General arguments about introduction of angular momentum to QM is that under a transformation of coordinates the x and y position operators mix (as it is usually written) $$\hat{x}' = \cos(\theta) \hat{x} - \sin(\theta)\hat{y}$$ $$\hat{y}' = \sin(\theta) \hat{x} - \cos(\theta)\hat{y}$$ The general conclusion is that the operator $\hat{L} = \hat{x}\hat{p_y} - \hat{y} \hat{p_x}$ is angular momentum and it generates rotations in the plane, $|x',y'⟩ = \exp\left(i\frac{\theta}{\hbar} \hat{L}\right)|x,y⟩$, where $|x',y'⟩$ are eigenstates of both $\hat{x}'$ and $\hat{y}'$. I find this very confusing because it mixes $\hat{x}, \hat{y}$ in one equation and they might possibly come from different Hilbert spaces (that need not be copies of the same thing).

In my QM course the lecturer said that we can think of the states depending on three real variables (as in Euclidean space) as living in a product of individual Hilbert spaces, $\mathcal{H}_{3d} = \mathcal{H} \otimes \mathcal{H} \otimes \mathcal{H}$ where pure position states are $|x,y,z⟩ = |x⟩ \otimes |y⟩ \otimes |z⟩ \in \mathcal{H}_{3d}$. What is the angular momentum in this notation? How does the unitary operator enacting rotations look like?

So far I have attempted to answer the questions in 2d. If this rotation is a symmetry of the system then we can write for some unitary $\hat{U}(\theta) \in \mathcal{H} \otimes \mathcal{H}$ by analogy to the previous relations $$\hat{x}' \otimes 1 = \hat{U}(\theta) \left(\hat{x} \otimes 1 \right)\hat{U}(\theta)^\dagger = \cos(\theta) \left( \hat{x} \otimes 1 \right) - \sin(\theta) \left( 1 \otimes \hat{y} \right)$$ $$1 \otimes \hat{y}' = U(\theta) \left(1 \otimes \hat{y} \right)U(\theta)^\dagger = \sin(\theta) \left( 1 \otimes \hat{y} \right) + \cos(\theta) \left( \hat{x} \otimes 1 \right)$$ and the eigenstates of $\hat{x}' \otimes \hat{y}'$ in terms of eigenstates of $\hat{x} \otimes \hat{y}$ are $|x'⟩\otimes |y'⟩=\hat{U}(\theta)|x⟩\otimes |y⟩$. Then by analogy to standard notation the angular momentum operator seems to be $$\hat{L} = \left(\hat{x}\otimes 1\right)\left(1 \otimes \hat{p}_y\right) - \left(1 \otimes \hat{y}\right) \left(\hat{p}_x\otimes 1\right)$$ and $\hat{U}$ seems to be $\hat{U}(\theta) = \exp(i \hat{L} \theta/\hbar)$. The conclusion is that both angular momentum and the unitary rotation operator are mixed operators (not a direct product of operators from original Hilbert spaces).

Is this correct? How does it look like in 3d? Is it $\hat{L} = (\hat{x}\otimes 1 \otimes 1, 1 \otimes \hat{y} \otimes 1, 1\otimes 1 \otimes \hat{z}) \times (\hat{p}_x \otimes 1 \otimes 1, 1 \otimes \hat{p}_y \otimes 1, 1\otimes 1 \otimes \hat{p}_z)$? How about $\hat{U}(\vec{n}, \phi)$ (in axis-angle convention or any other convenient for 3d rotations)?


2 Answers 2


You are correct if you really want to write it this way.

But I'd posit that this is needlessly complicated and does not add anything. Whenever you have position operators $x,y,z$ and momentum operators $p_x,p_y,p_z$ on the three Hilbert spaces fulfilling the canonical commutation relations, then by Stone-von Neumann your Hilbert spaces are all unitarily equivalent to the standard space of 1d wavefunctions $L^2(\mathbb{R})$ with the canonical representation of the position operator as multiplication and the momentum operator as differentiation, and therefore the total Hilbert space is equivalent to the standard space of 3d wavefunctions $L^2(\mathbb{R}^3)$ with the canonical representation of angular momentum as $L_i = \sum_{j,k} \epsilon_{ijk} x_j \partial_k$.

Regardless of what you do, once you have some representation for the $L_i$, the angle-axis rotation is just $$ U(\vec n, \phi) = \exp\left(\mathrm{i}\phi\vec n \cdot \vec L\right) = \exp\left(\mathrm{i}\phi\sum_i n_i L_i\right).$$


If you want to talk about this in such mathematical terms, it may be a good idea to have a look at Lie groups and Lie algebras.

I think it would help to decide, where does this angular momentum operator comes from, i.e. why does it enter the discussion at all? In classical phyiscs there is such a conserved quantity, but I always thought that trying to motivate its introduction in QM from CM was clumsy. Instead it may be better to focus on isotropy of space.

Let us try this. Let us say that you are working in isotropic space, thus all the observers, looking at your system from different directions, should observe similar effects (I am talking about co-variance rather than invariance here).

To make this tractable we introduce rotation operators $\hat{U}\left(\mathbf{\hat{n}}, \theta\right)$, that take in your state $|xyz\rangle$, as seen by one observer, and rotate it around some axis $\mathbf{\hat{n}}$, by angle $\theta$: $|\bar{x}\bar{y}\bar{z}\rangle=\hat{U}\left(\mathbf{\hat{n}}, \theta\right)|xyz\rangle$. Together these operators form the group of rotations SO(3). Now, note that we introduced how this group acts on your $\mathcal{H}_{3d}$ space allready, based on what we call to be a rotation, e.g.


The angular momentum operator arises as a Lie-algebra of your rotation group $\hat{U}\left(\mathbf{\hat{n}},\theta\right)=\exp\left(i\hat{L}_n \theta\right)$. The reason we call it angular momentum is because if the system is invariant under rotation around axis $\mathbf{\hat{n}}$ then classically there will be associated conserved angular momentum, and quantum mechanically you will have: $\hat{U}\left(\mathbf{\hat{n}},\theta\right)|xyz\rangle=\exp\left(i\lambda\right)|xyz\rangle$, where $\lambda\in\mathbb{R}$. If the system is isotropic, then Hamiltonian will be invariant under rotations, so the above-mentioned eigenstate will also be an eigenstate of the Hamiltonian. Thus if the system starts of in state $\hat{U}\left(\mathbf{\hat{n}},\theta\right)|xyz\rangle=\exp\left(i\lambda\right)|xyz\rangle$, it will remain in it. Commonly one then chooses to label the states of the system with the eigen-values of the Lie-algebra (so $|xyz\rangle\equiv|\dots,\lambda\rangle$). This is convinient since we know that such labels will not change in time, i.e. thery are conserved. Finally, we have a real-valued conserved quantity that arises as a result of isotropy, so we call it angular momentum.

My point here is that definition of angular momentum is more about rotations and isotropy than about classical angular momentum. For me this simplifies working with angular momentum. I hope it will also help you.


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