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QM: Work done due a time dependent perturbation

  1. Mar 6, 2015 #1
    1. The problem statement, all variables and given/known data
    A quantum particle of mass ##m## is bound in the ground state of the one-dimensional
    parabolic potential well ##\frac{K_0x^2}{2}## until time ##t=0##. Between time moments of ##t=0## and ##t=T## the stiffness of the spring is ramped-up as ##K(t) = K_0 +\frac{t}{T}[K_1-K_0)\quad 0\leq t\leq T\quad (3)## and it stays equal to ##K_1## afterwards, at ##t > T##. The overall change of the stiffness is small: ##|K1 − K0| ≪ K0##.
    (a) Does any work need to be done to exercise this ramp-up? Can this work be dependent
    on the duration T of the ramp-up and why?
    (b) If the answer to the latter question is positive, evaluate the needed work in the lowestorder
    of the perturbation theory.
    (c) Analyze the T-dependent part of the work, if any, and find out how much of the variation
    in the amount of work is achieved between the adiabatic (##T → ∞##) and nearly
    instantaneous (##T → 0##)limits?

    2. Relevant equations

    $$V(t)=\begin{cases}
    0;\quad -\infty<t<0\\
    \frac{tm(\omega_1^2-\omega_0^2)x^2}{2T};\quad 0\leq t\leq T\\
    \frac{m(\omega_1^2-\omega_0^2)x^2}{2};\quad T<t<\infty
    \end{cases}$$

    (Might need)

    $$P_{n\rightarrow k}=\frac{1}{\hbar^2\omega_{kn}^2}\Big|\int_{\infty}^{\infty}dt\dot{V}_{kn}e^{i\omega_{kn}t}\Big|^2$$

    Since $$ \lim_{t\rightarrow\infty}V(t)\neq 0$$

    where ##\omega_{kn}=\omega_{k}-\omega_{n}##

    3. The attempt at a solution

    a)

    Okay so yes, work needs to be done in order to exercise this ramp-up. My intuition says that the work will depend on ##T ##.

    Adiabatic case:

    As the perturbation is gradually applied, the state ##|\psi\rangle## will evolve with the Hamiltonian: in other words it will be a simultaneous eigen-state of the Hamiltonian. In this instance I'm sure the work done by the system is simply ##\langle H_f \rangle-\langle H_i \rangle##.

    Diabatic case:

    As the perturbation is applied instantaneously, the state ##|\psi\rangle## will remain the same since the system will have no time to adjust itself. ##|\psi\rangle## will become a linear combination of the eigen-states of the new Hamiltonian.

    (okay now here's where I lose my train of thought)

    Initially, the particle will jump between many different energy levels since ##|\psi\rangle## is a linear combination of states of the new Hamiltonian. Intuitively I feel that after sometime the system will establish itself and all will be well again: the particle will enter the new ground state. I'm not sure if I can use the definition of work that I used in the adiabatic case . Furthermore, Griffiths gives the example of the particle in the infinite well. Pull on the wall quickly enough and the probability density remains unchanged and there is no work done, but in this case by turning on ##V(t)## we are making the "well" narrower so I don't think you can make the same argument here. Is there a flaw in my logic that I'm just not seeing or something I'm missing?

    b)

    For this I'm guessing I need to evaluate ##W=\langle H_f \rangle-\langle H_i\rangle##


    c)

    Once I get parts a) and b) I can handle this part.

    Thanks for your help in advance!
     
    Last edited: Mar 6, 2015
  2. jcsd
  3. Mar 8, 2015 #2
    Nevermind everyone i figured it out:

    a)

    Yes, work must be done to "ramp it up". The work will depend on the value of ##T##:

    Diabatic case: If the perturbation is applied suddenly, then the system has no time to adjust and establish itself. The state would have to remain the same since the system has no time to respond. So the work done on the system is the work to carry the state from the old ground state energy to the new qround state energy plus the work to keep the particle confined to its initial state.

    Adiabatic case: If the perturbation is applied gradually, then the system has time to adjust and establish itself. So the state evolves along with the Hamiltonian. So by the end of the application of the perturbation, the particle will be in the new ground state of the new Hamiltonian. So the only work done on the system is that to carry the state from the old ground state energy to the new ground state energy.

    b)

    In order to look at the dependence of the work ##W## on ##T## I needed to look at ##\langle H\rangle=\hbar\omega_0\Big(n+\frac{1}{2}\Big)\rightarrow \langle H(t)=\frac{\hbar\omega(t)}{2}##(for ##n=0##)

    from:
    ##k(t)=k_0+\frac{t}{T}\Big(k_1-k_0\Big)\rightarrow \omega(t)=\sqrt{1+\frac{t}{T}\Big(\frac{\omega_1^2-\omega_0^2}{\omega_0^2}\Big)}##

    using:

    ##k_0>>|k_1-k_0|\rightarrow \omega_0^2>>|\omega_1^2-\omega_0^2|##

    ##\omega(t)\simeq \omega_0\Big[1+\frac{t}{T}\Big(\frac{\omega_1^2-\omega_0^2}{\omega_0^2}\Big)\Big]##

    Now I simply evaluate:

    ##
    W=\langle H_{t=t}\rangle-\langle H_{t=0}\rangle=\frac{\hbar\omega_0}{2}\Big[\frac{t}{T}\Big(\frac{\omega_1^2-\omega_0^2}{\omega_0^2}\Big)\Big]
    ##

    looking at the limits:

    Adiabatic Limit: ##T\rightarrow\infty## yields ##W_A\rightarrow 0##
    Diabatic Limit: ##T\rightarrow 0## yields ##W_D\rightarrow \infty##

    So ##W_D>W_A##
     
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