- #1
Konte
- 90
- 1
Hello everybody,
- In quantum mechanics, the state ## | \psi \rangle ## of a system that is in thermodynamic equilibrium can be expressed as a linear combination of its stationary states ## | \phi _n \rangle ## : $$ | \psi \rangle = \sum_n c_n | \phi _n \rangle $$
It permit us to express the mean value of energy as:
$$ \langle E \rangle _{\psi}= \sum_n E_n | c_n |^2 $$
- In other approach, one way to express the mean value of energy is by using the Boltzmann distribution. So my question is:
As the system is in thermodynamic equilibrium, is it allowed to think that Boltzmann distribution ## \frac{N_i}{N} = \frac{g_i e^{-\frac{E_i}{k_BT}}}{Z(T)} ## are equivalent to the ## | c_n |^2 ## ?
Thank you everybody.
Konte
- In quantum mechanics, the state ## | \psi \rangle ## of a system that is in thermodynamic equilibrium can be expressed as a linear combination of its stationary states ## | \phi _n \rangle ## : $$ | \psi \rangle = \sum_n c_n | \phi _n \rangle $$
It permit us to express the mean value of energy as:
$$ \langle E \rangle _{\psi}= \sum_n E_n | c_n |^2 $$
- In other approach, one way to express the mean value of energy is by using the Boltzmann distribution. So my question is:
As the system is in thermodynamic equilibrium, is it allowed to think that Boltzmann distribution ## \frac{N_i}{N} = \frac{g_i e^{-\frac{E_i}{k_BT}}}{Z(T)} ## are equivalent to the ## | c_n |^2 ## ?
Thank you everybody.
Konte