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Homework Help: Dimensionless form of the time-independent Schrodinger equation

  1. Aug 2, 2007 #1
    For a free particle, show that the time-independent Schrodinger equation can be written in dimensionless form as

    [tex] d^2\psi(z)/dz^2 = -\psi(z) [/tex].

    I do not see how you would get rid of the m (with units mass) in front of the del in the SE (or the other constants for that matter)...
     
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  3. Aug 2, 2007 #2

    Meir Achuz

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    Substitute z=[hbar/sqrt{2m}]x
     
  4. Aug 2, 2007 #3
    Sorry, this is probably a really basic question but if I have psi(x(z)), how do I get psi(z)?
     
  5. Aug 2, 2007 #4

    Dick

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    psi(z(x))=psi(z). They are the same thing. Meir is suggesting you change the variable you differentiate wrt, not the psi. Use the chain rule.
     
  6. Aug 3, 2007 #5
    Just so we are on the same page, the equation I am starting with is:

    [tex] -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi_E(x) = E \psi_E(x) [/tex]

    where (I thought) E was a constant not an operator.

    So, first, I do not see how that substitution would get rid of the constant E.

    Second, if I change variables I get

    [tex] -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi_E(v(x)) = E \psi_E(v(x)) [/tex]

    I cannot change the dx in the second derivative because that is part of an operator not a variable, right?

    Then if I use the chain rule one time, I get:

    d/dx psi(z(x)) = d/dx psi(z) * d/dx z(x) from which I do not see how you get a d/dz operator into the equation
     
  7. Aug 3, 2007 #6

    Meir Achuz

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    Sorry, it should have been sqrt{mE}.
    If z=ax, d/dx=ad/dz.
    Look at your calculus text.
     
  8. Aug 3, 2007 #7
    Sorry, I cannot find that rule in my calculus text. Does it have a name? I know how to substitute variables with integrals, but I have not seen it with derivatives. If you were going to do it with an integral it would be

    dz = a*x*dx but how do you get the derivative operator form? Sorry again, I should probably know this!
     
  9. Aug 3, 2007 #8

    nrqed

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    z = a x

    Now, [itex] \frac{d}{dx} = \frac{dz}{dx} \frac{d}{dz} = a \frac{d}{dz} [/itex]

    It's just the chain rule applied to a very simple case.
     
  10. Aug 3, 2007 #9
    The chain rule states that if a = f(b) and if b = f(c), then da/dc = da/db * db/dc.

    Your comment obviously makes sense with the Liebnex notation when you treat differentials like fractions, but I do not see how you get that directly from the chain rule. Apply, d/dx to both sides of the equation doesn't work...
     
    Last edited: Aug 3, 2007
  11. Aug 3, 2007 #10
    I think I see now. You can just replace d/dx with any dy/dx for any function y and you get the chain rule. Just a little too abstract for me!
     
  12. Aug 3, 2007 #11

    nrqed

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    To show how it comes from the chain rule, consider a function y(z) and z is a function of x itself z(x).
    Then the chain rule says

    dy/dx = dy/dz dz/dx

    Now let's say that z= ax. Then we have dz/dx = a (where a is a constant)

    So

    dy/dx = a dy/dz

    In this thread, you must think of the wavefunction psi as playing the role of y.

    Psi is initially a function of x. But then you imagine replacing all the x interms fo z in order to get a function psi(z). Plugging back in the Schrodinger equation, you then have to deal with calculating

    [itex] \frac{d \psi(z)}{dx} [/itex]. Using the rule above, you see that this is equal to [itex] a \frac{d \psi(z)}{dz} [/itex].
     
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