# Up and down spin are orthogonal, not antiparallel

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?

• Not Strictly Related : Understanding the Bloch sphere. Aug 2 at 8:47
• I like how this question is phrased and have often thought of it myself. I always thought this arose out of how an electron assumes 'states' depending on the orientation of the 'measurement'. That they are fundamentally different states (up and down). The orthogonality idea to me was just a way of emphasising that notion. Aug 2 at 22:13

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.

• Okay so although the spinors are orthogonal, the results from taking the spin operator on them are anti-parallel, right? And that gives the names up and down? Aug 2 at 19:59
• @user310291 Yes, exactly. Aug 2 at 20:10

The inner product of two spin vectors is not spatial overlap. Rather, $$|\langle \psi_1 | \psi_2 \rangle |^2$$ is the probability of measuring the state in $$| \psi_2 \rangle$$ if it is originally in $$|\psi_1 \rangle$$. Therefore, if a state is spin up, $$(1,0)$$, then you have $$0\%$$ chance of measuring it to be spin down, $$(0,1)$$.

An addendum to Thomas Fritsch's answer that resolves much of the initial confusion: you measure the eigenvalue of the operator acting on the state, not the eigenvector for the state. The relevant spin operator on the eigenvector clearly gives the desired anti-parallel eigenvalues, and the eigenvectors are simply vectors which behave in the desired manner, without an equally direct geometric interpretation (discussed here).

Some intuition for the orthogonality is: the z spin observed in a measurement must be +- 1/2, it cannot be zero. The orthogonality indicates that a body spinning upwards does not need to be expressed in a nontrivial sum of "up-ness" and down-ness." It is expressed solely as down-ness and has zero up-ness. This indicates orthogonality.