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The most common convention is to refer to qbits by the index of their significance, with the least-significant qbit having index $0$. This is cribbed from binary, where the significance index is the s …

All quantum operators must be unitary. Unitary means the conjugate-transpose of the operator is its inverse. In your case: $UU^{\dagger} = \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 1 & 0\\ 0 & 0 & 0 & …

These are called observables, which are unitary hermitian matrices whose eigenvectors are the possible outcomes of the measurement. For example, the observable of the standard computational basis is t …

If you use $\theta = \pi$, you get the following: $$ Rz(\pi)\begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{bmatrix} = \begin{bmatrix} e^{-i \pi/2} & 0 \\ 0 & e^{i \pi/2} \end{bmatrix} …

Here's the Rz matrix: $$ Rz(\theta) = \begin{bmatrix} e^{-i\theta/2} & 0 \\ 0 & e^{i\theta/2} \end{bmatrix} $$ As I understand it, Rz rotates around the Z axis on the Bloch sphere. Since $|+\rangle$ …

The tensor product is not a gate, but rather a way for us as humans to model the behavior of a quantum system. Whenever we're using multiple qbits, we can look at them in two ways: in their product st …

You say it's a problem that when using measurements your circuit is not reversible, but generating a truly random number is an inherently non-reversible operation. Consider that for an operation to be …

In a three-qubit system, it's easy to derive the CNOT operator when the control & target qubits are adjacent in significance - you just tensor the 2-bit CNOT operator with the identity matrix in the u …

Since the quantum teleportation circuit has three qbits, the matrix at each step is 8x8 and thus has 64 elements; this is pretty clunky to type out in its entirety, so I'll just walk you through step …