Letter to Editor Volume 1 Issue 2

Department of Physics, University of Eldoret, Kenya

**Correspondence:** Tanui PK, Department of Physics, University of Eldoret, P.O Box 1125−30100, Kenya, Tel +2547 2080 0446

Received: February 26, 2018 | Published: March 26, 2018

**Citation: **Namwetako JS, Tonui JK, Tanui PK. Solid 4 2 He and zero-point energy . Open Acc J Math Theor Phy. 2018;1(2):43 – 45 DOI: 10.15406/oajmtp.2018.01.00008

At very low temperatures, **${}_{2}{}^{4}He$
**becomes crystalline under high pressure of order of 65 Atmospheres or more. The
solid structure is hexagonal close packed (HCP) and the size of the crystal increases. **${}_{2}{}^{4}He$
, **being a quantum solid, is intrinsically restless with a large zero – point energy${\in}_{o}$
, which is of the order of 98 percent of the total energy, ${E}_{0}$
, in the crystalline state. The balance two percent will be the energy due to interparticle interaction. The total energy in the crystalline state can be calculated using the Gaussian form of the Lennard – Jones potential and the t – matrix formalism. The zero – point energy is calculated using Heisenberg’s uncertainty principle. If ${\in}_{o}$
is to be of the order of 98 percent of ${E}_{0}$
, the interparticle distance X turns out to be of the order of 1.5 Å, which is less than the hard sphere radius.

**Keywords****: **zero – point energy, $\lambda $
– transition, Hexagonal close pack, bose condensate

It is found that cooling ${}_{2}{}^{4}He$
liquid from ${T}_{\lambda}$
towards absolute zero temperature does not result in the solidification of ${}_{2}{}^{4}He$
liquid. However, ${}_{2}{}^{4}He$
becomes solid under large external pressure of the order of 25 Atmospheres or more. Helium was first solidified in Leiden by W. H. Keesom.^{1} Measurements on specific heat and thermal conductivity of solid Helium are given by H. R. Glyde.^{2} To study the properties of any interacting system, such as solid ${}_{2}{}^{4}He$
, the pair potential between the atoms has to be precisely known. Between the Helium atoms, the pair potential,^{3,4} $V\left(r\right)$
, is weakly attractive at large separation, $r\ge 3$
Å**, **with a minimum well depth $\in =10.95K$
. At close approach, $r\le \sigma =2.63$
Å, where hard – core radius is$\sigma $
. For$V(\sigma )={090}^{\circ}$
, $V(r)$
becomes steeply repulsive. The volume of solid${}_{2}{}^{4}He$
is 21.1cm3/mole. Since Helium is light, its thermal wavelength${\lambda}_{T}$
, is long. At$T=1.0K$
,${\lambda}_{T}\cong 10\AA $
for${}_{2}{}^{4}He$
. Helium is
thus difficult to localize. Attempts to localize it leads to a high kinetic or zero – point energy. Since$V\left(r\right)$
, is weak,

Helium does not solidify under attraction via$V\left(r\right)$
. Rather, it solidifies only under large external pressure as a
Consequence of the existence of the hard core of ${}_{2}{}^{4}He$
, much as billiard balls form a lattice under pressure. At higher pressure, ${}_{2}{}^{4}He$
gets compressed into close−packed (FCC and HCP) phases. At 4.9 k-bars, FCC${}_{2}{}^{4}He$
has a volume of 9.03cm3/mole and$\sigma =2.11A\xb0$
.

The degree of the quantumness in a solid is sometimes characterized by

$\Lambda =\frac{1}{2\pi}\frac{\lambda}{\sigma}=\frac{h}{\sigma {\left(2m\in \right)}^{\frac{1}{2}}}$ ……… (1)

Where$\lambda $ , is the de Boer parameter, which is the ratio of the de Broglie wavelength,$\lambda =\frac{h}{p}=\frac{h}{\sqrt{2m\in}}$ to the minimum separation of the atoms in the crystal which is$\sigma $ . If $\lambda $ is comparable to$\sigma $ , the solid is highly quantum. Moreover, the de Boer parameter $\Lambda $ and the Lindmann ratio$\delta $ are practically identical, and thus a large $\Lambda $ means quantum and very anharmonic vibration.In physics, a quantum solid is the type of solid that is intrinsically restless, in the sense that atoms continuously vibrate about their position and exchange places even at absolute zero temperature. The typical quantum solid, both in low density and high density, is the solid ${}_{2}{}^{4}He$
which is crystalline and is hexagonal – close – packed system. The atoms in a quantum crystalline solid are arranged in a regular array that may be characterized by a lattice of points. The atoms are held in the lattice by the inter – atomic bonds, meaning thereby that the atoms interact with each other, and there may be some specific form of interaction potential between the atoms. We now turn to the basic ideas and information that lead us to write the present manuscript. The properties of ${}_{2}^{4}He$
are dominated by the quantum – mechanical effect called zero – point motion which is the dynamical property of ${}_{2}^{4}He$
. Due to the low
atomic mass and weak inter–atomic forces${E}_{O}$
, atoms do not freeze even in the limit of zero temperature. Liquid${}_{2}^{4}He$
can be solidified only on the application of external pressure of 25bars or more, in addition to cooling. The effect allows the presence of delocalized vacancies in the crystalline solid that can condense into a coherent state at low temperatures.^{5} Even when it becomes solid, the liquid – solid transition is not well understood.

In this manuscript, we have concentrated on the physical characteristics of solid${}_{2}^{4}He$
, one is that ${}_{2}^{4}He$
crystals will be treated as quantum crystals, and the second is the calculation of the interaction energy between the ${}_{2}^{4}He$
atoms.
Since a quantum solid is intrinsically restless, the uncertainties in the values of position r and the momentum$p$
should be finite, and they are large in${}_{2}^{4}He$
because of its low mass. Hence, $\Delta r$
and$\Delta p$
should be large. At absolute
zero temperature,^{6,7 } the zero – point fluctuations of atoms in solid${}_{2}^{4}He$
are about 30 percent of the inter−atomic
spacing and as a consequence the zero – point energy represents roughly 98 percent of the total energy of the crystal.

This means that in the crystalline state$\left(T\to 0\right)$ , the interaction energy between the helium atoms will be roughly 2 percent of the total energy of the system. Thus, if ${E}_{o}$ is the total energy and${\in}_{1}$ is interaction energy, and${\in}_{o}$ is the zero – point energy, then,

${\in}_{1}=\frac{2}{100}{E}_{O}$

${\in}_{o}=\frac{98}{100}{E}_{O}$ ……………………… (3)

If X is the average inter–particle distance or inter−atomic spacing of the linear lattice, then the zero – point fluctuation$\Delta {X}_{o}$ of the atoms in solid helium is about 30 percent of$X$ . Hence,

$\Delta {X}_{0}=\frac{30}{100}X$
…………………… (4)
In the literature,^{7 }$X$
has two values, one is when$X\text{}=\text{}2.11\text{}\AA $
, and second when$X\text{}=\text{}3.67\text{}\AA $
. Both of these values will be used to get the numerical value for the zero – point – energy${\in}_{o}$
. Now, if $\Delta {X}_{o}$
is the uncertainty in the value of the position, and$\Delta {p}_{0}$
is the uncertainty in the value of the momentum, then according to Heisenberg’s uncertainty principle,

$\Delta {p}_{o}=\frac{\hslash}{\Delta {X}_{O}}$ ……………………… (5)

The zero−point−energy is then given by${\in}_{0}=\frac{{\left(\Delta {p}_{o}\right)}^{2}}{2m}=\frac{{\hslash}^{2}}{2m(\Delta {X}_{O})}$ ……… (6)

Where $m=6.46x{10}^{-24}g$ is the mass of the${}_{2}^{4}He$ atom. Eq. (6) shows that the value of zero–point–energy${\in}_{o}$ changes as$\Delta {X}_{o}$ , varies and the value of $\Delta {X}_{o}$ depend on the value of $X$ as shown in Eq. (4).Here we have quoted two extreme values for$X\text{}\left(X\text{}=\text{}2.11A\xb0\text{}and\text{}X\text{}=3.67A\xb0\right)$
. It is simple to calculate the value of $X$
that will lead to the value of ${\in}_{o}$
that will be 98 percent of the total energy${E}_{O}$
. Again there are two values for${E}_{O}$
in the literature;^{7}${E}_{o}=\text{}25K$
and ${E}_{o}=\text{}14K$
. In order that${\in}_{o}$
be 98 percent of ${E}_{O}$
,the value of $X$
will be the order of $1.7\text{}A\xb0$
which means that in the crystalline state the inter−particle distance between the atoms should be of the order of $1.7\text{}A\xb0$
. However, this needs to be verified experimentally, may be by using the neutron scattering method.

Our calculations bring out that if the zero−point−energy$\left({\in}_{o}\right)$ , is to be 98 percent of the total energy$\left({E}_{O}\right)$ , then the inter-particle distance between${}_{2}^{4}He$ atoms should be of the order of $1.5\text{}A\xb0$ which is less than the hard−core radius. More experimental and theoretical work needs to be done to determine the exact value of the fluctuation in position $\left(\Delta {X}_{o}\right)$ since this will determine${\in}_{o}$ , and we then calculate whether ${\in}_{o}$ is really 98 percent of${E}_{o}$ . In our next attempt, many−body theory will be used to calculate${\in}_{o}$ .

None

Author declares that there is no conflict of interest.** **

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©2018 Namwetako, et al . This is an open access article distributed under the terms of the Creative Commons Attribution License , which permits unrestricted use, distribution, and build upon your work non-commercially.

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