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Square of a wave function; way to understand

  1. Aug 20, 2011 #1
    1. The problem statement, all variables and given/known data
    In a book I found the following calculation without the way. How do you get from 1) to 2)?


    2. Relevant equations
    (1) [itex]\Psi(x,t)=\frac{A}{2\pi}\sqrt{\frac{\pi}{d²+i\frac{\hbar t}{2m}}} exp{\frac{-\frac{x²}{4}+id²k_{0}(x-\frac{k_{0}\hbar}{2m}t)}{d²+i\frac{\hbar t}{2m}}}[/itex]
    Code (Text):
    [itex]\Psi(x,t)=\frac{A}{2\pi}\sqrt{\frac{\pi}{d²+i\frac{\hbar t}{2m}}} exp{\frac{-\frac{x²}{4}+id²k_{0}(x-\frac{k_{0}\hbar}{2m}t)}{d²+i\frac{\hbar t}{2m}}}[/itex]
    (2) [itex]|\Psi(x,t)²|=\frac{A²}{4\pi\sqrt{d^{4}+\frac{\hbar^{2}t²}{4m²}}} exp{-\frac{(x-\frac{k_{0}\hbar}{m}t)^{2}}{2d²+\frac{\hbar^{2}t²}{2m²d²}}}[/itex]
    Code (Text):
    [itex]\Psi(x,t)²=\frac{A²}{4\pi\sqrt{d^{4}+\frac{\hbar^{2}t²}{4m²}}} exp{-\frac{(x-\frac{k_{0}\hbar}{m}t)^{2}}{2d²+\frac{\hbar^{2}t²}{2m²d²}}}[/itex]

    3. The attempt at a solution
    [itex](x-\frac{k_{0}\hbar}{m}t)^{2}=x²-2x\frac{\hbar k_{0}}{m}t+\frac{\hbar^{2} k_{0}^{2}}{m^{2}}t^{2}[/itex]
    [itex](d²+i\frac{\hbar t}{2m})^{2}=d^{4}+\frac{i \hbar t}{m}+\frac{i² \hbar^{2} t²}{4m²}[/itex]

    I have posted the code so that it is easier for you to help me.
     
    Last edited: Aug 21, 2011
  2. jcsd
  3. Aug 20, 2011 #2

    fzero

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    [itex]|\Psi|^2 = \Psi^* \Psi[/itex], it's not just the regular square. Compute [itex]\Psi^*[/itex] and then compute the product with [itex]\Psi[/itex]. You'll find that (2) is actually incorrect, since [itex]|\Psi|^2 [/itex] is real, not complex.
     
  4. Aug 20, 2011 #3

    ehild

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    Have you copied the formulae correctly? 2. is certainly wrong as it contains the imaginary unit i in spite of being an absolute value.

    Do not forget that (ex)2=e2x.

    ehild
     
  5. Aug 20, 2011 #4
    I am sorry, its without that i in 2) (I will correct it).

    You can find it here: http://books.google.de/books?id=wmY...er quantenmechanik&pg=PA8#v=onepage&q&f=false.
    Still, I do not know what "tricks" are used there. I know that (ex)2=e2x, but how do you get the [itex]\frac{i \hbar t}{m}[/itex] in front of the exp away?
    What about the next i² or the minus?
    You can't get it into the exp function, there would have to be a ln to do that.
    What do you do with the id²k_0 in the exponent?
    Why do you have a [itex]i\frac{\hbar^{2}t²}{2m²d²}[/itex] down in the exponent of the exp?
    And what do I do wrong with the tex so that it doesn't appear? [Edit: [/itex] instead of [\itex]]
     
    Last edited: Aug 21, 2011
  6. Aug 20, 2011 #5

    ehild

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    To get the magnitude-square of a complex number z you need to multiply it by its conjugate, z*. |z|^2=z z*. Conjugate means that all i-s are replaced by -i.
    If the complex number is written with real and imaginary parts, z=a+ib, |z|^2=z z*=(a+ib)(a-ib)=a^2+b^2.
    Given in the Euler form, z=r, |z|^2=(r)(re-iφ)=r^2

    The magnitude of a product or fraction is the product of magnitudes.

    |z1z2/z3|=|z1||z2|/|z3|

    Change the sign of all i-s to the opposite in Ψ: that is Ψ*, and multiply Ψ and Ψ*. You will see that all the imaginary terms cancel.
    Take care: the exponents add up.

    ehild
     
    Last edited: Aug 20, 2011
  7. Aug 21, 2011 #6
    Ok, I have tried it anew, but I can get only this far and some i stay.

    [tex]
    |\Psi^{2}|=\Psi^{*}\cdot\Psi=\frac{A}{2\pi}\sqrt{ \frac{\pi}{d^
    2 +i \frac{\hbar t}{2m} } } \cdot\exp\left\{ \frac{-\frac{x^{2}}{4}+idk_{0}\left(x-\frac{k_{0}\hbar}{2m}t\right)}{d^{2}+i\frac{\hbar t}{2m}}\right\} \cdot\frac{A}{2\pi}\sqrt{\frac{\pi}{d^{2}-i\frac{\hbar t}{2m}}}\cdot\exp\left\{ \frac{-\frac{x^{2}}{4}-idk_{0}\left(x-\frac{k_{0}\hbar}{2m}t\right)}{d^{2}-i\frac{\hbar t}{2m}}\right\} [/tex][tex]

    [1]=\frac{A^{2}}{4\pi\sqrt{d^{2}+\frac{\hbar^{2}t^{2}}{4m^{2}}}}\cdot\exp\left\{ \frac{-\frac{x^{2}}{4}+idk_{0}\left(x-\frac{k_{0}\hbar}{2m}t\right)}{d^{2}+i\frac{\hbar t}{2m}}+\frac{-\frac{x^{2}}{4}-idk_{0}\left(x-\frac{k_{0}\hbar}{2m}t\right)}{d^{2}-i\frac{\hbar t}{2m}}\right\} [/tex][tex]
    [2]=\frac{A^{2}}{4\pi\sqrt{d^{2}+\frac{\hbar^{2}t^{2}}{4m^{2}}}}\cdot\exp\left\{ \frac{\left(-\frac{x^{2}}{4}+idk_{0}\left(x-\frac{k_{0}\hbar}{2m}t\right)\right)\cdot\left(d^{2}-i\frac{\hbar t}{2m}\right)}{\left(d^{2}+i\frac{\hbar t}{2m}\right)\cdot\left(d^{2}-i\frac{\hbar t}{2m}\right)}+\frac{\left(-\frac{x^{2}}{4}-idk_{0}\left(x-\frac{k_{0}\hbar}{2m}t\right)\right)\cdot\left(d^{2}+i\frac{\hbar t}{2m}\right)}{\left(d^{2}-i\frac{\hbar t}{2m}\right)\cdot\left(d^{2}+i\frac{\hbar t}{2m}\right)}\right\} [/tex][tex]
    [3]=\frac{A^{2}}{4\pi\sqrt{d^{2}+\frac{\hbar^{2}t^{2}}{4m^{2}}}}\cdot\exp\left\{ \frac{\left(-\frac{x^{2}}{4}+idk_{0}\left(x-\frac{k_{0}\hbar}{2m}t\right)\right)\cdot\left(d^{2}-i\frac{\hbar t}{2m}\right)+\left(-\frac{x^{2}}{4}-idk_{0}\left(x-\frac{k_{0}\hbar}{2m}t\right)\right)\cdot\left(d^{2}+i\frac{\hbar t}{2m}\right)}{\left(d^{4}+\frac{\hbar^{2}t^{2}}{4m^{2}}\right)}\right\} [/tex][tex]
    [4] =\frac{A^{2}}{4\pi\sqrt{d^{2}+ \frac{\hbar^{2}t^{2}}{4m^{2}}}}\cdot\exp\left\{ \frac{\left[\left(-\frac{x^{2}}{4}+idk_{0}\left(x-\frac{k_{0}\hbar}{2m}t\right)\right)\cdot\left(-i\frac{\hbar t}{2m}\right)+\left(-\frac{x^{2}}{4}-idk_{0}\left(x-\frac{k_{0}\hbar}{2m}t\right)\right)\cdot\left(i \frac{\hbar t}{2m}\right)\right]\cdot d^{2}}{\left(d^{4}+\frac{\hbar^{2}t^{2}}{4m^{2}} \right )}\right\} [/tex][tex]
    [5] =\frac{A^{2}}{4\pi\sqrt{d^{2}+ \frac{\hbar^{2}t^{2}}{4m^{2}}}}\cdot\exp\left\{ \frac{\left[\left(-\frac{x^{2}}{4}+ idk_{0}x-idk_{0}\frac{k_{0}\hbar}{2m}t\right)\cdot\left(-i\frac{\hbar t}{2m}\right)+\left(-\frac{x^{2}}{4}-idk_{0}x+idk_{0}\frac{k_{0}\hbar}{2m}t\right)\cdot\left(i\frac{\hbar t}{2m}\right)\right]\cdot d^{2}}{\left(d^{4}+ \frac{\hbar^{2}t^{2}}{4m^{2}}\right)}\right\}
    [/tex][tex]
    [6] =\frac{A^{2}}{4\pi\sqrt{d^{2}+ \frac{\hbar^{2}t^{2}}{4m^{2}}}}\cdot\exp\left\{ \frac{\left[+\frac{x}{4}-idk_{0}x+idk_{0} \frac{k_{0}\hbar}{2m}t-\frac{x^{2}}{4}-idk_{0}x+idk_{0}\frac{k_{0}\hbar}{2m}t\right]\cdot\left[d^{2}+i\frac{\hbar t}{2m}\right]}{\left(d^{4}+\frac{\hbar^{2}t^{2}}{4m^{2}}\right)}\right\} [/tex][tex]
    [7]=\frac{A^{2}}{4\pi\sqrt{d^{2}+\frac{\hbar^{2}t^{2}}{4m^{2}}}}\cdot\exp\left\{ \frac{\left[2idk_{0}\left(-x+\frac{k_{0}\hbar}{2m}t\right)\right]\cdot\left[d^{2}+i\frac{\hbar t}{2m}\right]}{\left(d^{4}+\frac{\hbar^{2}t^{2}}{4m^{2}}\right)}\right\} [/tex]
     

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    Last edited: Aug 21, 2011
  8. Aug 21, 2011 #7

    BruceW

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    I think you went wrong on step 3 to 4 because [itex]d^2[/itex] is not a factor which you can take outside the brackets.
     
  9. Aug 21, 2011 #8

    Redbelly98

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    Agreed.

    Going back to Equation 3, look at the numerator of the exp{...} expression. It must be real, so all imaginary terms must cancel. You have two possible approaches at this point:

    1. Expand (multiply out) all terms in the numerator, and see the cancellation of the imaginary terms for yourself

    OR

    2. Perhaps you can see, by inspection of Equation 3, the imaginary terms in exp{...} will cancel. If so, you can just include the real terms when you expand the terms. Take note, [itex]imaginary \cdot imaginary = real[/itex], so it's not a matter of just dropping all imaginary terms and then doing the expansion. If this is not clear to you, proceed with Method 1 above.
     
  10. Aug 21, 2011 #9
    Ok, thanks, I have corrected that. Really, nice how the mixed i terms cancel. I get this far and now, how do I get to (2) from my first post?

    (3)[tex] =\frac{A^{2}}{4\pi\sqrt{d^{2}+\frac{\hbar^{2}t^{2}}{4m^{2}}}}\cdot\exp\left\{ \frac{\left(-\frac{x^{2}}{4}+idk_{0}\left(x-\frac{k_{0}\hbar}{2m}t\right)\right)\cdot\left(d^{2}-i\frac{\hbar t}{2m}\right)+\left(-\frac{x^{2}}{4}-idk_{0}\left(x-\frac{k_{0}\hbar}{2m}t\right)\right)\cdot\left(d^{2}+i\frac{\hbar t}{2m}\right)}{\left(d^{4}+\frac{\hbar^{2}t^{2}}{4m^{2}}\right)}\right\}
    [/tex](4)[tex] =\frac{A^{2}}{4\pi\sqrt{d^{2}+\frac{\hbar^{2}t^{2}}{4m^{2}}}}\cdot\exp\left\{ \frac{\left(-\frac{x^{2}}{4}d^{2}+id^{3}k_{0}\left(x-\frac{k_{0}\hbar}{2m}t\right)+\frac{x^{2}}{4}i \frac{\hbar t}{2m}-ii\frac{\hbar t}{2m}dk_{0}\left(x-\frac{k_{0}\hbar}{2m}t\right)\right)-\frac{x^{2}}{4}d^{2}-id^{3}k_{0}\left(x-\frac{k_{0}\hbar}{2m}t\right)-\frac{x^{2}}{4}i\frac{\hbar t}{2m}-ii\frac{\hbar t}{2m}dk_{0}\left(x-\frac{k_{0}\hbar}{2m}t\right)}{\left(d^{4}+\frac{\hbar^{2}t^{2}}{4m^{2}}\right)}\right\}
    [/tex](5)[tex] =\frac{A^{2}}{4\pi\sqrt{d^{2}+\frac{\hbar^{2}t^{2}}{4m^{2}}}}\cdot\exp\left\{ \frac{\left(-\frac{x^{2}}{2}d^{2}+2\frac{\hbar t}{2m}dk_{0}\left(x-\frac{k_{0}\hbar}{2m}t\right)\right)}{\left(d^{4}+\frac{\hbar^{2}t^{2}}{4m^{2}}\right)}\right\}
    [/tex](6)[tex] =\frac{A^{2}}{4\pi\sqrt{d^{2}+\frac{\hbar^{2}t^{2}}{4m^{2}}}}\cdot\exp\left\{ \frac{\left(-x^{2}d^{2}+2\frac{\hbar t}{m}dk_{0}\left(x-\frac{k_{0}\hbar}{2m}t\right)\right)}{\left(2d^{4}+\frac{\hbar^{2}t^{2}}{2m}\right)}\right\}
    [/tex](7)[tex]=\frac{A^{2}}{4\pi\sqrt{d^{2}+\frac{\hbar^{2}t^{2}}{4m^{2}}}}\cdot\exp\left\{ \frac{\left(-x^{2}d^{2}+2\frac{\hbar t}{m} dk_{0}x-\frac{\hbar^{2}t^{2}}{m^{2}} dk_{0}^{2}\right)}{\left(2d^{4}+\frac{\hbar^{2}t^{2}}{2m}\right)}\right\}[/tex]
     
  11. Aug 21, 2011 #10

    BruceW

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    You're incredibly close now. You just need to rewrite the function inside the exponential to make it the same as the one in (2)
     
  12. Aug 21, 2011 #11

    ehild

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    There is d^2 in the numerator of the exponent in the original wave function.

    [tex]-\frac{x^{2}}{4}+id^2k_{0}\left(x-\frac{k_{0}\hbar}{2m}t\right)[/tex]

    ehild
     
  13. Aug 22, 2011 #12
    [7][tex]
    =\frac{A^{2}}{4\pi\sqrt{d^{2}+\frac{\hbar^{2}t^{2}}{4m^{2}}}}\cdot\exp\left\{ \frac{\left(-x^{2}d^{2}+2\frac{\hbar t}{m}d^{2}k_{0}x-\frac{\hbar^{2}t^{2}}{m^{2}}d^2k_{0}^{2}\right)}{2d^{4}+\frac{\hbar^{2}t^{2}}{2m}}\right\}[/tex]
    [8][tex]
    \frac{A^{2}}{4\pi\sqrt{d^{2} +\frac{\hbar^{2}t^{2}}{4m^{2}}}} \cdot\exp\left\{ -\frac{\left(x^{2}-2 \frac{\hbar t}{m}k_{0}x+ \frac{\hbar^{2}t^{2}}{m^{2}}k_{0}^{2} \right)}{2d^{2} +\frac{\hbar^{2}t^{2}}{2md^{2}}} \right\}[/tex][9][tex]=\frac{A^{2}}{4\pi\sqrt{d^{2}+\frac{\hbar^{2}t^{2}}{4m^{2}}}}\cdot\exp\left\{ -\frac{ \left (x-\frac{\hbar k_{0}}{m}t\right)^{2}}{2d^{2}+\frac{\hbar^{2}t^{2}} {2md^{2}}} \right\}[/tex]

    Thanks everbody! I finally got it.
     
  14. Aug 22, 2011 #13

    BruceW

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    nice work. Its quite a long calculation to do.
     
  15. Aug 22, 2011 #14
    wow...

    may i enquire where in physics do you need such calculation?

    i have done a bit of Quantum mechanics but i don't recall such stuff
     
  16. Aug 22, 2011 #15
    In quantum mechanics, but it was not directly a task, just a way to find the Gaussian packet and read the expectation value for x and also the Delta_X², and since it was in the book I wanted to do it myself. But it seems there are easier ways to do this.
     
  17. Aug 22, 2011 #16
    wow ok lol isee thanks
     
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