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Significance of Commutation Relations

  1. Mar 14, 2010 #1
    I am aware that the commutation relation between conjugate variables shows that one quantity is the Fourier transform of the other, and so to imply the Heisenberg Uncertainty condition. So for example, the commutation relation between x, p (position and momentum respectively) leads to a non-zero value (i*h-bar).
    But what I am not clear about is what happens when you apply the commutator to a wave function (or in terms of an observation of a wave function). Let's take the example of angular momentum operators:

    [tex]\left[\hat{L}_z , \hat{L}^{2} \right][/tex]

    The fact that they commute, means they share no uncertainty relation between them. But what does this mean in terms of wavefunction and measurement? We can find the simultaneous eigenvalues of a wavefunction (what does this mean)? ... If I was to go and measure the angular momentum squared of (something?) and then the projected angular momentum on a given axis of the same (thing?), I would know these values to 100% certainty within, of course, the confines of experimental error. In terms of a visual experiment, does this mean I can find the angular momentum squared of, say, an electron, and -- " -- " -- with complete certainty? I ve been bombarded with theoretical frameworks of commutators, operators, wavefunctions and yet no appreciation of what is going on!

    I hope someone can help.
  2. jcsd
  3. Mar 14, 2010 #2


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    You've got it mostly correct AFAICS. Commutators are operators .. you apply them to wavefunctions the way that you would any other operator. If a commutator between two operators is zero, that means that the eigenfunctions of either operator are simultaneously eigenfunctions of the other .. i.e. the two operators share the same set of eigenfunctions. However, they will have different eigenvalues for the same eigenfunctions.

    You already know this in some sense if you have taken any chemistry classes ... remember the Pauli exclusion principle for atoms, which says that each electron must have 4 distinct quantum numbers? Those quantum numbers index eigenstates of a hypothetical one-electron atom, and all represent the different eigenvalues of 4 different operators, all of which commute for these eigenstates:

    the n quantum number indexes the total energy of the state, i.e. eigenvalue of the Hamiltonian operator

    the l quantum number indexes the orbital angular momentum of the state, i.e. eigenvalue of the L2 operator

    the ml quantum number indexes the projection of l on a space-fixed axis, i.e. eigenvalue of the Lz operator

    the ms quantum number indexes the projection of the electron spin on a space-fixed axis, i.e. eigenvalue of the Sz operator

    The eigenstates of the one-electron "hydrogen-like" atom are simultaneous eigenstates of all of those operators.*(see note below)

    Anyway, I hope this helps a bit ... feel free to ask more.

    *This is not true for multiple electron atoms ... the electron-electron interaction introduces additional terms to the Hamiltonian that mix the states together, but that does not destroy the utility of this indexing scheme for electrons in atoms.)
  4. Mar 14, 2010 #3
    What will really interest me is a simple (or maybe not) proof that the eigenfunction is the same or not depending on the operators used in the commutator. For example, I'd like to see the solution of the eigenfunction obtained for the total angular momentum squared and z-angular momentum operator and to see it is the same.
    But, what I'd really like to see is, the eigenfunction to be different (not simultaneous) for say the x-angular momentum operator and z-angular momentum operator.
  5. Mar 14, 2010 #4
    Just to split hairs, the eigenstates can be chosen to be simultaneous eigenstates of both operators. Not all eigenstates of L2 are eigenstates of Lz, for example.

    Sakurai has a really nice derivation for the commutation relations between the generators of rotations, that is, for angular momentum operators.
  6. Mar 14, 2010 #5
    So why do they (how can they!?) generalise the L2 & Lz commutation relations to equal 0?
  7. Mar 14, 2010 #6
    The fact that the operators can be chosen to have a common eigenbasis is not usually used to prove that they commute. The commutator [L2,Lz] can be calculated directly from the definition that the angular momentum operators are the generators of rotations. http://en.wikipedia.org/wiki/Angular_momentum#Angular_momentum_in_quantum_mechanics" , page 24, has a derivation for the fact that a complete commuting set of operators have a common eigenbasis. The book can be read for free in Google Books.
    Last edited by a moderator: Apr 24, 2017
  8. Mar 14, 2010 #7
    thanks for the citations. I am going out now but i'll read them when i get back.
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