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Common eigenstates

  1. Jan 10, 2014 #1
    If two operators commute my book says that "we can choose common eigenstates of the two." And I have seen it phrased like this in multiple other books.
    Does this mean that in general the eigenstates differ, but we can choose a set that is the same or what does it exactly mean in comparison to just saying that they have common eigenstates.
     
  2. jcsd
  3. Jan 10, 2014 #2

    dextercioby

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    There's no statement that if A and B (strongly) commute, then all eigenvectors of A are also eigenvectors of B and reverse. The statement is that if A and B are commuting compact operators, then you can choose a countable basis of the Hilbert space they act on from their common eigenvectors.
     
  4. Jan 10, 2014 #3

    DrClaude

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    Take the angular momentum operators ##l^2## and ##l_z##. The spherical harmonics ##Y_{l,m}## are eigenfunctions of both operators. But I can construct wave functions, such as
    $$
    \frac{1}{\sqrt{2}} \left( Y_{1,1} + Y_{1,-1} \right),
    $$
    which are eigenfunctions of ##l^2## but not ##l_z##, and conversely.

    So, if two operators commute, you can always find a complete basis set of states which are simultaneously eigenstates of both operators. But often, if you consider only one operator at a time, the eigenfunctions you calculate will not be eigenfunctions of the other operator.
     
  5. Jan 10, 2014 #4
    Okay. But the generic quantum problem is always. Find the eigenstates and eigenenergies of the hamiltonian and often this is solved by finding common eigenfunctions of H and an operator. But with what you are saying this is not all the eigenstates of the hamiltonian?
     
  6. Jan 10, 2014 #5

    DrClaude

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    No, that's not what I mean. Given a complete set of basis functions, if some eigenstates have degenerate (i.e., have the same eigenvalue), then you can always create a new complete set of basis functions by taking linear combinations of thoses degenerate eigenstates. If the Hamiltonian commutes with another operator, then there necessarily exist at least one way to make those linear combinations of degenerate eigenstates such that the resulting are eigenfunctions of both the Hamiltonian and the other operator.

    If there is no degeneracy, then the eigenstates are necessarily eigenfunctions of both commuting operators.
     
  7. Jan 10, 2014 #6
    Can you give an example? Maybe the kinetic energy operator and the hamiltonian for free electrons is an example, am I right?
     
  8. Jan 10, 2014 #7

    DrClaude

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    I just gave an example above.
     
  9. Jan 10, 2014 #8

    dextercioby

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    For a free particle, they are the same, right ?
     
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