Let's say I have a $1$-dimensional material with thermal expansion $\alpha$:

$$ \alpha l_0 = \frac{\Delta l}{\Delta T}$$

where $l$ is the length of the system and $T$ is it's temperature.

This is the same as saying:

$$ \frac{l}{l_0} - 1 = \alpha T$$

One can assume this to mean if there are $N$ particles the average spacing in a uniform material's particles is given by:

$$ \frac{\langle x_i - x_{i+1} \rangle}{l_0} - \frac{1}{N} = \frac{\alpha T}{N}$$

where $x_i$ is the $i$'th particle's position adjacent to the $i+1$ particle's position and the length of the system is given by $l \sim \langle x_1 - x_N \rangle $

Multiplying $N$ both sides:

$$ N \frac{\langle x_i - x_{i+1} \rangle}{l_0} - 1 = \alpha T$$

where $\langle x_1 \rangle \geq \langle x_2 \rangle \geq \langle x_3 \rangle \geq \langle x_4 \rangle \geq \dots $

Let's write everything we have gathered so far in terms of the density matrix:

$$ \rho = \sum_{ij} p_{ij} |\psi_i \rangle \langle \psi_j |$$

The temperature is given by:

$$ T= \Big( \frac{\partial S}{\partial U} \Big)_{N,V} =\Big(\frac{\partial \text{Tr}(\rho \ln\rho) }{ \partial \text{Tr}( \rho H)}\Big )_{N,V}$$

Where $H$ is the Hamiltonian. The average spacing is given by: $$\langle x_i - x_{i+1} \rangle =\text{Tr}(\rho(x_i - x_{i+1}))$$

Thus, we have a differential equation the density matrix must satisfy:

$$ \frac{N}{l_0} \text{Tr}(\rho(x_i - x_{i+1})) - 1 = \alpha \Big(\frac{\partial \text{Tr}(\rho \ln\rho) }{ \partial \text{Tr}( \rho H)}\Big )_{N,V}$$


Does such an approach already exist in the literature? What are some example Hamiltonian and density matrices which obeys this equation? Can a more explicit condition that the density matrix must obey be constructed (perhaps using $\text{Tr}(A \otimes B) = \text{Tr}(A) \text{Tr}(B) $)? Can this describe a strongly coupled system (my reason for asking about strongly coupled systems is that liquid mercury is a strongly coupled system)?

Edit: My attempt

Let us define $N/l_0 = \kappa$ as a constant.

$$ \kappa \text{Tr}(\rho(x_i - x_{i+1})) - 1 = \alpha \Big(\frac{\partial \text{Tr}(\rho \ln\rho) }{ \partial \text{Tr}( \rho H)}\Big )_{N,V}$$

Multiplying both sides with ${ \partial \text{Tr}( \rho H)}_{N,V}$:

$$ \kappa \text{Tr}(\rho(x_i - x_{i+1}) \partial \text{Tr}( \rho H)_{N,V} - \partial \text{Tr}( \rho H)_{N,V} = \alpha {\partial \text{Tr}(\rho \ln\rho)_{N,V} }$$

Using $A \partial B = \partial (AB) - B \partial A$:

$$ - \kappa \text{Tr}( \rho H) \partial \text{Tr}(\rho(x_i - x_{i+1}))_{N,V} - \partial \text{Tr}( \rho H)_{N,V} = \alpha \partial \text{Tr}(\rho \ln\rho)_{N,V} - \partial( \kappa \text{Tr}( \rho H) \text{Tr}(\rho(x_i - x_{i+1}))_{N,V} ) $$

Multiplying $- e^{ \kappa \text{Tr}( \rho(x_i - x_{i+1}))}$ both sides:

$$ \partial (e^{ \kappa \text{Tr}( \rho(x_i - x_{i+1}))} \text{Tr}( \rho H) )_{N,V}= (e^{ \kappa \text{Tr}( \rho(x_i - x_{i+1}))})\partial( \kappa \text{Tr}( \rho H) \text{Tr}(\rho(x_i - x_{i+1}))_{N,V} -\alpha ( e^{ \kappa \text{Tr}( \rho(x_i - x_{i+1}))}){\partial \text{Tr}(\rho \ln\rho)_{N,V} }$$

Integrating both sides and using a constant $c$:

$$ \text{Tr}( \rho H) = -\alpha e^{ -\kappa \text{Tr}( \rho(x_i - x_{i+1}))} \int ( e^{ \kappa \text{Tr}( \rho(x_i - x_{i+1}))}){\partial \text{Tr}(\rho \ln\rho)_{N,V} } + c e^{ \kappa \text{Tr}( \rho(x_i - x_{i+1}))} + e^{-\kappa \text{Tr}( \rho(x_i - x_{i+1}))} \int (e^{ \kappa \text{Tr}( \rho(x_i - x_{i+1}))})\partial( \kappa \text{Tr}( \rho H) \text{Tr}(\rho(x_i - x_{i+1}))_{N,V} $$

Now, we can substitute the L.H.S in the R.H.S and get some sort of series(?)

  • $\begingroup$ Just for clarification: $|\psi_i\rangle$ ist the full, $N$-parrticle, state or one-particle state at any of the sites? Are the single particles confined in 0-d (at any of the positions, $n\in\lbrace1,\ldots,N\rbrace$ which are constant, in some state $i$) or in 1-d (free, possibly interacting particles confined onsome interval with given one-particle states)? May the particles jump between sites? If so: bosons or fermions and what are onsite interactions? $\endgroup$ – denklo Nov 12 '18 at 8:24
  • $\begingroup$ Physically I was thinking of a $1$ dimensional thermometer (in some sense) ... So I was thinking of a Hamiltonian like $\hat H = p_1^2/2m + p_2^2 / 2m + \dots + V(x_1,x_2,x_3,\dots)$ where $V$ is the potential. Hence, $|\psi \rangle$ would be the full $N$ particle state .. I was thinking more along the lines of fermions .. $\endgroup$ – More Anonymous Nov 12 '18 at 8:55

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