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Mixed Vs Pure States

  1. Jan 24, 2009 #1
    Does the quote below (from a textbook) make sense to anyone? The context is that they are introducing mixed states, and the quote below is how they refer to the pure states discussed in previous chapters. Here is the quote:

    "In quantum mechanics, such maximum information is contained in a wavefunction that simultaneously diagonalizes a complete set of commuting operators relevant to the system."

    How does a wavefunction diagonalize an operator? I know some operators can be diagonalized with respect to certain basis sets of eigen functions, but I don't know what it means for a single wavefunction to diagonalize an operator. Maybe it should say the wave function is an eigenvector of every operator in a complete set of commuting operators?

    The book is "Introductory Quantum Mechanics" forth Ed. and the author is Liboff.

  2. jcsd
  3. Jan 25, 2009 #2
    Let me ask this again in a compact and maybe better-worded form:

    Does it make sense to say a (single) wave function diagonalizes a set of (many) operators? My textbook says this (see original post), but I think the author may be confused.
  4. Jan 25, 2009 #3
    When the author says that a wave function "diagonalizes a set of operators" he means that the state vector corresponding to that wave function is a simultaneous eigenvector of the operators in that set.

    Therefore the statement has content and can be interpreted, but it is also terribly sloppy to the point of being incorrect. This is a major problem with physics writing, and it gets worse as the topics become more advanced.
  5. Jan 25, 2009 #4
    That's also how I understand it. And I agree, current QM textbooks are junk.

    But what does this have to do with pure vs. mixed states? Maybe we are talking about density matrices...
  6. Jan 26, 2009 #5
    OK, thanks.

    The quote from the book occurred where Liboff was comparing pure states to mixed states, and it certainly reads as if he is saying a pure state is one which is an eigenvector of a complete set of commuting operators. If so, I believe again he is confused -- I believe what he is describing is a _DETERMINATE_ state with respect to these operators. The wording is so bad (and often wrong) that I really don't know what he is trying to say in a lot of these cases.
  7. Jan 26, 2009 #6


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    I don't have that book, so I don't know the larger context. But from the small extract
    you posted, I don't read it that way. Maybe if you post a larger extract I could try to say
    something more helpful.

    If you ever write any educational material yourself, you'll find that it's surprisingly
    hard to ensure that transmission from your mind to every reader's mind ends up clear.

    Moral: read multiple authors.
  8. Jan 26, 2009 #7
    The section is titled "Pure and Mixed States" and the author is discussing the differences. Sorry I don't have the book with me now, but he is definitely discussing differences between pure and mixed states surrounding this quote.

    I agree that it's not easy, but I have read many math/science books and cannot recall any with near the amount of incorrect statements, math errors, nonsensical statements (like the vector "diagonalizing a set of operators" above), and misleading statements as this one. If I didn't know better I would think someone just learning QM and linear algebra wrote this book.
  9. Jan 26, 2009 #8


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    That phrase should be understood in this way: the wavefunction determines a projection
    operator on the Hilbert space. Think of a ket [itex]|\psi\rangle[/itex] and the associated
    projection operator [itex]|\psi\rangle\langle\psi|[/itex]. The crucial point is that projection operators
    take you from the entire Hilbert space down to a subspace (in this case a very small subspace).
    When the "set of operators" you mentioned (I'll just call them "A" collectively) are restricted
    to this subspace, A acts diagonally -- meaning that A is equivalent to simple multiplication
    by a scalar.

    So that's what "diagonalizing an operator" means: when you restrict its domain to the
    particular subspace specified by that wavefunction, the operator is equivalent to scalar

    (This is actually a simple case of a deep mathematical result called the "Spectral Theorem".)

    It's the opposite, of course. People who know a great deal about a subject are not always
    the clearest teachers, having forgotten what it's like to be learning the subject for the
    first time.
  10. Jan 26, 2009 #9
    The projection operator for a vector v maps any other vector w to a multiple of v (Pw=cv), so the subspace you mention is just Span{v}. Saying that each operator in A acts diagonally on Span{v} is equivalent to saying that v is an eigenvector of these operators, which is exactly what we were saying above.

    Well, this is just an opinion, but I disagree strongly. I've read a few things from Feynmann, Einstein, Bell, ... and it's all very clear, concise, and accurate. When I've tried to understand fellow students in the classroom I got nowhere because of the inaccuracies, confused language, ....
  11. Jan 27, 2009 #10


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    Here we have an example of how what was in my mind failed to be transmitted into your mind
    without distortion. I said that not every expert makes a good teacher, but you seem to have
    heard "every expert is a bad teacher" -- which is not what I said. :-)

    Anyway, you seem to have the answer you were looking for. (If not, post a followup question.)
  12. Jan 31, 2009 #11
    Well I don't want to spend any more time on this, but for the record I said that when I see a poorly written book I assume it's written by someone that is not very familiar with the subject (poorlyExplained=>novice) and you said it's the opposite of that (poorlyExplained=>notNovice), so I said I disagree because I've observed plenty of poor explanations by novices new to a subject (not(poorlyExplained=>notNovice)).

    Anyway, the main issue is that everyone agrees that the book was trying to say that the wave function was an eigenvector of all the operators.
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