What are the possible polar orientations of the spin momentum vector for an electron?
We can begin our solution by recalling that electron spin has to do with the rotation of an electron about some access, not as an orbit like the Earth around the Sun, but as a spin like the Earth about its own axis or a spinning top. Given a particular axis of rotation, electron spin can occur in one of two ways: spin in one direction is called up spin, and spin the opposite way is called down spin. So an electron can spin up or down. When it spins up, the spin value is ℏ divided by two. And when it spins down, the 𝑠-value is the negative of that, negative ℏ divided by two.
Say that we had two coordinate axes, where the vertical axis is the rotation pole. If we were to start at the origin and mark off a distance of ℏ divided by two on the polar axis, that would give us the polar value of electrons with spin up. Likewise, if we were to mark off a distance of negative ℏ over two on the polar axis, that would give us the polar coordinate of a spin down electron. The question then becomes, where along the horizontal axis does the spin vector reach?
To solve that, we look up or recall the fact that the magnitude of the overall spin vector for the electron, capital 𝑆, equals the square root of three over two times ℏ. So for a spin up electron starting at the origin, points with a magnitude of the square root of three over two ℏ and with a polar coordinate of ℏ over two. Then for a spin down electron, that vector also has a magnitude of the square root of three over two ℏ, but this one has a polar coordinate of negative ℏ over two. So the polar orientation of a spin up electron is given by an angle that we’ll call 𝜃 sub 𝑢, and the polar orientation of a spin down electron is given by an angle we’ll call 𝜃 sub 𝑑.
To solve first for 𝜃 sub 𝑢, let’s consider the right triangle where 𝜃 sub 𝑢 is the bottom angle. The side adjacent to 𝜃 sub 𝑢 has a magnitude of ℏ divided by two and the triangle’s hypotenuse has a magnitude of the square root of three divided by two times ℏ. We see from this triangle that the cosine of 𝜃 sub 𝑢 equals ℏ over two divided by the square root of three over two times ℏ. In this expression, ℏ cancels as does the factor of one-half that appears in the numerator and denominator. This leaves us with one divided by the square root of three. Taking the arc cosine of both sides of this equation, we see that 𝜃 sub 𝑢 equals the inverse cosine of one divided by the square root of three. Entering this value on our calculator, we find that, to two significant figures, 𝜃 sub 𝑢 is 55 degrees.
If we now move on to solving for 𝜃 sub 𝑑, if we consider the right triangle representing the spin down electron, then we see that the angle at the top of the triangle is equal to 180 degrees minus 𝜃 sub 𝑑. We can write that the cosine of 180 degrees minus 𝜃 sub 𝑑 equals ℏ over two divided by the square root of three over two ℏ. Once again, ℏ over two cancels from the numerator and denominator, leaving us with one over the square root of three. Taking the inverse cosine of both sides, when we evaluate the right side, our cosine of one over the square root of three, we find as before that that rounds to 55 degrees. When we solve for 𝜃 sub 𝑑 by subtracting 55 degrees from 180 degrees, we find that 𝜃 sub 𝑑 is 125 degrees. That’s the polar orientation of the down spin momentum vector for an electron.