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1

An intuitive way to think about it is that $E[M]=E[X_1 \otimes Z_2]=E[X_1 \otimes \mathbb{1}]E[\mathbb{1} \otimes Z_2]$ If we only think about $E[\mathbb{1} \otimes Z_2]$, it is just the expectation value of $Z_2$ on the second qubit. Consider that our second Qubit in the entangled state $\frac{| 00\rangle + | 11\rangle}{\sqrt{2}}$ is measured to be \frac{+\... 1 Taking the last two terms of last expression you gave, we can do the following \begin{align} M \left(\frac{|00\rangle+|11\rangle}{\sqrt{2}}\right) &= X_1\otimes Z_2\left(\frac{|00\rangle+|11\rangle}{\sqrt{2}}\right) \\ &= \left(\frac{X_1|0\rangle \otimes Z_2|0\rangle+X_1|1\rangle \otimes Z_2|1\rangle}{\sqrt{2}}\right) \\ &= \left(\frac{|1\... 4 We know that \begin{gather} |0\rangle = \frac{|+\rangle+|-\rangle}{\sqrt{2}} \\ |1\rangle = \frac{|+\rangle-|-\rangle}{\sqrt{2}} \end{gather} $$Thus, we can rewrite the GHZ state as$$ \begin{align} |GHZ\rangle &= \frac{1}{\sqrt{2}}\left(|0\rangle|00\rangle+|1\rangle|11\rangle\right) \\ &=\frac{1}{2}\left(|+\rangle|00\rangle+|-\rangle|00\rangle+... 1 If you don't want to write your own function to do this then one way to do this is through qiskit pauli_measurement. For example: from qiskit import QuantumCircuit, QuantumRegister, ClassicalRegister from qiskit.quantum_info import Pauli from qiskit.aqua.operators.legacy import pauli_measurement qr = QuantumRegister(4) cr = ClassicalRegister(4) qc = ... 4 You can take only the keys of the returned dictionary: measurements = set(counts.keys()) print(measurements) If you want only one random key, you can just take the first one in the set: random_key = next(iter(a)) or store all the keys in a list and take a random one from here: from random import randint measurements_list = list(measurements) random_key = ... 4 Looking at the documentation for qiskit.result.Result.get_counts(), this doesn't seem to be a native option. A good alternative to this is the following: counts.get('1', 0) Which will return 0 if the key '1' is not present in the counts dictionary. You can loop through all possible\{0, 1\}^n$states and create a new dictionary with all the padded counts ... 4 Notice that it is somewhat a coincidence of that particular Bell state and choice of basis. The states$|0\rangle$and$|1\rangle$are in the$z$axis of the Bloch sphere and$|+\rangle$,$|-\rangle$are in the$x$-axis. The state you chose is a sum of products of single states that are the same, and it turns out that the same is true when you convert it to ... 2 One strong element of the intuition is related to the fact that it is maximally entangled. One definition of a pure state$|\psi\rangle$being maximally entangled is that the individual systems have density matrices $$\rho_A=\text{Tr}_B(|\psi\rangle\langle\psi|)=\frac{I}{d}$$ where$d$is the dimension of$A$'s Hilbert space. Now, one thing that the ... 2 Yet another derivation Applying a local unitary$U^A$on the first subsystem of a bipartite maximally entangled state$|\psi^{AB}\rangle$is equivalent to applying a possibly different unitary$V^B$on the second subsystem $$(U^A\otimes I)|\psi^{AB}\rangle = (I\otimes V^B)|\psi^{AB}\rangle\tag1.$$ In the specific case of the Bell state$(|00\rangle+|11\...

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The definition of information is context dependent and has different interpretations. See How to better define information forum (Physics Insights) As you did not provide a paper, I will assume a very standard context. In general when people talk about information loss in quantum mechanics due to collapse, it is related to irreversibility of collapse. If you ...

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If you measure a state $|\psi\rangle = a|0\rangle + b|1\rangle$, the post-measurement state will be either $|0\rangle$ or $|1\rangle$ with $|a|^2$ and $|b|^2$ probability, respectively. The important part here is that the values for $a$ and $b$ are lost in the post-measurement state and you end up with a basis state $\{|0\rangle, |1\rangle\}$ with an ...

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