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Two independent particles in an infinite square well

  1. May 3, 2015 #1

    Maylis

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    1. The problem statement, all variables and given/known data
    upload_2015-5-3_22-26-25.png

    2. Relevant equations
    upload_2015-5-3_22-26-39.png
    upload_2015-5-3_22-26-55.png

    3. The attempt at a solution
    a) For this part, I know for distinguishable particles, the expectation value of the square distance

    $$\langle (x_{1}^{2} - x_{2}^{2}) \rangle = \langle x^{2} \rangle_{2} + \langle x^{2} \rangle_{3} - 2 \langle x \rangle_{2} \langle x \rangle_{3}$$

    This just gets really ugly...
    $$\langle x^{2} \rangle_{2} = \int \psi_{2}^{*}x^{2}\psi_{2} dx = \frac {2}{a} \int x^{2} sin^{2}\Big (\frac {2 \pi}{a} x \Big ) dx $$
    Using the equation given in the problem (is that even right? It seems suspicious that it is independent of ##n##)
    $$\langle x^{2} \rangle_{2} = \frac {2}{a} \Big [ \frac {\pi^{3}}{6} - \frac {\pi}{4(\frac {2 \pi}{a})^{2}} \Big ] $$
    $$\langle x^{2} \rangle_{2} = \frac {2}{a} \Big [ \frac {\pi^{3}}{6} - \frac {a^{2}}{16 \pi} \Big ] $$
    $$ = \frac {\pi^{3}}{3a} - \frac {a}{8 \pi} $$

    Similarly for ##\langle x^{2} \rangle_{3}##
    $$\langle x^{2} \rangle_{3} = \frac {2}{a} \Big [ \frac {\pi^{3}}{6} - \frac {\pi}{4(\frac {3 \pi}{a})^{2}} \Big ]$$
    $$\langle x^{2} \rangle_{3} = \frac {2}{a} \Big [ \frac {\pi^{3}}{6} - \frac {a^{2}}{36 \pi} \Big ] $$
    $$=\frac {\pi^{3}}{3a} - \frac {a}{18 \pi}$$

    For ##\langle x \rangle_{2}##
    $$\langle x \rangle_{2} = \int \psi_{2}^{*}x \psi_{2} dx$$
    $$\langle x \rangle_{2} = \frac {2}{a} \int x sin^{2}(\frac {2 \pi}{a} x) dx $$
    From the equation given in the problem statement
    $$\langle x \rangle_{2} = \frac {2}{a} \frac {\pi^{2}}{4} = \frac {\pi^{2}}{2a} $$
    The same goes for ##\langle x \rangle_{3}##
    $$\langle x \rangle_{3} = \frac {2}{a} \int x sin^{2}(\frac {3 \pi}{a} x) dx$$
    $$\langle x \rangle_{3} = \frac {2}{a} \frac {\pi^{2}}{4} = \frac {\pi^{2}}{2a}$$

    $$\langle (x_{1}^{2} - x_{2}^{2}) \rangle = \langle x^{2} \rangle_{2} + \langle x^{2} \rangle_{3} - 2 \langle x \rangle_{2} \langle x \rangle_{3}$$
    $$\langle (x_{1}^{2} - x_{2}^{2}) \rangle = \Big (\frac {\pi^{3}}{3a} - \frac {a}{8 \pi} \Big ) + \Big (\frac {\pi^{3}}{3a} - \frac {a}{18 \pi} \Big ) - 2 \Big (\frac {\pi^{2}}{2a} \Big ) \Big ( \frac {\pi^{2}}{2a} \Big ) $$
    $$\langle (x_{1}^{2} - x_{2}^{2}) \rangle = \frac {2 \pi^{3}}{3a} - \frac {13a}{72 \pi} - \frac {\pi^{4}}{2a^{2}} $$
    ----------------------------------------------------------------------------------------------------------------------------------------
    (b) For this problem, I am basically looking at this webpage
    http://galileo.phys.virginia.edu/classes/252/symmetry/Symmetry.html

    symmetric:
    $$\psi_{23} (x_{1},x_{2}) = A sin \Big (\frac {2 \pi}{a} x_{1} \Big ) cos \Big (\frac {3 \pi}{a} x_{2} \Big ) $$
    Not sure how to get A, but anyways
    For symmetric
    $$ \psi_{23}^{S} = \frac {1}{\sqrt{2}} \Big [ sin \Big (\frac {2 \pi}{a} x_{1} \Big ) cos \Big (\frac {3 \pi}{a} x_{2} \Big ) +sin \Big (\frac {3 \pi}{a} x_{1} \Big ) cos \Big (\frac {2 \pi}{a} x_{2} \Big ) \Big ]$$

    For antisymmetric
    $$ \psi_{23}^{A} = \frac {1}{\sqrt{2}} \Big [ sin \Big (\frac {2 \pi}{a} x_{1} \Big ) cos \Big (\frac {3 \pi}{a} x_{2} \Big ) - sin \Big (\frac {3 \pi}{a} x_{1} \Big ) cos \Big (\frac {2 \pi}{a} x_{2} \Big ) \Big ] $$

    This is slightly different from my notes, which had two sine terms instead of a cosine term, and I am not sure if that is because this time it has electrons. I am just so confused...
    upload_2015-5-3_23-57-20.png
    upload_2015-5-3_23-57-38.png
     
    Last edited: May 3, 2015
  2. jcsd
  3. May 3, 2015 #2

    TSny

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    For part (a), note that your potential well extends from x = 0 to x = a. Thus, you will integrate from x = 0 to x = a. Note, though, that Griffiths' integrals extend from 0 to ##\pi##. So, a change of integration variable will be required if you want to use Griffiths' integrals.

    For part (b), the website that you link to assumes that the well extends from x = -L/2 to x = L/2. That causes some of the wavefunctions to be cosine functions. But, in your case where the well extends from x = 0 to x = a, all of the wavefunctions will be sine functions.
     
  4. May 3, 2015 #3

    vela

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    The answer you got in part (a) can't be correct because the units don't work out.

    You should review how to construct the symmetric and antisymmetric spatial wave functions. It's pretty straightforward how to do so. You shouldn't need to copy a result off of a webpage.
     
  5. May 3, 2015 #4

    Maylis

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    I just posted my new work since writing out all the latex and editing it until it is right can take over an hour. But for part (a), I got

    $$\langle (x_{1} - x_{2})^{2} \rangle = \frac {a^{2}}{27} - \frac {5a^{2}}{72 \pi^{2}}$$

    At least the units are correct for this answer
    ImageUploadedByPhysics Forums1430714949.829527.jpg

    (b) for part i. and ii.
    upload_2015-5-4_12-56-20.png

    part iii. looks mathematically brutal =(
     

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    Last edited: May 4, 2015
  6. May 4, 2015 #5

    Maylis

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    Is this the correct integral to evaluate for part iii? This looks brutal

    upload_2015-5-4_13-25-13.png
     

    Attached Files:

  7. May 4, 2015 #6

    TSny

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    I don't believe your answer is correct. Note, for example, that in calculating ##\left <x^2 \right >_2## your change of variable led to an integral from 0 to ##2 \pi##. But Griffiths' integrals are from 0 to ##\pi##.
     
  8. May 4, 2015 #7

    Maylis

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    True, but that's why I just multiplied by 2, because the function is even, so I should be able to just double my answer, and mulitplied by 3 for the integral from 0 to ##3 \pi##
     
  9. May 4, 2015 #8

    vela

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    According to Mathematica, the expectation value in part (i) works out to ##\frac{a^2}{6}\left(1-\frac{13}{12\pi^2}\right)##.
     
  10. May 4, 2015 #9

    Maylis

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    Then it has something to do with my limits of integration. I went to the prof's office hours and he said to just double what you would get integrating from 0 to pi since it is an even function. I assumed the same for tripling.
     
  11. May 4, 2015 #10

    TSny

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    Your prof must not have looked at your integral carefully. Changing Griffiths' limits of [0, ##\pi##] to [0, ## 2\pi##] will not double the result. A sketch of the integrand would help see that. I agree with vela's result (also using Mathematica).
     
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