# Quantum Hamiltonian and Classical Hamitonian Rotationally Invariant

From Shankar's QM book pg. 310, it was said that the quantum hamiltonian $$H$$ is rotationally invariant whenever the classical hamiltonian $$\mathcal{H}$$ is rotationally invariant.

For infinitesimal rotations, the classical variables $$(x,y,p_x,p_y)$$ transform as $$\bar{x}=x-y\epsilon$$ $$\bar{y}=y+x\epsilon$$ $$\bar{p}_x=p_x - p_y\epsilon$$ $$\bar{p_y}=p_y+p_x\epsilon$$ The classical hamiltonian $$\mathcal{H}$$ is rotationally invariant if $$H(\bar{x},\bar{y},\bar{p}_x,\bar{p}_y)=H(x,y,p_x,p_z).$$

The quantum operators $$(X,Y,P_x,P_y)$$ transform in a similar way: $$U^\dagger XU=X-Y\epsilon$$ $$U^\dagger Y U = X\epsilon + Y$$ $$U^\dagger P_x U=P_x - P_y\epsilon$$ $$U^\dagger P_y U=P_x\epsilon+ P_y$$

where $$U$$ is the infnitesimal rotation operator. The quantum hamiltonian is rotationally invariant if $$H(U^\dagger X U,U^\dagger Y U,U^\dagger P_x U,U^\dagger P_y U)=H(X,Y,P_x,P_y)$$.

Now since the operators $$X,P_x$$ and $$Y,P_y$$ do not commute like classical variables $$x,p_x$$ and $$y,p_y$$, how can we be sure that the quantum hamiltonian is invariant whenever the classical hamiltonian is?

For example consider that in the expansion of classical Hamiltonian $$\mathcal{H}(\bar{x},\bar{y},\bar{p}_x , \bar{p}_y)$$ we have two terms that cancel each other: $$\mathcal{H}(\bar{x},\bar{y},\bar{p}_x , \bar{p}_y)=...+xp_x-xp_x+...=\mathcal{H}(x,y,p_x,p_y)$$

How can we be sure that in the expansion of the quantum Hamiltonian $$H(U^\dagger X U,U^\dagger Y U,U^\dagger P_x U,U^\dagger P_y U)$$ these two terms will be $$H(U^\dagger X U,U^\dagger Y U,U^\dagger P_x U,U^\dagger P_y U)=...+XP_x-X P_x+...=H(X,Y,P_x,P_y)$$ which cancel instead of $$H(U^\dagger X U,U^\dagger Y U,U^\dagger P_x U,U^\dagger P_y U)= ... + XP_x - P_xX+...\neq H(X,Y,P_x,P_y)$$ which do not cancel?

You have hit upon a recondite problem of quantization, which rarely crops up in practical problems: A classical $$\cal H$$ has several, indeed, many different quantum Hs which have it as their common classical limit.
Of those, one chooses the ones that share the same symmetry (e.g., rotational) as $$\cal H$$, all things being equal, but this is not formally or theologically compulsory!
Here is an artificial simplistic toy example to Illusrtate the formal point. Consider two quantum Hermitian hamiltonians $$H_1= XP_y -YP_x ,\\ H_2= XP_y -YP_x +i\lambda X [X,P_x],$$ both of which have the same classical limit $${\cal H}=xp_y- yp_x,$$ which is rotationally invariant.
Your may see directly $$H_1$$ is rotationally invariant, but $$H_2$$ isn't. Usually, when in doubt, people choose $$H_1$$ in quantizing $${\cal H}$$, depending on context, as it, unlike its evil twin, $$H_2$$, inherits the rotational symmetry of the classical system. This is a discretionary choice.
Note that on p. 120 the author discusses complications related to first quantization - in particular the ambiguity of expressions such as $$xp$$, i.e. if it is to become $$XP_x$$ or $$P_xX$$. The rule is that such expressions are to become the symmetric sum: $$\frac{XP_x+P_xX}{2}$$.