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Simplifying an expression using limits

  1. Jan 16, 2016 #1
    1. The problem statement, all variables and given/known data
    Given that

    ##T = \frac{4E(V-E)}{4E(V-E)+V^2\sinh^2 (ka)}##

    Simplify the expression for T when a is large but not infinite, and again for the case when a tends to zero and the potential tends to infinity, such that ##d = Va## is a real constant. k is a constant, as is E (which is assumed to be less than V).


    2. Relevant equations


    3. The attempt at a solution
    To try and simplify it for the first case, I used the exponential definition of sinh, squared it and got that
    ##\sinh^2(x) = \frac{1}{4}(e^{2x}+e^{-2x}-2)##

    For large x the ##e^{-2x}## term would get small very quickly, especially compared to the other exponential term which would increase rapidly. So this becomes

    ##\sinh^2(x) = \frac{1}{4}(e^{2x}-2)##

    I don't know if that's what the question meant. It doesn't seem massively simpler. Is there something else that can be done to simplify the expression?

    And the second part, I don't know where to start! There should be some sort of constant ##d = Va##, but I thought ##sinh^2(0)=0##, which would mean that ##a## disappears. As far as I can see that expression simplifies to 1 for the second set of limits! Why is that wrong?
     
  2. jcsd
  3. Jan 16, 2016 #2
    What is the dominating term in the denominator?
    The limit is such that a tends to zero while V tends to infinity at the same time, so the limit of the term [itex]V^2 \sinh^{2}(ka)[/itex] cannot be done simply by setting a = 0 as you've done. A good starting point is to Taylor expand [itex]\sinh^{2}(ka)[/itex], and see where that gets you to, before taking the limits.
     
  4. Jan 16, 2016 #3

    Ray Vickson

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    Even better in the large-##a## case would be to write ##\sinh^2(x) \sim e^{2x}/4##, because ##e^{2x}## is overwhelmingly large compared to the 2. Also, if ##E(E-V)## remains constant, it is overwhelmingly small compared with the second term in the denominator, and that leads to even more simplification.
     
  5. Jan 16, 2016 #4

    vela

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    It would help to note that
    $$k^2 = \frac{2m(V-E)}{\hbar^2}.$$ Since ##ka## depends on both ##a## and ##V##, you need to first figure out how it behaves in the limit as ##a \to 0## and ##V \to \infty## before writing down an approximation for ##\sinh ka##. Start by writing ##V = d/a## and getting everything in terms of ##a##.
     
  6. Jan 16, 2016 #5
    Oh, I completely forgot about that! Thanks, I'll rewrite in terms of ##a## now.
     
  7. Jan 20, 2016 #6
    In terms of a, ##ka = \frac{\sqrt{2mad - 2ma^2E}}{\hbar}##.

    That still tends to zero, given d is a real constant. So ##sinh^2(ka)## tends to zero, since its taylor expansion is ##x^2 + \frac{x^4}{3} + \frac{2x^6}{45}##.

    Taking the ##V^2## term into account, how do I know what happens? I've never dealt with composite limits before, it kind of looks like V tends to infinity just as fast as the ##sinh^2## term tends to zero!
     
    Last edited by a moderator: Jan 20, 2016
  8. Jan 20, 2016 #7
    Sorry, misplaced # in the post above.

    In terms of a, ##ka = \frac{\sqrt{2mad - 2ma^2E}}{\hbar}##.

    That still tends to zero, given d is a real constant. So ##sinh^2(ka)## tends to zero, since its taylor expansion is ##x^2 + \frac{x^4}{3} + \frac{2x^6}{45}##.

    Taking the ##V^2## term into account, how do I know what happens? I've never dealt with composite limits before, it kind of looks like V tends to infinity just as fast as the ##sinh^2## term tends to zero!
     
  9. Jan 20, 2016 #8
    Is getting everything in terms of a equivalent to getting it in terms of V, or do both cases have to be considered?
     
  10. Jan 20, 2016 #9
    Why not try multiplying [itex]V^{2}[/itex] to your Taylor series expansion and see what you get? There'll be some simplifications that allow you to take the limit properly.
    [Hint: use Va = d]
     
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