Welcome to my reading club
Today we read the book "Thermodynamics in the quantum regime"
Quantum thermodynamics and High energy theory are very different
High energy physics has a lot of theories and logic since it is not yet fully connected to reality
A lot of reality facts have not yet entered the theory
But quantum thermodynamic is different
We know that quantum mechanics and classical statistics are well developed
Quantum mechanics is also used to build Quantum Field Theory
However, in the sense of thermodynamics, quantum mechanics cannot explain how small quantum systems form the large macro system and its features
We do not know these, thus the quantum thermodynamics is to connect the statistics of the classical and quantum regime
I choose this recent review book to read. It is very long, around 1000 pages. But it includes all.
I decide to make a small project to upload several discussions by time, and try to finish it
Now lets start
Today I will start from Chapter 1 section 2
1.2.1 discusses von Neumann formulism
Just a revision
First, an observable is an Hermitian operator
It works on a Hilbert space. Its eigenstates are orthonormal
Second, an event is presented by a projection operator, if you have a \psi state, it goes as a bra times a ket.
Third, quantum probability is presented by a density matrix.
By probability it means the state. Since we are treating ensembles, we have to use matrices.
Four, how to express probability
If you measure a state \psi, you have its operator P, a bra times a ket.
Trace over \rho times P
Five, expectation value is the trace of the observable times density matrix.
We knew these from QM, here we use matrices since the statistics studies mixed states
(1.3) is Schrondinger equation
(1.4) is its solution
When you have interaction, it is not commutative with free terms, then the U matrix has time ordering T
Then it discusses entropy
Two types entropy are discussed. von Neumann and Shannon entropy
Shannon entropy is the analogy of the Boltzmann entropy
Shannon entropy bases on an observable
Or to say the observable fixes some eigenstates
We sum the probability terms of the fixed eigenstates, pj times log pj
von Neumann entropy is different, if it has off diagonal elements, we have to diagonalize it
It is like, we find a set of probability for a new set of orthogonal bases
Then we sum the terms as shown with \lambda j
What are the two entropies' difference?
To discuss, you have to first have a density matrix
We produce a mixed ensemble from two pure states, which are not orthogonal in a representation of a certain set of eigenstate
For example, you produce a mixed state with z spin-up and x spin-up states. In the representation of spin-z, the density matrix is not diagonal
If you calculate entropy based on the measurement of spin-z, you can only measure the diagonal part
You get the Shannon entropy
If you want to calculate von Neumann entropy, you have to diagonalize it
These two gives different answer
You can prove Shannon entropy S_A >= von Neumann entropy
Shannon entropy over calculated information
von Neumann entropy is \rho transformation independent
(1.6) discuss equilibrium ensemble state
Gibbs ensemble in QM is the analogy of canonical ensemble in SM
Its probability distribution depends on H, with eigenvalues E
It says the Shannon entropy of H is the same with the von Neumann entropy of \rho, is that the \rho use the same eigenstates with H, as it is H's function
The next page there is a concept: The von Neumann entropy do not grow with Schrodinger evolution
Thus, the QM entropy cannot be used to explain the classical isolated system entropy increase
Thus, if you wish to find a criterion for QM entropy increase, you have to alter the entropy or look for other quantities
Alright, lets call it for today.
