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Calculating number of CNOT gates in Pauli evolution gate

First of all, when you say inbuilt, inbuilt to what? I assume you are talking about a particular compiler, like cirq or qiskit's (maybe you meant to ask this on Qiskit.SE?) The number of ...
Aditya Morolia's user avatar
4 votes
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Why is the Pauli Y gate eigenstate so hard to create?

In CSS codes, the reason this happens is because the Z observable is protected by the Z stabilizers and the X observable is protected by the X stabilizers. But the Y observable is protected by the ...
Craig Gidney's user avatar
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How to modify the quantum circuit to do superdense coding with the state $|00\rangle-|11\rangle$?

In order to encode two bits $x, z\in\{0,1\}$ using vanilla superdense coding protocol, we apply two Paulis to one of the qubits in a Bell pair \begin{align} |\beta_{zx}\rangle=(Z^zX^x\otimes I)|\beta_{...
Adam Zalcman's user avatar
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2 votes

How to modify the quantum circuit to do superdense coding with the state $|00\rangle-|11\rangle$?

It could be, yes, depending on whether you want the sender or the receiver to adjust their part of the protocol. Either of them applying a $Z$ gate to their half of the pair before proceeding with the ...
Mariia Mykhailova's user avatar
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Efficient way to calculate trace of product of Pauli string and matrix?

We have $\text{Tr}(PM)=\sum_{ij}P_{ij}M_{ij}$ and $P$ is very sparse: it has sparsity $\frac{1}{2^N}$. So if you store $P$ in a sparse format, you can compute $\text{Tr}(PM)$ efficiently by summing ...
Nichola's user avatar
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2 votes
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Commutation of $XX$ and $ZZ$ operators

TL;DR: The cause of the apparent paradox is degeneracy: each of the two operators has (infinitely) many eigenbases. One of the eigenbases is shared, but it does not contain $|{++}\rangle$. The shared ...
Adam Zalcman's user avatar
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2 votes

Commutation of $XX$ and $ZZ$ operators

Commuting operators share a set of eigenvectors that form a basis of the respective Hilbert space (in the finite dimensional case). Here are the 4 common eigenvectors of XX and ZZ that form a basis of ...
qubitzer's user avatar
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