Matrix Elements of x^3 Perturbation Theory

[henrik729]

Homework Statement

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.

Homework Equations

Equation (23.4) gives the matrix elements of x: x_n,n-1=x_n-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

x_n,n-1=x_n-1,n = n

I use the standard rule for matrix multiplication:
(AB)_n,m = SUM_r(A_n,r B_r,m)

The Attempt at a Solution

I get the following non-zero elements of x² using the above formulas
x²_n,n=x_m,m-1 x_n-1,n +x_n,n+1x_n+1,n = n²+ (n+1)²

x²_n,n-2 = x²_n-2,n = x_n,n-1 x_n-1,n-2 = n(n-1)

This gives next
x³_n,n-1=x_n,n-1x²_n-1,n-1 + x_n,n+1x²_n+1,n-1 = 3n³+2n

I can't find any flaw in this, yet in Landau the solution is given as 9n³

 

[vela]

You missed one of the non-zero terms of x², namely x²_n,n+2.

 

[henrik729]

isn't that one essentially the same as x²_n-2,n if I rename n --> n +2? (giving x²_n,n+2 = (n+1)(n+2) )

 

[vela]

You never said what x²_n-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?

 

[henrik729]

actually I did, I wrote it's equal to x²_n,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 going to try doing it that way later today.
But still, it would be nice to be able to solve it this way.

 

[vela]

actually I did, I wrote it's equal to x²_n,n-2 :)

Oops, I can't believe I repeatedly missed that!

 

[centaure]

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.

 

[centaure]

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).