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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]


    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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