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Schrodinger Equation

  1. Oct 1, 2006 #1
    By noting that the time dependance of the wave function is governed by the Schrodinger equation show that

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

    not sure where to start on this one actually...

    do i start by differentiatng <x>wrt x and then wrt t??

    then i get
    [tex] \frac{d}{dt} \frac{d <x>}{dx} = \frac{d}{dt} (\Psi^* x \Psi |_{-\infty}^{\infty}) [/tex]
    am i heading in teh right direction here? Or am i totally off??

    ve been thinkin a little more and i was thinking that maybe i should differentiate <x> wrt t and and since we know that
    [tex] m \frac{d <x>}{dt} = <p> [/tex]

    i can equate the two
    i then get
    [tex] m \left[\int_{-\infty}^{\infty} \Psi x \frac{d\Psi^*}{dt} + x \Psi^* \frac{d \Psi}{dt} dx \right] = <p> = \int_{-\infty}^{\infty} \Psi^* \left( -i \hbar \frac{\partial}{\partial x} \right) \Psi dx [/tex]

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

    is THis the right track??
    Last edited: Oct 2, 2006
  2. jcsd
  3. Oct 2, 2006 #2
    Are you sure your [itex] \frac{d}{dx} [/itex] in the first equation isn't [itex] \frac{d}{dt} [/itex] and some i is missing?
  4. Oct 2, 2006 #3


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    Where did you get the problem from...?
  5. Oct 2, 2006 #4


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    Staff: Mentor

    I was wondering the same thing.

    See - http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/scheq.html#c2

    And I thought of the same question as Dexter - where did this problem arise or what textbook did one find the problem in the OP?
  6. Oct 2, 2006 #5
    i corrected the problem yes it was d/dt in the problem (i posted two possible ways i was thinking of solving it ..)

    This Problem is from Chapter 3, Problem 9 of Introduction to Quantum Mechanics by A. C. Phillips
  7. Oct 2, 2006 #6
    Write down the derivative and think of the Schrödinger-Equation, then. ;) There's still an i (imaginary unit) missing, I think ...
    Last edited: Oct 2, 2006
  8. Oct 2, 2006 #7
    thje derivative with respect to time? or x??
  9. Oct 2, 2006 #8
    is this not it??
    [tex] \frac{d}{dt} \frac{d <x>}{dx} = \frac{d}{dt} (\Psi^* x \Psi |_{-\infty}^{\infty}) [/tex]

    or this??
    [tex] m \frac{d <x> }{dt} = m \left[\int_{-\infty}^{\infty} \Psi x \frac{d\Psi^*}{dt} + x \Psi^* \frac{d \Psi}{dt} dx \right] [/tex]

    thats the derivative of <x> wrt time in the second...
  10. Oct 2, 2006 #9
    No, you just have to take the derivative wrt time of the function [itex] (\Psi x \Psi^*) [/itex] and remember that [itex] \Psi [/itex] solves the Schrödinger-Equation (and [itex] \Psi^* [/itex] the "complex conjugate Schrödinger-Equation" ...)
  11. Oct 2, 2006 #10
    ok this gives
    [tex] \frac{d}{dt} (\Psi^* x \Psi) = \frac{d \Psi^*}{dt} x \Psi + \Psi^* \frac{dx}{dt} + \Psi^* x \frac{d \Psi}{dt} [/tex]

    x doesnt depend on time... so dx/dt = 0

    [tex] \frac{d}{dt} (\Psi^* x \Psi) = \frac{d \Psi^*}{dt} x \Psi + \Psi^* x \frac{d \Psi}{dt} [/tex]

    the time dependance of the wave function is governed by the Schrodinger equation...
    yesss [tex] H \Psi = i \hbar \frac{d \Psi}{dt} [/tex]

    for the conplex conjugate
    [tex] H \Psi^* = \hbar \frac{d \Psi}{dt} [/tex]
    im guessing this one... complex conjugate should just mean the negative sign??
    Last edited: Oct 2, 2006
  12. Oct 2, 2006 #11
    Ah, I think you were working with what I know as "Ehrenfest's Theorem". The Schrödinger Equation is this: [itex] i \hbar \frac{\partial}{\partial t} \Psi(\vec x, t) = \left(
    -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V(\vec x) \right) \Psi(\vec x, t) [/itex].
    Last edited: Oct 2, 2006
  13. Oct 2, 2006 #12
    Right, but you have conjugate it all and write out the Hamiltonian to see something. And don't forget about "i" just because it's not real, it will get mad if you do to often! :wink:
    Last edited: Oct 2, 2006
  14. Oct 2, 2006 #13
    so i have to ask.. why doesnt x depend on t?? Is this because we have made the assumption??
  15. Oct 2, 2006 #14
    conjugate it all... now i remember that a conplex conjugate you simply change the sign between the real and imaginary term.. but here its harder to see
    simply putting a negative sign in from the imaginary term would do it then?
    taht is

    [tex]- i \hbar \frac{\partial}{\partial t} \Psi(\vec x, t) = \left(
    -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V(\vec x) \right) \Psi(\vec x, t) [/tex]
  16. Oct 2, 2006 #15
    [itex] x [/itex] is an operator which is used to calculate the expectation value - the expectation value may change in time, but not the way it is calculated! If the expectation value changes, it is due to the wavefunction which changes in time. In fact, this is the perspective of the so-called "Schrödinger Picture", which usually is presented first in the lectures. There also is a "Heisenberg Picture" in which the operators change in time, but the wavefunction doesn't. So x doesn't depend on t because Mr. Schrödinger said "In MY picture we will put the time-dependence into the wavefunction, not the operators, for god's sake." (Actually, I don't think this was an historical event. :wink:)
  17. Oct 2, 2006 #16
    No, you have to conjugate [itex] \Psi [/itex] as well. Therefore [itex] \Psi \Rightarrow \Psi^* [/itex].
  18. Oct 2, 2006 #17
    so it bascially a general assumption?? If we were doing amtrix mechanics it would different then??

    How come it isnt taught at the undergrad level though??
  19. Oct 2, 2006 #18
    There always has to be time-dependence in some way. In fact, the Schrödinger and Heisenberg-Pictures are just different ways of writing down the same mathematics. There is a magical operator, call it U(t), which creates the time dependent wavefunctions out of the wavefunctions at t=0. So,

    [itex] \Psi(\vec x, t) = U(t,0) \Psi(\vec x, 0) [/itex]

    The time-dependent expectation value in bra-ket notation for, say, x would then be
    [itex] \langle \Psi(\vec x, t) | x | \Psi(\vec x, t) \rangle = \langle \Psi(\vec x, 0) | U^\dagger(t) x U (t) |\Psi(\vec x, 0) \rangle [/itex]

    So, in the Schrödinger Picture the time-independent operator is just x, but in the Heisenberg Picture, where wavefunctions are static are operators depend on time, it would be [itex] U^\dagger(t) x U(t) [/itex].
    Last edited: Oct 2, 2006
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