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QM Notation questions

  1. Sep 30, 2008 #1
    Ok, so I have some QM notation questions. I am not a physics student so I am not 100% sure what all the different symbols mean. I am working from Lie Algebras and Particle Physics by Georgi and he does not offer a key for all the symbols.

    Let A be a matrix representing a linear operator. Ok, what is the difference between A* and A^(dagger looking symbol).

    I've seen A* mean the conjugate transpose and A-dagger to be conjugate transpose. I've also seen A* to be transpose.

    So in terms of QM, what do these notations mean? What about when you are taking the trace of A, tr(A), but then you put tr(A)*??

    Thanks
     
  2. jcsd
  3. Sep 30, 2008 #2
    The star (*) is usually the complex conjugate: (a + bi)* = a - bi for all real a, b. It's not normally used on operators.

    The dagger is the conjugate transpose of a matrix. It is a generalization (of sorts) of the complex conjugate to operators on hilbert spaces. It's not usually used on scalars.
     
  4. Oct 2, 2008 #3
    Are you taking a class or just reading Georgi's book for fun? Because I took his class (and used his textbook), and it was, um, hard :wink:
     
  5. Oct 3, 2008 #4
    Ok, I'd like to start by saying that I'm in the same boat entirely as you! I don't understand any of this crazy mathsy language. But in my efforts to try to understand, I've grasped some rough concepts. I'd like to outline my understanding, though it's almost certainly wrong, so that someone can either say that I'm right (doubtful) or maybe correct where I'm going wrong.

    As I see it, every observable property in quantum mechanics, such as position, momentum, energy, angular momentum, can be found by applying what is known as a "self-adjoint operator" on a Hilbert space. This hilbert space is constructed from bases. I don't know if these bases are functions or unit vectors in the space or if these are equivalent. Anyway, these operators, which I've always had told to me are derivatives (like [tex]\mathbf{p}={\hbar\over i}\nabla[/tex]), have associated matrices. These matrices are Hermitian, that is to say the transpose of the matrix is the same as the complex conjugate of the matrix... Transpose is that funny dagger. Quoting wikipedia:-

    I don't understand this at all. Is this to say that the star means take the transpose of the matrix, then take the complex conjugate. Is this the same (for a Hermitian matrix) as taking the complex conjugate twice, which is the same as transposing the matrix twice? Am I understanding correctly?

    How is it that an operator has an associated matrix? Is it that the operators have eigenfunctions and eigenvalues in the same way as matrices have eigenvectors and eigenvalues. Thus a (wave?) function is a vector in Hilbert space? This is foggy at best. Is this all down to the two different formulations of QM by Schrodinger and Heisenberg? Can I look at the two formulations as the same? Or are they different beasts which I have to keep distinct?

    If someone could clear all this up, I'd love them forever! Also, how does all this relate to bras and kets? I've never been formally taught dirac notation and I've just moved to France where they use it everywhere. In England, they just avoided it, prefering operators and trying not to confuse us with the maths, just making sure that we understand the concepts... Help!

    http://en.wikipedia.org/wiki/Self-adjoint_operator
    http://en.wikipedia.org/wiki/Hermitian_matrix
     
    Last edited: Oct 3, 2008
  6. Oct 4, 2008 #5
    In order to find the conjugate transpose of the matrix, you take the complex conjugate of each element in the matrix, then flip the elements across the leading diagonal. So in the matrix you include in your post you take the complex conjugate of each element. Only the 2+i and 2-i terms are changed. The 2+i becomes 2-i and the 2-i becomes 2+1. The positions of these elements are then swapped giving you exactly what you had to begin with.

    Matrices that behave in this way are called Hermitian or Self-Adjoint.

    Operators do indeed have eigenfunctions and eigenvalues in exactly the same way as a matrix does. The eigenfunctions, however, are referred to as eigenstates and represent the possible states of the system - note that the system is not restricted to being in a single eigenstate though, it can be in a superposition of various eigenstates.

    A wave function is not a vector in a Hilbert Space, the wave function is typically less used in this formulation of Quantum Mechanics and you are correct that the difference is down dealing with different methods of treating Quantum Mechanical systems. It is through Hilbert Spaces and Dirac Notation that we can see the matrix formulation and the wavefunction formulation of Quantum Mechanics are actually equivalent.

    [tex] <x| \Psi> = \Psi (x) [/tex]

    Dirac notation simply allows you to easily represent the state of some system as a "ket" [tex] | \Psi > [/tex] is an example of such a ket. For every "ket" there is an associated "bra," such as [tex] < \Psi | [/tex], which is the conjugate transpose of the ket. Observables, what you can measure about the system, are expressed as linear hermitian operators acting upon the ket.

    I hope that helps to clear things up for you.
     
  7. Oct 4, 2008 #6

    George Jones

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    A wave function is a vector in the Hilbert space of (equivalence classes of) Lebesgue square-integrable functions.
     
  8. Oct 4, 2008 #7
    Hi Jason, I know you are an experienced math student, from this perspective you should think of [itex]A^{\dagger}[/itex] as the adjoint of [itex]A[/itex] i.e. the Reisz representation theorem establishes a dual correspondence between each operator in a Hilbert space and an operator in the dual of that space, the adjoint of the original operator. This is a generalization of the conjugate transpose for matrices.

    Not to start a debate over semantics, but there is a subtlety here at this level. Quoting Sakurai's Modern QM pg. 52:

    Note that the probability of being in a particular state is always given by the square of such an expansion coefficient in that basis (in this case position), not by the state vector itself (which does belong to a Hilbert space).
     
  9. Oct 4, 2008 #8

    George Jones

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    Yes, the mathematics of quantum mechanics, treated in a mathematically honest way, is quite subtle, and Sakurai hardly touches on these subtleties. See

    https://www.physicsforums.com/showthread.php?p=1883612#post1883612

    https://www.physicsforums.com/showthread.php?p=1889443#post1889443.

    I stand by my statement that a wave function is a vector in the Hilbert space of (equivalence classes of) Lebesgue square-integrable functions.
     
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