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QM: The time evolution of a Gaussian wave function

  1. Mar 19, 2017 #1
    Heads up, I only recently got into quantum mechanics and don't feel like I got a solid grasp on the material yet.

    1. The problem statement, all variables and given/known data

    Given is the wave function of a free particle in one dimension:
    \begin{equation}
    \psi(x,0) = \left( \frac{2}{\pi a^2} \right)^{1/4} e^{i k_0 x} e^{-x^2/a^2}
    \end{equation}
    I already calculated the Fourier-transform (necessary in another part):
    \begin{equation}
    g(k) = ( 2 \pi )^{-1/4} \sqrt{a} e^{-(k_0 - k)^2 a^2/4}
    \end{equation}
    Now I am tasked with calculating the time evolution ψ(x,t).

    2. Relevant equations

    We were given hints by tutors for using the time-operator:
    \begin{equation}
    |\psi (x, t)> = e^{-i / \hbar \hat{H} t } | \psi(x,0) >
    \end{equation}
    Thus, we would only need to find the eigenvalue of the hamiltonian:
    \begin{equation}
    \hat{H} | \psi(x,0)> = \lambda | \psi(x,0) >
    \end{equation}
    However, against my intuition of just putting in my original wave function and calculating away ( which yields a λ depending on x ), they told us to use the Fourier-transform of g(k) instead:

    \begin{equation}
    \hat{H} | \psi(x,0)> = \hat{H} \frac{1}{\sqrt{2 \pi}} \int g(k) e^{ikx} dk
    \end{equation}

    3. The attempt at a solution

    I tried calculating it the way they told us to, but the resulting integral seems endless and doesn't come to a head after hours of trying. Even if i were to finish it, it seems to me that the final term would still depend on x. I am also not that skilled in calculating eigenvalues, so allow me to ask: would it be fine if the eigenvalue for the Hamiltonian depends on x ( doesn't seem like it to me )? I'd also appreciate someone explaining to me why I'd have to use the double transformed function instead of the original one.

    Thanks in advance.
     
  2. jcsd
  3. Mar 19, 2017 #2

    vela

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    It doesn't make sense to write that since ##|\psi(x,0)\rangle## isn't an eigenstate of the Hamiltonian,.

    What are the eigenstates of the Hamiltonian? Once you identify those, it might make more sense why it's been suggested you use the Fourier transform.
     
  4. Mar 19, 2017 #3

    blue_leaf77

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    No, the initial wavefunction is clearly not an eigenfunction of free space Hamiltonian.
    Your starting point is the equation (3) but I would rewrite it in a more proper way.
    $$
    |\psi(t) \rangle = e^{iHt/\hbar} |\psi(t=0) \rangle
    $$
    where ##H = \frac{\hat p ^2}{2m}##. Since the Hamiltonian is a function of only momentum operator, it will be helpful to use closure relation in momentum space ##I = \int |p\rangle \langle p| dp## where ##I## is the identitty operator. Now insert it to the right of the time propagation operator to get
    $$
    |\psi(t) \rangle = \int e^{ip^2t/(2m\hbar)} |p\rangle \langle p|\psi(t=0) \rangle dp
    $$
    Now ##p## in the exponential factor is just a number not an operator as it was before. Since you want the state in position space, you need to project it with ##\langle x|## to get
    $$
    \langle x|\psi(t) \rangle = \int e^{ip^2t/(2m\hbar)} \langle x|p\rangle \langle p|\psi(t=0) \rangle dp
    $$
    You have calculated ##\langle p|\psi(t=0) \rangle## as ##g(k)## above (note that ##p=\hbar k##).

    Note: didn't realize vela has posted his answer while I was writing mine.
     
  5. Mar 20, 2017 #4
    Thanks for the replies, now it actually starts to make sense.

    Regarding your reply, blue_leaf77, I generally understand your approach but am wondering what exactly I have to write for ⟨x|p⟩. Would it be fine to just write it as the inner product
    \begin{equation}
    \langle x | p \rangle = \hat{X} \hat{P}
    \end{equation}
    since the particle only exists in one dimension?
     
  6. Mar 20, 2017 #5

    blue_leaf77

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    ⟨x|p⟩ is the momentum eigenstate projected into position space, in other words it's the eigenfunction of the operator ##-i\hbar \frac{d}{dx}##, shouldn't be too hard to calculate it or just make advantage of its popularity by searching in the internet.
     
  7. Mar 20, 2017 #6
    Ah, obviously. Thanks for the help, you cleared a lot of things up for me.
     
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