Eigenstates of Spin

Question

Why are the eigenstates of spin not functions? Is this because the spin, $s$, and magnetic quantum number, $m$, take discrete values? My textbook(see Griffiths' Introduction to quantum mechanics) in an earlier section used $Y_\ell ^m$ as the eigenfunction of the $L^2$ and $L_z$ operators. In a footnote, he says that $Y_\ell ^m$ was used instead of $\left|\ell \; m \right>$ because in the context of that section (angular momentum), a function seemed more natural.

Maybe I don't quite understand the difference between a function and a vector. A function is like an infinite-dimensional vector. So since $\left|s \; m_s \right>$ only has certain limited values of $s$ and $m_s$, it's a vector because the limited values imply it lives in a finite-dimensional space? As a side note, can a function be called a function if its domain is only a single point?

Answer 1

What you guessed is correct, the discrete values lead to a discrete number of eigenfunctions, but let me say the things a bit more precisely. The function $Y^m_ℓ$ for a GIVEN $ℓ$ and a GIVEN $m$ is indeed a FUNCTION, of the angles $θ$ and $φ$. However, for each $ℓ$ and for each $m$, we have ANOTHER function.

Now, let's take an example and see how we come to vectors. Let's take the hydrogen atom, the level $n = 2$. Let me disregard the spin, for simplicity. We have two possibilities for $ℓ$, i.e. $ℓ= 1$, and $ℓ= 0$. Now, for $ℓ= 0$ there is only one possible value of $m$, i.e., $m = 0$, while for $ℓ= 1$ we have 3 possibilities $m = -1$, $m = 0$, and $m = +1$. In all, we have 4 FUNCTIONS of $θ$ and $φ$, i.e. $Y^0$, $Y^1_{-1}$, $Y^1_0$, and $Y^1_1$ (for details see Laplace's spherical harmonics).

Sometimes in our calculi, it is convenient to represent them in a simplified way, in vector form, without specifying their dependence on the angles:

$$Y^0 = (1, 0, 0, 0),$$ $$Y^1_{-1} = (0, 1, 0, 0), \qquad Y^1_0 = (0, 0, 1, 0), \qquad Y^1_1 = (0, 0, 0, 1),$$

For instance, any wave-function suitable for the hydrogen level $n=2$ can be represented as a superposition of the four above vectors. I.e. in the space of wave-functions of the level $n = 2$, the four above vectors form a BASE.

To the difference of this example, the linear momentum admits CONTINUOUS, not discrete eigenvalues. Its eigenfunctions are of the form $e^{ipx/ħ}$, where the eigenvalue $p$ has continuous values. It is not possible to arrange the eigenfunctons as discrete vectors as we did in the previous case (for details, see Fourier transform).

Answer 2

As you might have already notice that spin $\vec s$ commutes with many variables, especially coordinates $\vec x$. So the Hilbert space of wave functions can be expressed as the tensor product of two Hilbert spaces $\mathscr S$ and $\mathscr X$ spanned by eignvectors of $\vec s$ and $\vec x$.

You should be familiar with $\mathscr X$ which is parametrized by $\vec x$. Similarly, $\mathscr S$ is parametrized by eigenvalues of $\vec s$, that is, $|s,m_s\rangle$. This is why $|s,m_s\rangle$ does not depend on $\vec x$.