# How can the Bloch sphere, built from one complex dimension, specify 2-complex dimensional Pauli spinors?

Two-component spinors can be identified with points on the surface of the Bloch Sphere. The Bloch sphere is constructed from the 1-complex-dimensional complex plane plus the point at infinity.

How does the Bloch Sphere represent two complex components?

It is probably most natural to identify the points of Bloch sphere with density matrices representing pure states of a spinor (two-level system). Arbitrary density matrix of a two-level system can be written as $$\hat{\rho} = \frac{1}{2} + \frac{1}{2}(\vec{\rho},\vec{\sigma}),$$ where $$\vec{\sigma} = (\sigma_x,\sigma_y,\sigma_z)$$, and $$\vec\rho$$ is a real 3-component vector with $$|\vec\rho| \le 1$$. The density matrix represents a pure state if and only if $$|\vec\rho| = 1$$, otherwise, the state is mixed.

Let me show how pure spinor states are mapped to a sphere. Arbitrary wavefunction of a two-level system can be written as $$|\psi\rangle = a|\uparrow\rangle + b |\downarrow\rangle,$$ where $$a$$ and $$b$$ are two complex numbers, as you wrote in the question. However, physical quantum states also

1. have norm unity, so $$|a|^2 + |b|^2 = 1$$
2. can be multiplied by arbitrary phase: $$|\psi\rangle \to e^{i\phi}|\psi\rangle$$.

So, the number of degrees of freedom representing the spinor reduces from $$4$$ (two arbitrary complex numbers) to $$2$$. The explicit parametrization can be chosen as $$a = \cos{\frac{\theta}{2}}e^{-i\phi/2},$$ $$b = \sin{\frac{\theta}{2}}e^{i\phi/2}.$$ The density matrix reads $$\rho = \left(\begin{matrix} a\\ b \end{matrix}\right) \begin{matrix} (a^* & b^*)\\ & \end{matrix}= \left(\begin{matrix} \cos^2\frac{\theta}{2} & \sin\frac{\theta}{2}\cos{\frac{\theta}{2}}e^{-i\phi}\\ \sin\frac{\theta}{2}\cos{\frac{\theta}{2}}e^{i\phi} & \cos^2\frac{\theta}{2} \end{matrix}\right) \\= \frac{1}{2} + (\vec{a},\vec{\sigma}),$$ where $$\vec{a} = \left( \begin{matrix}\sin{\theta}\cos{\phi},\\ \sin{\theta}\sin{\phi}, \\\cos\theta\end{matrix}\right).$$ We see that $$\vec{a}$$ is indeed a vector on a unit sphere, where $$\theta$$ and $$\phi$$ are polar angle and azimuthal angle respectively.

• Thank you for your clear answer, E. Incidentally, although your answer started from density matrices, it also surprised me that it confirmed the logic and correctness of an answer to the exact same question that I obtained first from ChatGPT4o. Commented Jul 4 at 18:08
• Well, I used ChatGPT in no way, I used Wikipedia instead :) I'm glad that ChatGPT also gave a good answer. Commented Jul 5 at 6:02
• Though one always has to watch out for erroneous LLM/ChatGPT replies, I've found it surprisingly helpful in some areas, especially in combination with Wikipedia: In medicine it clarified a radiologist's impenetrable MRI report, helping me negotiate and undergo a spine operation recently; and also led to an accepted answer elsewhere on Physics SE. According to the Economist magazine (June 29th 2024, p. 70), ChatGPT or similar is contributing to the writing of a significant percentage of academic publications nowadays - but not without risks... Commented Jul 6 at 12:34
• PS @Moderators - Although not involved directly in the answer above, I only became aware of the Physics-SE policy regarding no-use of LLMs subsequently. If it was announced publicly, I failed to notice it at the time. Mea culpa - mea maxima culpa Commented Jul 10 at 13:32
• LLMs answers should be based on good SE answers, not vice versa. Commented Jul 10 at 15:58

E.Anikin's answer explains lucidly how the spinor has two degrees of freedom. A simple way to look the map between the complex plane and the Bloch sphere it is that one complex dimension is equivalent to two real dimensions (one for each of the real and imaginary components). The two degrees of freedom in the complex plane translate to the two angles (equatorial and azimuthal) which act as coordinates on the Bloch sphere.

• Thank you for your intuitive answer CompassBearer. It seems surprising, at first, that a 2-complex component spinor can be represented by just two real numbers, until @E.Anikin explains the d.o.f. reduction. So, it's due to the combination of the normalisation (in effect defining the Bloch sphere's surface) and the ray-representation of |ψ⟩. Commented Jul 4 at 17:53