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A Is my solution of time-dependent Schrodinger equation right?

  1. Feb 4, 2017 #1
    The problem looks very simple. We have a time-dependent Hamiltonian:
    $$H(t) = B(t)H_0$$,
    where ##B(t)## is a numerical function, and matrix ##H_0## is time-indpendent.
    Let us consider:
    $$B(t) = \begin{cases}
    1,&\text{for $0\leq t\leq t_0$}\\
    A,&\text{for $t>t_0$.}
    \end{cases}$$
    Also, let us consider that ##H_0## has an eigenstate ##|n\rangle## :
    $$ H_0|n\rangle = \varepsilon_n |n\rangle$$.
    The problem is to find solution ##|\psi(t)\rangle## of the Schrodinger equation
    $$i\frac{\partial}{\partial t}|\psi(t)\rangle = H(t)|\psi(t)\rangle ,$$
    when ##|\psi(0)\rangle = |n\rangle##.

    My solution is:
    $$|\psi(t)\rangle = e^{ -i\varepsilon_n\int_0^t B(t')dt'}|n\rangle$$
    So, there are no transitions to other levels from the ##|n\rangle## state.

    Some people disagree with me because of "jump function makes exact solution not possible".

    Is my solution right or not?

    Thanks!
     
  2. jcsd
  3. Feb 4, 2017 #2

    Orodruin

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    Your solution is fine.

    What "people"?
     
  4. Feb 4, 2017 #3
    Thanks.

    Moderator note: links to external debate removed

    (It makes no sense to discuss on two different forums simultaneously. If you like, you can insert the arguments against your solution manually, such that we have a common basis for discussion without switching forums. However, since the author of these arguments isn't available to defend them, it likely will end nowhere.)
     
    Last edited by a moderator: Feb 4, 2017
  5. Feb 4, 2017 #4
    This is the "argument" against my solution:
     
  6. Feb 4, 2017 #5

    Orodruin

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    It is a simple matter of inserting your solution and checking that it satisfies the appropriate differential equation.

    In general when you have a Hamiltonian that does not commute at different times, you will not get the solution
    $$
    \lvert \psi(t)\rangle = e^{-i\int_0^t H(t') dt'} \lvert \psi(0)\rangle,
    $$
    but a time ordered exponential. In your case, the Hamiltonian at different times commute and your solution is fine. In this case, perturbation theory has nothing to do with it, the solution is exact - as can be checked by simply plugging it into the differential equation. However, if you do apply perturbation theory, the series is convergent everywhere (not only for small deviations) and converges to ... your solution.

    This is simply false. The Dyson series is the series of a simple matrix exponential in your case. I would strongly doubt the expertise of the person who wrote this.
     
  7. Feb 4, 2017 #6
    Orodruin: "..In your case, the Hamiltonian at different times commute.." and "..The Dyson series is the series of a simple matrix exponential in your case.."

    - Are You sure these two statements indeed take place in the "leap" time interval where even such quantity as Hamiltonian of quantum system - is undefined ? (Remark: we are talking here about B(t) as a true step-function not as something regular which can turn into step-function only in a certain limiting case.)
     
  8. Feb 4, 2017 #7

    Orodruin

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    Yes. There really is nothing strange going on here. The Hamiltonians at arbitrary times commute which makes te system's time evolution trivial. It is like solving an ordinary differential equation with the same type of inhomogeneity.

    You can chose to interpret the differential equation in a distributional sense at the jump if you must. It does not change the argumentations.
     
  9. Feb 4, 2017 #8
    Orodruin: "It is like solving an ordinary differential equation with the same type of inhomogeneity" - But on the leap (or "jump" ) time interval You cannot define any operator, meaning all related things such as matrices, commutation relations..
    Orodruin: "You can chose to interpret the differential equation in a distributional sense at the jump" - Please, specify what do You mean by this concerning this particular case?
     
  10. Feb 4, 2017 #9

    PeterDonis

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    What time interval is that? The Hamiltonian is well-defined for all times ##t##. It has a "jump discontinuity" in its value, but that doesn't mean there is any value or "interval" of ##t## where it is undefined.
     
  11. Feb 4, 2017 #10
    PeterDonis: "..The Hamiltonian is well-defined for all times t.." - It seems, if so, (and if definite causality law connects system's state at the moment "t" with its state at the moment "t+dt") then You should be able, at least, to write down such connection for Hamiltonian, or simpler, just for function B(t), i.e. the connection between B(t) and B(t+dt). Usually, one can do that by using differential i.e. in the form: B(t+dt)=B(t) + (dB(t)/dt)dt , but, how do You think, what meaning has quantity: (dB(t)/dt)dt - when t=t_0 ? - I think the differential of the function B(t) (and, hence, Hamiltonian) is undefined at the point t=t_0 since at this point derivative (dB(t)/dt) - diverges whereas the smallness of "dt", in general , isn't connected with the grade of (dB(t)/dt) divergence at t=t_0. That is why, I think, the infinitesimal interval "dt" "around" the value t=t_0 - represent the interval of function B(t) (and Hamiltonian) uncertainty.
     
  12. Feb 4, 2017 #11

    Orodruin

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    I am sorry, but what you are saying does not make much sense at all. ##B(t)## does not have to be differentiable in order to integrate it - it needs to be integrable. The important thing is not ##(dB/dt)dt## (even if that is perfectly well defined in a distributional sense), it is the integral of ##B(t)##.

    If you are unfamiliar with distributional derivatives, you probably should not be replying to an A-level thread. Regardless, they are not necessary here - the function ##B(t)## is integrable. In fact, it does not matter what value it takes at a single point with measure zero, what matters is its integral (or in the case of Hamiltonians that do not commute at different times, the time-ordered exponential).

    To be concrete, the time evolution of a general quantum system from time ##0## to ##t## is given by the unitary operator
    $$
    S(t,0) = \mathcal T e^{-i \int_0^t H(\tau) d\tau},
    $$
    where ##\mathcal T## denotes time ordering. In this case, time ordering is irrelevant since ##[H(\tau),H(\tau')] = 0## and therefore
    $$
    S(t,0) = e^{-iH_0 \int_0^t B(\tau)d\tau}.
    $$
    Note that the Hamiltonian's value at any particular time is not relevant for the time-evolution of the system, only its time ordered integral is. What you are saying is essentially the equivalent of claiming that
    $$
    \int_0^t B(\tau) d\tau
    $$
    does not have a definite value for ##t > t_0##.

    Also, please use the quote feature. Your posts are unnecessarily difficult to read.
     
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