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Homework Help: Matrix elements of x^3

  1. Feb 22, 2010 #1
    1. The problem statement, all variables and given/known data
    I'm working on problem 3 on page 136 in Landau and Lifgarbagez Quantum mechanics nonrelativistic theory.
    The problem is to determine the energy levels of an anharmonic linear oscillator with
    H=H0 + a*x3 + b*x4

    (where H0 is the hamiltonian of the harmonic oscillator).
    To solve this using pertubation theory, I need to find the matrix elements of x3 and x4. I have tried many times, but I don't get the same answer as Landau.

    2. Relevant equations
    Equation (23.4) gives the matrix elements of x: xn,n-1=xn-1,n=sqrt(nh/2mw).
    I have no problem getting the pre-factors right, so I simplify this expression by not writing explicitely the the constant term and the square root, so I get

    xn,n-1=xn-1,n = n

    I use the standard rule for matrix multiplication:
    (AB)n,m = SUMr(An,r Br,m)


    3. The attempt at a solution
    I get the following non-zero elements of x2 using the above formulas
    x2n,n=xm,m-1 xn-1,n +xn,n+1xn+1,n = n2+ (n+1)2

    x2n,n-2 = x2n-2,n = xn,n-1 xn-1,n-2 = n(n-1)

    This gives next
    x3n,n-1=xn,n-1x2n-1,n-1 + xn,n+1x2n+1,n-1 = 3n3+2n

    I can't find any flaw in this, yet in Landau the solution is given as 9n3
     
  2. jcsd
  3. Feb 22, 2010 #2

    vela

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    You missed one of the non-zero terms of x2, namely x2n,n+2.
     
  4. Feb 23, 2010 #3
    isn't that one essentially the same as x2n-2,n if I rename n --> n +2? (giving x2n,n+2 = (n+1)(n+2) )
     
  5. Feb 23, 2010 #4

    vela

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    You never said what x2n-2,n was in your OP.

    Just out of curiosity, is there a reason you're using matrices rather than calculating these results using the annihilation and creation operators?
     
  6. Feb 23, 2010 #5
    actually I did, I wrote it's equal to x2n,n-2 :)

    No good reason, I just didn't think about using creation and annihilation operators. I talked to some others today who also suggested doing that, so I'm gonna try doing it that way later today.
    But still, it would be nice to be able to solve it this way.
     
  7. Feb 23, 2010 #6

    vela

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    Oops, I can't believe I repeatedly missed that!
     
  8. Mar 29, 2012 #7
    For the matrix element (x^3)_(n-1,n), you have to consider the following 3 possible "transitions":

    1) (n-1) -> n -> (n+1) -> n
    2) (n-1) -> (n-2) -> (n-1) -> n
    3) (n-1) -> n -> (n-1) -> n

    The contribution to (x^3)_(n-1,n) are as follows (all of the following are multiplied by
    [hbar/(2 m w)]^(3/2) ):
    From 1), you get a contribution of Sqrt(n) (n+1),
    From 2), you get a contribution of Sqrt(n) (n-1),
    From 3), you get a contribution of Sqrt(n) (n).

    If you add them, you get Sqrt(n)(3n) = Sqrt(9 n^3).
    So (x^3)_(n-1,n) = [hbar/(2 m w)]^(3/2) Sqrt(9 n^3)
    = [hbar/(m w)]^(3/2) Sqrt[(9 n^3)/8] just like Landau and Lifgarbagez calculated in problem 3.
     
  9. Mar 29, 2012 #8
    Note that the way I have written the "transitions" above is translated in terms of the matrix elements x_(m,n) as follows (taking the example of 1) ):

    1) (n-1) -> n -> (n+1) -> n is x_(n-1,n) x_(n,n+1) x_(n+1,n)

    and the same for the other two.

    And for the last part of the problem, if you are going to calculate the 2nd approximation to E_n from the x^3 term using Eq (38.10):
    In addition to (i) (x^3)_(n-3,n) and (ii) (x^3)_(n-1,n), you will also need to calculate the matrix elements (iii) (x^3)_(n+3,n) and (iv) (x^3)_(n+1,n). Then you add the abs squared value (i.e. |V_(m,n)|^2 / (E0_n - E0_m) in 38.10) of (i)-(iv). Also take note that the denominator (i.e. (E0_n - E0_m) in 38.10) is negative for terms (iii) and (iv) and positive for terms (i) and (ii).
     
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