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If we look at a scattering process in the Schrödinger picture for a Hamiltonian $H = H_0(t) + V(t)$ where $H$ is independent of time (because we examine a theoretical situation after accelerating particles and turning off accelerators, so the system is isolated) and $V$ depends on time (because interaction term in general depends on distance between particles) and so will be $H_0$.

So the eigenstates of $H_0$ are dependent on time and I don't know how is their evolution? Of course their evolution as if they are state of a physical system is irrelevant. I want their evolution because of their dependence on $H_0(t)$.

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closed as unclear what you're asking by ACuriousMind, user36790, CuriousOne, Gert, garyp Jun 1 '16 at 0:56

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ I edited your question for grammar, but it is still not clear to me what you are asking. Why do you split a time-independent Hamiltonian into two time-dependent parts? What has this to do with "turning off accelerators"? Why are you interested in eigenstates of $H_0(t)$? It appears you are trying to do scattering theory/perturbation theory but have not properly understood it. $\endgroup$ – ACuriousMind May 30 '16 at 10:05
  • $\begingroup$ thanks. really I'm trying to understand lorentz invariant theory of scattering in schrodinger picture. (I turn off accelerators long before collision to have an isolated (but moving fast!) system of particles so I can have time independent hamiltonian $H$ for scattering.) I guess understanding eigenstates of time dependent operator $H_0$ (time dependent operator even in schrodinger picture!) will help me understanding S-matrix. $\endgroup$ – moshtaba May 30 '16 at 10:57
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The eigenstates of a time-independent Hamiltonian $$ H |\psi_j \rangle = E_j|\psi_j\rangle$$ have the usual rotating-phase time dependence in the Schrödinger picture: $$ |\psi_j(t)\rangle = |\psi_j(t_0)\rangle \cdot \exp(E_j(t-t_0)/i\hbar) $$ However, your formula indicates that $H_0$ and $V$ are time-dependent. So if the state vector is an eigenvector of $H_0(t)$ at one moment $t$, it will almost certainly not be an eigenstate of $H_0(t')$ at another moment $t'$ – because the operator is probably changing in a way that doesn't preserve the eigenvectors' being eigenvectors. For that reason, the simple "changing phase" Ansatz isn't a solution to the Schrödinger's equation.

For a generic Hamiltonian $H_0(t)$, there can't exist any state vector $|\psi(t)\rangle$ that simultaneously solves Schrödinger's equation (with the Hamilonian $H$ or even with $H_0(t)$); and that remains an eigenstate of $H_0(t)$ at each moment of time.

Because you say that $H$ is time-independent and $H_0,V$ seem to be time-dependent, it seems that all the formulae you are describing are in the Heisenberg picture, not Schrödinger's picture. In the Heisenberg picture, the state vector is independent of time (a constant function of $t$). In the Schrödinger's picture, the dependence of $H_0$ and $V$ on time would have to be "explicit" (the dynamical dependence is encoded in the evolution of the state vector in this picture) and in that case, it would be unreasonable for the explicitly time-dependent parts of $H_0,V$ to cancel in $H$.

In the Heisenberg picture, the dynamical time dependence is included in the evolution of the operators, $$H_0(t) = \exp(-Ht/i\hbar) H_0(0) \exp(Ht/i\hbar) $$ and similarly for $V$ instead of $H_0$ (and for any operator).

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  • $\begingroup$ about last line of the paragraph "Because you say ..." I think (am I wrong?) this experiment have time independent $H$ (because system is isolated). we can write $H=H_0+V$, now because $V$ is time-dependent ($V$ depends on distance between particles that depends on time) $H_0$ trivially become time dependent so that $H=H_0 +V$ remains true. $\endgroup$ – moshtaba May 30 '16 at 11:16
  • $\begingroup$ Sorry, in Schrödinger's picture, if you add an explicitly time-dependent perturbation $V(t)$, e.g. a variable external field, then be sure that the total Hamiltonian $H(t)$ will be time-dependent due to this explicit dependence, too. The total Hamiltonian is only time-independent for closed systems without any explicit dependence. $\endgroup$ – Luboš Motl May 30 '16 at 11:20
  • $\begingroup$ thanks, Let we accelerated two particles from left and right and then turn off accelerators, and examine the system of only this two particles in schrodinger picture . then $H$ is independent of time, is't it? if I split it into $H_0$ and $V$ then this two operators are time-independent?! what about intensity in interaction that is time dependent (distance dependent) that is coded in $V$?! thanks. $\endgroup$ – moshtaba May 30 '16 at 11:36
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    $\begingroup$ If (and when, e.g. for those values of $t$ when) $V$ explicitly depends on time, $H$ depends on time, too. I don't understand why we need 6 messages to communicate this self-evident point. In your original question, you said that $H$ was time-independent while $V$ was time dependent. I wrote that this is either implausible, or it's what happens in the Heisenberg picture. $\endgroup$ – Luboš Motl May 30 '16 at 12:05

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