Why do rotations of a multicomponent state function take this form?

Question

I am reading Leslie Ballentine's Quantum Mechanics, section 7.2, which is all about the explicit form of the Angular Momentum operators.

I understand how he gets the form for the single component state function, equation (7.18) which has the form $$\mathbf{R} \Psi(\mathbf{x}) = \Psi(R^{-1}(\mathbf{x})) $$ where $\mathbf{R}$ is given by $$ \mathbf{R}_n(\theta) = e^{i\theta \mathbf{\hat{n}} \cdot \mathbf{J}/\hbar} $$ He then identifies $\mathbf{J}$ with the orbital angular momentum operator $\mathbf{L}$. No problems there.

However, in the following section he claims that for a multicomponent state function we take the general form of (7.19) $$\mathbf{R} \begin{bmatrix} \Psi_1(\mathbf{x}) \\ \Psi_2(\mathbf{x}) \\ \vdots\end{bmatrix} = D \begin{bmatrix} \Psi_1(R^{-1}\mathbf{x}) \\ \Psi_2(R^{-1}\mathbf{x}) \\ \vdots\end{bmatrix}$$ where now we have, in addition to the coordinate transformation $R^{-1}(\mathbf{x})$, we also have a matrix $D$ that operates on the internal degrees of freedom --- which is to say it makes linear combinations of the components. Our $\mathbf{R}$ now takes the form (7.20) $$ \mathbf{R}_n(\theta) = e^{i\theta \mathbf{\hat{n}} \cdot \mathbf{L}/\hbar} D_n (\theta)$$

He then identifies $D$ with spin angular momentum so total angular momentum $\mathbf{J} = \mathbf{L} + \mathbf{S}$.

I still don't understand the reason why we need this $D$ matrix. Can someone explain to me what is going on here, specifically why the form (7.19) instead of (7.18)? Why does this matrix show up when we have a multicomponent state function?

Answer 1

Physically, $R$ is just a rotation. However, when you write $R \psi(x)$, the general result is $R \psi(x) = (R\psi) (R^{-1}x)$.

The $R^{-1}x$ takes in account that the rotation has an effect on the space-time coordinates, and it corresponds to orbital angular momentum , but you have to take in account that $\psi$ itself may be not a scalar (an invariant) under rotation.

If $\psi$ is a vector, you may write it in fact $\psi^\mu$, and then you have $(R\psi)^\mu = D(R){^\mu_\nu} \psi^\nu$, where $D(R)$ is the vectorial representation of $R$.

$\psi$ may be also be a spinor $\psi^\alpha$, so you will have to use the spinorial $D(R)^\alpha_\beta$ representation of $R$.

While you are considering the effect of rotations on internal degrees of freedom of fields, it corresponds in fact to spin angular momentum.

Answer 2

As you know, any operation, which can be continuous or discrete, in quantum mechanics is represented by an operator say $\hat{O}$. Under the operation, a physical quantity $\hat{X}$ transforms as $\hat{X}' = \hat{O}\hat{X}\hat{O}^{-1}$. In case $\hat{O}$ is continuous (forming a Lie group), one can often write it as $\hat{O} = e^{ir\hat{T}}$ (there is more general form which we do not discuss here), where $r$ is a continuous parameter and $\hat{T}$ another operator (generator of the Lie group).

Now let's consider three quantities which are supposed to transform as a vector: $\vec{V} = (V_x,V_y,V_z)$. In other words, under rotation (denoted by $\hat{R}(\theta)$, an operation that is continuous and characterised by a rotation angle $\theta$ along with the direction of rotation axis $\hat{n}$), it should transform like this (for simplicity, let $\hat{n}=(0,0,1)$): $$V'_x = \cos(\theta)V_x + \sin(\theta) V_y\\ V'_y = -\sin(\theta) V_x + \cos(\theta)V_y \\ V'_z = V_z,$$ where $V'_{x,y,z}$ denote the transformed components. On other hand, we have the following, $$V'_{x,y,z} = \hat{R}(\theta)~V_{x,y,z}~\hat{R}^{-1}(\theta),$$ which must agree with the above equations. This then puts a constraint on the form of $\hat{R}(\theta)$. To completely determine it, we have to the know the algebra defined by the commutators of $\vec{V}$. In the present case, we shall impose the following: $$\vec{V}\times\vec{V} = i\vec{V}.$$ Now, one can show that the $\hat{R}$ must be $$\hat{R}(\theta) = e^{i\theta\hat{n}\cdot\vec{V}},$$ regardless of the wave function under consideration.

In particular, we see that both angular momentum $\vec{L}$ and spin $\vec{s}$ (as well as their sum $\vec{J}$) satisfy the properties of $\vec{V}$. Thus, the rotation in terms of them must be $$\hat{R}(\theta) = e^{i\theta\hat{n}\cdot\vec{J}} = e^{i\theta\hat{n}\cdot\vec{L}}\cdot e^{i\theta\hat{n}\cdot\vec{s}}.$$ We have used the fact that $\vec{s}$ commutes with $\vec{L}$. In case you are not interested in spin components, you can ignore the second factor of the above expression.