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I understand linear transformations on the plane but cannot understand the Bloch sphere. How can a three dimensional sphere be generated by two linearly dependent vecotrs (the basis states 0 and 1)?

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    $\begingroup$ Hi Dimitris, I strongly advised you watch the lectures by Prof Shor in the course courses.edx.org/courses/course-v1:MITx+8.370.1x+1T2018/course . He explains everything you are asking. $\endgroup$ – Sam Palmer May 11 at 19:43
  • $\begingroup$ The bases the Bloch sphere is on is not that of the two states $|0\rangle$ and $|1\rangle$, the Bloch sphere is on the bases of the 3d spin states, $\sigma_x, \sigma_y, \sigma_z$. Also $|0\rangle$ and $|1\rangle$ are linear independent, otherwise they wouldn't form an orthonormal basis, this no such non trivial $\alpha,\beta \in \mathbb{R}$ s.t. $\alpha |0\rangle = \beta |0\rangle$ $\endgroup$ – Sam Palmer May 11 at 19:47
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    $\begingroup$ It's at the north pole. $\endgroup$ – tparker May 12 at 15:01
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If you read my previous answer here Cannot interpret transformations on the bloch sphere as matrix multiplications you can see that on the Bloch sphere $i|0\rangle$ lies on $|0\rangle$, because we can ignore the global phase here and the probability of being in state $|0\rangle$ is 1, when measuring you can't distinguish between the sign of $i^2$ and 1. But then what about $\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ and $\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)$ if we are saying we can ignore signs. We can only ignore signs for certain measurements, if you take a measurement of a single qubit gives you equal probability of being in $|0\rangle$ and $|1\rangle$, we can say the state of the qubit is in superposition, but we can't infer anything more. However these are still two distinct states as you can't take out a global phase to equate them, there is no number $\alpha$ where $\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) = \alpha\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)$. However $i|0\rangle = \alpha|0\rangle$ where $\alpha=i$, thus they are not distinct states up to a phase of $i$.

The basis of the Bloch sphere is not that of the two states $|0\rangle$ and $|1\rangle$, the Bloch sphere is on the basis of the 3 spin states, $\sigma_x, \sigma_y, \sigma_z$.

So if we start by assuming we can measure spin on one axis we get either $|0\rangle$ or $|1\rangle$ depending on where our measuring device is 'pointing' at, lets call this the z-axis. Now if we point our measuring device perpendicular to this axis, say the x-axis, we record either measuring $|0\rangle$ or $|1\rangle$ both with equal chance, i.e our measuring device is inbetween $|0\rangle$ and $|1\rangle$ of the Z-axis. Now If we point our device perpendicular to BOTH the x and z axis, we lay on the y-axis, but hold on, we only have one set of two real numbers for each state, so how can we REPRESENT a 3rd, we use an imaginary value to expand our state space! Now we can define space of the 3rd axis, so to summarise each of our axes lay on:

$Z$ has poles $|0\rangle$ and $|1\rangle$,

$X$ has poles $\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ and $\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)$

$Y$ has poles $\frac{1}{\sqrt{2}}(|0\rangle + i|1\rangle)$ and $\frac{1}{\sqrt{2}}(|0\rangle - i|1\rangle)$

Rather than getting bogged down with imaginary numbers, just remember that $i^2=-1$, and that in the complex space we can represent a real number or a complex number using $re^{i\theta} = r(\cos(\theta) + i\sin(\theta))$, and that real numbers are just where we have $i\sin(\theta)=0$. In fact, you may be wondering why we can saying by using $i$ we have a perpendicular axis, well, $\cos$ and $\sin$ form an orthogonal basis $\cos(\pi/2)=0$ , $\sin(\pi/2)=1$, hence we can can now see that we can form the 3rd orthogonal axis from 2 complex numbers by using the $i$ component, and in the case of $x$ and $z$ axis we just have 0 imaginary part.

Also $|0\rangle$ and $|1\rangle$ are linearly independent, otherwise they wouldn't form an orthonormal basis, there is no such non trivial $\alpha \in \mathbb{R}$ s.t. $\alpha |0\rangle = |1\rangle$

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