What is a "multicomponent state function" mathematically?

Question

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 here), 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) 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.

Answer 1

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?

I'm going to change the notation a bit to clarify my answer. Let $\big(\mathbf D_R\Psi\big)(\mathbf x)=\Psi(R^{-1}\mathbf x)$ be the action of the unitary operator corresponding to the rotation matrix $R$ (in the text, they use $\mathbf R$ for $\mathbf D_R$). Are you asking why $\mathbf D_R$ can't be defined via $\big(\mathbf D_R\Psi\big)(\mathbf x) = e^{i\phi}\Psi(R^{-1}\mathbf x)$?

If so, then the answer is that you could do that if you wanted to. If you did, then you would no longer have a genuine representation of the rotation group, since $$\mathbf D_{R_1} \mathbf D_{R_2} = e^{2i\phi}\Psi\big([R_1R_2]^{-1}\mathbf x\big) =e^{i\phi}\mathbf D_{R_1 R_2}\neq \mathbf D_{R_1 R_2}$$ Instead, you would have a projective representation which, as you say, is basically a representation "up to a phase." In principle, a projective representation is a perfectly good thing to work with, but carrying around those extra phase factors is a pain. If we can convert a projective representation to a genuine representation by trivially eliminating some pesky phase factors, there's absolutely no reason not to do so. Of course, there are some circumstances (specifically, systems with half-integer spins) in which genuine representations of $\mathrm{SO}(3)$ do not exist, and so we are forced to work with a projective representation of $\mathrm{SO}(3)$ (which can be expressed as a genuine representation of its double cover $\mathrm{SU}(2)$).

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(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"?

Yes - an $N$-component state function is an element of $L^2(\mathbb R^d) \otimes \mathbb C^N$. Loosely speaking, the $\mathbf D$ written in the text is the $N\times N$ matrix which acts on the $\mathbb C^N$ part. In practice, such systems arise when we endow our particles with spin - a spin-$s$ particle is described using the Hilbert space $L^2(\mathbb R^d) \otimes \mathbb C^{2s+1}$.

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.

Under the action of a generic operator, each component of the state function transforms like a scalar and the components are mixed together by the action of the matrix $\mathbf D$. The full rotation operator can be written as the composition of these two actions (strictly speaking, this is the tensor product of operators which act on the $L^2(\mathbb R^d)$ space and the $\mathbb C^N$ space respectively).

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From Hall's Quantum Theory for Mathematicians, pp. 383-384:

The simplest possibility for the Hilbert space of a single particle is the Hilbert space $L^2(\mathbb R^3)$, which certainly carries an (ordinary) unitary action of $\mathrm{SO}(3)$, as we have been discussing in this chapter. Based on various experimental observations, however, physicists have proposed a modification to the Hilbert space for an individual particle that incorporates “internal degrees of freedom.” The proposal is that for each type of particle, the quantum Hilbert space should be of the form $L^2(\mathbb R^3)\otimes V$, where $V$ is a finite-dimensional Hilbert space that carries an irreducible projective (emphasis mine) unitary representation of $\mathrm{SO}(3)$. Here $\otimes$ is the Hilbert tensor product (Appendix A.4.5). The (projective) action of $\mathrm{SO}(3)$ on $V$ describes the action of the rotation group on the internal degrees of freedom of the particle.

[...]

Summary 17.20 (Spin) Each type of particle has a “spin” $\ell$, which is a non-negative integer or half-integer. The Hilbert space for such a particle is $L^2(\mathbb R^3)\otimes V_\ell$, where $V_\ell$ is an irreducible projective representation of $\mathrm{SO}(3)$ of dimension $2\ell + 1$.

Answer 2

1. You can derive the action of the rotation operator using its explicit form $R_z=-i\hbar \frac{\partial }{\partial \theta}$. We don't have a choice on the explicit form of the rotation operator, as it's given by canonical quantisation.

You can see that exponentiating this and applying it to a vector $f(r,\theta)$ will translate the angular position by an angle $\theta _0$, because the exponentiation gives a Taylor series. No extra phase factors show up.

2. This is because the two rotation matrices commute, So you can flip their order. They commute because they operator on different labels of the vector components. The components of the vector $\psi (x,y,s)$ are specified by three labels : $x$ and $y$ are continuous labels, while $s$ (=$1,2$) is a discrete label. The first rotation matrix $R_z$ only operates on the $x$, $y$ labels while the $D$ operator only acts on the $s$ label.

Specifically, the vector space is spanned by the basis $|x,y\rangle \otimes |s\rangle$, $-\infty <x,y<\infty$, $s=1,2$. Here's the action of $R_z$ on a basis vector:

$$R_z (|x,y \rangle \otimes |s\rangle) = |R(x,y)\rangle \otimes |s\rangle$$

$R(x,y)$ refers to the $(x,y)$ co-ordinates after rotation. You can see that the action of $R_z$ on a basis vector leaves the value of $s$ unchanged. Similarly, the action of $D$ on a basis vector leaves $x,y$ unchanged:

$$D(|x,y \rangle \otimes |s\rangle )= |x,y\rangle \otimes (\alpha |1\rangle + \beta |2\rangle)$$

where $D|s\rangle = \alpha |1\rangle + \beta|2\rangle$.

From this, it follows that the commutator of the matrices $R_z$ and $D$ is zero. You can prove this by applying $R_zD - DR_z$ on any basis vector.