# [Physics] Up and down spin are orthogonal, not antiparallel

quantum mechanicsquantum-spinspinorsvectors

In conventional coordinate systems (anything you solve a simple Newtonian mechanics problem with), up and down are + and – z. A vector pointing up and a vector pointing down are anti-parallel.

But in qm, we have up and down spinors making and orthonormal basis. These basis vectors are also called positive and negative z spin angular momentum. I understand the math for how spinors like (1,0) and (0,1) are orthogonal. I also see how they can be expressed as superpositions of x and y spinors, using complex numbers such that a two-component spinor can represent quantities in 3 dimensions. (this seems a little like what I have studied about symmetry groups like SU(1), so if that is relevant in the solution, I appreciate a discussion, but if it is unrelated, please do not bother correcting any huge mistakes in this sentence because I did not fully try to study it on my own yet).

My question is this: what is the intuition for saying that spin up and down are orthogonal?

You need to distinguish the orthogonality in spinor space ($$\mathbb{C}^2$$) from orthogonality in vector space ($$\mathbb{R}^3$$). The spaces are different, and therefore scalar product and orthogonality in these spaces have entirely different meanings.

Example:

The two spinors $$|\uparrow\rangle=\begin{pmatrix}1\\0\end{pmatrix}$$ and $$|\downarrow\rangle=\begin{pmatrix}0\\1\end{pmatrix}$$ are orthogonal to each other because their scalar product is zero: $$\langle\uparrow|\downarrow\rangle=0$$

The expectation values of the spin vector $$\vec{S}=\frac{\hbar}{2}\vec{\sigma}$$ (where $$\vec{\sigma}$$ is the Pauli vector $$\vec{\sigma}=\sigma_x\hat{x}+\sigma_y\hat{y}+\sigma_z\hat{z}$$) for these two spinors are:

$$\vec{S}_\uparrow = \langle\uparrow|\vec{S}|\uparrow\rangle = \frac{\hbar}{2}\langle\uparrow|\vec{\sigma}|\uparrow\rangle = + \frac{\hbar}{2} \hat{z}$$ and $$\vec{S}_\downarrow = \langle\downarrow|\vec{S}|\downarrow\rangle = \frac{\hbar}{2}\langle\downarrow|\vec{\sigma}|\downarrow\rangle = - \frac{\hbar}{2} \hat{z}$$

These two vectors are antiparallel to each other. Their scalar product $$\vec{S}_\uparrow \cdot \vec{S}_\downarrow$$ is not zero.