# When Did the Universe Turn From a Quantum System to a Multi-particle System?

For completeness here is a timeline of the universe using the standard model of particle physics and General Relativity ( and thermodynamics etc etc)Take your choice going backwards in time.The universe we have now is after the transparency of light. Particles as we study them now, existed after quark confinementThe energy of the universe appeared at the time of quantum gravity in this timeline. The coalescence into particles as we know them, after quark confinement.In the quantum gravity era is the beginning of the Universe and the Big Bang model has a fast inflationary period (needed to fit the cosmic microwave background radiation data) as seen in the plot . It is not involved in present day particle creation which happened much later.

1. Misconception about the expectation of a quantum system

Nothing is actually wrong.The quantum answer includes \$c_2^*c_1\$ term which depends on the phase difference between \$c_1\$ and \$c_2\$. But the phase difference is something that would vary randomly with even the tiniest coupling between states 1 and 2. And thermodynamics requires such a coupling to exist to be applied. So the stat mech answer "averages over" things that very randomly in thermal systems. Since that phase varies, \$\$langle c_2^*c_1

angle = 0\$\$ averages out to zero, and the only terms that do not average out to zero in thermal equilibrium are the terms that both stat mech and quantum preserve. The quantum answer applies when the states have no interaction, in which case it is not correct to apply stat mech because stat mech gives the answer "after we have reached thermal equilibrium" and the smaller the interaction between 1 and 2 is, the longer it takes to reach equilibrium, and when the interaction is vanishingly small, the time to thermal equilibrium approaches infinity.

2. Is the presence of Planck's constant in a formula sufficient to deduce that it pertains to a quantum system?

As long as Plank's constant has a value above zero you can deduce it pertains to quantum physics, if you let the constant reach zero, then and only then, you can be dealing with classical physics, the pre-quantum physics

3. Does observation in quantum theories always imply interaction (affecting quantum system with photons, electromagnetic fields, etc.)?

In elementary particle physics interactions means an exchange of four momenta between incoming and out going particles, where observable are just the incoming and outgoing particles , and the effect of the four momentum exchange is measured by using energy momentum and angular momentum conservation, and a number of quantum number conservations. In this field observation and interaction is the same. One observes the effects of interaction. But physics is not just elementary particle physics. It is concerned with complex systems, like protons and neutrons, nuclei, atoms, lattices etc. At the level of nuclei for example, there are continuous interactions between the protons and the neutrons, which are not observable directly. Levels of mathematics have to be used, so in this sense observations are a subset of interactions. In my Zenon example why do physicists speak of just observation, not the effects of interaction of their (quantum) system with ultraviolet waves of certain frequency, amplitude, etcBecause the "interaction" is not one of the fundamental ones, but a complicated mathematically synergy of UV photons gives a behavior that is emergent from the fundamental electromagnetic interaction which involves one photon at a time. It is a great multiplicity of fundamental interactions, and thus they stick to "observations".Simple quantum mechanical models have been developed for complex systems as for example the phonons , and then one might have a simple interaction model and be back in a framework where interaction and observation can be the sides of the same coin.

4. Use of operators in a time-dependent Hamiltonian quantum system

\$a_1\$ is not hermitian so be careful about its eigenvectors. Most importantly, the definition of the creation and destruction operators involves the frequency \$omega\$ so the creation operator on both sides of \$t=0\$ will not be the same so the harmonic oscillator eigenstates, i.e. the number states, will not be the same. This is evident because the length scale for the number states depends also on \$omega\$.In particular, the change of scale is equivalent to a squeezing transformation so the vacuum state for \$t_2\$ will not be the vacuum state for \$t_1\$. You will need to expand \$vert 0_2

angle\$ in terms of \$t_1\$-states in order compute the action of \$a_1\$, i.e. given \$lambda_1=sqrtfracmomega_1hbar\$ \$\$ psi_n(x)=frac1sqrt2^n n! left(fraclambda^2_1pi

ight)^1/4e^-lambda_1^2 x^2/2 H_n(lambda_1 x) \$\$ and \$\$ phi_n(x)=frac1sqrt2^n n! left(fraclambda^2_2pi

ight)^1/4e^-lambda_2^2 x^2/2 H_n(lambda_2 x) \$\$ the simplest way to proceed is to write \$\$ vert n_2

angle =sum_n_1 vert n_1

angle langle n_1vert n_2

angle \$\$ with \$\$ langle n_1vert n_2

angle =int_-infty^infty phi_n_2(lambda_2 x) phi_n_1(lambda_1 x) \$\$ Thus, for instance, \$\$ langle 0_1vert 0_2

angle= sqrtfrac2lambda_1lambda_2lambda_1^2lambda_2^2, , qquad langle 2_1vert 0_2

angle = fracsqrtlambda_1lambda_2(lambda_1^2-lambda_2^2)(lambda_1^2lambda_2^2)^3/2 \$\$

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