# Where is $i|0\rangle$ located on the bloch sphere?

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)?

• 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. – Sam Palmer May 11 at 19:43
• 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. $\alpha |0\rangle = \beta |0\rangle$ – Sam Palmer May 11 at 19:47
• It's at the north pole. – tparker May 12 at 15:01

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. $$\alpha |0\rangle = |1\rangle$$