1
vote
0answers
107 views

How to understand the matrix behind a Hamiltonian?

thanks to the answers I received to my previous questions, I could derive correctly an elegant partition function for my problem which resembles a second quantized model taking the particles to be ...
1
vote
0answers
45 views

Ascertaining a mathematical equality to derive a partition function

we have an equation like this: $$\mathcal N(x)=\sum_{q=1}^\infty (\psi(x,q) \log(q)) \qquad (1)$$ while $\psi(x)$ is the function for some oscillations (may contain complex part), $x\in \Bbb R$ and ...
2
votes
0answers
107 views

Number theoretical function applied in physics? [closed]

I have a series of number theoretic phenomena (mathematics) that I can describe exactly by the superpositions or linear combination of the below function (I know it is an inverse Fourier type). Does ...
4
votes
1answer
393 views

How do I calculate the probability that the oscillator is in a certain state using partition function?

So let's say I have a single ($N=1$) quantum harmonic oscillator and the energy is determined by $E_n = (n + 1/2) \cdot \hbar \omega$ (where $n$ is the quantum number and $n$ = $0, 1, 2, \ldots$) ...
0
votes
2answers
70 views

What is the minimal set of expectation values I need in a statistical model?

At least if $\vec v$ is really only a one dimensional parameter, measuring all the moments $\langle v^n \rangle_f$ seems to give me all the information to compute $\langle A \rangle_f$ with $A(v)$ ...
1
vote
1answer
73 views

semiclassical exact expression (in one dimension only)

let be $ N(x)= \sum_{n} H(x-E_{n}) $ the eingenvalue 'staircase' function and let be a system so $ V(x)=V(-x)$ and $ V^{-1}(x)=\sqrt \pi \frac{d^{1/2}}{dx^{1/2}} N(x) $ then would it be true that ...
1
vote
1answer
155 views

Quantum Stat-Mech Proof of an Inequality for the Partition Function

I have the following problem that I was unable to solve for class, but I had a couple first steps that I started with that I am unable to finish. I know I can't get this since it's already been ...