Ballentine on the "multicomponent state function"

[EE18]

I have just finished reading Ballentine Chapter 7.2 and I am positively baffled, perhaps because Ballentine is being sloppy for the first time. I attach the discussion in Ballentine at the end of this post if it helps, though I hope my writing will be independent thereof. This question is intimately connected to this one, though it goes a little further I think.

Earlier (in Chapter 3), Ballentine argued on the grounds of our universe (in the low speed limit investigated in this book) obeys Galilean symmetries. That is (jumping off from ACuriousMind's answer ), for any Galilean transformation $\tau$ there is a corresponding unitary operator $U(\tau)$ such that, if we began in a state $\lvert \psi \rangle$ then we have that $U(\tau)\lvert \psi \rangle$ is the corresponding state (in the sense of preserving certain reasonable requirements which Ballentine sets out on page 63 (not pictured here) (for example ACuriousMind's $O$ would be our $\tau$ and would be applied to e.g. the coordinates of our transformed function if we projected into coordinate space.

Now for the particular case of rotations ($\tau = R$) we have that $U(R) = \exp(-i \hat{\textbf{n}} \cdot \textbf{J})$ (incidentally, Ballentine drops the minus signs that he includes earlier in the table on page 69 -- again not pictured but this is conventional -- why? Is this an erratum? He does it multiple times later so perhaps not, but also he seems to choose a minus sign in front of $J_z$ just at the end of Case (i) so perhaps not). Now for my two questions:

(1) I follow (again it follows exactly from ACuriousMind's answer here) that (7.18) is correct (since $\psi$ is a scalar), but is it really strictly speaking a scalar? Why isn't (7.18) correct up to a phase? Shouldn't we be allowed to insert some $e^{i\phi}$ in (7.18) seeing as (7.18) is just the statement that states must be invariant under rotation, and a state is indeed equivalent up to a phase?

(2) The main question is what on earth a "multicomponent state function" is. Ballentine introduces this with zero definition. Does "multicomponent state function" perhaps mean "state from Hilbert space which is built out of tensor products of other Hilbert spaces"? Now with this stated, I follow the general pattern of (7.19) -- in particular, objects like vectors/tensors/spinors transform more generally than do scalars and the operator $D$ captures this. But I'm afraid I don't then follow how (7.20) obtains, and perhaps this is because I can't make sense of what kind of mathematical object a "multicomponent state function" is. Is Ballentine in going from (7.19) using the following formal (not rigorous) development, where we "expand the objects inside the multicomponent state function" after choosing a rotation about $z$ as Ballentine did before?
$$\textbf{R}\begin{bmatrix} \psi_1(\textbf{x}) \\ \psi_2(\textbf{x}) \\ \dots\end{bmatrix} = D\begin{bmatrix} \psi_1(R^{-1}\textbf{x}) \\ \psi_2(R^{-1}\textbf{x}) \\ \dots\end{bmatrix} \stackrel{?}{=} D\begin{bmatrix} \psi_1(\textbf{x}) + \epsilon\left(y\frac{\partial}{\partial x} - x\frac{\partial}{\partial y} \right)\psi_1\\ \psi_2(\textbf{x}) + \epsilon\left(y\frac{\partial}{\partial x} - x\frac{\partial}{\partial y} \right)\psi_2 \\ \dots\end{bmatrix}\stackrel{??}{=} D\begin{bmatrix} e^{-i \epsilon\hat{\textbf{n}} \cdot \textbf{L}}\psi_1(\textbf{x}) \\ e^{-i \epsilon\hat{\textbf{n}} \cdot \textbf{L}}\psi_2(\textbf{x}) \\ \dots\end{bmatrix} \\ \stackrel{???}{=} De^{-i \epsilon\hat{\textbf{n}} \cdot \textbf{L}}\begin{bmatrix} \psi_1(\textbf{x}) \\ \psi_2(\textbf{x}) \\ \dots\end{bmatrix}$$

from which we conclude indeed that $\textbf{R} = De^{-i \epsilon \hat{\textbf{n}} \cdot \textbf{L}}$? None of the manipulations I've done make any sense based on any math I've ever learned but they formally "get me there". I'm hoping people can shed light here.

 

[PeterDonis]

Shouldn't we be allowed to insert some $e^{i \phi}$ in (7.18) seeing as (7.18) is just the statement that states must be invariant under rotation

No, that's not what 7.18 is. 7.18 is an expression for what a rotation transformation does to a state. 7.17, the equation for the rotation operator $\mathbf{R}$, already is a phase; that's obvious just from looking at 7.17. You can't just insert an arbitrary phase in addition to that.

The main question is what on earth a "multicomponent state function" is.

It's a state function of a system with internal degrees of freedom, i.e., degrees of freedom in addition to the ones captured by the configuration space label $\mathbf{x}$. The most common example of this in QM is the intrinsic spin of elementary particles, which is the case Ballentine discusses here.

Is Ballentine in going from (7.19) using the following formal (not rigorous) development, where we "expand the objects inside the multicomponent state function" after choosing a rotation about as Ballentine did before?

No. The "rotation" expressed by the "internal" matrix $D$ is not a rotation in ordinary space. It's a "rotation" in the space spanned by the internal degrees of freedom. The only relationship it has to ordinary space is that, in order to choose a basis for the "spin space"--the Hilbert space of the internal degrees of freedom involved with spin--you have to pick a direction in ordinary space, corresponding to choosing a particular axis about which to measure the spin. That's what 7.21 is describing: the unit vector $\mathbf{\hat{n}}$ is the chosen spin axis, and the vector $\mathbf{S}$ is the vector containing the unitary operators (matrices) that describe measuring spin about the $x$, $y$, and $z$ axes. The "rotation" $D$ then describes what happens to all of these operators when you rotate the ordinary space coordinate axes by an angle $\theta$--which of course also rotates the basis vectors you are using to describe the spin operators.