General properties of helium droplets are introduced and the ripplon and phonon excitation spectrum established. The ripplon and phonon thermal properties are calculated in detail using methods derived in previous chapters. The angular momentum distributions vs. excitation energy are calculated.


Angular Momentum Droplet Size Level Density Helium Atom Canonical Partition Function 
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Helium is the lightest of the rare gases. As a rare gas it is chemically inactive and is only bound to other atoms, other helium atoms included, by weak and non-directional forces. The potential between two rare gas atoms of the same type is often described by a sum of two high powers of the reciprocal of the interatomic (internuclear, to be exact) separation. The most common is called the Lennard-Jones potential and has powers −12 for the repulsive part and −6 for the attractive part;
$$ V(r) = 4\varepsilon \biggl( \biggl(\frac{\sigma}{r} \biggr)^{12}- \biggl(\frac{\sigma}{r} \biggr)^6 \biggr). $$
The potential is shown in Fig. 10.1. The potential reproduces the long distance behavior of the interaction of two neutral particles that varies as 1/r 6. This distance dependence is a general behavior for neutral particles and is due to the polarizability induced by quantum mechanical fluctuations in the electronic charge distributions. The 1/r 12 term represents the strong repulsion between the atoms at short distances, but the precise power has a less fundamental origin. The value of σ and ε give the equilibrium distance and binding energy. The equilibrium distance, r 0, is found as
$$ \frac{dV}{dr} = 0\quad \Rightarrow\quad r_0 = 2^{\frac{1}{6}} \sigma = 1.122\sigma, $$
and the binding energy
$$ -V(r_0) = \varepsilon. $$
This energy is the classical (ħ=0) energy at the bottom of the two-body potential in Eq. (10.1). The Lennard-Jones parameters for the rare gases are given in Table 10.1.
Fig. 10.1

The potential energy of two atoms that interact with the Lennard-Jones potential

Table 10.1

Lennard-Jones parameters of the rare gases. The masses are rounded averages over the isotopic compositions







ε (K)






σ (Å)






m (u)






The main reason to single out helium for a special treatment is that the interaction with other helium atoms is weak due to the strong binding of electrons inside the atom. Combined with the small mass of the atom, this causes so large a quantum delocalization in the positions of the atoms that they are not bound to lattice points. This prevents liquid helium from crystallizing at low temperatures, as all other elements do (at atmospheric pressure, above 24 atmospheres it does crystallize). The low temperature phase is a quantum liquid. Precisely what type of quantum liquid is determined by the isotope. Helium has two stable isotopes, 3He and 4He, of which the first is a fermion and very rare, with a natural relative abundance of 1.4⋅10−6. Most interest is therefore focused on the bosonic 4He, and we will deal exclusively with that species here.

One can get a semi-quantitative understanding of the fact that helium doesn’t crystallize (but not of the properties of the liquid, which is a much more involved and interesting problem) by calculating the zero point energy of the vibrational motion of the dimer with the potential Eq. (10.1). Approximating the bottom of the potential with a harmonic oscillator, the vibrational frequency is
$$ \mu \omega^2 = \frac{d^2 V}{dr^2}\bigg\vert_{r=r_0} = 62\cdot 2^{2/3} \frac{ \varepsilon}{\sigma^2}, $$
with the reduced mass μ=m/2. We then have the ratio of the zero point energy to the dimer binding energy
$$ \frac{\frac{1}{2}\hbar \omega}{\varepsilon} = \sqrt{\frac{62}{2^{1/3}}} \frac{\hbar}{\sigma \sqrt{m\varepsilon}} \approx 7.01 \frac{\hbar}{\sigma \sqrt{m\varepsilon}}. $$
Values of this ratio for all noble gases are given in Table 10.2. Clearly He stands out with a zero point motion energy that exceeds the binding energy. It should not be inferred from this simple calculation that the helium dimer does not exist. It does, just about, but it has a very long bond, about 50 Å, and is bound by the minute but finite energy of 10−7 eV, or 10−3 K. For comparison, the radius R of a liquid He drop composed of N atoms and with size independent density D is where m a is the mass of the atom and the density is D=0.145 g/cm2 at 0.4 K.
Table 10.2

The zero point energy of the rare gas dimers relative to the Lennard-Jones energy













Small helium droplets will be described with the liquid drop model. It is worth keeping a few facts in mind before we blindly apply this description to helium droplets. First, the density of the droplet is not a step function. Given the delocalization of helium atoms in the liquid, one must expect that also the surface densities vary over a finite length. Scattering experiments show that the atomic density varies from 90 % of the central value to 10 % over a distance of 6 Å of the surface. The surface diffuseness can also be calculated theoretically with Density Functional Theory. It is found to be 7.0 Å, for the bulk surface and slightly larger for finite sizes, defined in the same way as the experimental value. This renders the dimensions of droplets partly a matter of definition. With the radius of r N =2.22 Å N 1/3 from Eq. (10.6), the ratio of the surface diffuseness to the radius is 0.68 for N=100, 0.15 for N=104, and 0.03 for N=106. It would be a conservative estimate to set these ratios equal to the relative error in the energy of the surface modes we will calculate below, but the numbers do tell us that the simple estimate for the quantum energy of the excitations needs to be taken with a grain of salt.

The bosonic nature of 4He combined with the delocalization means that liquid helium may be described as a Bose-Einstein condensate (BEC) at sufficiently low temperatures, although it should be mentioned that this classification has been questioned. In any case, a phase transition occurs at 2.17 K for bulk liquid helium, called the lambda transition after the similarity of the heat capacity vs. temperature around the transition temperature with the Greek letter. Below this temperature the liquid has a number of interesting properties that have been studied intensely. One is that the liquid is superfluid, i.e. moves with no friction. This also holds for interactions with foreign objects, provided the relative speed of the object and the helium does not exceed a value called the Landau critical velocity, and which is 60 m/s in the bulk.

For bulk helium the binding energy per atom is 0.86 meV, or 10 K. In a beam of freely evaporating helium droplets the temperature is a small fraction of this energy. In Chap.  6 it was found to be on the order of 3–4 % for typical clusters, and the value is similar for helium. Evaporative cooling therefore generates internal energies corresponding to temperatures of ≈0.4 K and is used to thermalize impurities to this temperature. A temperature of 0.4 K is routinely obtained in conventional cryostats for pieces of macroscopic matter, but is an otherwise unreachable temperature for gas phase molecule and cluster beams. The temperature is well below the lambda temperature of 2.17 K, and for all but the smallest droplets one must expect that they will be superfluid. And indeed, some molecules have been observed to rotate freely inside helium droplets with rotational constants close to the vacuum values, which is only possible for superfluid droplets.

Experimentally, droplets are formed during expansion of pre-cooled He gas with a stagnation pressure, i.e. the pressure in the source, close to the saturated vapor pressure. The method has the potential to produce droplets containing very large numbers of He atoms. After production the droplets can be passed through a chamber where they pick up one or more molecules or atoms that can aggregate in the droplets to form clusters. Studies, for example of spectroscopic nature, can then be performed downstream in a molecular beam machine. The pickup chamber may be heated to high temperatures to get a workable vapor pressure of the dopants without destroying the droplets, because helium is almost transparent to black body radiation at even very high temperatures.

Absorption of photons by dopants are monitored by the loss of mass in the droplet, caused by the relaxation of the excitation energy and dissipation into the droplet which results in He atoms boiling off. The method has a high sensitivity, because the absorption of a photon with even a small energy will cause the loss of a large number of He atoms. An infrared photon with an energy of 500 cm−1, say, equal to 62 meV, will cause the loss of typically 500 cm−1/0.86 meV=72 atoms from the droplet. Cycling through this process several times can give a mass loss that can be measured experimentally.

10.1 The Excitation Spectrum

The calculation of the thermal properties of a helium droplet is greatly facilitated by the fact that the two most important contributions are separable to a good approximation. The effects of dopants is less known and we will consider the pure helium droplet.

The ground state of a droplet is spherical because that shape reduces the surface energy to the minimum. Waves (or ripples) on the surface of this sphere are the lowest energy excitations of the droplet shape. The quantized excitations are called riplons, and they come with an associated angular momentum, ≥2. The spectrum starts at angular momentum 2 because zero angular momentum is spherically symmetric and does not represent a wave, and a putative unit angular momentum wave would correspond to a displacement of the whole droplet and is therefore a translation.

The quantum energies of the ripplons are
$$ E_{\ell} = \hbar \omega_0 \bigl(\ell(\ell-1) (\ell+2) \bigr)^{1/2}. $$
Each mode has a degeneracy of 2+1. The frequency ω 0 is
$$ \omega _0 = \sqrt {\frac{\sigma _t }{DR^3}}, $$
where σ t is the surface tension, D the mass density, and R the droplet radius. If we use the assumption of constant density (see Chap.  7), disregarding its shortcomings mentioned above, the denominator in the square root is equal to 3m a N/4π, where m a is the mass of the atom and N is the number of atoms in the droplet. This gives, with σ t =3.54⋅10−4 N/m,
$$ \omega_0 = \sqrt {\frac{4\pi \sigma _t }{3m_a N}}, $$
$$ \hbar \omega_0 \approx 3.6~\mathrm{K} N^{-1/2}. $$
The only other type of excitation relevant for us are phonons. They are quantized sound waves, as one can hear form the name, and involve also atoms in the interior of the droplet. They arise as solutions to the wave equation that describes compression waves over the whole volume of the droplet. The dispersion relation relates frequencies to wave vectors,
$$ \omega_{n,\ell } = ck_{n,\ell } $$
where c is the speed of sound and k is the wavenumber.1 Hence phonon quantum energies are:
$$ \varepsilon_{n,\ell } = \hbar ck_{n,\ell}. $$
The wavenumber k n, is determined by the boundary condition at the surface. For a spherical particle and a free surface, the k’s are given by
$$ k_{n,\ell }= a'_{n,\ell}/R, $$
where \(a'_{n,\ell}\) is the n’th root of the derivative of the spherical Bessel function, \(j'_{\ell}\).
It is convenient to express these energies in scaled units. We could use the highest frequency in the spectrum, the Debye frequency, as the energy scale. The Debye temperature for phonons in liquid 4He is ≈20 K, which is a very high energy in this connection (bulk helium boils at 4 K under atmospheric pressure and we will be looking at droplets below 1 K). We will therefore express phonon energies in units of the energy of the longest wavelength, where the 0.40 K values for the density (0.145 g/cm2) and the speed of sound (238 m/s) were used. This energy scale decreases slower than that of the ripplons with N. Nevertheless, phonons turn out to be more important that ripplons for large sizes, because the number of phonons are proportional to N, whereas the number of ripplons is proportional to N 2/3. We will return to a quantitative estimate below, when numbers are available.

Finally, it should be mentioned that a third type of excitations exist, called rotons. We will not consider these.

10.2 Ripplon Thermal Properties

He droplets are almost always observed freely flying through a molecular beam machine. The most important ensemble for helium droplets is therefore the microcanonical ensemble, which means that we need to calculate the level density of the species to describe their thermal properties. In a previous chapter we have developed different formulae to calculate level densities directly from both the vibrational spectra, with the Beyer-Swinehart algorithm, and a inversion of the canonical partition function. We want a closed analytical form and use the latter procedure here.

We begin with a calculation of the leading order contribution from the ripplon excitation to provide a quick and portable result, followed by a more accurate calculation where higher order terms are included. The calculation requires that we have the canonical partition function. The ripplon elementary excitations are bosons and we therefore sum over all possible ‘occupation numbers’, which in this case are excitation energies:
$$ \ln Z = - \sum_{\ell =2}^{\ell_{max}} (2\ell+1) \ln \bigl( 1-e^{-\beta\varepsilon_{\ell}} \bigr). $$
To leading order, we can replace the sum in Eq. (10.15) by an integral from zero to infinity, and approximate the energy eigenvalues (10.7) by ε ħω 0 3/2. The integral then evaluates to
$$ \ln Z = \varGamma \biggl( \frac{7}{3} \biggr) \zeta \biggl(\frac{7}{3} \biggr) \beta^{ - 4/3} = 1.685 (\beta \hbar \omega_0 )^{ - 4/3}, $$
where ζ is the Riemann zeta function and the energy scale was reintroduced in the last equality. The energy is then
$$ E = -\frac{\partial \ln Z}{\partial \beta} = 2.25 \frac{T^{7/3}}{ ( \hbar \omega_0 )^{4/3}} = 0.407N^{2/3} \frac{T^{7/3}}{\mathrm{K}^{4/3}}. $$
The temperature and entropy are expressed in terms of the excitation energy and use of Eq. ( 3.45) gives
$$ \rho(E) = \frac{e^S}{\sqrt { - 2\pi (\partial E/\partial \beta )}}, $$
where S=βE+lnZ is the entropy. We then get the leading order ripplon level density
$$ \rho_{rip} (E) \approx 0.31 \biggl( \frac{E}{\hbar \omega_0} \biggr)^{ - 5/7} \exp \biggl(2.48 \biggl(\frac{E}{\hbar \omega_0} \biggr)^{4/7} \biggr). $$

This expression is sufficient for most purposes, but we have the tools to calculate higher order contributions and it is worth doing, both to check if the corrections are indeed small and because it is so typical for these kinds of problems that it is worth giving another example. The calculation is lengthy, and will not be shown in all detail.

The first improvement is to use the relation between the microcanonical energy and temperature from Chap.  3:
$$ E = - \partial \ln Z /\partial \beta - \beta ^{ - 1}, $$
where Z is the canonical partition function at the microcanonical temperature β −1. The next is to evaluate the partition function in Eq. (10.15) more accurately by using the Euler-Maclaurin formula. We use the first three terms in the expansion: The upper limit of integration of the first term has been set to infinity. The actual value is on the order of max ≈2πR/λ min ≈2πR/(2d), where λ is the wavelength and d is the interatomic distance. In the liquid drop approximation (R=N 1/3 d/2) one has max πN 1/3/2. In view of Eqs. (10.7, 10.8) this yields a size-independent ripplon Debye temperature of ε max ≈7.1 K. Using this value to estimate the error in lnZ, we find that the neglected terms are on the order of (βε max /4−7 max /6)exp(−βε max ). For T=1 K this is a relative contribution to ln(Z) of less than 10−2/N 1/3 which can be ignored.
In the third step the relation between energy and angular momentum, Eq. (10.7), is inverted to express the angular momentum as a function of the energy. This allows a calculation of the integral in Eq. (10.21) by substitution. Finally, with the expansion in βE 2 of the exponential exp(−βE 2), we get the result
$$ \ln Z \approx c_2 \beta^{-4/3} +\frac{c_1}{3} \beta^{-2/3} -\frac{349}{96}+\frac{7}{3}\ln(2\sqrt{2}\beta). $$
The numerical constants c 1=1.917 and c 2=1.685 that appear in this expression are given by
$$ c_n = \sum_{j=1}^{\infty} j^{-2n/3-1} \int_0^{\infty} x^{2n/3} e^{-x}dx = \zeta(2n/3+1)\varGamma(2n/3+1). $$
Inserting the numerical values of c 1 and c 2 into Eq. (10.22) and reintroducing the physical energy scale gives
$$ \ln Z = -\beta E +S = 1.685 \biggl(\frac{T}{\hbar \omega_0} \biggr)^{4/3} + 0.639 \biggl(\frac{T}{\hbar \omega_0} \biggr)^{2/3} - \frac{{349}}{{96}} - \frac{7}{3}\ln (2\sqrt{2} T/\hbar\omega_0 ). $$
The first term coincides with Eq. (10.16), as it should. The other terms vary slower with N than the leading order term and are finite-size corrections.
The result in Eq. (10.24) is a very good approximation to the exact result. The comparison with a numerical summation of the partition function for ripplons in Eq. (10.15) is shown in Fig. 10.2. Already at temperatures where T is equal to the lowest excitation energy \(\varepsilon_{2}=\hbar \omega_{0} \sqrt{6}\) is the free energy well represented by the above expression. At higher energies the agreement improves monotonically.
Fig. 10.2

The (negative of) the ripplon free energies, calculated with Eq. (10.24) (dashed line) and the summation in Eq. (10.15) (full line) which is exact apart from setting the upper summation limit to infinity. The dotted line is the first term of Eq. (10.24). The lowest excitation energy is indicated as ε 2. The data were published in K. Hansen, M.D. Johnson and V.V. Kresin, Phys. Rev. B 76 (2007) 235424

We proceed by finding the caloric curve, or energy-temperature relation from Eq. (10.22) or Eq. (10.24) for the free energy and Eq. (10.20): If we invert this relation and rewrite it slightly as
$$ T = \biggl(\frac{1}{2.247} \biggr)^{3/7} \biggl(E ( \hbar\omega_0)^{4/3} -0.426 T^{5/3}(\hbar \omega_0)^{2/3} +\frac{10}{3} T (\hbar \omega_0)^{4/3} \biggr)^{3/7}, $$
we can calculate T(E) by iterative approximations of the right hand side. The result is that T=E 3/7(ħω 0)4/7, times a power series in (ħω 0/E)2/7 which we need not write down here.
We now have all quantities that enter Eq. (10.18) and after using that F=ETS, inserting Eqs. (10.24), (10.25) and β(E) we get, keeping the leading three terms in the exponential, the total ripplon level density
$$ \rho_{rip} (E) = 0.205 (\hbar\omega_0)^{5/7} E^{-12/7} \exp \bigl( 2.48 (E/\hbar\omega_0 )^{4/7} + 0.507 (E/\hbar\omega_0 )^{2/7} \bigr), $$
where the energy scale is still given by ħω 0=3.6 K N −1/2.

The calculation above gives the level density for a given total energy. It is also possible to calculate the angular momentum-specified level density. This function is relevant because excitations come with angular momentum and angular momentum is a conserved quantity.

One important fact to note here is the degeneracy of the ripplon energy levels, 2J+1. This is the usual quantum mechanical degeneracy of a state with angular momentum J. This degeneracy signals that we cannot use a completely classical calculation, even for large quantum numbers. A degeneracy proportional to J gives a canonical thermal rotational energy of T in the high temperature limit. In contrast, a classical situation would give a degeneracy factor proportional to J 2, corresponding to the rotational degeneracy of a stiff body (see Eq. ( 7.29)). The canonical thermal energy of this is 3T/2 (see Chap.  2). Obviously, the difference between quantum and classical rotations has consequences for the angular momentum-resolved level density.

We can add the projections of the angular momenta, m, from all the individual degrees of freedom. The sum of thermal averages is zero:
$$ \langle M_{rip} \rangle \equiv \sum_{\ell=2}^{\infty} \langle m_{\ell}\rangle = 0, $$
because there is no preferred direction in space and therefore all the contributions from different angular momenta, , are zero, 〈m 〉=0.
The sum of squares is not zero (obviously). It is calculated as the sum over all ’s and all projections of the angular momentum, m. A single of these states with energy kE is populated with the probability exp(−kβE )/Z, where Z −1=1−exp(−βE ), and the length of the angular momentum projection is km where −m. For a given this gives
$$ \bigl\langle m_{\ell}^2 \bigr\rangle = \sum _{m=-\ell}^{\ell} \sum_{n=0}^{\infty} Z^{-1} m^2 n^2 e^{-n\beta E_{\ell}} = \biggl( \frac{2}{3} \ell^3 + \ell^2 + \ell/3 \biggr) \bigl(1-e^{-\beta E_{\ell}}\bigr) \sum_{n=0}^{\infty} n^2 e^{-n\beta E_{\ell}}. $$
The sum over m 2 will be represented by the leading term 2 3/3 in the following. The sum over n can be done with the result
$$ \bigl\langle m_{\ell}^2 \bigr\rangle = \frac{2}{3} \ell^3 \biggl[ \frac{e^{-\beta E_{\ell}}}{ 1-e^{-\beta E_{\ell}}} + 2 \frac{e^{-2\beta E_{\ell}}}{(1-e^{-\beta E_{\ell}})^2} \biggr]. $$
The total variance for the whole system of ripplons is then the sum of contributions from all :
$$ \bigl\langle M_{rip}^2 \bigr\rangle = \sum_{\ell=2}^{\infty} \bigl\langle m_{\ell}^2 \bigr\rangle. $$
We can approximate the sum over with an integral and set the lower integration limit to zero, which is acceptable for high temperatures. With the substitution u=βħω 0 3/2 we get
$$ \bigl\langle M_{rip}^2 \bigr\rangle \approx \frac{4}{9} \biggl( \frac{T}{\hbar \omega_0} \biggr)^{8/3} \int _0^{\infty} u^{5/3} \biggl( \frac{e^{-u}}{1- e^{-u}} +2 \frac{e^{-2u}}{(1- e^{-u})^2} \biggr) du. $$
The first integral is calculated by expanding 1/(1−e u ) in e u and integrating term by term, which gives the product Γ(8/3)ζ(8/3). For the last integral we expand the denominator as
$$ \frac{1}{(1-e^{-\beta E_{\ell}})^2} = \sum_{n=0}^{\infty} \sum _{k=0}^{\infty} e^{-(n+k)\beta E_{\ell}} = \sum _{n=0}^{\infty} (n+1)e^{-n\beta E_{\ell}}. $$
Inserting this into Eq. (10.32) one gets after some reshuffling of sums that the integral is IΓ(8/3)[2ζ(5/3)−ζ(8/3)]=1.981… and thus
$$ \bigl\langle M_{rip}^2 \bigr\rangle \approx \frac{4}{9} \biggl( \frac{T}{\hbar \omega_0} \biggr)^{8/3} I = 1.981\ldots \biggl( \frac{T}{\hbar \omega_0} \biggr)^{8/3} = 0.785\ldots \biggl( \frac{E}{\hbar \omega_0} \biggr)^{8/7}, $$
where the leading order term in the caloric curve, Eq. (10.26), was used. With ħω 0=3.6N −1/2 K, this can be expressed as
$$ \bigl\langle M_{rip}^2 \bigr\rangle = 5.65 \cdot 10^{-3} N^{4/3} \biggl(\frac{T}{0.4~\mathrm{K}} \biggr)^{8/3}. $$
This is the mean of the square of the total angular momentum projection on a fixed axis. We still need to find the distribution of this quantity, and do so by invoking the Central Limit Theorem which says that if enough stochastic variables are added, the sum will be normally distributed with a variance which is the sum of the variances of the individual terms. This is exactly what we have calculated for M, and we therefore have the (normalized) distribution of the projection of the total angular momentum:
$$ \rho(E,M) = \rho(E) \frac{1}{\sqrt{2\pi} \sigma_M} e^{-M^2/2\sigma_M^2}, $$
with \(\sigma_{M}^{2} = \langle M_{rip}^{2} \rangle\) given in Eq. (10.34).
This distribution can be used to find the distribution of the total angular momentum, J. The number of states of a given M, denoted by n(M), is the number of states of a given angular momentum, n(J), with the projection M. For every J, there is one projection with the value M, provided JM. For J<M there is zero. In other words:
$$ n(M) = \sum_{J=M}^{\infty} n(J). $$
Taking the difference of this relation for M and M+1 we get
$$ n(M) - n(M+1) = \sum_{J=M}^{\infty} n(J)- \sum_{J=M+1}^{\infty} n(J) = n(J=M). $$
Replacing the difference on the left hand side with a derivative, and the number of states, n, with the density of states, both of which are permissible procedures because the spacing of J is unity, one gets
$$ \rho(E,J) = -\frac{\partial \rho(E,M)}{\partial M} \bigg\vert_{M=J+1/2} = \rho(E) \frac{J+1/2}{\sqrt{2\pi}\sigma_M^3} e^{-(J+1/2)^2/2\sigma_M^2}. $$
With the size dependence of the energy scale from Eq. (10.10), the calculated σ M and setting T=0.4 K, the angular momentum distributions for different droplet sizes are found to be
$$ \rho(0.4\ \mathrm{K},J) \propto N^{-2} (J+1/2 ) \exp \bigl(-88 (J+1/2 )^2 N^{-4/3} \bigr). $$
The distribution is shown in Fig. 10.3 for the droplet size N=104. On inspection of Eq. (10.40) one sees that the distribution scales and has the same form for different sizes, with a scale factor that varies as N 2/3. The maximum of the distribution is reached at \(J = N^{2/3}/\sqrt{88 \cdot 2} = 0.075N^{2/3}\). Through σ M , the distribution scales with temperature as T 2/3.
Fig. 10.3

The angular momentum distribution from the ripplon degrees of freedom at a temperature of 0.4 K for N=104. The maximum for other droplet sizes varies as N 2/3, as described in the text

10.3 Molecular Beam Temperatures

We will now estimate the previously announced temperature for droplets flying freely through vacuum. The derivation will include only the ripplon thermal properties because these are the most important for droplets that are not extremely big. The droplets cool by evaporating atoms and is an example of the ensembles treated in detail Chap.  6. Following this derivation we set k=C/tG 2, where t is the time since the droplets started free flight. We know the evaporation rate constant of helium atoms from a droplet from Eq. ( 5.11). With zero electronic degeneracy and a geometric cross section it is
$$ k(E) = \frac{m}{\pi^2\hbar^3} \pi r_1^2 N^{2/3} T_d^2 \frac{\rho_{N-1}(E-E_a)}{\rho_N(E)} = \nu \frac{\rho_{N-1}(E-E_a)}{\rho_N(E)}, $$
where r 1=2.22 Å and E a =10 K are the atomic radius and the evaporative activation energy, respectively, and T d is the product (daughter) temperature. With T d =0.4 K the pre-exponential factor is calculated to
$$ \nu \equiv \frac{m}{\pi^2\hbar^3} \pi r_1^2 N^{2/3} T_d^2 = 1.8\cdot 10^9 N^{2/3}~\mathrm{s}^{-1}. $$
The level density of the product, N−1, is close enough to that of droplet N to initially ignore the difference, i.e. we set ρ N−1(E)≈ρ N (E) (see Exercise 10.1), and the ratio of level densities is then ρ N−1(EE a )/ρ N (E)≈ρ N (EE a )/ρ N (E). The leading order finite heat bath correction is E a /2C (see Chap.  3). With E a =10 K and the heat capacity from the ripplons, we get a value relative to the temperature of
$$ \frac{E_a}{2CT} = 5.3 N^{-2/3} T^{-7/3}\ \mathrm{K}^{7/3}, $$
which is 45N −2/3=(300/N)2/3 for T=0.4 K. This all adds up to a rate constant of
$$ k(E) = 1.8\cdot 10^9 N^{2/3}\ \mathrm{s}^{-1} \exp \biggl(-\frac{E_a}{T (1- (\frac{300}{N} )^{2/3} )} \biggr). $$
For precise estimates, N needs to be above 104 before the finite heat bath correction can be ignored. Above that number the beam temperature, which is at the same time approximately both parent and daughter temperature, is
$$ T = \frac{E_a}{\ln (1.8\cdot 10^9 N^{2/3}\ \mathrm{s}^{-1} t )}. $$
The precise value depends weakly on the droplet size and on the observation time. The droplets are neutral, their speed thermal and typical flight times after creation will be between 100 μs and 10 ms, say. Sizes are not easy to measure for neutral particles but a good deal of effort has been devoted to this question and there are tools available that will give mean sizes with reasonable certainty. If we limit ourself to numbers between 104 and 108 atoms, we get for the extreme cases that T=0.55 K (for 100 μs, N=104) and T=0.35 K (for 10 ms, N=108). These numbers are a lot closer to each other than one may have guessed, given the wide range of the parameters, but they are not identical. The value of 0.36 K often quoted in the literature refers to relatively large droplets measured after a relatively long time.

10.4 Phonon Level Density

The leading-order behavior of the phonon density of states can be determined using the expression for the Debye heat capacity of bulk phonons calculated in the chapter on vibrational thermal properties (Eq. ( 4.44)), with some modifications. If the droplet is superfluid, which experimentally seems to be the case when it flies through vacuum, it can only support waves in the direction of the propagation. The two perpendicular modes are not supported by a superfluid, and the low temperature Debye heat capacity in Eq. ( 4.44) must therefore be divided by a factor 3:
$$ \frac{\partial E}{\partial T} = C_{bulk} \approx \frac{2 \pi^2 V}{15} \biggl(\frac{T}{\hbar c} \biggr)^3 = \frac{8\pi^3}{45} R^3 \biggl(\frac{T}{\hbar c} \biggr)^3= \frac{8\pi^6}{45} \biggl( \frac{T}{\tilde{\varepsilon}} \biggr)^3, $$
where c is the speed of sound, and the energy scale, \(\tilde{\varepsilon} = \hbar c\pi /R\) from Eq. (10.14) was used. The experimental data for bulk helium, where practically all modes are phonons, agree very well with this equation, as shown in Fig. 10.4. We find the energy content by integration with respect to T:
$$ E = V \int_0^{T} C_{bulk}\bigl(T'\bigr)dT' = \frac{2\pi^6}{45} \frac{T^4}{\tilde{\varepsilon}^3} =0.92 N^{2/3} \frac{T^4}{\mathrm{K}^3}. $$
With the caloric curve in Eq. (10.47) we can find the entropy in terms of the excitation energy by integration with the factor 1/T. From the general result in Eq. (10.18), the level density is found to be
$$ \rho_{ph} (E) \approx A \varepsilon^{-3/8} E^{-5/8} \exp \biggl( 3.409 \biggl( \frac{E}{\tilde{\varepsilon}} \biggr)^{3/4} \biggr), $$
where A is 0.32.
Fig. 10.4

The experimentally measured 4He bulk heat capacity at the saturated vapor pressure (circles) and the predicted 1/3 Debye heat capacity, calculated with Eq. (10.46) and the experimentally measured density of 0.1451 g/cm3 and the speed of sound c=238 m/s. There are no adjustable parameters in the theoretical curve

This is the first estimate of the level density, and depending on the size of the particle it may be precise enough. Higher order contributions are mainly generated by the deviations of the roots of the Bessel functions from an equidistant spectrum. The calculation of the thermal properties of such a system seems like a very difficult problem, because approximations for roots of these functions are usually calculated for asymptotically large quantum numbers, and asymptotically large in that sense may still be larger than relevant for our high temperature expansions.

Fortunately there exists an expansion of the smoothed density of states in a finite cavity. When applying these corrections, which we will not derive, we get the heat capacity (per unit volume)
$$ C = C_{bulk} + \frac{9\zeta(3)}{4\pi}\frac{1}{\hbar^2 c^2 } \frac{T^2}{R} + \frac{1}{6}\frac{1}{\hbar c}\frac{T}{R^2}. $$
As one can see from the R dependence, the corrections can be understood as a surface and a curvature term. The entropy and the caloric curve can be found by integration as above and the rest of the procedure is similar to the one used for ripplons. We have gone through this procedure a couple of times and just give the result: This expression is accurate enough to warrant a comparison with the level density calculated with the Beyer-Swinehart algorithm (see Chap.  4) and the roots of the Bessel functions. We must expect that there are corrections that are unaccounted for because the argument of the exponential is an expansion in E 1/4, and the fourth term in Eq. (10.49), which we don’t know, will be constant and contribute to the pre-exponential. The numerical calculation gives a leading order correction of exp(−0.62(E/ε)0.2). An effective value of A≈0.05 can be used for phonon energies below 400 \(\tilde{\varepsilon}\).

Phonons also carry angular momentum, and one of the two indices of the spherical Bessel functions gives the angular momentum (the other counts the number of nodes in the radial motion). For the calculation of the phonon angular momentum-resolved density of states we will be satisfied with the leading order contribution.

For the free surface boundary condition, the roots of the Bessel functions are asymptotically (n+/2−3/4)π≈(n+/2)π. If we use the phonon energy scale in Eq. (10.14), the quantum energies are thus n+/2. In analogy with the calculation for ripplons, we sum the squares of projections of angular momentum as
$$ \bigl\langle M_{ph}^2 \bigr\rangle = \sum_{n=0}^{\infty} \sum_{l=0}^{\infty} \sum_{m=-\ell}^{m=\ell} \frac{ \sum_{k=0}^{\infty} m^2 k^2 e^{-k\beta E_{n,\ell}}}{\sum_{k=0}^{\infty} e^{-k\beta E_{n,\ell}}}. $$
There is one more summation here compared with the ripplon sum, because the radial nodes are also summed over, consistent with the three dimensional nature of phonens vs. the two dimensional nature of ripplons. The sum over m 2 gives 2 3/3, as before, and the sum over k is also identical to the ripplon case. We therefore have
$$ \bigl\langle M_{ph}^2 \bigr\rangle \approx \frac{2}{3} \sum_{n=0}^{\infty} \sum _{l=0}^{2n} \ell^3 \biggl[ \frac{e^{-\beta E_{n,\ell}}}{1-e^{-\beta E_{n,\ell}}} +2\frac{e^{-2\beta E_{n,\ell}}}{(1-e^{-\beta E_{n,\ell}})^2} \biggr]. $$
We perform one of the remaining two sums over constant energy surfaces in the n, plane. For the first term in Eq. (10.52), this gives an approximate value of
$$ \frac{2}{3} \sum_{n=0}^{\infty} \frac{e^{-\beta E_{n,0}}}{1- e^{-\beta E_{n,0}}} \sum_{l=0}^{2n} \ell^3 \approx \frac{8}{3} \sum_{n=0}^{\infty} \frac{e^{-\beta E_{n,0}}}{1- e^{-\beta E_{n,0}}} n^4 \approx \frac{8}{3} \int _{0}^{\infty} \frac{e^{-\beta n}}{1- e^{-\beta n}} n^4 dn. $$
This is calculated with the same method used for ripplons; expand the integrand in exp(−βn), interchange integration and summation and shuffle some limits of the sums. The second term is calculated analogously. We get
$$ \bigl\langle M_{ph}^2 \bigr\rangle \approx \frac{8}{3} {\rm\varGamma}(5) \bigl(2\zeta(4)-\zeta(5) \bigr)T^5 = 72.2 \biggl(\frac{T}{\tilde{\varepsilon}} \biggr)^5, $$
where the energy scale has been reintroduced. Inserting the value of \(\overline{\varepsilon}\), the mean square M is
$$ \bigl\langle M_{ph}^2 \bigr\rangle \approx 6.3 \cdot 10^{-6} N^{5/3} \biggl( \frac{T}{{\rm K}} \biggr)^5, $$
$$ \bigl\langle M_{ph}^2 \bigr\rangle \approx 6.5 \cdot 10^{-8} N^{5/3}, $$
for T=0.4 K.

Once \(\langle M_{ph}^{2} \rangle\) is calculated, the distributions are derived following the procedure for ripplons in all details, and Eq. (10.39) can be used also for phonons, simply by replacing σ M with the phonon value.

10.5 Thermal Properties of the Combined Excitation Spectrum

The ratio of the leading orders of the ripplon and the phonon excitation energies is
$$ \frac{E_{ph}}{E_{rip}} = \frac{\frac{2\pi^6}{15}\frac{T^4}{\tilde{\varepsilon}^3}}{ 2.25 \frac{T^{7/3}}{(\hbar \omega_0)^{4/3}}} = 0.018 N^{1/3} \biggl( \frac{T}{\mathrm{K}} \biggr)^{5/3}. $$
For T=0.4 K we have equal contributions when N=2⋅107. Below this fairly large but not outrageously oversized droplet, the excitation energy of the surface modes (ripplons) exceed that of the bulk modes (phonons).
When the droplet size approaches the crossover size, it will have both ripplon and phonon oscillations excited at the same time and the thermal properties will be determined by the total level density obtained by combining the ripplon and phonon functions. One can calculate the joint level density as the convolution,
$$ \rho_{total} (E) = \int_0^{E} \rho_{rip}(E-\epsilon) \rho_{ph}(\epsilon) d\epsilon. $$
As it turns out, this is not trivial to do with the same precision with which the two input functions are known. Instead of performing the convolution, we can calculate the thermal properties of the system at the energy partition where the joint distribution, i.e. the integrand in Eq. (10.58), has its maximum. This energy corresponds to identical temperatures in the two subsystems (see Chap.  1) and makes microcanonical thermal properties such as energy, entropy and heat capacity of the two subsystems additive,
$$ E = E_{rip} + E_{ph} = 2.247 \frac{T^{7/3}}{(\hbar \omega_0)^{4/3}} + \frac{2\pi^6}{45}\frac{T^4}{\tilde{\varepsilon}^3} +\cdots. $$
In the limit where the phonons carry only a relatively small amount of energy, the convoluted level density in Eq. (10.58) can be performed approximately in a procedure similar to the one used to define the microcanonical temperature in Chap.  1. The result is that the total level density is the product of the ripplon level density and the phonon canonical partition function;
$$ \rho_{total} (E) \approx \rho_{rip}(E) Z_{ph}\bigl(T(E)\bigr), $$
where T is the microcanonical temperature of the ripplons. The simplest way to calculate the phonon partition function here is to start with the level density and multiply with the Boltzmann factor and the preexponential factor as given in Eq. ( 3.45). The calculation of the expression is left as an exercise to the reader.
By the same logic as used in Eq. (10.60) one can calculate the level density of the joint system when the phonons dominate the thermal properties. The result is symmetric in the subscripts relative to Eq. (10.60);
$$ \rho_{total} (E) \approx \rho_{ph}(E) Z_{rip}\bigl(T(E)\bigr). $$
Also the angular momenta of the two different types of excitation add. The ratio of the ripplon and phonon contributions to the width of the angular momentum distributions is, to leading order,
$$ \frac{\sigma_{ph}}{\sigma_{rip}} \equiv \sqrt{\frac{\langle M_{ph}^2 \rangle}{\langle M_{rip}^2 \rangle}} \sim 0.0098 ( T/{\rm K} )^{7/6} N^{1/6}. $$
This ratio is small compared to unity up to very large droplet sizes. For our 0.4 K favorite example, unity is reached for a droplet size with a radius of 19 μm! For smaller sizes than this, one can ignore the contribution from phonons in the angular momentum distribution.

10.6 Exercises


Show that, for ripplon level densities,
$$ \rho_{N-1}(E) \approx \rho_N(E) \exp \biggl(- \frac{\alpha}{N^{1/3} (\frac{T}{3.6\ \mathrm{K}})} \biggr), $$
where α is on the order of unity.


Show that the distribution in Eq. (10.39) is normalized.


Complete the steps between Eq. (10.52) and Eq. (10.54).


Calculate the joint ripplon-phonon level density both for situation when ripplons carry most of the excitation energy and when phonons do. Insert the scaling factors to get an explicit expression.


  1. 1.

    The term wavenumber doubles as the length of the wave vector, as here, and an energy unit.

Copyright information

© Springer Science+Business Media Dordrecht 2013

Authors and Affiliations

  • Klavs Hansen
    • 1
  1. 1.Department of PhysicsUniversity of GothenburgGothenburgSweden

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