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WavemechanicsofaparticletrappedinanimpenetrablesphericalcavityARTICLEJULY2014

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Wave mechanics of a particle trapped in animpenetrable spherical cavity

Yatendra S. Jain

Department of PhysicsNorth-eastern Hill University, Shillong, 793022, India

Abstract

Wave mechanics of a particle in an impenetrable spherical cavity is criticallyanalysed to conclude correct values of the ground state momentum and energy of theparticle, viz., k10 = pi/D and E10 = h

2/8mD2 (not k10 = pi/R and E10 = h2/8mR2

as concluded in the literature). The conclusion is supported by experiments andseveral points of logic and physical realities.

Key words: Quantum particle, spherical cavity.

cwith author

Retired Professor, Communication address profysjain@gmail.com

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1. Introduction

Wave mechanics of a particle trapped in an impenetrable spherical cavity (Fig.1(A)) isthe simplest of all problems related to the motion of a particle in a spherically symmetricpotential V (r). These problems are described by 3-D Schrodiger equation written in polarcoordinates, r, and as,

h2

2m

1

r2

[(

r

(r2

r

)+

1

sin

(sin

)+

1

sin2

2

2

)]+ (E V (r)) = 0, (1)

with = (r, , ), m = mass of the particle, h(= h/2) being reduced Plancks constantand E = eigen energy. However, the approach of solving this problem, as reported brieflyin Section 2 and with details in [1, 2]), misses certain physical realities and concludeserroneous results; this is confirmed by the fact that the results do not agree with relatedexperimental observations (Section 4). Evidently, we do not have correct understanding ofthe quantum states of the particle trapped in spherical cavity for the reasons concluded inSection 5. However. in this chapter, we avoid these reasons to conclude error free resultsshowing good agreement with experiments.

2. Salient points of the approach

The eigen energy solutions for a particle constrained to move under the influence of aspherically symmetric V (r) can be obtained simply by solving the radial part of its 3-DSchrodinger equation (Eqn.(1)), i.e.,

h2

2m

1

r2d

dr

(r2dRl(r)

dr

)+

[El

(V (r) +

h2

2m

l(l + 1)

r2

)]Rl(r) = 0. (2)

Here Rl(r) is the eigen function of a quantum state of energy El with l (=0, 1, 2, 3, ...)being a quantum number which signifies that the particle in the state has an angular

momentum,l(l + 1)h/2.

As V (r), in our problem, is defined by

V (r < R) = 0 and V (r R) =, (3)

we use V (r) = 0 to arrange Eqn.(2) as

d2Rl(r)

dr2+2

r

dRl(r)

dr+

[2m

h2El l(l + 1)

r2

]Rl(r) = 0, (4)

which has the form of standard Bessel equation. Accordingly, Rl(r) can be expressed as

Rl(r) = Ajl(kr) +Bnl(kr) (5)

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where jl(kr) and nl(kr) (two mutually independent Bessel functions) are the solutionsof Eqn.(4) with k representing momentum wave vector of the particle and A and Bbeing constants whose values depend on the physical realities of the problem. In variancewith the nature of jl(kr), which remains finite at all r, nl(kr) diverges to at r = 0.Consequently, it is argued [1, 2] that nl(kr) is a physically unacceptable function for ourproblem for which we must have B = 0 in Eqn.(5). This renders

Rl(r) = Ajl(kr). (6)

Simplifying Eqn.(2) further by using,

Rl(r) =l(r)

r, (7)

we obtain,h2

2m

d2l(r)

dr2+

[El

(V (r) +

h2

2m

l(l + 1)

r2

)]l(r) = 0, (8)

which indicates that our problem of solving Eqn.2 gets reduced to solving a Schrodingerequation which represents simple 1-D motion of a particle along a radial line of thespherical cavity in a modified potential,

V (r) = V (r) + h/2m)l(l + 1)/r2. (9)

The added term (h/2m)l(l+1)/r2, obviously, represents the source of centrifugal force onthe particle. Use of Eqn.(7) in Eqn.(6), renders

l(r) = rRl(r) = Arjl(kr) (10)

Evidently, all l(r) not only vanish at r = 0 (or kr = 0), they also have zero values of eachjl(kr) at several points of non-zero kr (counted by an integer index (n = 1, 2, 3...) whichare defined by knlr = nl (with nl being an integer or non-integer number). On theother hand all l(r) have to have their zero value at r = R as a result of V (r = R) =(Eqn.3); using this as a boundary condition, we have

knlR = nl or knl =nl

R(11)

which gives

Enl =h2k2nl2m

=2nlh

2

8mR2(12)

as eigen energies of different quantum states of the particle.

3 : l = 0 states

A l = 0 state (known as sstate) is a state, where particle moves along any one ofthe infinite many radial lines of the spherical cavity, as depicted in Fig.1B(i) where it is

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shown to be moving along the solid line, although, in principle, it has equal probabilityto move along any other radial line (two shown in the figure or many not shown in thefigure). For a better perception of different l states we also draw notional sketches ofclosed paths representing l 6= 0 states (such as p, d, f, .. etc.) in Fig.1C whichdepicts the particle moving on closed polygons. It may, however, be clear that real orbitscould be different and smoother than these polygons; their use in the depiction is intendedonly to highlight the difference of l 6= 0 states.

In order to determine kn0 and En0 of l = 0 states (sstates) of the trapped particle,it is instructive to rewrite Eqn.(8) by putting l = 0 and V (r) = V (r < R) = 0, -relevantto our problem. We have,

h2

2m

d20(r)

dr2+ E00(r) = 0. (13)

which shows that all s-states of the particle should be exactly identical to the statesof a particle trapped in 1-D box of size D = 2R (diameter of the cavity cf., Fig.1B(i)and Fig.1B(ii)) because two walls of infinite potential on a radial line are located at aseparation D and the particle sees zero potential at all points between r = R to r = +R.Accordingly, kn0 and En0 of these states should be given by

kn0 =n

Dand En0 =

n2h2

8mD2(14)

indicating that the ground state of the particle identified with n = 1 has

k10 =

Dand E10 =

h2

8mD2(15)

Note that integer value of n is consistent with the fact that zeroes of j0(kr) = sin(kr)/krtoo occurs at kr = n (where n = 1, 2, 3, ... is an integer) which indicates that n0 for thes =states is an integer number.

Interestingly, the approach of finding relations for kn0 and En0 (as used in the litera-ture) does not make any argument to conclude that the size of the equivalent 1D box isD. Also, it asks no question as to why the state function of a particle moving across theradial line of the sphere [i.e. along the length of an equivalent 1D box (cf., Fig.1B(i) andB(ii))] should vanish at the mid point of 1D box as a matter of a boundary condition asused in the approach. Without any such argument or question,it uses n0 = n directly inEqns.(11 and 12) and concludes,

kn0 =n

Rand En0 =

n2h2

8mR2(16)

which renders

k10 =

Rand E10 =

h2

8mR2(17)

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To facilitate further discussion in relation to these results, we identify Eqn.(14) + Eqn.(15)as Set-I and Eqn.(16)+Eqn.(17) as Set-II. It is clear that only one of these sets representsthe correct result. However, since experimental observation (as analysed in Section 4)reveals that results of Set-II concluded in the literature are erroneous.

4 : Comparison with experiments

4.1 Electric mobility of electron bubble

Although there are several real systems which represent a quantum particle trappedin a spherical cavity, however, the best example is an electron bubble (EB), -an electronoccupying its ground state in a self created spherical cavity in liquid 4He as depictedin Fig.2A. The bubble of radius R (cf., Fig.2B representing an expanded picture of thebubble shown in Fig.2A)) is exclusively occupied by an electron at its ground state. An EBneither represents a quantum particle in the spherical cavity placed with a scatterer likean atom at the center of the cavity (cf., Fig.2C) nor with a spherical cavity having anotherconcentric sphere of impenetrable walls as depicted in Fig.2D(i); it rather represents aparticle in a spherical cavity (cf., Fig.2D(ii)) which is exclusively occupied by the quantumparticle. We note that, in the absence of an external pressure, the energy of an EB isgiven by [3]

Eb =A

(R)2+BR2 (18)

where A = h2/8m and B = 4 [with being the surface tension of liquid 4He]), respec-tively, represent zero-point energy of the electron and surface energy of the bubble; here = 2 and 1, respectively, refers to E10 of Set-I and Set-II. We use Eqn.(18) to find R ofan EB for its lowest possible energy by using R(Eb) = 0. This renders

R = (A/B)1/41

1/2(19)

which reveals that the said R of EB would be smaller by a factor of 1/2 = 1/1.42 if

we use E10 from Set-I in place of that from Set-II. To decide which E10 is correct, wecompare experimentally observed mobility (exp) of an EB measured under the influencean electric field E with theoretically value [4]

theo =vdE

=e

6R, (20)

which depends on R. In Eqn.(20), vd is drift velocity of an EB, e is electron charge,and is viscosity of liquid 4He. Enss and Hunklinger [page 67], make their observationon how exp and theo compare, by stating, the prefactor 1/6 must be replaced with1/4 to obtain a quantitative agreement which means that exp is higher than theo bya factor of 1.5 or the real R of EB in liquid Helium-4 is shorter (by a factor of 1/1.5)

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than that obtained by using E10 of Set-II concluded in the literature. In what followsfrom Eqn.(19), R estimated by E10 of Set-I is expected to be shorter by a factor of 1/1.42which falls reasonably close to 1/1.5 (a correction suggested by Enss and Hunklinger[4] forthe quantitative agreement between theo and exp. Evidently, exp supports the accuracyof E10 of Set-I (Eqn.(15) or, in other words, it questions the accuracy of E10 of Set-II(Eqn.(17)).

4.2 Experimental Density of liquid Helium and E10 of a4He-atom

Based on experimentally known density () of liquid 4He available in [5], we find thatE10 determined separately by using Eqn.15(Set-I) and Eqn.17(Set-II) fall around 4.2 and17 K, respectively. To this effect, we assume that each atom occupies a spherical cavityof diameter D and determine its value by using D = v1/3 = (m/)1/3 where m is mass ofa He-atom. Each atom in liquid 4He has a potential energy Vo 21K and a latent heatL 7K [4,6] which indicates that its kinetic energy is about 14 K at the boiling pointof the liquid [5]. Since atoms at T 6= 0 have finite amount of kinetic energy due to theirthermal motions, E10 is expected to much lower than 14K. This is corroborated by thefact that liquid 4He becomes superfluid (He-II) at 2.17 K and He-II has very little amountof kinetic energy due to thermal motions [5] indicating that E10 of a

4He atom can notbe higher than the maximum possible kinetic energy, 3kBT 3x2.17 = 6.5K. Since thisagrees closely with E10 4.2 estimated from Eqn.(15), it is evident that Eqns.(16 and17) (Set-II) which render E10 17 K have errors.

5. Other indicators of errors in set-II (Eqn.(16/17)

1. In what follows from Section 2, kn0 and En0 of Set-I (Eqn.14/15) are lower thanthose of set-II (Eqn.16/17) by a factor of 1/2 and 1/4, respectively. Since Eqn.(14/15) isconsistent with the physical reality that the particle in its sstates, moving along a radialline does not see any potential at r = 0, it is evident that R0(r) (sstate wave functions,Eqn.6) should not be zero at r = 0 (as a matter of a condition); however, since Set-II isobtained by using R0(r = 0) = 0, Eqns.(16 and 17) have a source of an error.

2. Fig.3 depicts four cavities of different shapes and symmetries, -say, spherical, cubical,rectangular, and irregular. A particle, trapped in each of these cavities, experiences zeropotential at all points within the boundaries of their impenetrable walls (whether the pointis the center of a cavity or away from it); this implies that the particle motion at all thesepoints can be described by

h2

2m2 (r) + E(r) = 0 (21)

with (r) = 1Vexp [ik.r] which can also be written as

(x, y, z) =1Vexp [i(kxx+ kyy + kzz)]. (22)

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Here k is particle momentum related corresponding energy E (Eqn.21) through

E =h2

2m[k2x + k

2

y + k2

z ] =h2

2mk2 (23)

with kx, ky and kz being the magnitudes of kx, ky, and kz representing three mutuallyperpendicular components of k. Although, the boundary condition, that (r) (Eqn.22)should be zero at every point on the impenetrable walls of every cavity, is expected tomodify the plane wave nature of Psi(r), however, there is no reason for which (r) shouldbe zero (as a matter of necessary condition) at any point where potential (V (x, y, z)) iszero. While this premise is honored in determining the eigen value solutions for theparticle trapped in a cubical or rectangular or irregular shape cavity, it is not honored inconcluding set-II (Eqn.(16/17) for a spherical cavity; if (x, y, z) expressed in cartesiancoordinates is not expected to vanish at a point (say, at the center of the cavity, (0,0,0)),how it can be expected to vanish at the same point just because the same is expressedas (r, , ) in polar coordinates. This, obviously, indicates a flaw with what kn0 and En0concluded by Set-II (Eqn.16/17).

3. Huang and Yang [7] used Eqn.(2) for a particle trapped in the space between twoconcentric impenetrable spherical surfaces of radii R and a (with R > a) (shown inFig.2D(i)). Using the boundary condition, Rl=0(r = R) = 0 and Rl=0(r = a) = 0, theyobtained,

R0(r) = Csin k(r a)

krfor a < r < R and Rl=o(r a) = 0, (24)

where C stands for normalization constant and

kn0 =n

(R a) , and En0 =n2h2

8m(R a)2 with n = 1, 2, 3, ... (25)

Using a = (an infinitesimally small radius of inner sphere, we have

kn0 =n

R n

R, as R >> n = 1, 2, 3, ..., (26)

En0 =n2h2

8mR2(27)

and

Rl=0(r) = Bsin [k(r )]

kr B sin kr

krat r > and Rl=0(r ) = 0 (28)

for the particle in a cavity which has an infinitely small size impenetrable sphere which,obviously, serves as a source of scattering potential at r = 0. We call this as Cavity-(i)(cf., Fig.2D(i)) to distinguish it from Cavity-(ii) (cf., Fig.2D(ii)) which has no such source

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of potential at any r < R. Since literature for Cavity-ii concludes Set-II(Eqn.16/17) andwe find sames values (Eqn.27) for Cavity-(i), we have,

En0(i) = En0(ii), (29)

which contradicts the normal expectation, that two different physical systems are expectedto have some difference in their energy spectrum. Again, this analysis indicates thatEn0(ii) is erroneous.

4. In different types of scattering, which are described in terms Rl(r) (Eqn.(5) and/orEqn.(6)), one uses a beam of particles, moving along the z-axis as a plane wave (r) =A exp(ikz), towards the scatterer (a source of spherically symmetric V (r) at r = 0) whichscatters the beam particles in all possible directions as spherical waves (exp(ikr)/r asdepicted in Fig.4(A). The theory of the phenomenon finds that the scattered particlesmoving as spherical waves (see concentric circles in, see Figs.4(A-D)), which at large r,where V (r) can be presumed to be zero and l(l + 1)/r2 has vanishingly small value, arerepresented by Rl(r) = Ajl(r) or exp(ikr)/r] and rightly assumed to have zero value atr = 0 because the scatterer and scattered particles are not expected to occupy commoncoordinates. However, the possibility of transformation of a plane wave (exp(ik.r)) into spherical waves is questionable if there is no scatterer at r = 0 (cf., Fig.4(C)). Henceit appears that the approach, which concludes Set-II(Eqn.(16/17)) for Cavity-ii (Fig.4(Cand D)), presumes that : (i) the plane wave representing the trapped particle too has itsscattering from the center of the cavity r = 0 and (ii) the spherical waves, (Rl(r)), sogenerated have zero value at this point. However, both these assumptions are unjustifiedsince they ignore two important physical realities: (i) Rl(r) is not expected to vanish atr = 0 (as matter of boundary condition) at a point where V (r) = 0, and (ii) a simple planewave is not expected to get transformed into spherical waves from a point which is notoccupied by a scatterer as a source of non-zero V (r), as this violates the basic principleof cause and effect which means that nothing happens without a cause.

If scattering of the particle is believed to occur in Cavity-ii, it is expected to occur whenthe particle strikes a point on the boundary wall of the cavity (cf., Fig.4(D). Evidently,the origin (r = 0) of scattering of the particle in Cavity-ii is located at its boundarywall, -not at its centre. Naturally, Rl(r) generated in this process and travelling along theradial line (i.e., a swave) is bound to have its zero at two points on the inner surfaceof the cavity, -separated, obviously, by D (not by R). Once this reality is accepted forswaves, it becomes clear that l 6= 0 states (p, d, f, .... etc., where particle followsnon-radial paths, as shown in Fig.1(C) can also be believed to have

knl =nl

Dand Enl =

2nlh2

8mD2(30)

It may be mentioned that arguments (3) and (4) have been reported in our earlier report[8] on this issue.

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6. Conclusion

Analysing the wave mechanics of a particle trapped in a spherical cavity of diameterD = 2R, we find that the correct values of energy and momentum of the particle in itsl = 0 states are En0 = n

2h2/8mD2 and kn0 = n/D. This is supported unequivocally byexperimental observations related to the mobility of an electron bubble in liquid Helium-4and zero point energy of an atom in liquid 4He. In what follows, we find reasons for whichEn0 = n

2h2/8mR2 and kn0 = n/R, concluded in the literature, have a source of error andit is for this reason that these values are inconsistent with experiments. A generalisationof our analysis also concludes that for l 6= 0 states one has Enl = 2nlh2/8mD2 andknl = nl/D. This study is expected to help in having a better understanding of thequantum states of a particle trapped in Cavity-(i) and Cavity-(ii).

Acknowledgment: Author is thankful to Drs. S. Dey, Simanta Chutia and Mr. J. P.Gewali for useful interaction.

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References

[1] S. Flugge, Practical Quantum Mechanics, (Springer-Verlag, New York, 1974).

[2] G. Aruldas, quantum Mechanics, (Prentice-Hall of India, Delhi, 2002).

[3] H. Maris and S. Balibar, Physics Today, (Feb 2000), pp 29-34.

[4] C. Enss and S. Hunklinger, Low Temperature Physics, Springer Verlag, Berlin,(2005).

[5] J. Wilks, The properties of Liquid and Solid Helium. Clarendon Press, Oxford (1967).

[6] K.H. Benneman and J.B. Ketterson, (eds.): The Physics of Liquid and Solid Helium,Part-I. Wiley, New York (1976).

[7] K. Huang and C. N. Yang, Phys. Rev. 105, 767 (1957).

[8] S. Dey and Y. S. Jain, On the wave mechanics of a particle in two different impene-trable spherical cavities, arXiv:1002.4308.

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Figure 1: Depiction of: (A) a particle in a spherical cavity, (B) motion of the particlealong a radial line in its l = 0 state (or sstate and its equivalence to the motion motionof a particle trapped in 1D box of size D), and (C) the motion of the particles along non-radial closed paths representing l 6= 0 states. While we chose closed polygons to highlightthe difference among different l 6= 0 states, the real orbits could be more complex andsmooth.

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Figure 2: Depiction of: (A) an electron bubble, (B) expanded picture of the bubblemarked with its radius R, (C) particle trapped in a cavity located with a source (say anatom with spherically symmetric structure), (D-(i)) a particle trapped in a space betweentwo concentric spheres of impenetrable walls of radius R and a and (D-(ii)) a particletrapped in spherical cavity having V (r < R) = 0, -a true depiction of the case of presentstudy. D-(i) and D-(ii) are, respectively, identified as Cavity-(i) and Cavity-(ii) whena = , -infinitely small size

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Figure 3: particle trapped in (A) spherical, (B) cubical, (C) rectangular, (D) irregularcavities having different shapes and size. Particle in the left and right side depictions is,respectively, seen at the mid point and off mid point of cavities.

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Figure 4: (A) A beam of particles represented by exp (ikz) is scattered from a sphericallysymmetric scatterer (dark black) presumably placed at the center r = 0 of a sphericalshell (2D representation by dashed circle); the gray colour circle represents the extendedregion of the potential V (r). Spherically symmetric scattered waves exp(ikr)/r (whichcan also be described as Rl(r) (Red circles) as shown in (B) where these waves also meetthe impenetrable boundary walls of spherical cavity, (C) Should spherical waves emergeeven if there is no scatterer at the center of spherical cavity ? (D) Depiction of the factthat, in a case when cavity has no scatterer at its center, the real scattering/ reflection ofa particle takes place at a point, -any where at the inner surface of the spherical cavity,hence r = 0 of a Rl(r) lies at such points.

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