking to prepare the ground state $...


1

In principle, Bob here just has to guess the $2\times 2$ matrix $\sigma$. If he starts with any parametric state $\sigma(\alpha,\beta)$ with $\alpha,\beta\in\mathbb{C}$ and measures the outcome Tr$(M\sigma)$ with the post measurement state $\sigma '=M\rho M^\dagger/\text{Tr}(M\rho)$, he receives a number and he has to tune $\alpha,\beta$ to come close to ...


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