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Questions tagged [time-evolution]

The quantum mechanical time evolution operator governs how observables and/or states evolve during finite time steps, and is always unitary. Use this tag for questions about the time evolution operator, or the different equations of motion in the Schrödinger/Heisenberg/Dirac pictures. For time-independent Hamiltonians, the time evolution operator is simply exp(-iHt).

3 votes
1 answer
55 views

Do the phases of the wavefunctions for non-interacting quantum systems evolve faster if those systems are conceived of as components within some larger system rather than each being treated separately?...
benjimin's user avatar
  • 1,744
-4 votes
4 answers
117 views

In standard quantum mechanics, the formalism successfully predicts transition probabilities between quantum states, but the transition itself is often treated as instantaneous or left without a ...
Joyanta Biswas's user avatar
4 votes
1 answer
100 views

I am reading Shaul Mukamel's Principles of Nonlinear Spectroscopy. When talking about time evolution in Liouville space, he says in page 61 that $$\mathscr{U}_{jk,j^\prime k^\prime}(t,t_0)=U_{jj^\...
余力圣's user avatar
8 votes
1 answer
218 views

I am investigating the dynamics of two interacting bosonic modes with self-Kerr and cross-Kerr nonlinearities. The Heisenberg equations of motion are: $$ \frac{d\hat a}{dt} = - i\omega_a \hat a + \...
Rob's user avatar
  • 181
-1 votes
1 answer
110 views

As we know, instantaneous eigenstates of energy are the states in which the wavefunction collapses instantly after we measure energy. And we also know that instantaneous eigenstates don't satisfy the ...
S K's user avatar
  • 181
2 votes
2 answers
315 views

Sorry about the stupid title. I have no idea how to summary the following content into a concise title. This is related to exercise 5.8 of Cohen-Tannoudji's QM textbook. Let us consider a 1D harmonic ...
MrBetadine's user avatar
5 votes
1 answer
302 views

A very standard and widely discussed fact in quantum mechanics is that pure quantum states are only defined up to an overall phase factor.$^1$ That is, the state vectors $|\psi\rangle$ and $e^{i \...
tparker's user avatar
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5 votes
1 answer
78 views

I am reading a derivation of the interaction picture in subsection 7.2.4 from Schwartz's QFT book, and I am confused about how the interaction-picture potential $V_I(t)$ is being defined and used. At ...
Maj's user avatar
  • 51
4 votes
1 answer
130 views

When the potential is time varying,there are no energy (total energy) eigenstates.So what energies values we obtain when we measure the energy of a particle in such a potential?Does "not having ...
S K's user avatar
  • 181
3 votes
0 answers
97 views

Newton's Laws can be presented as a statement regarding the acceleration and time evolution of objects: $$\ddot {\bf x} = F(\dot {\bf x},{\bf x}) $$ In other words, the time evolution of a physical ...
Nathaniel Bubis's user avatar
5 votes
3 answers
499 views

Given two bosonic operators $A$, $B$ (in the Heisenberg picture) in a QFT, the time-ordered product of $A$ and $B$ is defined as $$ T\{A(t_1)B(t_2)\}=\theta(t_1-t_2)A(t_1)B(t_2)+\theta(t_2-t_1)B(t_2)A(...
Hezaraki 's user avatar
1 vote
0 answers
116 views

On Chapter 7 of Fetter & Walecka, the authors prove Dyson formula for the (imaginary time) propagator $U(t,t_{0}) = e^{H_{0}t_{0}}e^{-H(t_{0}-t)}e^{-H_{0}t}$, where I am ommitting the $\hbar$'s. ...
InMathweTrust's user avatar
1 vote
0 answers
175 views

Follow-up question to How to deal with explicit time dependence in the Heisenberg picture? Time evolution of an operator A (in Heisenberg picture) is given as: $$\frac{\mathrm{d}\hat{A}_H}{\mathrm{d}t}...
DarkMIR4GE's user avatar
1 vote
0 answers
59 views

This might be a conceptual question more than a technical one, however it would be very useful to see different conceptual approaches to address this problem physically/mathematically. The question ...
BrightNeutrino's user avatar
5 votes
1 answer
199 views

In the Schrödinger, Heisenberg, and interaction pictures, the time evolution of an operator $A$ is defined differently. In the $\textbf{Heisenberg picture}:$ \begin{equation} A_H(t) = e^{i(H_0 + V)t} ...
seeker's user avatar
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