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Stationary States: Differential Equations Issues

  1. Jun 24, 2011 #1
    So in Griffith's (ed. 2 page 37) there's an equation that says that
    **pretend the h's are h-bars...I don't know Latex very well**


    Since in this simplified case V where is a function of x alone he says that each side is equal to a constant but I'm still trying to figure out why.

    I can see that the LHS is a function of t alone because he made the wave function separable and the [itex]\varphi[/itex] is a function of t and the [itex]\psi[/itex] is a function of x but I'm not sure why it's important. I see that if we set the equation equal to a constant the rest of the math works out nicely but I can't see what allows us to do this.

    If we had say [itex]\frac{dy}{dx}[/itex]=[itex]\frac{dz}{dt}[/itex] and we set the whole thing equal to c, how would I know that y(x) = cx and z(t) = ct (they're the same constant right?) so y(x)/x=z(t)/t ?

    Any direction would be great and thank you for your time,
  2. jcsd
  3. Jun 24, 2011 #2


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    Staff: Mentor

    h-bar in Latex is \hbar (surprise! :surprised).

    The equation has to be true for any value of x and any value of t.

    First pick a value of t and use it to evaluate the LHS. This gives you a number. No matter what value you use for x, the RHS must evaluate to that number. So the RHS is constant.

    Now start over. Pick a value of x and use it to evaluate the RHS... (see if you can fill in the rest.)
  4. Jun 24, 2011 #3


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    Can you see another possibility? Remember that mathematically, a variable is free to take any value in its domain. So if the LHS is only a function of t, and the RHS is only a function of x, can you find any function f(x,t) that is NOT a constant, yet still satisfies both sides of the equation independently?

    In other words, what you are given is:


    and I am asking you to do is find any f(x,t) that is NOT a constant and satisfies BOTH




    for ALL allowable values of x and t.
  5. Jun 24, 2011 #4
    haha thank you for that

    I think I see what you mean now.
  6. Jun 24, 2011 #5
    That makes a whole lot more sense!

    thank you very much for your time,
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