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I Why not describing the radiatation of atom spectrum by TDSE

  1. May 30, 2017 #1
    The radiation of an atom was interpreted by time-independent schrodinger equation:electron was transformed from high energy level state to lower and emit a photon.


    Could we treat this process through a wavefunction ##{\psi}(t)##? Before emiting,the system's wavefunction is ##{\psi}(0)## and after emiting photon,it is ##{\psi}(t_0)##.
    ##{\psi}(t)## is constrained by time-dependent schrodinger equation and contain all information of the system.


    Is there any papers incorporate photon emiting in wavefunction as well as the method to define ##{\psi}(t=0)##?
     
  2. jcsd
  3. May 30, 2017 #2

    blue_leaf77

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    You are definitely mistaken. The formal approach to describe radiative transitions in atoms is in fact through the use time-dependent perturbation theory. When one takes EM radiation into account in the Schroedinger equation, e.g. for the purpose of deriving the transition rate, the time-dependent Schroedinger equation is inevitable. Upon employing certain approximation it's possible to reduce this problem to a perturbation treatment.
    You determine it yourself, depending on where you want the electron to start from before the transition. However, it's usually assumed to be one of the energy eigenstates.
     
  4. May 30, 2017 #3

    hilbert2

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    Stimulated emission can be modeled with the nonrelativistic Schrödinger equation by adding an oscillating term in the potential energy to represent the EM field, as in the equation

    ##i\hbar\frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2\mu}\nabla^2 \Psi (x,y,z,t) - \frac{e^2}{4\pi \epsilon_0 r}\Psi (x,y,z,t) + C\sin (\omega t)x\Psi (x,y,z,t)##

    but spontaneous emission must be handled with quantum electrodynamics (by using perturbation theory, too, but there it's more difficult than the nonrelativistic case).
     
  5. May 30, 2017 #4
    Thanks.I check my book and I am mistaken.But I am still confused.
    As the book said,
    ##{\phi}_n## is the eigenstate of time-independent schrodinger equation with eigenvalue ##E_n##.
    Adding the time-dependent perturbation ##{\hat{H}}^{'}(t)##, wave function becomes ##\Psi=\sum\limits_{m=0}^{\infty}a_m(t){\phi}_m\exp{(-\frac{i}{\hbar}{E_m}t)}##,and the ##a_m(t)## was interpreted as the probability for the wave function transform from inital ##{\phi}_n\exp{(-\frac{i}{\hbar}E_nt)}## to final ##{\phi}_m\exp{(-\frac{i}{\hbar}E_mt)}##.The initial function and final function satisfied time-independent schrodinger eqution.Why not just choose the ##\Psi## as the final function?


    Why the oscillating term was very large far away from the system? Is it more reasonable to divide ##\sin(\omega t)## by ##x##?
    With which kind of perturbation potential,we could distinguish adsorption or emission?
     
  6. May 30, 2017 #5

    hilbert2

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    A potential energy that grows in linear proportion to ##x## means that there's a constant force towards negative ##x##-direction everywhere. This is a reasonable approximation if the beam of light (which causes the transitions), as well as its wavelength, are much larger than a single atom.

    In solving the spontaneous emission problem the whole system is different and can contain an undetermined number of photons besides the emitting/absorbing atom or molecule.
     
    Last edited: May 30, 2017
  7. May 30, 2017 #6

    blue_leaf77

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    One usually is interested in finding out the probability of the atom being found in some other eigenstate different from the initial one after the transition because this is directly related to the transition rate between the initial state and the final one. The transition rate is a quantity that can be fairly easily measured in experiments.
     
  8. May 30, 2017 #7
    Thanks!
    I means with the perturbation added, the atom in ##{\phi}_n## with time evolving to be fould in ##\Psi=\sum\limits_{m=0}^{\infty}a_m(t){\phi}_m\exp{(-\frac{i}{\hbar}{E_m}t)}##,this is a mixture of many eigenstates.However,as the book said,the atom would be found at another eigenstate in the end.The mixture of many eigenstates seems "collapsing" to an eigenstate,I know this word--collapsing,but I think the "collapsing" must be a process control by time-dependent schrodinger eqution.

    This is my reason :
    If we in momentum representation,atom state "collapsing" to an eigenstate function ,which may be a different function with the eigenstate fucntion in position representation.The final wavefunction collapse to positon eigenstate or momentum eigenstate seems not right.
     
    Last edited: May 30, 2017
  9. May 30, 2017 #8

    vanhees71

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