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Wave Function in Lorent tranform

  1. Jun 21, 2008 #1
    how to tranform wave function(x,t) to same coordinate wave function(x',t')
    with Lorentz Tranformation (please show all the calculus).

    and why we know the Lorentz tranform can do this function to use in every inertial frames.
     

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  3. Jun 21, 2008 #2

    Fredrik

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    It does seem to be possible to make sense of [itex]\psi(x,t)=\psi'(x',t')[/itex] using some relativistic quantum field theory, but is that really what you're looking for?

    The equation you posted looks like an attempt to find a relativistic version of the Schrödinger equation by making the substitutions

    [tex]E\rightarrow i\hbar\frac d{dt}, p\rightarrow -i\hbar\frac d{dx}[/itex]

    in the relativistic formula [itex]E^2=p^2c^2+m^2c^4[/itex] (with m=0) instead of in the non-relativistic formula [itex]E=p^2/2m[/itex]. If you do it in the non-relativistic equation, the result is the Schrödinger eqauation. It might seem plausible that you'd get a relativistic wave equation if you start with the relativistic expression for the energy instead.

    However, this doesn't work. The "psi" isn't going to be a wave function. It's a classical field, which needs to be "quantized" to a quantum field before we can do anything with it.

    Here's how I would make sense of [itex]\psi(x)=\psi'(x')[/itex] (where x now represents all the coordinates including time), if we assume that the "wave function" represents a one-particle state in the quantum field theory of a (not necessarily massless) non-interacting scalar field:

    [tex]\psi(x)=\langle 0|\phi(x)|\psi\rangle=\langle 0|U(\Lambda)^\dagger U(\Lambda)\phi(x)U(\Lambda)^\dagger U(\Lambda)|\psi\rangle=\langle 0|\phi(\Lambda x)|\psi'\rangle=\psi'(x')[/tex]
     
  4. Jun 21, 2008 #3
    I don't think the OP is using the term 'wave function' in a quantum mechanical sense, since the attached thumbnail is the classical wave equation.

    The definition of the lorentz transformation is such that psi(x,t) = psi(x',t') i.e. events occurring at x,t are seen as occurring at x',t' in the other inertial frame.
     
  5. Jun 21, 2008 #4

    Fredrik

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    D'oh. Well, at least it would have been a good answer in the quantum mechanics forum. :smile:
     
  6. Jun 22, 2008 #5
    thank you for answer

    My mean is how to tranform change the function(x,t) to function(x',t')
    My teacher tell me it can do by calculus
    He sample a start is use chain rule differential calculus

    dU/dx' = (dU/dx)(dx/dx') + (dU/dt)(dt/dx')

    and I no idea
    why in the end ddU/dx'x' = (1/c^2)(ddU/dt't')

    please show me if it true.
     

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  7. Jun 22, 2008 #6

    Mentz114

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    Start with the Lorentz transformation

    [tex]x' = \gamma x - \beta\gamma t[/tex]
    [tex]t' = \gamma t - \beta\gamma x[/tex]

    now calculate

    [tex] \frac{dx}{dx'} , \frac{dt}{dt'}[/tex]

    and apply

    [tex]\frac{d\Psi}{dx'} = \frac{d\Psi}{dx}\frac{dx}{dx'} + \frac{d\Psi}{dt}\frac{dt}{dx'}[/tex]

    then differentiate again. Repeat for t'. I think that ought to work, but I haven't time to do the whole calculation.
     
    Last edited: Jun 22, 2008
  8. Jun 22, 2008 #7
    Thank you for answer

    Now I can do it ,thank you for all thinking

    if you want see answer i can post but you will wait a many time.

    if you want see please tell me.
     
  9. Jun 23, 2008 #8
    wave function and LT

    You can see a transparent approach in
    C.Moller
    The Theory of Relativity
    Clarendon Press 1952 p.56 Ch.23
     
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