Why can we say that the cross product of $\hat{x}$ and $\hat{p}$ will give the quantum operator $\hat{L}$?

Question

The second postulate of quantum mechanics says that all physical observables are described by their corresponding operators acting on quantum states. I assume this is why we can change from classical $x$ and $p$ to quantum operators. But I'm not logically convinced we can simply say that their cross products (for angular momentum) will also produce the quantum operator for angular momentum. Is there a way to prove that cross products (or other operations) of quantum operators will always give a new viable Hermitian operator corresponding to a physical observable? Or is this just part of a postulate?

Edit: Nobody is in disagreement that theoretical physics consists of theories that must be shown to be good models. To give a simple example to help clarify my question. $F=ma$ is a postulate that is a great model for everyday physics problems that freshman do. But we can use definitions such as $F = qE$ to find the acceleration of a particle in an electric field by making the substitution $qE = ma \rightarrow a = qE/m$. If this did not predict the acceleration of a particle in a field, but it did predict the acceleration of a ball rolling down a hill, then we know that there is a hole somewhere that we are missing theoretically.

My original question is: starting with the postulate of classical x and p corresponding to quantum operators $\hat{x}$ and $\hat{p}$ can we show derive that classical $\vec{L} = \vec{r} \times\vec{p}$ will also correspond to quantum operator $\hat{L}$ = $\hat{x} \times \hat{p}$?

Answer 1

It is straightforward to show that the orbital angular momentum is $L = x\times p$ just from the idea that the angular momentum operator $L$ must represent infinitesimal rotations. It is exactly the same logic that shows that the momentum operator is $-\mathrm{i}\hbar\partial_x.$

Obviously, if $R_x(\phi)$ is a "rotation operator" around the $x$-axis, then $R(\phi)\lvert \vec r\rangle = \lvert R_x(\phi)\vec r\rangle$, where the term inside the ket is the classical rotation of the position vector $\vec r$. Hence

$$ \langle \vec r\vert R_x(\phi) \vert \psi\rangle = \langle R_x(\phi)^{-1}\vec r\vert \psi\rangle = \psi(R_x(\phi)^{-1}\vec r)$$

and a classical infinitesimal rotation is $R_x(\phi)^{-1}\vec r = r- \phi r_y \hat{z}+ \phi r_z \hat{y} +\mathcal{O}(\phi^2)$ so this is

$$ \psi(R_x(\phi)\vec r) = \psi(\vec r) + \phi r_z(\partial_y\psi)(\vec r) - \phi r_y (\partial_z \psi) (\vec r) + \mathcal{O}(\phi^2) $$

and like for the translation operator we conclude that infinitesimal rotations around the $i$-th axis are generated by $r_y\partial_z - r_z\partial_y$ in the position basis, which generalises without effort to the generic statement that $L = r\times p$ generates rotations in general.

(There may be some sign errors somewhere in there but they don't change the form of the argument)

Answer 2

What makes quantities in physics important is that they are useful. What makes them useful is the equations that they obey. In classical physics, the angular momentum (defined in terms of components as $L_{i}=\epsilon_{ijk}x_{j}p_{k}$) turns out to be useful in a lot of situations. It transforms like a (pseduo-)vector, and—even more importantly—in some fairly natural systems it is conserved. A particularly well known example is that, for a particle moving in a central potential $V(\vec{x})$, $\vec{L}$ is conserved, according to the classical equations of motion. (The conservation of $\vec{L}$ may be demonstrated in many ways, such as via the fact that $\{L_{i},H\}=0$, where $\{\cdot,\cdot\}$ is the Poisson bracket and $H$ is the Hamiltonian.)

With the transition to quantum mechanics, it will initially be an open question whether the operator with components $\hat{L}_{i}=\epsilon_{ijk}\hat{x}_{j}\hat{p}_{k}$ will still be the "right" expression for the angular momentum. That $\hat{L}_{i}$, so defined, must at least be part of the angular momentum in the quantum theory, because the classical and quantum expressions must correspond in the classical limit (taking $\hbar\rightarrow0$). However, there is no a priori reason why there cannot be additional intrinsically quantum mechanical contributions to the angular momentum, $$\hat{L}_{i}=\epsilon_{ijk}\hat{x}_{j}\hat{p}_{k}+\hat{M}_{i},$$ where the operator $\hat{M}$ is at least $\mathcal{O}(\hbar)$. What ultimately decides which expression for the angular momentum is the "right" one is which expression is most useful. It turns out that in nonrelativistic quantum mechanics, the naively chosen $\hat{L}_{i}=\epsilon_{ijk}\hat{x}_{j}\hat{p}_{k}$ is still a conserved quantity, since it commutes with the Hamiltonian, $[L_{i},H]=0$; this is directly analogous to the classical theory. The fact that $\hat{L}^{2}$ and $\hat{L}_{z}$ commute with $H$ is something that we take advantage of when solving the problem of a particle in central potential. Those two operators can be diagonalized simultaneously with $H$, and their eigenvalue are consequently "good" quantum numbers that can be used as part of the parameterization of the eigenvalue and eigenstate spectrum. So this definition of the angular momentum operator turns out to be very useful, and that is why we use it in nonrelativistic quantum mechanics.

However (although it is not usually discussed in these terms), the situation changes in relativistic quantum mechanics. For the Dirac equation with a central potential, it turns out that $\hat{L}_{i}=\epsilon_{ijk}\hat{x}_{j}\hat{p}_{k}$ does not commute with the full Hamiltonian. Instead, it behooves us to define a different quantity to be the angular momentum, $$\hat{J}_{i}=\epsilon_{ijk}\hat{x}_{j}\hat{p}_{k}+\frac{\hbar}{2}\hat{\Sigma}_{i}.$$ This, of course, has exactly the form hypothesized above, with the intrinsically quantum contribution $\hat{M}_{i}$ now being nonzero. The operator $\hat{M}_{i}=\frac{\hbar}{2}\hat{\Sigma}_{i}$ is not constructed out of the position and momentum variables; instead it is a matrix operator, acting in the space of Dirac spinors. By convention, we still call the first term, $\epsilon_{ijk}\hat{x}_{j}\hat{p}_{k}$ the "orbital angular momentum" component $\hat{L}_{i}$; the new term is the "spin angular momentum." However, it is the total angular momentum $\hat{J}_{i}$ which is conserved and so ultimately the most important (labeling the energy eigenstates, for example).

Answer 3

Up to some domain issues, the product of two hermitian operators is hermitian if they commute so $$ L_x=yp_z-z p_y $$ does become a hermitian operator. Next, under Dirac quantization, the quantum commutator should be $i\hbar \times$(the classical Poisson bracket) and indeed we have $$ [\hat L_x,\hat L_y]=i\hbar \widehat{\{L_x,L_y\}}_{PB} \tag{1} $$ so $\hat L_x\to \hat y\hat p_z-\hat z \hat p_y$ is certainly the simplest operator that satisfies the basic requirements of the theory. You can get (1) to hold by adding (for instance) some scalar function $g_i$ to each $\hat L_i$ so that $[\hat L_i,g_j]=0 \forall i,j$ but it's not clear what has been gained, especially as there's no obvious experimental justification for such a scalar. Alternatively, you could think that the RHS of (1) is the start of some series in $\hbar$ with the next term very small but again there's no evidence of this.

Answer 4

There is no proof behind it, really it's just a postulate. It seemed to work for other quantities - think for example the Hamiltonian being

$$H = \frac{p^2}{2m} + V(x)$$

where $p$ is the momentum operator and $x$ is the position operator.

So, it could lead to incorrect predictions a priori for sure. But, in experiments measuring energy and in your case angular momentum, that postulate worked so far.

There are a lot of assumptions that bring you where you are. Even to assume that experiments should be predicted through observable operators at all, is an assumption - which has worked so far but may not work forever.

I've always wanted to try to frame Quantum Mechanics by starting from experiment, rather than making theoretical assumptions. This kind of stuff still bugs me. But of course the theory doesn't follow directly from experiment so it would just be "inspiration" for the theoretical assumptions rather than proof.

Edit: Yes, the form of $L$ follows from the commutation relations or the fact that it is an infinitesimal generator of rotations. But this just extends the radius of circular logic. Those three are all equivalent assumptions, and the point is that regardless of which one you choose as a postulate in your theory, we are already at "the bottom", the base assumptions which are just taken for granted as a starting point in QM. Sure you can justify any one of those three by any of the others, but that doesn't bring you anywhere. So I don't see a sense in using them as justification.