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I shot schrodinger's cat and then I ate it

  1. Oct 18, 2005 #1
    I'm currently reading up about solving the schrodinger equation for a one-dimensional particle in a one-dimensional box with an infinite level of potential outside the box, and I have a *slight* problem (no, really, it is very slight)

    I've got to the equation where the Wavefunction = Acos(2*x*pi/lambda - omega*t). And the text I'm working on says that using the planck relationship omega is equal to energy? ey!:grumpy: I was under the impression that omega equalled angular velocity (not that I have a clue why a wave would have angular velocity, but that's another story).

    Besides, I'm not entirely sure how that equation is in the form it is - the wave equation arrived at by modelling a travelling 1-d wave comes to Asin(2*pi/lambda(x ± vt). Firstly, how does that get to a cos expression (is that just coming from using different initial conditions where in the first case x = A at t = 0 (well, psi, or whatever) and x = 0 at t = 0 in the second? That leaves the omega*t bit of the first equation - I'm assuming that is equivalent to 2*pi*vt/lambda from the traveling wave equation - how do those two relate?

    And finally where did the minus sign go from the second equation?

    Blech, sorry for the convoluted post, I'm not really this confused! honest!:uhh:

    Cheers,
    Just some guy
     
  2. jcsd
  3. Oct 18, 2005 #2

    ZapperZ

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    You really have not provide sufficient information for anyone (at least for me) to accurately address this specific problem. You present a rather specific form of the answer or end result, but you left out important info on the nature of the problem.

    [tex] E = hf[/tex]

    But [tex] f = \frac{\omega}{2 \pi}[/tex]

    So [tex] E = \frac{h \omega}{2 \pi}[/tex]

    or [tex] E = \hbar \omega[/tex]

    This means that [tex]\omega[/tex] is equivalent to energy except for a planck constant.

    This is where you left out quite a bit of info. The form of the solution to the wavefunction depends on where you choose x=0. If you choose it to be at one of the walls, then your solution will always be of the sine form due to the boundary condition. But if you choose to be at the center of the well, then it will be an alternating sine and cosine functions.

    I'm a bit confused when you indicated that the wave equation was "... arrived at by modelling a travelling 1-d wave...". Did you not solve this from the Schrodinger equation? If you did, then there is no "modelling" of any kind. It's just solving the differential equation via brute force and that's that.

    Zz.
     
  4. Oct 18, 2005 #3
    Well, the text I'm reading got the equation from solving the wave equation for an ideal string: d^2y/dx^2 = (p/t)d^2y/dt^2, to which the solution is y(x,t) = Asin((2pi/lambda)(x±vt))
     
  5. Oct 18, 2005 #4

    ZapperZ

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    Yes, now look at the time-independent schrodinger equation for V=0, and see if you see the connection. Then look at the boundary conditions.

    Zz.
     
  6. Oct 18, 2005 #5
    you mean -h-bar^2/2m*((d^2)psi/dx^2) = E*psi? Sorry, I don't see the relation - does h-bar/2m relate to p/t?
     
  7. Oct 18, 2005 #6

    ZapperZ

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    Er.. I goofed. I should have said time DEPENDENT schrodinger equation.

    Don't pay attention to the h-bar, etc... those are just constants. Pay attention to the FORM of the differentials.

    Zz.
     
  8. Oct 20, 2005 #7
    ok, so the partial derivatives are sort of equivalent - I don't get the connection though.

    Basically I don't get how one reaches the wavefunction for an electron (=Acos((2*pi*x/lambda) * (omega*t)) from the expression for a travelling wave in one dimension - y(x,t) = Asin(2*pi/lambda)*(x±vt)

    The opening text says "Starting with the expression for a traveling wave in one dimension, the connection can be made to the Schrodinger equation." - I just don't get how :grumpy:
     
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