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Observables and eigenvalues

  1. Jun 27, 2011 #1
    i have been trying to learn a bit of quantum mechanics,this is some thing that has been bothering me ,
    if the states of a system can be expressed as vectors in the Hilbert space,what is the motivation behind saying that physical observables can be given by operators?even then ,how can we say that the values they take are given by their eigenvalues?can this be proved somehow or reasoned out?
     
  2. jcsd
  3. Jun 27, 2011 #2

    Matterwave

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    This is a postulate of QM, and as a postulate, it is not proven.
     
  4. Jun 27, 2011 #3
    if its a postulate then what motivates it?
     
  5. Jun 27, 2011 #4
    the motivation is purely mathematical. in Hilbert space, you can think of physical observables as matrix transformations. it so happens that a matrix transformation has a set of eigenvalues and eigenfunctions defined for it mathematically. these then corresponde to the possible values of the observable and the set of eigenstates of the system
     
  6. Jun 27, 2011 #5
    Hi Ardie,

    Can you give a a physical example of matrix transofrmation or how it may represent a physical process other than observables? Thanks
     
  7. Jun 27, 2011 #6
    for example the hamiltonian in 1 dimension is a 2by1 matrix, that acts on a 1by1 vector (scalar psi) to give the scalar value of observable energy of the system (E)
     
  8. Jun 27, 2011 #7
    the act of transformation by a matrix is simply multiplying the vector by a matrix. you do this when u operate on psi with d/dx or d/dt or multiply etc, all these operations can be encoded into a matrix relation, and hence the analogy to Hilbert space.
     
  9. Jun 27, 2011 #8
    I think the motivation came from the historic development of QM as the matrix mechanics of Heisenberg, Born and Jordan. Namely, the canonical Poisson bracket of two "observables" [itex]f(q, p)[/itex] and [itex]g(q, p)[/itex]:
    [tex]
    \left\{f, g\right\} \equiv \sum_{k = 1}^{s}{\left(\frac{\partial f}{\partial q_{k}} \, \frac{\partial g}{\partial p_{k}} - \frac{\partial f}{\partial p_{k}} \, \frac{\partial g}{\partial q_{k}}\right)}
    [/tex]
    "goes over to" a commutator:
    [tex]
    -\frac{i}{\hbar} \, \left[F, G\right]
    [/tex]
    The simplest objects that do not commute are matrices. That is why the observables are correspondent to matrices and the new mechanics was called matrix mechanics.

    The equation of evolution for a classical observable [itex]f(q, p)[/itex] is given by a Poisson bracket:
    [tex]
    \frac{d f}{d t} = \left\{f, H\right\}
    [/tex]
    where [itex]H = H(q, p)[/itex] is the Hamiltonian of the system, should go over to the Heisenberg equation of motion:
    [tex]
    \frac{d F}{d t} = -\frac{i}{\hbar} \, \left[F, H\right]
    [/tex]

    Every matrix can be written as:
    [tex]
    F(t) = U^{-1}(t) \, f \, U(t)
    [/tex]
    where [itex]f[/itex] is a diagonal matrix with the eigenvalues as the diagonal elements and the time evolution is given by the evolution matrix [itex]U(t)[/itex]. The above equation of motion is then equivalent to:
    [tex]
    \frac{d U(t)}{d t} = -\frac{i}{\hbar} \, H \, U(t), \ \frac{d U^{-1}(t)}{d t} = \frac{i}{\hbar} \, U^{-1}(t) \, H
    [/tex]
     
    Last edited: Jun 27, 2011
  10. Jun 27, 2011 #9
    Thank you ardie.
     
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