1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Work Check: Wavefunction Normalisation

  1. Dec 11, 2017 #1
    1. The problem statement, all variables and given/known data
    Find relation between real normalisation constants ##B_1## and ##B_2## for the following wavefunction,
    \Psi_k =\sum_{k=1,2} \frac{B_k}{\sqrt{4\sigma ^2 + 2it}} \exp (ip_k (x - \frac{p_k}{2}t) - \frac{(x - p_k t)^2}{4\sigma ^2 + 2it})

    The working is rather long so thanks in advance for helping, it is greatly appreciated.

    2. Relevant equations

    3. The attempt at a solution

    Normalisation requires ##\int_{\infty} dx |\Psi (x,0)|^2 = 1##, and
    $$\Psi_k (x, 0) =\sum_{k=1,2} \frac{B_k}{2\sigma} \exp (ip_k (x) - \frac{(x)^2}{4\sigma ^2})$$
    Knowing that,
    $$|\Psi_k|^2 = |B_1 \psi_1|^2 + |B_2 \psi_2|^2 + (B_2 \psi_2)^* (B_1 \psi_1) + (B_2 \psi_2)(B_1 \psi_1)^*$$
    $$|B_k \psi_k|^2 = \frac{B_k ^2}{4\sigma^2}\exp (-\frac{x^2}{2\sigma ^2})$$
    $$(B_2 \psi_2)(B_1 \psi_1)^* = [\frac{B_1}{2\sigma} \exp (-ip_1 (x) - \frac{(x)^2}{4\sigma ^2})][\frac{B_2}{2\sigma} \exp (ip_2 (x) - \frac{(x)^2}{4\sigma ^2})]$$
    $$= \frac{B_1 B_2 }{4\sigma^2} \exp [-i(p_1 - p_2)x - \frac{x^2}{2\sigma^2}]$$
    $$(B_2 \psi_2)^* (B_1 \psi_1) = \frac{B_1 B_2 }{4\sigma^2} \exp [i(p_1 - p_2)x - \frac{x^2}{2\sigma^2}]$$
    Before computing the integral, I simplified by letting ##\frac{B_1 B_2}{4\sigma ^2} = \phi##, ##\frac{1}{2\sigma ^2 } = g## and ##p_1 - p_2 = U##
    This gave,
    $$(B_2 \psi_2)(B_1 \psi_1)^* = \phi \exp (-gx^2 - iUx)$$
    $$(B_2 \psi_2)^*(B_1 \psi_1) = \phi \exp (-gx^2 + iUx)$$
    The expression for the integral is,
    $$\int_{\infty} dx |\Psi (x,0)|^2 = 1 =\int_{\infty} dx [\frac{B_1 ^2 + B_2 ^2}{4\sigma^2}\exp (-gx^2) + \phi \exp (-gx^2 - iUx) + \phi \exp (-gx^2 + iUx)] $$
    The integral of the first integrand can be computed with the Gaussian integral result which gives,
    $$\frac{B_1 ^2 + B_2 ^2}{4\sigma ^2} \sqrt{\frac{\pi}{g}} = \frac{B_1 ^2 + B_2 ^2}{4\sigma ^2}\sqrt{2\pi \sigma ^2} = \frac{\sqrt{2\pi} (B_1 ^2 + B_2 ^2)}{4\sigma}$$
    The integral of the second integrand, which is also equal to that of the third, was done by completing the square,
    $$\int_{\infty} dx \phi \exp (-gx^2 - iUx) = \phi e^{\frac{-U^2}{4g}} \int_{\infty} dx \exp (-[\sqrt{g}x + \frac{iU}{2\sqrt{g}}]^2)$$
    I used ##dy = \sqrt{g} dx## and rewrote this as,
    $$\frac{\phi e^{\frac{-U^2}{4g}}}{\sqrt{g}} \int_{\infty} dx \exp (-y^2) = \frac{\phi e^{\frac{-U^2}{4g}}}{\sqrt{g}} \sqrt{\pi} $$
    Substituting for ##g, U, \phi## then gives,
    $$\frac{\sqrt{2\pi} B_1 B_2}{4\sigma} \exp (-\frac{(p_1 - p_2)^2 \sigma ^2}{2})$$

    Finally, summing up all 3 integrals of all 3 terms gives,

    $$1 =\frac{\sqrt{2\pi} (B_1 ^2 + B_2 ^2)}{4\sigma} + \frac{\sqrt{2\pi} B_1 B_2}{2\sigma} \exp (-\frac{(p_1 - p_2)^2 \sigma ^2}{2})
  2. jcsd
  3. Dec 11, 2017 #2


    User Avatar
    Gold Member

    Shouldn't this be moved to the Advanced Physics Homework forum?
  4. Dec 11, 2017 #3
    I thought that this was pretty basic stuff (in the context of everything else), and decided to post it here. Apologies if I should have posted this in the other thread.
  5. Dec 11, 2017 #4


    User Avatar
    Gold Member

    Its not about you not posting in the correct forum. I think you might just get some better and faster answers in the advanced physics forum if what you're studying is wave functions and normalization.
  6. Dec 11, 2017 #5
    Could anybody take the time to assist? Many thanks!
  7. Dec 12, 2017 #6

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Letting ##B_1## and ##B_2## be possibly complex, I used Maple and obtained
    $$\int_R |\Psi(x,0)|^2 \, dx = \frac{\sqrt{2 \pi}}{4 \sigma} ( |B_1|^2 + |B_2|^2)
    + \frac{\sqrt{2 \pi}}{4 \sigma} \exp \left( -\frac{1}{2} \sigma^2 (p_1 - p_2)^2 \right) ( B_1 \bar{B_2} + B_2 \bar{B_1})$$
    This agrees with your answer if we assume the ##B_i## are real.
  8. Dec 13, 2017 #7

    Thank you!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted