Eur. J. Phys. 17 (1996) 19–24. Printed in the UK 19
Relativistic particle in a box
P Alberto , C Fiolhais andVMSGil † † ‡ Departamento de F´ısica da Universidade de Coimbra, P-3000 Coimbra, Portugal †Departamento de Qu´ımica da Universidade de Coimbra, P-3000 Coimbra, Portugal ‡ Received 3 August 1995
Abstract. The problem of a relativistic spin 1/2 particle Resumo. O problema de uma part´ıcula de spin 1/2 confinada confined to a one-dimensional box is solved in a way that por uma caixa a uma dimensao˜ e´ resolvido de uma maneira resembles closely the solution of the well known muito semelhante a` da resolu¸cao˜ do problema de uma quantum-mechanical textbook problem of a non-relativistic part´ıcula no-relativista numa caixa referido em muitos livros particle in a box. The energy levels and probability density introdutorios´ de Mecanicaˆ Quntaica.ˆ Os n´ıveis de energia e a are computed and compared with the non-relativistic case. densidade de probabilidade sao˜ calculados e comparados com os valores nao-relativistas.˜
1. Introduction quantization rule for the wavenumber, k nπ/L, n 1, 2,... . Outside the box the wavefunction= vanishes= Energy quantization in atoms and molecules plays an which means that the derivative of the wavefunction is essential role in the physical sciences. discontinuous at the well walls (z 0 and z L). This From which theoretical arguments does the quanti- is related to the fact that the potential= has a infinite= jump zation of energy comes from? The axioms of quantum at the well walls. It is worth stressing this point as we theory were built to explain that phenomenon. A simple move later to the relativistic case. The energy levels of example of application of these rules, which has great equation (1) are given by pedagogical value, is the study of a particle in a one- 2 2 2 2 2 2 dimensional box, i.e., an infinite one-dimensional square p h¯ k h¯ n π En (3) well. Indeed, in many introductory courses of physics = 2m = 2m = 2mL2 or chemistry the study of electronic wavefunctions in where p hk¯ nπ/L is the quantized momentum. atoms and molecules is made by analogy with a parti- In a similar= = fashion, we propose to solve the Dirac cle in a box, without having to solve the more involved equation for a particle in a box, emphasizing the role Schrodinger¨ equation for systems ruled by the Coulomb of the boundary conditions in the solution of the wave potential. In this way one is able to introduce the con- equation in this paper ourselves. At the same time, we cepts of energy quantization and orbitals for atoms and will get the solutions in a way which closely resembles molecules without being lost in the mathematical de- the non-relativistic approach. The method we shall tails of solving the Schrodinger¨ equation for a central use set ourselves apart from either the one-dimensional potential. Dirac equation approach [3, 4] and the calculations of We note that for obtaining quantization it is not so Greiner [5] both using a finite square well. By using much the type of differential equation which must be the three-dimensional Dirac with a one-dimensional solved but the boundary condition which must be obeyed potential well in the z-axis, we are able to include by the solution: a particle confined to a finite region spin in the wavefunction without increasing much the of space does have discrete energy levels. In order to mathematical burden. The crucial points are the use solve the time-independent Schrodinger¨ equation for a of a Lorentz scalar potential, which avoids the Klein free non-relativistic particle of mass m inside a one- paradox problem, and boundary conditions which assure dimensional box of length L in the z-axis, the continuity of the probability current rather than that h¯ 2 d2ψ of wavefunction itself. In this way we are able to avoid Eψ (1) the problems referred in [3–5] when the depth of the 2 − 2m dz = potential goes to infinity. which is the stationary wave (0 z L) In the next section we solve the Dirac equation in an one-dimensional infinite square well, leaving the ψ(z) C sin(kz) (2) = mathematical details for an appendix. In the conclusions one has to impose the boundary conditions ψ(0) we present some comment on our solution and compare ψ(L) 0. These same conditions give rise to the= it with other approaches found in the literature. =
0143-0807/96/010019+06$19.50 c 1996 IOP Publishing Ltd & The European Physical Society
20 P Alberto et al
2. Solution of the Dirac equation in an one-dimensional infinite square well m(z)
Let us consider the time-independent relativistic M equation for the wavefunction of a free electron of mass m moving along the z direction. This Dirac equation E can be written as m Hψ (α p c βmc2)ψ Eψ (4) IIIIII ˆ = z ˆz + = where H α p c βmc2 is the Dirac energy operator O L z ˆ z z (Hamiltonian),= ˆp + i¯h d is the z component of the ˆz =− dz momentum operator and αz and β the matrices Figure 1. Plot of the mass as function of position m(z) showing the three different zones I, II, III in which the 0 σz I 0 αz β . (5) solution of the Dirac is evaluated. Eventually we take the = σz 0 = 0 I limit M . − →∞ Here, σz isa2 2 Pauli matrix and I is the 2 2 unit matrix. × × well. By analogy with the non-relativistic case, we We wish now to consider the solutions of (4) for a could proceed by adding to the Dirac Hamiltonian in free particle moving in the direction of the positive z- (4) a potential V(z) such that axis with momentumhk ¯ . These can be taken from a relativistic quantum mechanics textbook (for example, V z 0 0 [1]), giving the following normalized wavefunction V(z) 00
χ ψ ik0z I(z) A e− σzhk¯ 0c = −E Mc2 χ + χ χ ψ (z) B eikz Ce ikz II σzhkc¯ χ − σzhkc¯ χ = E mc2 + E− mc2 (12) + + χ ψ ik0z III(z) De σzhk¯ 0c = E Mc2 χ + Figure 2. Graphical solution of (16) when L = h¯ /(mc). The functions tan(kL) and kL are plotted. The values 2 2 2 2 where k0 E /c M c /h¯ , and A, B, C and D are − = − of kL for which the two curves intersect are the solutions constants. of the equation tan(kL) = kL. In the appendixp we compute the solutions in the case − M with suitable boundary conditions (the outward flux→∞ of probability at walls is zero). The result is that which is exactly the non-relativistic result (10) apart ψ ψ ψ from a normalization factor and the spinor χ. I(z) and III(z) vanish identically, and II(z) is given by In the appendix we show that the wavenumber k is provided by the transcendental equation 2 cos kz δ χ − 2 ψ (z) B iδ/2 2P hk¯ II e (13) tan(kL) . (17) = δ 2 2iP sin kz σzχ = P 1 =−mc − 2 − where From the discrete set of values of k that satisfy (17) equation we get the discrete energy eigenvalues E 2P hkc¯ = δ arctan P . (14) h¯ 2c2k2 m2c4. We can check that for non-relativistic = P 2 1 = E mc2 − + momenta+ tan(kL) 0 and kL nπ, n 1, 2,..., p The need for a special boundary condition instead of recovering the non-relativistic∼ result.∼ The size= L of the simply requiring that the wavefunction be continuous box provides a criterion to decide whether a particle at the walls is explained in the appendix. Actually, inside a well is relativistic or not, at least for the the wavefunction turns out to be discontinuous at the first energy eigenvalues. The distance scale is set boundaries of the box, as can be seen from (13) This is by the length L0 h/(mc)¯ , which is the Compton not surprising if one realizes that the Dirac equation is = wavelength of the particle divided by 2π. For L L0 a first order differential equation with a infinite jump in or smaller the particle is relativistic and we must apply∼ the potential in this case. In the non-relativistic particle the relation (17). This length is about 0.004 A˚ for in a box, we have a second-order differential equation, an electron. Figure 2 shows the graphic solution of leading to a discontinuity in the first derivative of the (17) when L L .IfLis much bigger than L , wavefunction. 0 0 the slope of the= line approaches zero, in which case For non-relativistic momenta,hk ¯ mc and P 0 so that ∼ the two curves cross at kL π, 2π, 3π, ,..., corresponding to the non-relativistic= behaviour. Figure 0 δ arctan π. (15) 3 shows the values of kL/π for boxes with L L0, ∼ 1 = L 10L and L 100L , in comparison with= the − = 0 = 0 Setting δ π and P 0 in (13) we get non-relativistic case. One sees that a particle inside a = = box with L 100L is already non-relativistic, at least = 0 π for the lower spectrum. It is clear from Figure 4 that iπ/2 2 cos kz χ ψ (z) B e − 2 the relativistic levels are lower than the corresponding II = 0 ! (16) non-relativistic ones. 2 sin(kz)χ It is interesting to note that the relativistic probability B eiπ/2 = 0 density constructed from the wavefunction (13) has no 22 P Alberto et al
This quantity is never zero for k 0 (if k 0 the wavefunction is zero inside the well)6= because= 0 < P<1. This could only happen if P 0, i.e., in the non-relativistic limit. Figure 5 shows→ the densities corresponding to the first three levels of a particle in a box with sizes L0,10L0 and 100L0. The latter case is non-relativistic and can be taken as a reference, showing the characteristic nodes of a stationary wave in a closed box.
3. Conclusions
We have analysed the relativistic particle in a one- dimensional box and compared it with the non- relativistic case. The quantized momentum values and corresponding energies emerge as solutions of a simple transcendental equation, in contrast with the Figure 3. Solutions of (16) for boxes of sizes L = L , 0 more involved case of the relativistic particle in the L =10L0 and L = 100L0, together with the non-relativistic case (for which kL = nπ). three dimensional Coulomb problem. We have checked that both the spectrum and the wavefunctions go to their non-relativistic values as the size of the box grows. The basic scale is set by the length L0 which is related to the Compton wavelength for the particle in the box. The probability density shows no nodes in a relativistic regime (L L0) but its minima approach zero as the box becomes∼ larger. Finally, let us compare the boundary conditions we have used with other approaches. For instance, (17) similar in form to the one obtained by Greiner [5]. However, one cannot make a correspondence between the two equations because of the way the potential well is defined in [5] and the problems that the approach used has when the depth of the well is allowed to go to infinity. In that work the wavefunction is required to be continuous at the walls of the well. One should also mention a textbook treatment of the same problem [7] in which two components of the wavefunction were required to be continuous at the box walls, while the other two were not. We think that there is no Figure 4. Relativistic spectrum of kinetic energies for the reason to treat the components of the wavefunction same box sizes of figure 3 differently, especially when one is far from the non- 2 2 4 2 (Ekin = (hkc¯ ) + m c mc ) in comparison with the relativistic regime. However, if one would demand that − 2 2 non-relativistic spectrum (Ekin = (nh¯ π) /(2mL )). The all components be continuous and also to be zero outside valuesp shown are the logarithm of the kinetic energy in units of mc2 for the lowest nine levels. the box, the wavefunction would be identically zero everywhere. This means that the treatment of [7] is in fact restricted to the non-relativistic regime. On the ψ ψ zeros. Indeed, if we compute II(z)† II(z) we obtain other hand, the continuity of the wavefunction, which was crucial for solving the Schrodinger¨ equation (a second order equation), does not need to be imposed for 2 2 δ ψ (z)†ψ (z) 4 B cos kz solving the Dirac equation (a first order equation), with II II = | | − 2 a mass going to infinity outside a certain region. This allows one to impose a boundary condition in which the 2 2 δ P sin kz (18) flux of probability is continuous at the box walls but the + − 2 wavefunction is not. By doing that, one is still able to δ guarantee the fulfillment of the continuity equation for 4 B 2 1 (P 2 1) sin2 kz . = | | + − − 2 the probability density. Relativistic particle in a box 23
Figure 5. Plot of the probability density ψ†(z)ψ(z) normalized to unity for the first three levels of a relativistic particle in boxes of sizes L = L0, L =10L0 and L = 100L0. The latter case is almost identical with the non-relativistic limit, as it is apparent from Figs. 4 and 5. Note that the density is discontinuous at the walls in the relativistic regime.
Appendix the wave vector k p/h¯ , this condition just states that E the probability current=E density at z 0 and z L is In this appendix we present the mathematical formalism equal to the value of ψψ at those points.= Applying= now ψ ¯ leading to (17). (20) to II(z) at z 0 gives the condition Let us start by considering the solutions of the Dirac = iP 1 equation given by (12). Since we are going to let C B − (22) 2 = iP 1 M , we have E To calculate the eigenvalues for the energy we have References to consider the boundary condition (20) at z L. This results in the condition = [1] Bjorken J D and Drell S D 1964 Relativistic Quantum ikL ikL ikL ikL Mechanics (New York: McGraw-Hill) i(B e C e− )P B e C e− (27) − − = + [2] Chodos A et al 1974 Phys. Rev. D 9 3471 which, after collecting terms and using (22) yields Chodos A et al 1974 Phys. Rev. D 10 2599; Thomas A W 1984 Adv. Nucl. Phys. 13 1 ikL 2 ikL 2 e (iP 1) e− (iP 1) [3] Coulter B L and Adler C G 1971 Am. J. Phys. 39 305 + = − [4] CoutinhoFABet al 1988 Am. J. Phys. 56 904 ikL ikL (28) e e− 2iP [5] Greiner W 1990 Relativistic Quantum Mechanics − . eikL e ikL = P 2 1 (Berlin: Springer) + − − [6] Itzykson C and Zuber J 1980 Quantum Field Theory Using (23), we write finally (New York: McGraw-Hill) 2P hk¯ [7] Sherwin C W 1959 Introduction to Quantum Mechanics tan(kL) . (29) (New York: Henry Holt) = P 2 1 =−mc −