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Schrodinger Equation in a representation

  1. May 31, 2012 #1
    The general evolution of a ket [itex]|\psi\rangle[/itex] is according to
    [itex]-i\hbar\frac{d}{dt}|\psi\rangle=H|\psi\rangle[/itex]
    without specifying a representation.

    From this equation, how can you simply get a equation in a certain representation [itex]F[/itex] as below:
    [itex]-i\hbar\frac{\partial}{\partial t}\langle f|\psi\rangle=\langle f|H|\psi\rangle[/itex] ?

    doesn't it need the validity of
    [tex]\langle f|\frac{d}{dt}|\psi\rangle=\frac{∂}{∂t}\langle f|\psi\rangle[/tex]
    ?

    does this always hold for any ket and bra in a Hilbert space and its dual space?
     
  2. jcsd
  3. Jun 1, 2012 #2

    dextercioby

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    As long as f is time independent, yes. This is true in the Schroedinger picture for example, where f can be x or p_x and as operators these are time independent.
     
  4. Jun 1, 2012 #3
    thank you for your reply, but I really think some condition is required.

    for example, look upon [itex]\langle f|[/itex] (independent of [itex]t[/itex]) as a eigenbra of [itex]F[/itex] whose eigenbras are continuous in [itex]f[/itex]. and [itex]\langle f|[/itex] acts on [itex]|\psi\rangle[/itex] as an linear functional.
    that is:
    [itex]\langle f|\psi\rangle=\psi (f,t)=\int^{+\infty}_{-\infty}\phi^*(x,f)\psi (x,t)dx[/itex]

    so, [itex]\langle f|\frac{d}{dt}|\psi\rangle=\frac{∂}{∂t}\langle f|\psi\rangle[/itex] in this case means:

    [itex]\int^{+\infty}_{-\infty}\phi^*(x,f)\frac{\partial}{\partial t}\psi (x,t)dx=\frac{\partial}{\partial t}\int^{+\infty}_{-\infty}\phi^*(x,f)\psi (x,t)dx[/itex]

    I mean, doesn't this equation (exchange of derivative and improper integral) require some condition, such as the uniform convergence of the improper integral[itex]\int^{+\infty}_{-\infty}[/itex]?
     
  5. Jun 1, 2012 #4

    strangerep

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    One way to deal with this more rigorously is to work in the context of so-called "rigged Hilbert space" (Gel'fand triples) -- if you're not familiar with these terms, think "generalized functions" or "distributions". Derivatives are then typically interpreted as some kind of "weak derivative".
    http://en.wikipedia.org/wiki/Weak_derivative
     
  6. Jun 8, 2012 #5
    The integral here is a Lebesgue integral, not a Riemann integral, so I think interchange of derivative and integral is allowed.
     
  7. Jun 8, 2012 #6
    lugita15, I know little about Lebesgue integral.
    Be it a Lebesgue integral, interchange of derivative and integral is always allowed ?
    I read somewhere that Lebesgue integral is a generalization of Riemann integral, then if the interchange in Riemann integral does not hold for a certain integral, will it hold in Lebesgue integral ?
     
  8. Jun 9, 2012 #7

    strangerep

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  9. Jun 9, 2012 #8
    thank you.
     
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