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Free Particle and the Schroedinger Equation

  1. Oct 31, 2011 #1
    1. The problem statement, all variables and given/known data

    Consider the time-dependent one-dimensional Schroedinger Equation for the free particle, i.e. let the Potential be [itex]V(x)=0[/itex]. Consider a wave packet, i.e.
    [itex]\psi(x,t)=\int_{-\infty}^{\infty}=A(k)\exp[i(kx-\omega(k)t]dk[/itex].
    Consider especially the Amplitude distribution
    [itex]A(k)=\alpha\cdot\exp[-(k-k_0)^2\cdot d_0^2]\exp[-ikx_0][/itex]
    (then it says "Gaussian Wave Packet").

    For t=0, [itex]\psi(x,0)[/itex] is given by a simple expression. Calculate this [itex]\psi(x,0)[/itex] and sketch the associated probabilty density [itex]\rho(x,t=0)=|\psi(x,0)|^2[/itex]. What must α be, so that the wave function is normalized, i.e. [itex]\int_{-\infty}^{\infty} \rho(x,t=0)dx=1[/itex]? What is the physical meaning behind "the parameters" [itex]k_0, x_0, d_0[/itex]?

    Hint: [itex]\int_{-\infty}^{\infty} \exp[-(s+c)^2]ds=\sqrt{\pi}[/itex] for all [itex]c\in[/itex] ℂ

    2. Relevant equations

    Schroedinger's Equation

    3. The attempt at a solution
    Well, this is a big one. First of all I tried the final question. We were taught that there is a "k-Space" and a "position space" in which the probability density is a gaussian bell curve. x0 should be the highest point of the curve in the position space, whereas k0 is the corresponding amount in the k-space. Also, when I look at the lecture notes, k0 is a vector which corresponds to a wave with a certain momentum. Somehow I don't think that this explaination is too good. What do you think?

    Now for the first question. I tried to insert the amplitude distribution into the wave function and then tried to combine the exponential functions. That looked like this:

    [itex]\psi(x,0)=\int_{-\infty}^{\infty}{\alpha\cdot\exp[-(k-k_0)^2\cdot d_0^2]\exp[-ikx_0]\exp[i(kx)]dk}[/itex]

    [itex]\psi(x,0)=\int_{-\infty}^{\infty}{\alpha\cdot\exp[-(k^2-2kk_0+k_0^2)\cdot d_0^2]\exp[-ikx_0]\exp[i(kx)]dk}[/itex]

    [itex]\psi(x,0)=\int_{-\infty}^{\infty}{\alpha\cdot\exp[k^2d_0^2+2kk_0d_0^2+k_0^2d_0^2]\exp[-ikx_0]\exp[i(kx)]dk}[/itex]

    And finally arrived at
    [itex]\psi(x,0)=\int_{-\infty}^{\infty}{\alpha\cdot\exp[k^2d_0^2+2kk_0d_0^2+k_0^2d_0^2-ikx_0+ikx]dk}[/itex]

    I was told that I would now have to use a completing the square to bring that into a form [itex]Ak^2+Bk+C[/itex], so that I could use the "hint" given in the task. But I got some kind of mental block there. Could someone please help me?

    Final question: Is it true that when we look for α, all that has to be done is to set the Integral of ρ multiplied by α equal to 1 and α then is like 1/ρ ?
     
  2. jcsd
  3. Oct 31, 2011 #2
    An addendum, just received an E-Mail from the professor, who said that due to many people having difficulties, he would give the solution for ψ, and he said it looks like this:

    [itex]\psi(x,0)=\frac{1}{\pi^{1/4}\cdot 2^{1/4}\cdot d_0^{1/2}}\exp[i\cdot k_0(x-x_0)]\exp[-\frac{(x-x_0)^2}{4d_0^2}][/itex]

    When I use that, should the probability density be:

    [itex]\rho(x,0)=\frac{1}{\sqrt{2\pi}d}\exp[2ik(x-x_0)]\exp[-\frac{(x-x_0)^2}{2d^2}][/itex]

    ?
     
  4. Oct 31, 2011 #3

    vela

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    Use the substitution k' = k-k0 first. That'll simplify the algebra quite a bit.
    You simply set the integral of ρ equal to 1 and solve for [itex]\alpha[/itex].

    Almost. The probability density isn't ψ2; it's |ψ|2*ψ.
     
  5. Oct 31, 2011 #4
    Ah, thank you! When I take the complex conjugate of ψ for taking the square, then I get

    [itex]\rho(x,t=0)=\frac{1}{\sqrt{2\pi}d}\exp[-\frac{(x-x_0)^2}{2d^2}][/itex].

    Looks like a gaussian bell curve to me from the form of the equation. Could that be it?
     
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