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Derivative of complex exponential differs by a sign

  1. Sep 30, 2015 #1
    I know this is probably the least of my worries at the moment but my quantum textbook solves ##\frac{\mathrm{d}\phi (t) }{\mathrm{d} t}=\frac{iC}{h}\phi (t) ## as ##\phi (t) = e^{-i(\frac{C}{h})t}##. Is this not off by a sign? Its really bugging me.
     
  2. jcsd
  3. Sep 30, 2015 #2
    And the h is really h bar but I don't know the latex code for it
     
  4. Sep 30, 2015 #3

    Mark44

    Staff: Mentor

    I don't see how they got the minus sign. Can you post a picture of the textbook page?

    \hbar -- ##\hbar##
     
  5. Sep 30, 2015 #4
    again
     

    Attached Files:

  6. Sep 30, 2015 #5

    Geofleur

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    Write ## i\hbar\frac{1}{\phi}\frac{d\phi}{dt} = C ## as ## \frac{d}{dt}\ln\phi = \frac{C}{i\hbar} ##. Now multiply the RHS by one in disguise as ## \frac{i}{i}\frac{C}{i\hbar} = \frac{iC}{i^2\hbar} = -\frac{iC}{\hbar} ##. We get ## \frac{d}{dt}\ln\phi = -\frac{iC}{\hbar} ##. Integrate both sides to get ## \ln\phi = -\frac{iCt}{\hbar} + A ##, where ## A ## is some constant. Finally, exponentiate both sides to get ## \phi = Be^{-\frac{iCt}{\hbar}} ##, where ## B ## is a constant.
     
  7. Oct 1, 2015 #6

    Mark44

    Staff: Mentor

    I don't think the problem was how to solve the differential equation, but rather how the text went from one DE to another. In the image of post #4, the text has
    @Summer95, I believe the text is being a little sloppy here -- C in the left equation isn't the same as C in the right equation; the two differ by a sign. That's what I think is going on here.
     
  8. Oct 1, 2015 #7

    Ssnow

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    If the ## C## is the same the second equation needs a minus in front the right side ... ( in any case the disjunction is true and also the solution given)
     
  9. Oct 1, 2015 #8

    Geofleur

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    Oh I see! Oops... :oops:
     
  10. Oct 1, 2015 #9
    ok I see now, of course. thank you!
     
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