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For a time dependent wavefunction, are the instantaneous probability densities meaningful? (The question applies for instances or more generally short lengths of time that are not multiples of the period.)

What experiment could demonstrate the existence of a time dependent probability density?

Can an isolated system be described by a time dependent wavefunction? How would this not violate conservation of energy?

I see the meaning of the time averaged probability density. Is the time dependence just a statistical construct?

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Hello Praxeo, Welcome to Physics.SE. Please try to ask questions only related to the topic (of yours) or consider a revision to focus on your question. The problem (for now) is, you're asking lot of questions... –  Waffle's Crazy Peanut Nov 15 '12 at 16:38
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2 Answers

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1) Why do you believe that instantaneous probability densities are not meaningful?

2) Essentially any non-stationary state for which you need to compute time-dependent wavefunctions: e.g. chemical reaction dynamics, particle scattering, etc.

3) Yes, the time dependant Schrödinger equation applies to isolated systems.

4) By definition energy is conserved in an isolated system. Moreover, the Schrödinger equation conserves energy because the generator of time translations is the Hamiltonian and this commutes with itself $[H,H]=0$, i.e. energy is conserved. For isolated systems, the Hamiltonian is time-independent (explicitly) and the time-dependent wavefunction $\Psi$ has the well-known form $\Psi = \Phi e^{-iEt/\hbar}$, with $E$ the energy of the isolated system.

5) I do not understand the question.

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In (4), one needs a further condition that the Hamiltonian is itself time-independent, $\frac{\partial H}{\partial t} = 0$. –  Stan Liou Nov 16 '12 at 9:34
    
As well, one has to distinguish the energy certainty from the energy conservation. –  Vladimir Kalitvianski Nov 16 '12 at 15:17
    
@StanLiou: Conservation, by definition, implies zero production $d_iH/dt=0$. If the Hamiltonian has explicit time dependence then the equation of motion contains a 'flow' term $d_eH/dt$ but the production term continues being zero. –  juanrga Nov 16 '12 at 18:28
    
@VladimirKalitvianski: Not sure what do you mean, but the conservation law $[H,H]=0$ is independent of the kind of quantum state. –  juanrga Nov 16 '12 at 18:32
    
It's certainly correct and tautologous to say that energy is conserved in an isolated system. But if your last sentence were correct, all quantum systems whatever would conserve energy, because $[H,H] = 0$ is an exact identity and $H$ is always the generator of time translation. Hence, I expected the point of $[H,H] = 0$ to be a reference to $\frac{dA}{dt} = \frac{\partial A}{\partial t} + \frac{1}{i\hbar}[A,H]$ in the Heisenberg picture or analogous expectations in the Schrödinger picture. In the Lagrangian formalism, energy through Noether's theorem also needs no explicit time dependence. –  Stan Liou Nov 16 '12 at 19:13
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Yes, $|\psi(t)|^2$ is an instantaneous probability density.

Passage of a wave packet can be experimentally observed.

An isolated system can be in a superposition of different energy eigenfunctions. It does not violate the energy conservation law because initially the system is not in an eigenstate - it has some energy uncertainty at $t=0$. This uncertainty evolves as any other uncertainty.

EDIT: Let us make a superposition of two states: $$\psi(t)=c_1\psi_1(x)e^{-iE_1 t}+c_2\psi_2(x)e^{-iE_2 t}.$$ It means that we can find in experiment the system in state 1 with probability $|c_1|^2$ and in state 2 with probability $|c_2|^2$. The system is free and this is due to coefficients $c_1$ and $c_2$ being constant in time (occupation numbers do not depend on time).

Measuring the system energy will give sometimes $E_1$ and sometimes $E_2$, with the same probabilities. So initially and later on the system does not have a certain energy. The state $H\psi$ depends on time as $$H\psi=c_1 E_1 \psi_1(x)e^{-iE_1 t}+c_2 E_2 \psi_2(x)e^{-iE_2 t}.$$ It is not an eigenstate of the Hamiltonian, so the time derivative $\partial\psi/\partial t$ is not proportional to $\psi$.

The Hamiltonian expectation value, however, does not depend on time: $$\langle\psi|H|\psi\rangle = |c_1|^2 E_1 + |c_2|^2 E_2 = const.$$ In other words, it is the energy expectation value that conserves, not the energy. The latter is undefined, uncertain in this free state.

You invoke the "energy conservation law" $dH/dt=0$ which is an operator relationship. If the system has a certain energy $E_n$ in the initial state, this value remains the system energy in later moments, so your "conservation law" may be cast in a form $dE(t)/dt=0$ that means $E=const=E(0)=E_n$.

But if the system does not have a certain energy at the initial state $\psi(0)$, then there is no $E(0)$ to conserve and your operator relationship turns into conservation of the expectation value.

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