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Identical particles and separating the Schrodinger equation

  1. Mar 6, 2017 #1
    1. The problem statement, all variables and given/known data
    Two identical particles, each of mass m, move in one dimension in the potential
    $$V = \frac{1}{2}A(x_1^2+x_2^2)+ \frac{1}{2}B(x_1-x_2)^2$$
    where A and B are positive constants and ##x_1## and ##x_2## denote the positions of the particles.
    a) Show that the Schrodinger equation is separable in the variables ##x_1+x_2## and ##x_1-x_2##. Find the eigenvalues and the corresponding eigenfunctions.
    b)Discuss the symmetry of the eigenfunctions with respect to particle exchange.

    2. Relevant equations
    Schrodinger equation:
    $$HΨ = EΨ$$
    Hamiltonian for identical particles
    $$H = \frac{-ħ}{2m}\frac{d^2}{dx_1^2}-\frac{-ħ}{2m}\frac{d^2}{dx_2^2}+\frac{1}{2}A(x_1^2+x_2^2)+ \frac{1}{2}B(x_1-x_2)^2$$

    3. The attempt at a solution
    I plugged in the Hamiltonian into the Schrodinger equation, and separated variables by setting ##Ψ=X(x_1+x_2)Y(x_1-x_2)##, replaced Ψ with the product and divided by ##XY##. I end up with something I can't really solve. I'm mostly not sure if my separation method is correct, I'm confused on what it means by 'separable in the variables ##x_1+x_2## and ##x_1-x_2##' rather than just ##x_1## and ##x_2## for example. Thanks
     
  2. jcsd
  3. Mar 6, 2017 #2

    DrClaude

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    Staff: Mentor

    Define two new variables, say ##R \equiv x_1+x_2## and ##r \equiv x_1-x_2## and try and rewrite the Hamiltonian in terms of ##R## and ##r##.
     
  4. Mar 6, 2017 #3
    Thanks for the reply
    Rewriting the Hamiltonian in terms of those new variables, I get
    $$H = \frac{-ħ^2}{2m}\frac{d^2}{dx_1^2}-\frac{-ħ^2}{2m}\frac{d^2}{dx_2^2}+\frac{1}{2}A(R^2-2x_1x_2)+ \frac{1}{2}Br^2$$
    This seems incorrect though because of the factor of ##-2x_1x_2## which is due to ##R^2 = x_1^2 + x_2^2+2x_1x_2##. Also, how would I go about changing the terms ##\frac{d^2}{dx_1^2}## and ##\frac{d^2}{dx_2^2}## in terms of R and r?
     
  5. Mar 6, 2017 #4

    DrClaude

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    Staff: Mentor

    There is also a factor ##x_1x_2## that can appear from ##r^2##.

    You have to use the multivariable chain rule.
     
  6. Mar 6, 2017 #5
    Oh right! So I got
    $$H = \frac{-ħ^2}{2m}\frac{d^2}{dx_1^2}-\frac{-ħ^2}{2m}\frac{d^2}{dx_2^2}+\frac{1}{2}A(\frac{R^2}{2}+\frac{r^2}{2})+ \frac{1}{2}Br^2$$
    for the Hamiltonian.
    For the ##\frac{d^2}{dx_1^2}## part I get that:
    $$ \frac{d^2}{dx_1^2}=\frac{d^2}{dR^2}(\frac{dR}{dx_1})^2=\frac{d^2}{dR^2}$$
    but I could have also used ##r## so am I just allowed to pick one thereby combining the two derivative terms (since the answer is the same for ##\frac{d^2}{dx_2^2}##) or am I doing something wrong?
     
  7. Mar 7, 2017 #6

    DrClaude

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    Staff: Mentor

    The ##\frac{d^2}{dx_1^2}## operator will act on a function of ##R## and ##r##, which both depend on ##x_1##.

    Have a look at https://www.math.hmc.edu/calculus/tutorials/multichainrule/
     
  8. Mar 7, 2017 #7
    Oh I see, I think I got it. Thanks!
     
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