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chrysaor4
chrysaor4
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We actually do use latitude and longitude to describe qubits, because one very common way of writing the state is of a qubit is

$$ \begin{align} \cos{(\theta/2)}|0\rangle + e^{i \phi}\sin{(\theta/2)} |1\rangle \end{align} $$

where $\theta$ is the angle formed by the Bloch vector and the $z$-axis (so technically it is the complement of the latitude, which is measured up from the equator instead of down from the North pole), and $\phi$ is the longitude with respect to the $x$-axis. This is the most general form for a pure quantum state, which you can think of as having only quantum uncertainty (outcome is undefined before measurement), but no classical uncertainty (outcome is well-defined but unknown to the observer). The "pair of probabilities" referred to in those videos is likely the probability that one obtains either $|0\rangle$ or $|1\rangle$ upon measurement - knowing these only determines the magnitude of the vector's projection onto the $z$-axis (i.e. its latitude), but does not determine its longitude, so those probabilities alone are not sufficient to specify a pure quantum state of the above form.

We actually do use latitude and longitude to describe qubits, because one very common way of writing the state is of a qubit is

$$ \begin{align} \cos{(\theta/2)}|0\rangle + e^{i \phi}\sin{(\theta/2)} |1\rangle \end{align} $$

where $\theta$ is the angle formed by the Bloch vector and the $z$-axis (so technically it is the complement of the latitude, which is measured up from the equator instead of down from the North pole), and $\phi$ is the longitude with respect to the $x$-axis. This is the most general form for a pure quantum state, which you can think of as having only quantum uncertainty (outcome is undefined before measurement), but no classical uncertainty (outcome is well-defined but unknown to the observer). The "pair of probabilities" referred to in those videos is likely the probability that one obtains either $|0\rangle$ or $|1\rangle$ upon measurement - knowing these only determines the magnitude of the vector's projection onto the $z$-axis (i.e. its latitude), but does not determine its longitude, so those probabilities alone are not sufficient to specify a pure quantum state of the above form.

We actually do use latitude and longitude to describe qubits, because one very common way of writing the state is of a qubit is

$$ \begin{align} \cos{(\theta/2)}|0\rangle + e^{i \phi}\sin{(\theta/2)} |1\rangle \end{align} $$

where $\theta$ is the angle formed by the Bloch vector and the $z$-axis (so technically it is the complement of the latitude, which is measured up from the equator instead of down from the North pole), and $\phi$ is the longitude with respect to the $x$-axis. This is the most general form for a pure quantum state, which you can think of as having only quantum uncertainty (outcome is undefined before measurement), but no classical uncertainty (outcome is well-defined but unknown to the observer). The "pair of probabilities" referred to in those videos is likely the probability that one obtains either $|0\rangle$ or $|1\rangle$ upon measurement - knowing these only determines the vector's projection onto the $z$-axis (i.e. its latitude), but does not determine its longitude, so those probabilities alone are not sufficient to specify a pure quantum state of the above form.

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chrysaor4
chrysaor4
  • 1.4k
  • 7
  • 9

We actually do use latitude and longitude to describe qubits, because one very common way of writing the state is of a qubit is

$$ \begin{align} \cos{(\theta/2)}|0\rangle + e^{i \phi}\sin{(\theta/2)} |1\rangle \end{align} $$

where $\theta$ is the angle formed by the Bloch vector and the $z$-axis (so technically it is the complement of the latitude, which is measured up from the equator instead of down from the North pole), and $\phi$ is the longitude with respect to the $x$-axis. This is the most general form for a pure quantum state, which you can think of as having only quantum uncertainty (outcome is undefined before measurement), but no classical uncertainty (outcome is well-defined but unknown to the observer). The "pair of probabilities" referred to in those videos is likely the probability that one obtains either $|0\rangle$ or $|1\rangle$ upon measurement - knowing these only determines the magnitude of the vector's projection onto the $z$-axis (i.e. its latitude), but does not determine its longitude, so those probabilities alone are not sufficient to specify a pure quantum state of the above form.