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When a qubits is in a quantum state it can be measured as $[0\rangle$ or $[1\rangle$. Then why does the Bloch sphere have these two states on antipodal sides of the spheres? If I want to plot the position of an object and their velocity, then I don't put position on one side of the axis and velocity on the other side. I would put them on different axis.

I read an explanation here, but you get some practical issues when you use the Bloch sphere. These different quantum states are projected on the same place on the Bloch sphere:

  • $[0\rangle$
  • $i[0\rangle$
  • $-[0\rangle$
  • $-i[0\rangle$

Why can't the real probabilities be projected like this? enter image description here

Then in three dimensions you can add an axis $i[0\rangle$ and in four dimensions you add an axis $i[1\rangle$.

jng224
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user3635700
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    It is your gut feeling that tells you that since the zero and one state are orthogonal as states, they should be orthogonal on the Bloch sphere as well. But orthogonality as states is just a completely different concept, which confusingly shares the same name as 'pictorial' orthogonality. More deeper, this fact is also closely related to spinors. – Kenneth Goodenough Oct 18 '21 at 20:27
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    Note that $|0\rangle$ and $i|0\rangle$ are the same state, so they should be represented as the same point in the Bloch sphere. More precisely, the Bloch sphere is a representation of all possible density matrix states of a qubit, where the pure states, which are of the form $|\psi\rangle\langle\psi|$ for some state $|\psi\rangle$, are on the surface of the sphere, and all the states inside the sphere are mixed states (i.e., non-trivial convex combinations of pure states). – march Oct 18 '21 at 20:51
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    The point is, the Bloch sphere is a representation of all possible distinct states of a qubit, and so you don't want different points representing the same state, as is indicated in your diagram. – march Oct 18 '21 at 20:52
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    The analogy with position and velocity breaks down here: $|0\rangle$ and $|1\rangle$ are two possibilities for the same property, so they should correspond to two opposite values of position. Note that this physically may correspond to the direction in which an microscopic magnet is pointing, so a sphere is the best way to depict that – Quantum Mechanic Oct 18 '21 at 21:37
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    Related : Understanding the Bloch sphere. – Frobenius Oct 18 '21 at 22:51
  • Got a bit confused when people stated that some states are equivalent, but now I understand it has something to do with multiplication between complex numbers. For any complex number $x$ you have that $i\cdot x$, $-\cdot x$, $-i\cdot x$ all have the same magnitude. When multiplied with some other complex number $y$, you get 4 new complex numbers that all have the same magnitude as $x\cdot y$. And the magnitude of two complex numbers solely determines the probability of what to measure from a qubit. So quantum states like $[1>$ and $-i[1>$ are indeed equivalent. Thanks, I think get it now! – user3635700 Oct 19 '21 at 20:57
  • If you are more familiar with optics, it might help to think of the Poincare sphere, where orthogonal polarizations in the lab frame correspond to opposite points on this sphere instead of orthogonal axes – Rol Oct 26 '21 at 06:36

1 Answers1

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This is a very good question that many people ask when they learn about the Bloch sphere. There are two important reasons

Physical reason

The most famous two state system in quantum mechanics is the spin 1/2 particle. When you draw the spin 1/2 states on the Bloch sphere something very nice happens, the points on the Bloch sphere correspond to the physical directions of the states. For example the $|\uparrow\,\rangle$ state corresponds to a point that points in the $+z$ direction. Similarly the state $|x+\rangle=\tfrac{ 1}{\sqrt{2}}(|\uparrow\,\rangle+|\downarrow\,\rangle)$ points in the $+x$ direction on the Bloch sphere. To be more precise for some state $|\psi\rangle$ the vector of expectation values given by $\langle\psi|\vec S|\psi\rangle=\langle \vec S\rangle =(\langle S_x\rangle,\langle S_y\rangle,\langle S_z\rangle)^T$ points in the same direction as the Bloch vector. See also this picture:

enter image description here

Mathematical reason

Another reason that the Bloch representation is good is that states which are physically the same correspond to the same point on the Bloch sphere. This is also mentioned in the comments. In your diagram $|0\rangle$ and $-|0\rangle$ are antipodal points but they represent the same state up to an arbitrary phase. Remember that $|\psi\rangle$ and $e^{i\alpha}|\psi\rangle$ represent the same state. The Bloch sphere intuitively leaves out any global phase. To represent the full state on a Bloch sphere you would have to specify an additional phase for any point. In Steane's "Introduction to Spinors" this is represented by drawing a state as a little flag. The direction of the flag gives the point on the Bloch sphere and the angle that flag makes gives the global phase. So in this representation the states $|0\rangle$ and $-|0\rangle$ both point in the z-direction but the first one has its flag rotated to $\alpha=0$ and the second one has it rotated to $\alpha=\pi$ radians. Interestingly when you rotate a spin 1/2 particle by $360^\circ$ its state will pick up a minus sign and you would have to rotate it by another $360^\circ$ before it becomes exactly the same state.

enter image description here

AccidentalTaylorExpansion
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