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# Perturbed harmonic oscillator

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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.

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#### Solution Preview

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]

where:

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 = ...

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