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Particle in a 1D potential V(x)

  1. Jan 13, 2016 #1
    1. The problem statement, all variables and given/known data
    There's a particle moving in a 1D potential V(x) with mass m. The particle's normalised wavefunction is ψ(x,t). Use the time dependent Schrodinger equation to show that ##\frac{\partial{\rho}}{\partial{t}} + \frac{\partial{j}}{\partial{x}} = 0##
    Where
    ##j(x,t) = -\frac{i\hbar}{2m}(\psi^{*} \frac{\partial{\psi}}{\partial{x}} - \psi \frac{\partial{\psi}}{\partial{x}})##

    I also have to show that j(x,t) is real. All I know about j is that it has to be equal to the magnitude of ψ(x,t)##^{2}##.

    2. Relevant equations
    Time dependent Schrodinger equation:
    ##i\hbar \frac{\partial{\psi}}{\partial{t}} = -\frac{\hbar^{2}}{2m} \frac{\partial^{2}{\psi}}{\partial{x^2}} + V(x) \psi##

    The probability density ρ = |ψ(x,t)|^2 or ψψ*.

    3. The attempt at a solution
    I'm having a bit of trouble with the calculus element. Pretty sure I'm differentiating wrong AND integrating wrong, but anyway, here's what I got:
    First I thought I'd calculate ##\frac{\partial{j}}{\partial{x}}## since j is given. I got
    ##-\frac{i\hbar}{2m}(\psi^{*}\frac{\partial^{2}{\psi}}{\partial{x^2}}+\frac{\partial{\psi}}{\partial{x^2}}\psi^{*}-\psi\frac{\partial^{2}{\psi^{*}}}{\partial{x^2}}+\frac{\partial{\psi}}{\partial{x}}\frac{\partial{\psi}}{\partial{x}})##.

    Then I thought I would solve the Schrodinger equation. But (and I know this isn't exactly a good reason) the question is only worth five marks! Which makes me think that I might not have to solve the Schrodinger equation. And I wouldn't actually know how to solve it anyway...

    So is my first calculation right? And do I need to solve the Schrodinger equation in order to answer this question? If I do... how do I do it??
     
    Last edited: Jan 13, 2016
  2. jcsd
  3. Jan 13, 2016 #2
    Why do you think solving the Schrodinger equation would be useful? Besides, it is not possible to solve the Schrodinger equation in closed form for general [itex]V(x)[/itex]

    There is a mistake in your expression for the current density. There should be a complex conjugation on the second term: [tex]j(x,t) \equiv -\frac{i\hbar}{2m} \left(\psi^{*} \partial_{x} \psi - \psi \partial_{x} \psi^{*} \right)[/tex]
    Calculating [itex]\partial_{x} j(x,t)[/itex] is a good place to start - do correct the complex conjugation mistakes though.
    Have you tried calculating [itex]\partial_{t} \rho[/itex]?
     
  4. Jan 13, 2016 #3
    So it represents current density! Interesting. OK, I'll sort out the conjugation mistakes - if I add the *, should it be correctly differentiated though?

    I have tried calculating ρ for a start, but pretty unsuccessfully. I think I broke some rules along the way, while trying to solve the schrodinger equation, the potential was still in the equation at the end. I'll post where I got to...
     
  5. Jan 13, 2016 #4
    So corrected version for partial derivative of j: ##-\frac{i\hbar}{2m}(\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}}-\frac{\partial{\psi}}{\partial{x}}\frac{\partial{\psi^{*}}}{\partial{x}})##
     
  6. Jan 13, 2016 #5
    As for ##\rho##, actually I know my method was wrong. I made ∂Ψ/∂t the subject and then integrated. I don't know how to solve the Schrodinger equation, so I can't find ##\rho##.
     
  7. Jan 13, 2016 #6
    You don't need to solve the Schrodinger equation to "find" [itex]\rho[/itex].
    [tex]\partial_{t} (\psi^{*}\psi) = \psi \partial_{t} \psi^{*} + \psi^{*} \partial_{t} \psi[/tex]
    Now, use the Schrodinger equation to replace [itex]\partial_{t} \psi^{*} [/itex] and [itex]\partial_{t} \psi[/itex]
     
  8. Jan 15, 2016 #7
    Got it, thank you!
     
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