# Aren't all spin states (up, down, left, right, in, out) orthogonal?

Hopefully someone can clear up a basic misconception I am having about the nature of spin state vectors.

According to the book i am reading, The basis vectors of up/down spin are orthogonal to each other. Same goes for in/out and left/right. I understand this because if the spin is measured as up it precludes it from being down. Thus , the inner product of the two basis vector, is 0.

But where I'm confused is, aren't all spin states orthogonal? Cant there only be one spin at a time? As in, if you have an apparatus oriented along the z-axis, and measure a +1 (up spin), then doesnt that preclude the spin from being left, right, in, or out, unless you shift the apparatus/remeasure?

I understand that if you were to measure along the x or y axis at this point you would have a 50/50 chance of getting an up or down spin, whereas if you measured via the negative z axis you would have 100% chance of getting a down spin.

So then what does orthogonality represent in spin systems? My guess is that after the initial apparatus direction is set, and the original spin value is measured, orthogonality means having a 0% chance of achieving that same measured spin after adjusting the apparatus along a different axis and remeasuring. Thus if you measured spin +1 with the apparatus in the +z direction, and then shifted the apparatus to -z to perform a second measurement, you would have a 0% chance of it being +1, making it orthogonal. But if you shifted the apparatus along the +/- x/y axis, you would have a 50% chance of getting the same measured spin?

Is my reasoning correct, or way off base?

• Your question is a little confusing. What statistics you will get in a spin experiment depends on how you prepare your spin system, not just on how you align the measurement. Normally we leave the directions of the axis of the measurement alone and work out the dynamic of the spin, instead, since that's where the actual physics is. – CuriousOne Oct 28 '15 at 0:43
• Notice that we use a vector to represent spin, but "orthogonal" means that the inner-product $\langle\text{up}|\text{right}\rangle$ between the two states is zero, and that inner-product is not the dot product between vectors. Otherwise up would not be orthogonal to down. – dmckee --- ex-moderator kitten Oct 28 '15 at 0:55
• The vectors $(0,1)$ and $(1,0)$ are orthogonal, and they form a basis for ${\mathbb R}^2$. Does it follow that any two vectors in ${\mathbb R}^2$ are orthogonal? – WillO Oct 28 '15 at 1:01

You can think of the spin state of an electron as represented by a vector $(\alpha,\beta)$. Depending on how you set things up, "Up" might be represented by $(1,0)$, "Down" by $(0,1)$, "Left" by $(1,1)$, and "Right" by $(-1,1)$. Up is orthogonal to Down, and Left is orthogonal to Right, but Up is not orthogonal to Left.
Angular momentum in quantum mechanics in general works like this: the total is measured by $L^2 = \hbar^2 \ell (\ell+1)$ whereas the projection along any axis is measured by $L_z = \hbar~m$ between $-\ell \le m \le \ell.$ Both $\ell$ and $m$ are simultaneously measurable (i.e. the $L^2$ and $L_z$ operators commute), and they must be spaced by integers but they don't actually have to be integers.
For a spin-$\frac12$ system, $\ell = 1/2$ and the allowed values for $L_z$ are $m = -1/2$ and $m = +1/2.$ These are "up" and "down" in theory.
However, pay very very careful attention, because $L^2$ is actually $\frac34 \hbar^2$, therefore the total angular momentum is definitely $\hbar\sqrt{3/4},$ or about $0.866~\hbar.$ Only $0.5$ of this is pointed in the $z$ direction in the $m = +1/2$ state.
This is because the $m=+1/2$ state consists of a quantum superposition of nonzero $L_x, L_y$ values with a constant $L_z$ value such that $L^2 = L_x^2 + L_y^2 + L_z^2$ is constant. When you are in the definitely "spin-up" state there is a 50/50 chance of seeing you in the "spin-left" or "spin-right" state. They are not orthogonal states in the Hilbert space, even though they are orthogonal directions in 3d space!
Really what has happened is that you were talking about two spaces and the word "orthogonal" means two different things in each, resolving the confusion. Yes, two wavefunctions which are orthogonal in phase space have zero overlap: you cannot measure the one state as being "in the other state" when you measure it. And that's not the same as being orthogonal in 3D space because the quantum spin-up state has $x$ and $y$ components to its angular momentum.