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Derivation of the velocity of an expectation value

  1. May 31, 2010 #1
    1. The problem statement, all variables and given/known data
    I am trying to derive for myself the velocity of the expectation value from the information given, specifically that

    [tex]<x> = \int_{-\infty}^{\infty}x|\Psi (x,t)|^2 dt[/tex] (1)

    Eq (1) can be transformed into,

    [tex]\frac{d<x>}{dt} = -\frac{i\hbar}{m}\int\Psi^*\frac{\partial\Psi}{\partial x}dx[/tex] (2)

    I couldn't come up with Eq. (2). Below is my work.

    2. Relevant Equations.

    Schr[tex]\ddot{o}[/tex]dinger Equation
    [tex]i\hbar\frac{\partial\Psi}{\partial t}=-\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2}+V\Psi[/tex]

    3. The attempt at a solution

    Taking the derivative of both sides of Eq. (1) results in,

    [tex] \frac{d<x>}{dt} = \int x\frac{\partial}{\partial t}|\Psi|^2 dx[/tex] (3)

    As [tex]\Psi^2 = \Psi^*\Psi[/tex] taking the derivative of the r.h.s of Eq.(3) results in,

    [tex]\int x\left(\Psi^*\frac{\partial\Psi}{\partial t}+\Psi\frac{\partial\Psi^*}{\partial t}\right)[/tex] (4)

    Manipulating the Schr[tex]\ddot{o}[/tex]dinger Equation to get [tex]\frac{\partial\Psi}{\partial T}[/tex]

    [tex]-\frac{\partial\Psi}{\partial t} = \frac{\hbar}{2im}{\partial^2\Psi}{\partial x^2}-\frac{V\Psi}{i\hbar}[/tex] (5)

    Multiplying both sides of Eq. (5) by [tex]i^2[/tex],

    [tex]\frac{\partial\Psi}{\partial t} = \frac{i\hbar}{2m}\frac{\partial^2\Psi}{\partial x^2}-\frac{iV\Psi}{\hbar}[/tex] (6)

    Taking the complex conjugate of Eq. (6),

    [tex]\frac{\partial\Psi^*}{\partial t} = -\frac{i\hbar}{2m}\frac{\partial^2\Psi^*}{\partial x^2}+\frac{iV\Psi^*}{\hbar}[/tex](7)

    Substituting Eq. (6) & (7) into Eq. (4) results in,

    [tex]=\Psi^*\frac{i\hbar}{2m}\frac{\partial^2\Psi}{\partial x^2}-V\frac{i\Psi^*\Psi}{\hbar}-\Psi\frac{i\hbar}{2m}\frac{\partial^2\Psi^*}{\partial x^2}+V\frac{i\Psi^*\Psi}{\hbar}[/tex]

    [tex]=\Psi^*\frac{i\hbar}{2m}\frac{\partial^2\Psi}{\partial x^2}-\Psi\frac{i\hbar}{2m}\frac{\partial^2\Psi^*}{\partial x^2}[/tex]

    [tex]=\frac{i\hbar}{2m}\left(\Psi^*\frac{\partial^2\Psi}{\partial x^2}-\Psi\frac{\partial^2\Psi^*}{\partial x^2}\right)[/tex](8)

    Eq. (8) is equivalent to,

    [tex]\frac{i\hbar}{2m}\left[\frac{\partial}{\partial x}\left(\Psi^*\frac{\partial\Psi}{\partial x}-\Psi\frac{\partial\Psi^*}{\partial x}\right)\right][/tex](9)

    (The inside can be confirmed just by taking the derivative, which I'm going to leave out for brevities sake)

    So, the velocity of the expectation value now becomes,

    [tex]\frac{d<x>}{dt} = \frac{i\hbar}{2m}\int x\frac{\partial}{\partial x}\left(\Psi^*\frac{\partial\Psi}{\partial x}-\Psi\frac{\partial\Psi^*}{\partial x}\right)dx[/tex](10)

    Taking the constant out I solved the integrand by using integration by parts.

    Let [tex]u=x[/tex]
    [tex]dv=\frac{\partial}{\partial x}\left(\Psi^*\frac{\partial\Psi}{\partial x}-\Psi\frac{\partial\Psi^*}{\partial x}\right)dx.[/tex]

    Then, [tex]du=dx[/tex]
    [tex]v=\left(\Psi^*\frac{\partial\Psi}{\partial x}-\Psi\frac{\partial\Psi^*}{\partial x}\right)[/tex]

    Substituting this into [tex]\int udv = \int uv-\int vdu[/tex]

    [tex]x\left(\Psi^*\frac{\partial\Psi}{\partial x}-\Psi\frac{\partial\Psi^*}{\partial x}\right)\bigg|_{\infty}^{\infty}-\int \left(\Psi^*\frac{\partial\Psi}{\partial x}-\Psi\frac{\partial\Psi^*}{\partial x}\right)dx[/tex] (11)

    Before the integral that term goes to 0 because (as I'm told, I don't actually know how to prove this) [tex]\Psi[/tex] goes to zero must quicker then x goes to infinity. So the velocity becomes,

    [tex]\frac{d<x>}{dt} = -\frac{i\hbar}{2m}\int \left(\Psi^*\frac{\partial\Psi}{\partial x}-\Psi\frac{\partial\Psi^*}{\partial x}\right)dx[/tex](12)

    I transformed,

    [tex]\Psi\frac{\partial\Psi^*}{\partial x}=\frac{\partial\Psi^*\Psi}{\partial x}-\Psi^*\frac{\partial\Psi}{\partial x}[/tex](13)

    Substituting Eq (13) into Eq (12),

    [tex]\frac{d<x>}{dt} = -\frac{i\hbar}{2m}\int \left(\Psi^*\frac{\partial\Psi}{\partial x}-\frac{\partial\Psi^*\Psi}{\partial x}+\Psi^*\frac{\partial\Psi}{\partial x}\right)dx[/tex]

    [tex]=\frac{d<x>}{dt} = -\frac{i\hbar}{2m}\int \left(2\Psi^*\frac{\partial\Psi}{\partial x}-\frac{\partial\Psi^*\Psi}{\partial x}\right)dx[/tex](14)

    Eq.(14) Becomes

    [tex]\frac{d<x>}{dt} = -\frac{i\hbar}{m}\int \Psi^*\frac{\partial\Psi}{\partial x}dx+\frac{i\hbar}{2m}\int\frac{\partial\Psi^*\Psi}{\partial x}dx[/tex](15)

    The left part of the r.h.s. of Eq.(15) is what I want but I don't know how to eliminate the right portion of it.

    Anyone have any ideas??

    Thanks in advance!

  2. jcsd
  3. May 31, 2010 #2
    Eqn. (15) Looks incorrect. You should only have the first term on the rhs.

    Of course, the second term is an integral of a derivative of [itex] |\Psi|^{2} = \Psi^{\ast} \, \Psi[/itex], so it can be integrated. But, the wavefunction disappears at infinity, so the contribution from that term is zero.
  4. May 31, 2010 #3
    So, I can assume again that [tex]\Psi^2\Psi[/tex] goes to zero as it goes to infinity? If that's the case then thanks!! =)
  5. May 31, 2010 #4
    Of course. This is a necessary (but not sufficient) condition for the probability normalization integral:

    \int_{-\infty}^{\infty}{|\Psi(x, t)|^{2} \, dx} = 1

    to converge.
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