For simple harmonic oscillator H = p^2/(2m) + (K/2)x^2 turn on
the potential V(x) = cx^4
Calculate the first order correction to the ground state energy.
Prove that the first order energy correction vanishes for all n = odd integer excited state levels.© BrainMass Inc. brainmass.com March 4, 2021, 8:06 pm ad1c9bdddf
I'm going to use the following notation:
If A is an operator, then C(A) denotes its Hermitian conjugate (usually denoted as A-dagger in books).
Let's write the Hamiltonian as:
H = p^2/(2m) + 1/2 m omega^2 x^2
(omega = sqrt(K/m) )
This can be written in terms of the annihilation and creation operators as:
H = h-bar omega [C(a) a + 1/2]
a = Squareroot[m omega/(2 h-bar)](x + i/(m omega) p)
C(a) = Squareroot[m omega/(2 h-bar)](x - i/(m omega) p)
The commutation relation is:
[a, C(a)] = 1
These operators act on the eigenstates n> of the Hamiltonian as follows:
C(a)|n> = Squareroot[n+1]|n+1>
a|n> = Squareroot[n]|n-1> for n > 0, if n = 0 then:
a|0> = 0
Expressing x and p in terms of a and C(a) gives:
x = ...
A detailed solution is given.