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4D Schrodinger equation

  1. Sep 28, 2014 #1
    This is a rather naive question concerning the dimension of the schrodinger equation. If the Schrodinger eqation can be wrtiten in a three dimensional form using the laplacian operator can it be writen in a 4d version. I understand that the schrodinger equation shows the developement of the state of a particle in time... as I said naive!

    s3d1.gif
     
  2. jcsd
  3. Sep 28, 2014 #2

    Orodruin

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    The answer you are looking for would depend on what you consider your four dimensions to be. Is it the space-time of special relativity? Four spatial dimensions? (Say, four particles moving in 1 dimension) Something completely different?
     
  4. Sep 28, 2014 #3
    I meant a wave something along the lines of psi(x,y,z,t). Would that give me the evolution of the particles state in a 5 dimensional system???
     
  5. Sep 28, 2014 #4

    jtbell

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    Do you mean the time-dependent Schrödinger equation (TDSE)?
    $$-\frac{\hbar^2}{2m} \left( \frac{\partial^2 \Psi}{\partial x^2} + \frac{\partial^2 \Psi}{\partial y^2} + \frac{\partial^2 \Psi}{\partial z^2} \right) \Psi(x,y,z,t) + U(x,y,z,t) \Psi(x,y,z,t) = i \hbar \frac {\partial}{\partial t} \Psi(x,y,z,t)$$

    When V is independent of t, that is, U(x,y,z,t) = U(x,y,z), the solutions of the TDSE are of the form ##\psi(x,y,z)e^{-iEt/\hbar}##, where ##\psi(x,y,z)## satisfies the time-independent Schrödinger equation (TISE), which you gave in your first post.
     
  6. Sep 28, 2014 #5
    Yep...I mean can one expand the laplacian operator so that one has the deriviativ of x,y,z,t???
     
  7. Sep 28, 2014 #6

    Orodruin

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    The Klein-Gordon equation ##(\partial_t^2 - \nabla^2)\phi = 0## appears in relativistic quantum mechanics and involves the d'Alembert operator - which in some sense is the relativistic equivalent of the Laplace operator. However, note the difference in sign between the tempotal and spatial derivatives.

    In the non-relativistic limit, the KG equation gives back the Schrödinger behaviour.
     
  8. Sep 28, 2014 #7
    Something like this?

    (d^2/dt^2+d^2/dx+d^2/dy^2+d^2/dz^2)-(h/2pi)/2m psi(x,y,z,t)+V(x,y,z,t)psi(x,y,z,t)=?

    instead of

    upload_2014-9-28_17-35-41.png
     
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