Generalised Ramachandran pairing interaction in helium-4 and helium-3 superfluids

Abstract

We make use of the ionisation energy theory, Ramachandran interaction theory and the energy-level spacing renormalisation group technique to extend the Drude Hamiltonian to derive the Ramachandran pairing mechanism. This particular mechanism is exploited to explain the notorious discontinuous jumps in the specific heat data at critical points in both He-4 and He-3 superfluids. The well-known spin states (due to Balian–Werthamer and Anderson–Brinkman–Morel theories) and the Leggett’s spontaneously broken spin-orbit symmetry are shown to enhance Ramachandran attraction between two He-3 atoms without the need to invoke the spin-induced pairing or the phonon-mediated Cooper pairing mechanism in superconductors. In addition, we shall show physically that the spin-exchange mechanism can neither cause pairing between electrons nor between He-3 atoms.

Appendix

1.1 Derivation of the Ramachandran attraction: Eq. (1)

In order to derive eq. (1), we need to first derive some relevant equations which are needed to understand the existence of the so-called Ramachandran attraction. We start from the one-dimensional Drude model Hamiltonian for a two-atom system (isolated), which is given by [52, 53]

$$\begin{aligned}&H = H_\mathrm{O} + H_\mathrm{I}, \end{aligned}$$

(16)

$$\begin{aligned}&H_\mathrm{O} = \frac{p_1^2}{2m} + \frac{1}{2}Qr^2_1 + \frac{p_2^2}{2m} + \frac{1}{2}Qr^2_2, \end{aligned}$$

(17)

$$\begin{aligned}&H_\mathrm{I} = \frac{1}{4\pi \epsilon _0}\bigg [\frac{(e)(e)}{R} + \frac{(-e)(e)}{R + r_2}\nonumber \\&\qquad \quad + \frac{(-e)(e)}{R - r_1} + \frac{(-e)(-e)}{R - r_1 + r_2}\bigg ], \end{aligned}$$

(18)

where $p_{1,2}$ and $Q_{1} = Q_{2} = Q$ are the electrons momenta and spring’s (interaction potential) constant, respectively, and R is the distance between two neutral atoms (interatomic distance). The subscripts 1 and 2 are used to identify electrons 1 and 2, both bounded to nuclei 1 and 2, respectively. The ordinary Hamiltonian without interaction, $H_\mathrm{O}$, consists of two kinetic energy terms and two semiclassical harmonic-oscillator type el–ion potential terms. These potential terms refer to the el–ion interactions within their respective atoms (see figure 1 in ref. [32]). In contrast, the interaction Hamiltonian, $H_\mathrm{I}$, captures all the Coulomb interactions that may exist between the two interacting atoms.

Assuming that the polarisation of electrons obey harmonic oscillation with very weak interaction, we can impose these conditions, namely, $|r_1| \ll R$ and $|r_2| \ll R$, which means $|Rr_1 - Rr_2| \approx |R^2r_1 - R^2r_2| \approx 0$ and $R^3 \gg |Rr_1r_2|$, and therefore, we can readily rearrange $H_\mathrm{I}$ to read,

$$\begin{aligned} H_\mathrm{I} \cong \frac{e^2r_1r_2}{2\pi \epsilon _0R^3}. \end{aligned}$$

(19)

Subsequently, we can further rearrange eqs (17) and (19) by using the following variables [52]:

$$\begin{aligned} r_{\pm } = \frac{1}{\sqrt{2}}\big (r_1 \pm r_2\big ) \end{aligned}$$

(20)

and

$$\begin{aligned} p_{\pm } = \frac{1}{\sqrt{2}}\big (p_1 \pm p_2\big ) \end{aligned}$$

(21)

to arrive at

$$\begin{aligned} H= & {} \bigg [\frac{1}{2m}p^2_{+} + \frac{1}{2}\bigg (Q - \frac{e^2}{2\pi \epsilon _0R^3}\bigg )r^2_+\bigg ] \nonumber \\&+ \bigg [\frac{1}{2m}p^2_{-} + \frac{1}{2}\bigg (Q + \frac{e^2}{2\pi \epsilon _0R^3}\bigg )r^2_-\bigg ]. \end{aligned}$$

(22)

Equation (22) is in a suitable form to be compared with eq. (23),

$$\begin{aligned} -\frac{\hbar ^2}{2m}\frac{\mathrm{d}^2\psi }{\mathrm{d}x^2} + \frac{1}{2}m\omega ^2x^2\psi = \hat{H}\psi = E\psi , \end{aligned}$$

(23)

(see the kinetic and potential energy operators on the left-hand side of eq. (23)). In doing so, we should be able to infer the ground-state energy from the Drude Hamiltonian given in eq. (22), which is

$$\begin{aligned} E= & {} \frac{1}{2}\hbar \big (\omega _+ + \omega _-\big ), \end{aligned}$$

(24)

$$\begin{aligned}= & {} \frac{1}{2}\hbar \bigg [\sqrt{\frac{Q - (e^2/2\pi \epsilon _0R^3)}{m}} \nonumber \\&\quad + \sqrt{\frac{Q + (e^2/2\pi \epsilon _0R^3)}{m}}\bigg ]. \end{aligned}$$

(25)

We shall not work out the algebraic rearrangements here because the complete solution is already available in ref. [52], and the rearrangements are straightforward. Anyway, it should be clear that m is the electron mass, while the interaction constant Q is different for different atoms. Before we move on to derive the Ramachandran interaction potential operator, let us prove the following theorem:

Theorem 1

The coordinates, $\mathbf {{r}}_1$ and $\mathbf {r}_2$ are vectors such that we have aligned them to obtain maximum electronic polarisation along the x-axis where $\mathbf {r}_1 \rightarrow r_1$ and $\mathbf {r}_2 \rightarrow r_2$. Therefore, eq. (22) also applies to real (three-dimensional) atoms such that the polarisation has been confined to a particular direction (one-dimensional).

Proof

First, imagine that there are two polarisable spherical atoms. Subsequently, suppose the electron coordinate, $\mathbf{r} _1$ (from sphere 1) has been aligned along the x-axis, so that $|\mathbf{r} _1| = r_1$. The induced polarisation caused by sphere 1 due to $\mathbf{r} _1$ shall affect $\mathbf{r} _2$ giving rise to

$$\begin{aligned} |\mathbf{r} _2| = r_2(\theta ,\phi ), \end{aligned}$$

(26)

where

$$\begin{aligned} \theta< \pi /4,~ \phi < \pi /4, \end{aligned}$$

(27)

$\theta$ is the angle in the xy-plane and $\phi$ denotes the angle in the xz-plane. It is easily verified that if $\theta \ge \pi /4$ and $\phi \ge \pi /4$, then there is an additional induced polarisation affecting $|\mathbf{r} _2| = r_2(\theta ,\phi )$. If such an additional polarisation does exist, then we can take it into account by realigning the coordinate, $\mathbf{r} _2$ (from sphere 2) with respect to this new $\mathbf{r} _1'$. In this ‘additional-polarisation’ case, $|\mathbf{r} _1'| = r_1'$, which then leads to

$$\begin{aligned} |\mathbf{r} _2'| = r_2'(\theta ,\phi ), \end{aligned}$$

(28)

where the angles remain the same, for example, $\theta < \pi /4$ and $\phi < \pi /4$. This means that,

$$\begin{aligned} \theta \in (0,\pi /4],~\phi \in (0,\pi /4]. \end{aligned}$$

(29)

For one-dimensional polarisation cases, we always have $\theta = 0 = \phi$, which is the case for eq. (22), while for two-dimensional polarisation cases, one has $\theta \in (0,\pi /4]$ and $\phi = 0$. $\square$

1.2 Renormalised Van der Waals interaction

Before one can derive the Ramachandran interaction potential operator, we need to first derive the standard vdW interaction operator, which can be obtained from the ground-state energy E (see eq. (25)) and from the ground-state energy $E'$ when $H_\mathrm{I} = 0$. This is important to observe and acknowledge the fundamental difference between vdW and Ramachandran attraction. It is trivial to derive $E'$ because $H_\mathrm{I} = 0$. Hence, eq. (22) reduces to

$$\begin{aligned} H' = \bigg [\frac{1}{2m}p^2_{+} + \frac{1}{2}Qr^2_+\bigg ] + \bigg [\frac{1}{2m}p^2_{-} + \frac{1}{2}Qr^2_-\bigg ]. \end{aligned}$$

(30)

Subsequently, eq. (30) directly leads us to

$$\begin{aligned} E' = \hbar \omega ' = \hbar \sqrt{\frac{Q}{m}}. \end{aligned}$$

(31)

After imposing the condition,

$$\begin{aligned} Q \gg \frac{e^2}{2\pi \epsilon _0R^3} \end{aligned}$$

(32)

in eq. (25), we can find the standard vdW interaction potential,

$$\begin{aligned} V_\mathrm{vdW} = E - E' \cong -\frac{\hbar }{8m^2\omega '^3}\bigg (\frac{e^2}{2\pi \epsilon _0}\bigg )^2\frac{1}{R^6}. \end{aligned}$$

(33)

The negative sign in eq. (33) implies that $V_\mathrm{vdW}$ is an attractive interaction, and also, it is weak due to $1/R^6$ dependence. From eq. (34),

$$\begin{aligned} Q^\mathrm{dressed} = Q\exp {[\lambda (\xi - E_\mathrm{F}^0)]}, \end{aligned}$$

(34)

the renormalised Q,

$$\begin{aligned} \tilde{Q} = Q\exp [\lambda \xi ], \end{aligned}$$

(35)

where we prefer to denote any renormalised parameter with a tilde. Recall here that $\tilde{Q}$ (from eq. (35)) replaces the standard Q given in $H_\mathrm{O}$ to evaluate the changes to Q when we replace one atom with another. Moreover, we (as required by the Drude Hamiltonian) do not allow any electronic wave function overlapping, in which, the said overlapping shall lead to quantum phase transition.

Subsequently, we need to renormalise eq. (33), which can be done by first substituting all Q in eq. (25) with eq. (35), which leaves us with

$$\begin{aligned}&\tilde{E} = \frac{1}{2}\hbar \bigg [\sqrt{\frac{\tilde{Q} - (e^2/2\pi \epsilon _0R^3)}{m}}\nonumber \\&\qquad + \sqrt{\frac{\tilde{Q} + (e^2/2\pi \epsilon _0R^3)}{m}}\bigg ]. \end{aligned}$$

(36)

Here, any variable found to wear a tilde is a renormalised parameter. We now expand eq. (36) using the series,

$$\begin{aligned} \sqrt{1 + (\pm x)} = 1 \pm \frac{x}{2} - \frac{x^2}{8} \pm \frac{x^3}{16} - \cdots \end{aligned}$$

(37)

to arrive at

$$\begin{aligned} \tilde{E}= & {} \frac{1}{2}\hbar \bigg [\sqrt{\frac{\tilde{Q}}{m} - \frac{e^2/2\pi \epsilon _0R^3}{m}} + \sqrt{\frac{\tilde{Q}}{m} + \frac{e^2/2\pi \epsilon _0R^3}{m}}\bigg ], \end{aligned}$$

(38)

$$\begin{aligned}= & {} \frac{1}{2}\hbar \sqrt{\frac{\tilde{Q}}{m}}\bigg [\sqrt{1 + \bigg (-\frac{e^2/2\pi \epsilon _0R^3}{\tilde{Q}}\bigg )}\nonumber \\&+ \sqrt{1 + \bigg (\frac{e^2/2\pi \epsilon _0R^3}{\tilde{Q}}\bigg )}\bigg ]. \end{aligned}$$

(39)

Substituting

$$\begin{aligned} {\tilde{\omega }}' = \sqrt{\frac{\tilde{Q}}{m}}, \end{aligned}$$

(40)

into eq. (39) and noting that,

$$\begin{aligned} x = \mp \frac{e^2}{2\pi \epsilon _0R^3\tilde{Q}}, \end{aligned}$$

(41)

we can get

$$\begin{aligned} \tilde{E}= & {} \frac{1}{2}\hbar {\tilde{\omega }}'\bigg \{\bigg [1 + \frac{1}{2}\bigg (-\frac{e^2}{2\pi \epsilon _0R^3\tilde{Q}}\bigg ) \nonumber \\&- \frac{1}{8}\bigg (-\frac{e^2}{2\pi \epsilon _0R^3\tilde{Q}}\bigg )^2\nonumber \\&+ \frac{1}{16}\bigg (-\frac{e^2}{2\pi \epsilon _0R^3\tilde{Q}}\bigg )^3 - \cdots \bigg ] \nonumber \\&+ \bigg [1 + \frac{1}{2}\bigg (\frac{e^2}{2\pi \epsilon _0R^3\tilde{Q}}\bigg ) - \frac{1}{8}\bigg (\frac{e^2}{2\pi \epsilon _0R^3\tilde{Q}}\bigg )^2\nonumber \\&+ \frac{1}{16}\bigg (\frac{e^2}{2\pi \epsilon _0R^3\tilde{Q}}\bigg )^3 - \cdots \bigg ]\bigg \}. \end{aligned}$$

(42)

It is clear from eq. (42) that all the terms with odd powers ($x^{1,3,5,\ldots }$) cancel out, and the terms with only even powers, $x^{2,4,6,\ldots }$ survive. Moreover, we should also observe that

$$\begin{aligned} \bigg |\pm \frac{e^2}{2\pi \epsilon _0R^3\tilde{Q}}\bigg | \le 1. \end{aligned}$$

(43)

Consequently, we just need to impose the condition,

$$\begin{aligned} Q\exp [\lambda \xi ] \gg \frac{e^2}{2\pi \epsilon _0R^3}, \end{aligned}$$

(44)

which means,

$$\begin{aligned} \bigg (\frac{e^2}{2\pi \epsilon _0R^3}\bigg )^{\texttt {n}}\tilde{Q}^{-\texttt {n}} \approx 0, \end{aligned}$$

(45)

for all $\texttt {n} \ge 4$ where $\texttt {n} \in \mathbb {N}_\mathrm{even}$. Here, $\mathbb {N}_\mathrm{even}$ is the set of even natural numbers. After imposing the said condition,

$$\begin{aligned} \tilde{E}= & {} \frac{1}{2}\hbar {\tilde{\omega }}'\bigg \{\bigg [1 - \frac{1}{8}\bigg (-\frac{e^2}{2\pi \epsilon _0R^3\tilde{Q}}\bigg )^2\bigg ]\nonumber \\&+ \bigg [1 - \frac{1}{8}\bigg (\frac{e^2}{2\pi \epsilon _0R^3\tilde{Q}}\bigg )^2\bigg ]\bigg \}, \end{aligned}$$

(46)

$$\begin{aligned}= & {} \frac{1}{2}\hbar {\tilde{\omega }}'\bigg [2 - \frac{1}{4\tilde{Q}^2}\bigg (\frac{e^2}{2\pi \epsilon _0R^3}\bigg )^2\bigg ]. \end{aligned}$$

(47)

To see if there is any (repulsive or attractive) interaction between the two identical and neutral atoms as a result of $H_\mathrm{I}$, we need to subtract $\hbar {\tilde{\omega }}'$ (see eq. (31)) from eq. (47) to obtain

$$\begin{aligned} \tilde{E}(\xi ) - \hbar {\tilde{\omega }}'= & {} \hbar {\tilde{\omega }}'\bigg \{\frac{1}{2}\bigg [2 - \frac{1}{4\tilde{Q}^2}\bigg (\frac{e^2}{2\pi \epsilon _0R^3}\bigg )^2\bigg ] - 1\bigg \} \nonumber \\= & {} \hbar {\tilde{\omega }}'\bigg [-\frac{1}{8\tilde{Q}^2} \bigg (\frac{e^2}{2\pi \epsilon _0R^3}\bigg )^2\bigg ]. \end{aligned}$$

(48)

We now substitute $\tilde{Q}^2 = \big ({\tilde{\omega }}'\big )^4m^2$ and then followed by the substitution, ${\tilde{\omega }}' = \omega '\exp {[\frac{1}{2}\lambda \xi ]}$ into eq. (48) to arrive at,

$$\begin{aligned} \tilde{V}_\mathrm{vdW}(\xi )= & {} \tilde{E}(\xi ) - \hbar {\tilde{\omega }}' \nonumber \\= & {} \bigg \{-\frac{\hbar }{8m^2(\omega ')^3}\bigg (\frac{e^2}{2\pi \epsilon _0}\bigg )^2 \frac{1}{R^6}\bigg \}\nonumber \\&\times \exp \bigg [-\frac{3}{2}\lambda \xi \bigg ], \end{aligned}$$

(49)

$$\begin{aligned}= & {} V_\mathrm{vdW}\exp \bigg [-\frac{3}{2}\lambda \xi \bigg ]. \end{aligned}$$

(50)

The negative sign in eq. (49) implies that there is an attractive interaction, and this is the van der Waals attraction in its renormalised form. The standard (or unrenormalised) vdW formula is given in the curly bracket. Regardless of whether $V_\mathrm{vdW}$ is renormalised or not, eq. (49) is only valid for weakly interacting systems due to the condition given in eq. (44). Apparently, $\hbar {\tilde{\omega }}'$ is the renormalised ground-state energy $\tilde{E}'(\xi )$ when $H_\mathrm{I} = 0$. The compatibility of eq. (49) with eq. (33) is obvious, while the additional physics that can be extracted from eq. (49) is as follows: for a given interatomic distance, R, we should have

$$\begin{aligned} \tilde{V}_\mathrm{vdW}(\xi ) \rightarrow 0,~\mathrm{when}~ \xi \rightarrow \infty , \end{aligned}$$

(51)

where eq. (51) implies that the valence electrons are literally not polarisable, which is as it should be. In contrast, $\tilde{V}_\mathrm{vdW}(\xi )$ gets a maximum value if $\xi$ is permitted to reach a physically allowable minimum value. Here, we did not take the minimum value to read, $\xi \rightarrow 0$ because this is physically not acceptable for atomic systems where all electrons in such systems are bound to their respective nucleus. Thus $\xi \ne 0$ is always valid. In solids however, $\xi = 0$ is permitted entirely for a different reason. In particular, we should note that there is such a thing as many-body potential giving rise to metallic bond that keeps all the electrons bounded within a given solid. Even though these electrons are collectively kept bounded within the solid, they can be effectively free (not bounded to any particular ion or group of ions) to form Fermi metals, for which $\xi = 0$ or $\xi$ is an irrelevant constant.

1.3 Ramachandran interaction theory and analysis

The essence of the Ramachandran interaction is the existence of a stronger attraction between two atoms, stronger than the vdW attraction. As pointed out above, even the renormalised version of the standard vdW attraction (see eq. (49)) is too weak due to this $1/R^6$ dependence. In this weak interaction case, the induced electronic polarisation in each atom is either too small or, if the polarisation is relatively large, then the polarisation is still approximately isotropic (see figure 2 in ref. [32]). The condition that assures this weak interaction and isotropic polarisation is given in eq. (44). Basically, there are three types of polarisation between interacting atoms. The first is the weakly interacting atom, before or after the atom is exposed to an external disturbance, namely, intense laser or high temperatures. Such exposure leads to an isotropically polarised atom, hence weakly interacting. On the other hand, strongly interacting atoms can give rise to anisotropic polarisation and asymmetric polarisation. However, an additional interaction is activated due to anisotropic polarisation. The said additional interaction is nothing but the stronger el–el Coulomb repulsion between the anisotropically polarised atoms.

On the contrary, if two atoms are of different types (atom-1 $\ne$ atom-2) such that one of the atoms is easily-polarisable, while the other is a less polarisable atom, then the atoms are asymmetrically polarised. In this case, the el–el Coulomb repulsion is weaker than anisotropically polarised atoms, for a given R. If the two atoms are identical, then, we definitely need external assistances (high temperature, pressure and concentration that can lead to violent atomic collisions) to form the transition state. But to complete the chemical reaction from the transition state to the product, one still needs strong attraction that can only be made available by a Coulomb-type attraction, which leads to wave function transformation that induces chemical reaction. Therefore, identical atoms eventually require stronger attraction, in addition to other external help, to proceed further from the transition state to the final product. In contrast, the asymmetrically polarised atoms do not explicitly need any external ‘help’, and the reaction between dissimilar atoms can be spontaneous.

Theorem 2

For the Ramachandran attraction of the first type, asymmetric and anisotropic polarisation can be induced with a stronger vdW attraction between dissimilar and identical atoms, respectively, regardless of whether $R \gg r_1 + r_2$ or $R > r_1 + r_2$ where $r_1$ and $r_2$ denote the radii of atom-1 and atom-2, respectively. Here, strong attraction can be imposed manually by requiring,

$$\begin{aligned} Q\exp {[\lambda \xi ]} = \tilde{Q} = \frac{e^2}{2\pi \epsilon _0R^3}, \end{aligned}$$

(52)

instead of eq. (44). Equation (44) is the condition for weak attraction, used earlier to derive the vdW attraction. Here, eq. (52) is valid even if $R \gg r_1 + r_2$ where $R \gg r_1 + r_2$ is a necessary condition so as to make sure eq. (19) is also valid. This condition (eq. (52)) gives us the interatomic Ramachandran attraction of the first type,

$$\begin{aligned} \tilde{V}^{\mathrm{I}}_{\mathrm{Ramachandran}}(\xi ) = \hbar \omega '\bigg (\frac{1}{\sqrt{2}} - 1\bigg )\exp {\bigg [\frac{1}{2}\lambda \xi \bigg ]}. \end{aligned}$$

(53)

Equation (53) imposes a large anisotropic polarisation, much larger than what is allowed by the standard vdW attraction. The new condition allows a large interatomic interaction energy even if $\xi$ is small (see eq. (53)). Here, eq. (53) is the analytic relation that permits $\xi$ to have the smallest value – in other words, $\xi = 0$ is forbidden as demanded by all the physical atoms found in the Periodic Table of Chemical Elements.

Proof

The interatomic Ramachandran attraction of the first type gives rise to the transformation

$$\begin{aligned} \exp {\bigg [-\frac{3}{2}\lambda \xi \bigg ]} \rightarrow \exp {\bigg [\frac{1}{2}\lambda \xi \bigg ]}, \end{aligned}$$

(54)

which can be confirmed by comparing eq. (49) with eq. (53), whereas the interatomic Ramachandran attraction of the second type gets rid of the el–el Coulomb repulsion by activating $\tilde{V}^{\mathrm{el-ion}}_{\mathrm{Coulomb}}(\xi )$. From now on, we shall drop the term ‘interatomic’ for convenience because of course we are exclusively dealing with interatomic interaction in the atomic systems. The physics of the standard renormalised vdW attraction (eq. (49)) relies on the fact that polarisation is small due to weak interaction such that $\xi$ is not even allowed to reach zero where $\xi \ne 0$ because atomic systems cannot permit $\xi = 0$, and therefore, $0 \ll \xi < \infty$ (see eq. (44)), which is indeed a condition for small polarisation.

For large polarisation however, we need $0 < \xi \ll \infty$, and rightly so, the transformation (eq. (54)) is automatically activated in our derivation of the Ramachandran attraction of the first type. Note that, the large polarisation here refers to the polarisation of electron-1 due to atom-2 and vice versa, and that is why we need larger $\xi$ to enhance the Ramachandran attraction. In other words, the Ramachandran attraction is activated when the atomic polarisation is large in the presence of large ionisation energy, which can only be true for atoms that interact strongly such that this strong interaction is not due to temperature.

The said large polarisation can be imposed by invoking eq. (52) that is in agreement with $0 < \xi \ll \infty$. Consequently, we can readily make use of eq. (52), which enforces eq. (36) to reduce to a mathematically simpler form (for strongly interacting atoms),

$$\begin{aligned}&\tilde{E}(\xi )' = \frac{1}{2}\hbar \bigg [0 + \sqrt{\frac{2{\tilde{k}}}{m}}\bigg ], \end{aligned}$$

(55)

but physically, it is difficult to understand eq. (55) within the standard vdW formulation due to eq. (54). But never mind, the Ramachandran attraction (type I) in the presence of large polarisation (and large ionisation energy) is as follows:

$$\begin{aligned} \tilde{V}^{\mathrm{I}}_{\mathrm{Ramachandran}}(\xi )= & {} \tilde{E}(\xi )' - \hbar {\tilde{\omega }}' \nonumber \\= & {} \frac{1}{2}\hbar \sqrt{\frac{2\tilde{Q}}{m}} - \hbar {\tilde{\omega }}'. \end{aligned}$$

(56)

Again, noting that $\tilde{Q} = Q\exp {[\lambda \xi ]}$ and ${\tilde{\omega }}' = \omega '\exp {[\frac{1}{2}\lambda \xi ]}$, we can simplify eq. (56) as

$$\begin{aligned}&\tilde{V}^{\mathrm{I}}_{\mathrm{Ramachandran}}(\xi ) = \hbar \omega '\bigg (\frac{1}{\sqrt{2}} - 1\bigg )\exp {\bigg [\frac{1}{2}\lambda \xi \bigg ]}, \end{aligned}$$

(57)

which is eq. (53). $\square$

1.4 Quantitative comparison in support of Ramachandran attraction

We have achieved our original objectives by first proving that the Ramachandran attraction between He atoms is responsible for the specific heat capacity to jump up discontinuously at the critical point during cooling. This is only possible with the formation of Ramachandran pairs between two He atoms, be they He-4 or He-3, which automatically proves why BEC does not imply superfluidity. In particular, only a cusp-like feature is observed if the transition is induced by pure BEC. Apart from that, we also have proved that there are four possibilities for the A-phase to B-phase transition in the specific heat capacity during cooling for the He-3 superfluid. To choose the correct microscopic physics for this particular transition, we have to rely on the specific heat experimental data, whether there is a cusp or a discontinuous jump and whether the jump is upward or downward during cooling.

Despite the above facts, it is always worthwhile to provide some numerical comparison in support of our theory presented thus far, especially in support of the Ramachandran attraction, at least indirectly. From the experimental data reported by Morton et al [49], we know that $\xi _{{\mathrm{He}:4}}> \xi _{\mathrm{He}:3}$ when $\xi _{{\mathrm{He}:4}} - \xi _{\mathrm{He}:3} = 313,520$ MHz. As a consequence, from the Ramachandran attraction formula (see eq. (1)) between neutral atoms, we can readily deduce that Ramachandran attraction is stronger for He-4 pairs than for He-3 pairs, which means that the superfluid critical temperature for liquid He-4 should be higher to form superfluidity. This is indeed the case from the specific heat data reported in ref. [47] for He-4 (at saturated He vapour) and in ref. [41] for He-3 (at 33.4 bar), even though both measurements were carried out at different pressures. Note that no estimate is available for the ground-state energies for He-4 or He-3 pairs and that there are two unknowns ($\omega _{\mathrm{pair}}$ and $\xi _{\mathrm{pair}}$) and only one equation, eq. (1), for each system.

When we say that the Ramachandran attraction is stronger than that of the vdW force, it means for strongly interacting atoms, the strongest Ramachandran attraction for atomic He or for any particular system does not imply that the attraction of one system should be of the same order or magnitude in other systems. Hence, such numerical comparison (between different systems), even by means of inequalities, would be misleading.

What we can say here is that when we consider the attraction between any strongly correlated atoms, the Ramachandran attraction is always stronger than that of the vdW force, which is indeed the case from the above derivation. For example, it is worth noting that we can substitute eq. (52) into eq. (53) so as to derive the Ramachandran attraction as a function of R, i.e., $V^{\mathrm{I}}_{\mathrm{Ramachandran}}(R)$, which is given by

$$\begin{aligned}&V^{\mathrm{I}}_{\mathrm{Ramachandran}}(R) = \frac{e\hbar }{\sqrt{2m\pi \epsilon _0}}\bigg (\frac{1}{\sqrt{2}} - 1\bigg )\frac{1}{R^{3/2}}, \end{aligned}$$

(58)

where the effect of $\xi$ is replaced by R, the distance between two atoms. We can estimate the value for $V^{\mathrm{I}}_{\mathrm{Ramachandran}}(R)$ between two atomic helium from the above formula, which is about 1 eV or between $1.3 \times 10^{-19}$ J (0.8 eV) and $1.9 \times 10^{-19}$ J (1.2 eV) when R is between 300 pm and 240 pm, respectively. Here, the van der Waals separation between two helium atoms is 280 pm. We stress that the stated range for $V^{\mathrm{I}}_{\mathrm{Ramachandran}}(R)$ between two atomic helium differs for different interacting chemical element pairs as pointed out in the above paragraph.

Our estimate makes physical sense (as anticipated) because it is larger than that of the van der Waals highest value, which is about 0.04 eV. However, our estimate is still rather crude and sensitive to R where the choice of R should be large due to electron–electron repulsion (between helium atoms), which will reduce the magnitude of the above estimate. Nevertheless, $V^{\mathrm{I}}_{\mathrm{Ramachandran}}$ by definition, can be larger than all other weak intermolecular interactions such as van der Waals, hydrogen bonds, ionic interaction and hydrophobic interaction because the next step after $V^{\mathrm{I}}_{\mathrm{Ramachandran}}$ is the type-II Ramachandran attraction that can induce chemical reaction.

In addition, we also have discussed the proper numerical comparison in ref. [54] to evaluate the stronger Ramachandran attraction in physical chemistry to study the microscopic physics of hydration energy for different cations. Hence, only for weakly interacting atoms, the Ramachandran attraction is not activated (that can be deduced from the above derivation where eq. (52) is not activated, instead, eq. (32) is activated), and in this case, the usual vdW attraction plays its part, which is indeed true for weakly interacting gases. If we take $V^{\mathrm{R}}_{\mathrm{Type:I}}(\xi ) \propto k_{{\mathrm{B}}}T_{\mathrm{C},\lambda }$, which is reasonable, then there are two factors that have led to $T_{\mathrm{C}} < T_{\lambda }$, namely, $\xi _{\mathrm{He}:3} < \xi _{{\mathrm{He}:4}}$ and the obstacle due to spin-half effect.

Of course, one can argue that the said lower critical temperature for He-3 is entirely due to the effect of spin-$\frac{1}{2}$ nuclei as discussed in the earlier sections. If this is the case, then we do not know what is the interaction responsible for superfluidity in He-4. Moreover, we have to assume that the spin interaction is responsible for He-3 pairing despite the fact that this is not possible physically as discussed using eq. (14). It is unfortunate that we only have two superfluids, namely, He-4 and He-3, and therefore unambiguous numerical or quantitative analysis in support of Ramachandran attraction can only come from novel measurements, which can be used to verify whether He-4 pairing does occur in He-4 superfluids. It is worth noting that the Ramachandran attraction is not responsible or is not activated for water at different temperatures (between 600 and 250 K) and pressures (between 0.1 and 100 MPa) [55].