Entropy before and after collapse of the wavefunction and interpretation?

Seems like it might be pretty rudimentary but I want to see if my thinking is on the right track as well as what the result means.

The question is, is the entropy of the collapse of a wavefunction or entangled system more before the collapse? My reasoning is that as Entropy is a measure of the "unpredictability" of information content (in this context), the more we know the less entropy there is. So for example if we have N fair coin tosses in which we don't know the outcome, we would have N bits of unknown information as $ \sum_{i = 0;N} 2^N = N$ but the first we flip and once the answer is known we would have an entropy of N-1 as we don't know those N-1 states. Once all is known we have an entropy of 0. Of course we would deal with a continuous function for waves, but am I on the right track of thought? If so, then how does the second law of thermodynamics fit in in that the Universe is supposed to go to a higher state of entropy?

Just an aside question, but is the wavefunction collapse the same as the collapse of an entangled system? Seems that the wavefunction itself is just a superposition of several different eigenstates, is it safe to say that these eigenstates are "entangled"?

Answer

The question is, is the entropy of the collapse of a wavefunction or entangled system more before the collapse? My reasoning is that as Entropy is a measure of the "unpredictability" of information content (in this context), the more we know the less entropy there is.

I will use the statistical definition of physical entropy, which has been shown to coincide with the thermodynamical at the limits.

In Boltzmann's definition, entropy is a measure of the number of possible microscopic states (or microstates) of a system in thermodynamic equilibrium, consistent with its macroscopic thermodynamic properties (or macrostate). To understand what microstates and macrostates are, consider the example of a gas in a container. At a microscopic level, the gas consists of a vast number of freely moving atoms, which occasionally collide with one another and with the walls of the container. The microstate of the system is a description of the positions and momenta of all the atoms. In principle, all the physical properties of the system are determined by its microstate. However, because the number of atoms is so large, the motion of individual atoms is mostly irrelevant to the behavior of the system as a whole. Provided the system is in thermodynamic equilibrium, the system can be adequately described by a handful of macroscopic quantities, called "thermodynamic variables": the total energy E, volume V, pressure P, temperature T, and so forth. The macrostate of the system is a description of its thermodynamic variables.

kB is Boltzmann's constant and Omega is the number of microstates consistent with the given macrostate.

When we go to the quantum mechanical framework :

The quantum state of a system can be expressed as a superposition of "basis" states, which can be chosen to be energy eigenstates (i.e. eigenstates of the quantum Hamiltonian). Usually, the quantum states are discrete, even though there may be an infinite number of them. For a system with some specified energy E, one takes Ω to be the number of energy eigenstates within a macroscopically small energy range between E and E + δE. In the thermodynamical limit, the specific entropy becomes independent on the choice of δE.

Collapse of the wavefunction means a quantum mechanical interaction that will change the energy E , an extra microstate. So after the "collapse" there is one more microstate introduced into the statistical entropy , (maybe a photon has gone off). Entropy is not defined for one particle, in physics. It is as statistical phenomenon of accumulated behavior of quantum dominated interactions to large distances where the quantization is averaged out into the thermodynamic variables, in this case S, entropy. So collapse/interaction increases entropy.

You ask :

Just an aside question, but is the wavefunction collapse the same as the collapse of an entangled system? Seems that the wavefunction itself is just a superposition of several different eigenstates, is it safe to say that these eigenstates are "entangled"?

Quantum mechanical solutions for physical boundary conditions will give wavefunctions, whose square give the probability of interaction .

These wave functions may describe many body systems, but they are coherent in space and time for all the variables describing the system. Coherent means that all the phases entering the problem are theoretically known for each particle , a single wave function describes the system .

Entanglement means that one coherent wavefunction is describing a system.

Collapse is a way of saying "interaction" or "measurement". Before "collapse" we only have probability distributions from the state function. After "collapse" an interaction has taken place that gives us one value measured in that probability distribution. To get the probability distribution we have to do many identical setup "collapses"/measurements so as to determine the square of the wavefunction before "collapse".

Yes, if one has all the boundary conditions and one known state wave function then you can say that the eigenstates are entangled. Entanglement is a verbose way of talking about state functions.