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Absolute value of exponentials being multiplied

  1. Feb 26, 2015 #1

    Maylis

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    I know this is an elementary question, but it has been some time since I multiplied exponentials, and with imaginary terms combined with absolute values, things get muddled up so easy that I want to clear this up

    So if I have

    $$ \Psi (x,t) = c_{1} \psi_{1} e^{- \frac {i E_{1}}{\hbar} t} + c_{2} \psi_{2} e^{- \frac {i E_{2}}{\hbar} t} $$

    I want to find ##\mid \Psi (x,t) \mid^{2}##

    Well, I suppose I will do ##\mid \Psi (x,t) \mid \mid \Psi (x,t) \mid##. My first problem is actually determining the absolute value of this quantity.

    $$ \mid c_{1} \psi_{1} e^{- \frac {i E_{1}}{\hbar} t} + c_{2} \psi_{2} e^{- \frac {i E_{2}}{\hbar} t} \mid $$

    But how to analyze this?
     
  2. jcsd
  3. Feb 26, 2015 #2
    Your definition of ##|\Psi(x,t)|^2## is not correct. Since ##\Psi## is complex, you need to evaluate ##\Psi^*\Psi##.
     
  4. Feb 26, 2015 #3

    Maylis

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    I know this is dumb, but for a real ##a##, why is ##e^{-ia}*e^{ia} = 1##?? I don't know how to show this.
     
  5. Feb 26, 2015 #4
    Short of just giving you the answer, think about the rules of multiplying terms with powers.
     
  6. Feb 26, 2015 #5

    Maylis

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    oh yeah, duh. I just get freaked out when I see complex numbers and absolute values :cry:
     
  7. Feb 26, 2015 #6

    Maylis

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    Okay, well I am almost good, but one part I haven't figured out.

    I get the answer to be (using Euler's Identity)

    $$c_{1}^{2} \psi_{1}^{2} + 2c_{1}c_{2} \psi_{1} \psi_{2} \Big [cos \Big ( \frac {E_{2}t}{\hbar} - \frac {E_{1}t}{\hbar} \Big ) + i \hspace {0.02 in} sin \Big ( \frac {E_{2}t}{\hbar} - \frac {E_{1}t}{\hbar} \Big ) \Big ] + c_{2}^{2} \psi_{2}^{2} $$

    However, the answer has the term with ##i \hspace {0.02 in} sin (\theta)## missing. How did that term get cancelled?
     
  8. Feb 28, 2015 #7

    Maylis

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    I'll just cut to the chase and post the example/solution, how was the term with the imaginary number and sine cancelled??

    upload_2015-2-28_15-26-35.png
     
  9. Mar 2, 2015 #8
    First off, and I know it's always bold to accuse a textbook of making errors, but the author seems to have missed a ##t## in the ##c_2## exponentials on the penultimate line.

    To see the logic, explicitly multiply out the cross terms, remembering the rules for multiplying powers that we discussed earlier. Then use Euler's identity as given in the book, and remember that ##cos(\theta) = cos(-\theta)## and ##sin(\theta) = -sin(-\theta)##.
     
  10. Mar 3, 2015 #9

    Maylis

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    Does that mean my solution is correct?
     
  11. Mar 3, 2015 #10
    No the sine terms should still cancel.
     
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