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I If operators commute then eigenstates are shared

  1. Apr 5, 2016 #1
    I'm confused about the statement that if operators commute then eigenstates are shared.

    My main confusion is this one:

    ##L^2## commutes with ##L_i##. Then these two share eigenstates. But ##L_x## and ##L_y## do not commute, so they don't share eigenstates. Isn't this violating some type of transitivity? Namely, if ##[L^2,L_i]=0## then exactly what accounts for ##L_x## and ##L_y## not sharing eigenstates (considering they share eigenstates with ##L^2##)?

    A mathematical explanation of this would be much appreciated.

    If my confusion is not well understood please ask for clarification.

    Many thanks.
     
  2. jcsd
  3. Apr 5, 2016 #2

    blue_leaf77

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    In a vector space of fixed value of ##L##, the operator ##L^2## is a multiple of the identity. Therefore, every operator in this space, regardless of its hermicity, must commute with it. But obviously not every operator commutes with one another in this space.
    I think it can also be explained in terms of the (ir)reducibility of the space. The set of operators ##\{L_i: i=1,2,3\}## is irreducible, therefore any operator which commutes with all operators in this set must be a multiple of the identity. One of such operators is ##L^2##, therefore it must be a multiple of identity.
     
    Last edited: Apr 5, 2016
  4. Apr 6, 2016 #3
    L2's eigenvalues are degenerate; each has one eigenstate it shares with either Lz, Lx or Ly. So all of Lz's eigenstates belong also to L2, none to Lx or Ly.

    Each eigenstate can be written |l,m> where l is the angular momentum quantum number and m, the magnetic quantum number. l belongs to L^2, m belongs to either z, x, or y. Conventional is to use z. So when you use L2 and Lz to get a basis of commuting eigenstates you get the ones where m represents the z quantum number. Strictly it should be subscripted z. With this basis, Lz is diagonal but not Lx or Ly. If you used Lx or Ly, you'd get the other groups of L2's degenerate eigenstates. Any of the 3 groups spans the space.
     
  5. Apr 6, 2016 #4

    stevendaryl

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    To say that [itex]L^2[/itex] and [itex]L_x[/itex] share eigenstates just means that there is (at least one) state that is an eigenstate of both operators. It doesn't mean that every eigenstate of one operator is an eigenstate of the other operator.

    So since [itex]L^2[/itex] and [itex]L_x[/itex] commute, there is a state [itex]| l, l_x \rangle[/itex] that is an eigenstate of [itex]L^2[/itex], and is also an eigenstate of [itex]L_x[/itex]. Similarly, there is a state [itex]|l, l_y\rangle[/itex] that is an eigenstate of [itex]L^2[/itex] and [itex]L_y[/itex]. But [itex]|l, l_x\rangle[/itex] is not an eigenstate of [itex]L_y[/itex], and [itex]|l, l_y\rangle[/itex] is not an eigenstate of [itex]L_x[/itex].
     
  6. Apr 6, 2016 #5

    PeroK

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    Although, generally, it tends to be proved or assumed that they share a complete set of eigenstates (that span the Hilbert Space). Just having one eigenstate in common wouldn't lead to very much!
     
  7. Apr 6, 2016 #6
    They must share a complete set spanning the space. They can't commute only on one subspace not on another
     
  8. Apr 6, 2016 #7

    PeroK

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    That doesn't follow. Every operator commutes with the identity operator, but not every operator has a complete set of eigenvectors.

    For two commuting Hermitian operators, however, it can be proved that they share an eigenbasis. But, it's not a one-line proof.
     
  9. Apr 6, 2016 #8
    I though we were talking about observables, which always span the space, whether shared with another commuting observable or not.

    But, what's an example of a (non-observable?) operator that doesn't have a complete set of eigenvectors?
     
    Last edited: Apr 6, 2016
  10. Apr 6, 2016 #9
    To put it another way if the operator is positive semidefinite Hermitian it has complete set of eigenvectors. The proof I learned actually constructs the basis. You choose any eigenstate as your first then construct another from the rest of the space using ... I forget the theorem, maybe Chebyshev? It guarantees there's at least one more orthogonal eigenvector. You keep going until you have N of them. There are quicker non-constructive proofs.

    Physically this corresponds to the fact that we always find an observable has a definite real value. If we couldn't define a basis in some part of the overall space, there must be experimental conditions where we see an observable superposed in two states.

    If you allow any sort of operator then you're right, it doesn't have to be diagonalizable.

    Yes, every operator commutes with the Identity. Physically that means you can first do nothing, then measure; or measure, then do nothing; and get the same result either way.
     
  11. Apr 7, 2016 #10
    secur: do you know somewhere I can study this proof you mention in post 9?

    The proofs I've seen show the existence of at least one. But as someone already mentioned here, it is generally assumed the shared eigenstates form a complete basis, so just showing the existence of one is pretty shallow. Does anyone know of a book that covers this in detail?
     
  12. Apr 7, 2016 #11

    PeroK

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    You can find all this online these days. For example:

    http://ocw.mit.edu/courses/physics/...study-materials/MIT8_04S13_OnCommEigenbas.pdf
     
  13. Apr 7, 2016 #12
    hi david, If I recall correctly it was in my functional analysis textbook, "Real Analysis" by Royden and Fitzpatrick. But I could be wrong, it's been more than 40 years. I'll look through the books I have lying around, let you know if I find it. It's simple enough, I might remember or figure out how it goes.

    That ref proves that commuting observables share all complete bases except for degenerate eigenvalues - exactly the case we have here. Every eigenvector of L2 is degenerate with 3 eigenvectors, one for each of x, y, z, as I mentioned in my first post.

    But the proof that davidbenari is asking about is different: it proves that a (single) Hermitian operator must span the space. (That's the point under dispute.) Although it's got to be online I don't know where.

    - no that's not an assumption but a fact. PeroK's reference proves it mathematically, but physically it's even more obvious. If it weren't true there would be experimental observations showing two observables commute but other experiments showing they don't. For example if you measure the positions of two separate particles one way, it doesn't matter which order you do it in. But if you measure them some other way, it does matter. Physically it just doesn't make sense.
     
  14. Apr 7, 2016 #13
    david, the proof appears on p. 39 of "Principles of Quantum Mechanics" by Ramamurti Shankar. I can recommend the book as very easy to read, since it's only for first-year grad students at Harvard (not noted for difficult STEM courses!).

    Shankar also gives an example of an operator that doesn't span the space:

    4 1
    -1 2

    Has one eigenvalue, 3, and one eigenvector, [a -a].

    The example can be thought of as a diagonal matrix plus a rotation matrix (or, in QM, you'd be looking at a Hermitian plus a Unitary matrix / operator).

    Shankar doesn't mention it of course, but such a matrix can represent the following physical situation. Take some item with two bars coming out to the sides, like a bicycle's handlebars. Holding the front wheel of the bike steady with your knees, pull the left handlebar towards you and push the other away. Then the stress tensor in the 2-dim horizontal plane through the pylon has just one eigenvalue, somewhat counterclockwise from the axis along the bike. To the left of it, the stress vectors are all being pulled back; to the right, pushed forward; none of them will be another eigenvector, you see.
     
  15. Apr 7, 2016 #14
    Thanks secur! I'll check it out.
     
  16. Apr 9, 2016 #15

    George Jones

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    davidbenari, maybe we should think a little more concretely about what blue_leaf77 and stevendaryl wrote. Consider spin 1/2 (so I only have to type in 2x2 matrices!) with hbar set to one.

    $$S^2 = \frac{3}{4} \begin{pmatrix}
    1 & 0\\
    0 & 1
    \end{pmatrix}, \quad
    S_z = \frac{1}{2} \begin{pmatrix}
    1 & 0\\
    0 & -1
    \end{pmatrix}, \quad
    S_x = \frac{1}{2} \begin{pmatrix}
    0& 1\\
    1 & 0
    \end{pmatrix}
    $$

    Because ##S^2## is a multiple of the identity, any non-zero state in an eigenstate of ##S^2## .

    The eigenstates of ##S_z## are

    $$
    \begin{pmatrix}
    1 \\
    0
    \end{pmatrix}, \quad
    \begin{pmatrix}
    0 \\
    1
    \end{pmatrix},
    $$

    which are also eigenstates of ##S^2##.

    The (normaiized) eigenstates of ##S_x## are

    $$\frac{1}{\sqrt{2}}
    \begin{pmatrix}
    1 \\
    1
    \end{pmatrix}, \quad
    \frac{1}{\sqrt{2}}\begin{pmatrix}
    1\\
    -1
    \end{pmatrix},
    $$

    which are also eigenstates of ##S^2##. ##S_z## and ##S_x##, however, have no common eigenstates.
     
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