Debye function

The Debye functions are closely related to the polylogarithm.

They have the series expansion[1]

${\displaystyle D_{n}(x)=1-{\frac {n}{2(n+1)}}x+n\sum _{k=1}^{\infty }{\frac {B_{2k}}{(2k+n)(2k)!}}x^{2k},\quad |x|<2\pi ,\ n\geq 1,}$

where ${\displaystyle B_{n}}$ is the n-th Bernoulli number.

${\displaystyle \lim _{x\to 0}D_{n}(x)=1.}$

If ${\displaystyle \Gamma }$ is the gamma function and ${\displaystyle \zeta }$ is the Riemann zeta function, then, for ${\displaystyle x\gg 0}$,

${\displaystyle D_{n}(x)={\frac {n}{x^{n}}}\int _{0}^{x}{\frac {t^{n}\,dt}{e^{t}-1}}\sim {\frac {n}{x^{n}}}\Gamma (n+1)\zeta (n+1),\qquad \operatorname {Re} n>0,}$[2]

The derivative obeys the relation

${\displaystyle xD_{n}^{\prime }(x)=n(B(x)-D_{n}(x)),}$

where ${\displaystyle B(x)=x/(e^{x}-1)}$ is the Bernoulli function.

The Debye model has a density of vibrational states

${\displaystyle g_{\rm {D}}(\omega )={\frac {9\omega ^{2}}{\omega _{\rm {D}}^{3}}}}$ for ${\displaystyle 0\leq \omega \leq \omega _{\rm {D}}}$

with the Debye frequency ω_D.

Inserting g into the internal energy

${\displaystyle U=\int _{0}^{\infty }d\omega \,g(\omega )\,\hbar \omega \,n(\omega )}$

with the Bose–Einstein distribution

${\displaystyle n(\omega )={\frac {1}{\exp(\hbar \omega /k_{\rm {B}}T)-1}}}$.

one obtains

${\displaystyle U=3k_{\rm {B}}T\,D_{3}(\hbar \omega _{\rm {D}}/k_{\rm {B}}T)}$.

The heat capacity is the derivative thereof.

The intensity of X-ray diffraction or neutron diffraction at wavenumber q is given by the Debye-Waller factor or the Lamb-Mössbauer factor. For isotropic systems it takes the form

${\displaystyle \exp(-2W(q))=\exp(-q^{2}\langle u_{x}^{2}\rangle }$).

In this expression, the mean squared displacement refers to just once Cartesian component u_x of the vector u that describes the displacement of atoms from their equilibrium positions. Assuming harmonicity and developing into normal modes,[3] one obtains

${\displaystyle 2W(q)={\frac {\hbar ^{2}q^{2}}{6Mk_{\rm {B}}T}}\int _{0}^{\infty }d\omega {\frac {k_{\rm {B}}T}{\hbar \omega }}g(\omega )\coth {\frac {\hbar \omega }{2k_{\rm {B}}T}}={\frac {\hbar ^{2}q^{2}}{6Mk_{\rm {B}}T}}\int _{0}^{\infty }d\omega {\frac {k_{\rm {B}}T}{\hbar \omega }}g(\omega )\left[{\frac {2}{\exp(\hbar \omega /k_{\rm {B}}T)-1}}+1\right].}$

Inserting the density of states from the Debye model, one obtains

${\displaystyle 2W(q)={\frac {3}{2}}{\frac {\hbar ^{2}q^{2}}{M\hbar \omega _{\rm {D}}}}\left[2\left({\frac {k_{\rm {B}}T}{\hbar \omega _{\rm {D}}}}\right)D_{1}\left({\frac {\hbar \omega _{\rm {D}}}{k_{\rm {B}}T}}\right)+{\frac {1}{2}}\right]}$.

From the above power series expansion of ${\displaystyle D_{1}}$ follows that the mean square displacement at high temperatures is linear in temperature

${\displaystyle 2W(q)={\frac {3k_{\rm {B}}Tq^{2}}{M\omega _{\rm {D}}^{2}}}}$.

The absence of ${\displaystyle \hbar }$ indicates that this is a classical result. Because ${\displaystyle D_{1}(x)}$ goes to zero for ${\displaystyle x\to \infty }$ it follows that for ${\displaystyle T=0}$

${\displaystyle 2W(q)={\frac {3}{4}}{\frac {\hbar ^{2}q^{2}}{M\hbar \omega _{\rm {D}}}}}$ (zero-point motion).