Does Commuting with the Hamiltonian Guarantee Conservation of Angular Momentum?

[latentcorpse]

Just reading about how if an operator commutes with the Hamiltonian of the system then its corresponding obserable will be a conserved quantity.

my notes say that if \hat{L} commutes with the hamiltonian then the angular momentum will be conserved.

this kind of makes sense but surely it's going to depend on more specifics such as the form of the hamiltonian, no?

or is there a way to prove this is always true?

 

[quZz]

If a system is isotropic (for example closed system) then its Hamiltonian will always commute with L since L is a generator of an infinitesimal rotation

 

[latentcorpse]

could you explain a bit more fully please?

 

[quZz]

OK. But let me be a bit lazy and not mark vectors in bold =) hope you'll know where it should be.
The space is known to be isotropic, that is if you rotate a closed system as a whole in inertial reference frame then with the same initial conditions you should get the same results of experiments, or to put in another way all directions in space are equivalent for the system.

Consider a closed system. Suppose it is described by the wave function \psi(r_1}, ...). Each vectorr_a represents position of a-th particle. Now if you rotate your system to some infinitesimally small angel over some direction: \delta\varphi then each coordinate vector will change:
\delta r_a = \delta \varphi \times r_a.
It is very important that vector \delta\varphi is the same for all a since you rotate the system as a whole. Now follow me:
\psi(r_1 + \delta r_1, ...) = \psi(r_1, ...) + \sum_a \delta r_a \nabla_a \psi(r_1, ...) = \psi + \sum_a (\delta \varphi \times r_a) \nabla_a \psi = \left(\hat{1} + \delta \varphi \sum_a r_a \times \nabla_a \right) \psi
You see that operator \hat{1} + \delta \varphi \sum_a r_a \times \nabla_a represents an infinitesimal rotation. Parameters of this operation are components of \delta \varphi. Operator standing near a parameter and divided by imaginary i is known as a generator (in math terminology you sometimes don't have to divide by i).
So generator of a rotation is operator of angular momentum
\hat{L} = -i \sum_a r_a \times \nabla_a = \sum_a r_a \times \hat{p_a}.

Now recall what I've written in the beginning and let \hat{H} be Hamiltonian of the system. The symmetry of rotations mean that if you first rotate and then let the system evolve with time (following Schroedinger equation) or you first let it evolve and then rotate you should get the same result:
\left(\hat{1} + i \delta \varphi \:\hat{L} \right) \hat{H}\psi = <br /> \hat{H}\left(\hat{1} + i \delta \varphi \: \hat{L} \right) \psi<br />.
Now as you expand both sides and compare (taking into account that \delta \varphi is arbitrary) you get
[\hat{H}, \hat{L}] = 0.

You can use the same considerations to derive conservation of momentum as a consequence of homogeneity of space (i.e. symmetry under parallel translations).
The same ideas apply in classical mechanics and are known as Noether's theorem.