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Symmetry and Hamiltonian

  1. Oct 2, 2006 #1
    Why do operators representing some symmetry commute with the Hamiltonian?
     
  2. jcsd
  3. Oct 2, 2006 #2
    This is a very deep question, and worthy of discussion. There are a few ways that you can look at this, but I'm going to look at it from the standpoint of Noether's Theorem, which states, more or less, that every continuous symmetry corresponds to a conserved quantity of the motion. This also applies to classical systems, or field theories, since the proof (as I've seen it) only relies on the presence of a lagrangian for the system and a generator of some transformation that leaves the action invariant.

    As a specific example, let's look at the case of angular momentum, because it's the simplest non-trivial one. In the case of the coulomb potential, for example, the hamiltonian is given by

    [tex]
    H = \frak{\mathbf{p}^2}{2m} + \frac{q_1 q_2}{r}
    [/tex]

    The hamiltonian above is invariant under rotation (let's pretend we're in rectangular coordinates for now), and to see this, notice simply that [tex]\mathbf{p}^2[/tex] and [tex]1/r[/tex] are both functions of the magnitude of the vector, which is invariant under rotation, as we know from linear algebra. Thus, any generator of rotation must correspond to a conserved quantity. In both the classical and quantum case, this is the angular momentum, as we interpret the angular momentum as the generator of rotations.

    So in this case we find that the angular momentum is a conserved quantity. However, because of the structure of angular momentum, we have to choose [tex]L^2[/tex] and one of the angular momentum components, usually [tex]L_z[/tex], to be conserved.

    Why do these commute with the hamiltonian? Well, we know that, for example, [tex]L_z[/tex] must be a conserved quantity, so the time evolution of its state kets must be zero. From the Heisenberg equations of motion, we have that

    [tex]
    \dot{L_z} = \frac{i}{\hbar} [ L_z, H ]
    [/tex]

    Knowing that the time evolution of the operator is zero, we have that the z component of the angular momentum commutes with the hamiltonian. A similar exercise arises for [tex]L^2[/tex]. This is also why the angular momentum eigenstates and energy eigenstates are product states of the two.

    The fundamental physics to keep in mind, however, is Noether's Theorem, and that if you find some symmetry transformation of the hamiltonian/lagrangian that leaves the equations of motion invariant, then you've managed to find a conserved quantity.
     
    Last edited: Oct 2, 2006
  4. Oct 2, 2006 #3
    My answer seems lame now! I was going to say that symmetries of the hamiltonian are eigenstates and so they are just multiplication by a constant...
     
  5. Oct 2, 2006 #4
    Okay, so basically quantummechanically symmetries give rise to conserved operators (as classically they also give rise to conserved quantities) and via the Heisenberg equation of motion this implies the operator commutes with the Hamiltonian.

    Thanks a lot!
     
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