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Schrodinger Equation

  1. Sep 28, 2009 #1
    Dear fellow quantum enthusiasts,

    My mathematical shape is a bit disastrous while I'm following a Quantum Physics course.
    I use several handbooks, and the image below is an excerpt from the David Griffiths Handbook, introduction to quantum mechanics.I put some notes aside, what I don't understand.And bye the way, these aren't the only things I dont understand, bu I start with this.For those who have the textbook, this is a piece of chapter 1.4 Normalization. Can someone clear things up?
    schrodinger%25unclear.JPG
     
  2. jcsd
  3. Sep 28, 2009 #2

    alxm

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    1.21 - you have a function of x and t, f(x,t). You integrate that function over all x, so you're left with a function of t alone, F(t). The derivative of F(t) with respect to t is an ordinary derivative, since it's a single-valued function. When you move it inside the integral (which is okay, since it is independent of t), you're dealing with the derivative of f(x,t) with respect to t, which is a partial derivative because it's a two-valued function. That's what it says.

    [tex]\psi[/tex] and [tex]\psi^*[/tex] are functions, not constants.

    1.25: Note that [tex]\frac{\partial}{\partial x}[\psi^*\frac{\partial\psi}{\partial x} - \psi\frac{\partial\psi^*}{\partial x}] = (\frac{\partial\psi^*}{\partial x}\frac{\partial\psi}{\partial x} + \psi^*\frac{\partial^2\psi}{\partial x^2}) - (\frac{\partial\psi}{\partial x}\frac{\partial\psi^*}{\partial x} + \psi\frac{\partial^2\psi^*}{\partial x^2}) = \psi^*\frac{\partial^2\psi}{\partial x^2} - \psi\frac{\partial^2\psi^*}{\partial x^2}[/tex]
    (chain rule)

    1.26: The complex conjugate of zero is zero. If [tex]\psi[/tex] goes to zero then [tex]\psi^*[/tex] goes to zero. And the partial derivatives don't matter - anything multiplied by zero is zero.

    (Okay this is slightly hand-waving; a math hard-liner might point out that it's a limit, so that it doesn't necessarily go to zero if the derivative diverges, but we happen to know it doesn't.)
     
    Last edited: Sep 28, 2009
  4. Sep 28, 2009 #3

    CompuChip

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    (1.25): There are no additional assumptions here, it is just a rewriting. Try working out the derivative using the product rule.

    (1.26): If [itex]\Psi[/itex] vanishes somewhere (like at plus or minus infinity) then clearly so does its complex conjugate, and so does [itex]\Psi[/itex] multiplied by anything (in particular, by [itex]\partial\Psi/\partial x[/itex])
     
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