- #1

EE18

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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.