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    Harmonic Oscillator and Wave Packets

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    Consider the 1-D harmonic oscillator, with Hamiltonian H=p²/2m+½mw²x²=hw(a¬a+1/2), and with energy eigenstates H|n}=En|n}=hw(n+1/2)|n}. The eigenvectors of the annihilation operator a are known as coherent states: a|z}=z|z}, where z is in general a complex number (a is not Hermitian, so z is not necessarily real). Take |z} to be normalized: {z|z}=1.

    a) Find an expression for |z} as a linear combination of the eneergy eigenstates |n}. Make sure it is normalized. Calculate {z1|z2}.
    b) If the oscillator is in the state |z} at the instant that the oscillator energy is measured, what is the probability of obtaining the result En? What is the expectation value for the measured energy, {z|H|z}?
    c) Show that for this state, change in x*change in p = h/2, so that the state is a minimum uncertainty wave packet.

    Please see the attached for the proper format and notations.

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    https://brainmass.com/physics/energy/harmonic-oscillator-wave-packets-31256

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    Solution Preview

    Please see the attached file for the adequate notation.

    From the problem, we have a|z> = z|z>

    Suppose, |z> = ?n fn (z) |n> and a|z> = ?n fn (z) ?n |n-1>

    z|z> = ?n fn (z) ?n |n-1>
    z ?n fn (z) |n> = ?n fn (z) ?n |n-1>

    So, we have the following:
    z?n fn (z) = fn+1 (z) ?(n+1)
    fn+1 (z) = (z/?(n+1)) fn (z) ...

    Solution Summary

    The answer provides an expression for the complex number z, given the energy eigenstates within the harmonic oscillator. Using probabilities and the Hamiltonian equation, this expression is determined, alongside the expectation value for the measured energy.

    $2.49

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