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I am trying to develop an intuition for how to tell if a certain variable (linear momentum, angular momentum, $L^2$, and parity) will be constants of motion in a particular potential.

I know the definition of a constant of motion: operator can not be a function of time, and it must commute with the Hamiltonian. However, if I have a particle, I want to develop symmetry arguments for determining whether it will commute:

(1) If I can rotate a particle around the $x,y,z$ axis, and the potential on the particle remains the same then angular momentum remains a constant of motion

(2) If I can move $x,y,z$ direction, and have the potential remain the same on the particle, then regular momentum remains a constant of motion.

(3) if moving the particle to $-x,-y,-z$ does not change the effecting potential, parity will be a constant of motion.

Hence, for a central potential I quickly realize that angular momentum will be constants of motion, but not linear momentum. For a particle in a uniform electric field in the z direction, $z$ angular momentum will be a constant of motion, but not linear or other angular components.

I am having difficulties applying this argument to a square potential. If I move within the box, the potential remains the same. If I move out of the box, the potential is different. Would I then conclude that there are no constants of motion for a square potential? How exactly can I show this with $[H, V(x)]$?

If this argument makes sense please let me know!

Dr. Ernesto Chinchilla
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I think only parity and energy are constants of the motion for the particle in a box (and functions thereof). Importantly, you need to be able to make all translations (rotations) for linear (angular) momentum to be conserved. Note that if we translate by more than the length of the box, the potential changes, and hence the potential is not invariant under all translations, and hence linear momentum is not conserved. Etc.

Showing that $[H,V(x)]$ for the particle in a box is unfortunately difficult, if not impossible (I believe that there are explicit forms for the potential that allows you to do this, provided you're okay with taking derivatives of delta functions and such). However, that doesn't show you that the system would be invariant under translations! The generator of translations is the momentum operator, so a system will be invariant under translations if it commutes with the momentum operator, in which case momentum will be a good quantum number, i.e., it will be conserved. It is really not obvious to see why momentum doesn't work here. Basically, the particle-in-a-box problem is not a good one, because of the pathologies associated with the infinite boundaries (among other things.

It might be worth considering a slightly more complicated (but more nicely behaved) system like the harmonic oscillator, for which $V(x) = \frac{1}{2}m\omega^2 x^2$. You can then show explicitly that the momentum operator $\frac{\hbar}{i}\frac{\partial}{\partial x}$ does not commute with the Hamiltonian, which shows that it is not a conserved quantity and also (although not as obviously why) that the system is not translationally symmetric (although you can see that from the fact that the potential is not translationally invariant).

All of this falls under the heading of (1) the quantum mechanical version of Noether's theorem that relates conserved quantities to quantities that commute with the Hamiltonian and (2) Lie theory (Lie groups and Lie algebras), which in part relates symmetries (like translational or rotational symmetries) to physical quantities (like momentum and angular momentum) that generate those symmetries.

march
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