# How can a spin totally aligned along the $x$ axis have uncertain $z$ component?

Let's work with the eigenvectors $$|x+ \rangle = [\sqrt{1/2}, \sqrt{1/2}] \text{ and } |z+\rangle = [1, 0].$$ Suppose I measure $S_x$ and find our state vector is $|x+\rangle$. If spin is angular momentum, that means the vector points in the positive $x$ direction.

If I then measure $S_z$ I will find it either in $|z+\rangle$ or $|z-\rangle$. This seems contradictory. We just determined the vector is aligned with the x axis but now find it is aligned with the $z$ axis.

Should we not really think of spin angular momentum as a vector pointing along an axis and if not how should we think of it?

• @knzhou: Thanks. I really have no idea what I'm doing. I just learned a tiny amount of quantum mechanics from a freshman level book. Can you explain what a spinor is and how it corrects my confusion above? – user7348 Aug 12 '16 at 21:57
• I posted an answer that gives a heuristic way of understanding this. The other answer is the more "official" take you'll find in textbooks. – knzhou Aug 12 '16 at 22:24
• "Should we not really think of spin angular momentum as a vector pointing along an axis" Yes, we shouldn't. "and if not how should we think of it?" It's an operator-valued vector, the operator acts in spinor space. – Robin Ekman Aug 13 '16 at 5:06

Spin is not described by a three-component vector, it's described by a two-component spinor. The correspondence between the two is subtle and nonintuitive; generally, it's a bad idea to try to think of quantum spin as vectorial at all.

You've found the sharpest reason why: a spin with definite $z$ component doesn't have a definite $x$ component. There's an intuitive way to see this, but to do it, we have to temporarily leave quantum mechanics and do some high school geometry.

Consider vectors in the plane. We can decompose them into $x$ and $y$ components; call the $x$ and $y$ unit vectors $|x \rangle$ and $|y \rangle$. If we know that a vector is pointing along the $|x \rangle$ direction, we know for sure that it has zero component along the $|y \rangle$ direction.

We can also consider the rotated basis $$|+\rangle = \frac{|x\rangle + |y\rangle}{\sqrt{2}}, \quad |-\rangle = \frac{|x\rangle - |y\rangle}{\sqrt{2}}.$$ Again, if we know that a vector is along the $|+\rangle$ direction, it has zero component along the $|-\rangle$ direction. However, a $|+\rangle$ vector is not definitely either $|x \rangle$ or $|y\rangle$, it's a superposition of both. The questions "$|x\rangle$ or $|y\rangle$" and "$|+\rangle$ or $|-\rangle"$ do not simultaneously have definite answers, because the bases are incompatible. This is an example of the uncertainty principle, as I explain further here.

Now let's return to quantum mechanics. All of the weirdness of quantum mechanics comes from the fact that bases which intuitively look compatible (thinking on a naive classical level) are not compatible in the abstract Hilbert space.

The state space of a spin 1/2 particle is two-dimensional, so we can keep working with the plane. Since the spin is 1/2, all angles are shrunk by a factor of two. For example, the states $|z, +\rangle$ and $|z, -\rangle$ are 180 degrees apart in real space, so they're only 90 degrees apart in spinor space. That means they form an orthogonal basis just like $|x \rangle$ and $|y \rangle$ did above.

Now, the states $|z, + \rangle$ and $|x, + \rangle$ are 90 degrees apart in real space, so they're 45 degrees apart in spinor space. That is, we have the correspondence $$(|z, + \rangle, |z, -\rangle) \leftrightarrow (|x\rangle, |y \rangle), \quad (|x, + \rangle, |x, -\rangle) \leftrightarrow (|+\rangle, |- \rangle)$$ Translating our conclusions from the plane, if we know a vector is $|x, +\rangle$, it is a superposition (in fact an equal superposition) of both $|z, + \rangle$ and $|z, -\rangle$. And the questions "$|z, + \rangle$ or $|z, - \rangle$" and "$|x, + \rangle$ or $|x, - \rangle$" do not have simultaneously have definite answers.

So, the answer to your question doesn't require any high-powered math: a spin along the $x$ axis has uncertain $z$ component for exactly the same reason that traveling northeast is a combination of both north and east. The tricky part is seeing that these two properties are one and the same, which results from the "angle halving" of spin 1/2.

• Spin is angular momentum. Angular momentum is a vector. Therefore spin is a vector. What is wrong with this reasoning? – user7348 Aug 13 '16 at 0:20
• @user7348 The part "angular momentum is a vector". That's only true for one particular kind of angular momentum, i.e. the kind you learned about in mechanics. – knzhou Aug 13 '16 at 0:21
• I think I'm just getting way ahead of myself. I haven't even started reading Griffith's text on QM. Do the commutation relations indicate uncertainty and if they do then uncertainty in what? The very statement implies uncertainty in something that exists. It seems like we're saying if spin is (z, +) then there is uncertainty in spin x. It also seems like we're saying it doesn't exist at all. – user7348 Aug 13 '16 at 0:29
• @user7348 Translate it into the language I used. If you're walking northeast, then are you walking north, or you walking east? Which one of the two is it? There's no definite answer. It's both at the same time. And comparing spin along z and x is the same thing. – knzhou Aug 13 '16 at 1:10
• OK, but the state vector only has a probabilistic interpretation. It doesn't really mean that a particle in |+x> = sqrt(1/2)|+z> + sqrt(1/2)|-z> is in both +z and -z at the same time. What it means is that if the state vector acts on an a z oriented eigenstate say +z it gives the amplitude whose square is the probability of being in the +z state. e.g. |<+z|+x>|^2 = 1/2. The state vector has a probabilistic interpretation in proper quantum mechanics. Then a correction to your analogy would be a man who is either walking N or E and the squared amplitude gives the probability. – user7348 Aug 13 '16 at 2:31

You should indeed not think of quantum mechanical angular momentum as being a vector pointing anywhere. Due to the commutation relations $$[S_i,S_j] = \mathrm{i}\epsilon_{ijk}S_k$$ you can't expect that an arbitrary eigenstate of $S_x$ is also one of $S_y$ or $S_z$ - the operators $S_x,S_y,S_z$ do not commute and therefore cannot share an eigenbasis (though they may share some eigenvactors).

In fact, if one examines the eigenstates, one finds that the states that are eigenstates of one of the components of angular momentum are completely scrambled in the other two components- the $x_+$-state has equal probability to have "up" or "down" for $S_y$ and $S_z$.

Now, that you find it contradictory that we may find a state to have spin aligned with the z-axis if we determined it to have spin aligned with the x-axis before suggests you are still thinking classically. In quantum mechanics a measurement changes the state. It is indeed classically absurd to get these results because classically one merely "looks" at a state to measure it - one doesn't change it. In quantum mechanics, this does not work anymore - every measurement is an interaction, so if you "measure whether the spin is aligned with the z-axis", you are really carrying out an interaction with the state that will force it to occupy one of the two possible eigenstates of $S_z$.

In particular, it is not possible to measure all three components of angular momentum simultaneously. The components do not commute, you can only ever ask whether spin is aligned with one particular direction, but you cannot determine a definite vector of angular momentum as you could in classical mechanics. At least one component of the angular momentum vector will always have uncertainty.

• Suppose you measure it in (z, +) then you can say you "changed it to (z, +)" and at that moment it is a vector pointing along +z? Then you measure along x axis and say you get (x, +). Then you would say it is a vector that points along +x? – user7348 Aug 13 '16 at 0:36
• @user7348: No, you cannot say that. As I say, since the components of angular momentum do not commute, it is not good to think of it as a single, definite vector pointing anywhere. – ACuriousMind Aug 13 '16 at 10:39

Notwithstanding all this discussion about spin in the context of quantum mechanics, it is perhaps worth noting that spin also exists in the context of classical theories. The polarization of a classical optical field also represents spin angular momentum. It also obeys the commutation relation $$[S_i,S_j]=i\epsilon_{ijk}S_k ,$$ because this commutation relation is not restricted to quantum mechanics. It comes from the Lie algebra of the SU(2) Lie group, which governs the properties of rotations in three-dimensions.

So if one were to make the same measurements in the $x$ and $z$ bases, one would find the same funny situation that they are not orthogonal. However, these classical measurements differ from the quantum equivalent in that they do not collapse the state. Therefore one can make multiple measurements.

Another way to understand the orthogonality of the $x$ and $z$ bases is to consider the configuration space of all such spin states. In quantum mechanics it is called the Bloch sphere. In optics it is called the Poincare sphere. Apart from the difference in their names, they are identical. Any two points located at geometrically opposite points on the sphere are orthogonal. Hence, for instance $|x,+\rangle$ and $|x,-\rangle$ are orthogonal and therefore represent a basis.