Periodic Boundary Conditions. Classical Limit We have found the geometry of rectangular box with non-penetrable walls to be quite convenient for statistical applications. There is, however, even better setup. A disadvantage of non-penetrable walls is that the momentum is not a good quantum number, and we thus cannot straightforwardly introduce a distribution over momenta (velocities). The way out is to introduce periodic boundary conditions (PBC). We start with 1D case which easily generalizes to any dimension. Instead of 1D well of the length L, consider a ring of the same length. The SchrÄodingerequation does not change and reads: ¹h2 ¡ Ã00(x) = EÃ(x) ; x 2 [0;L] ; (1) 2m but the boundary conditions are di®erent. Now we have the conditions of periodicity: Ã(0) = Ã(L) ;Ã0(0) = Ã0(L) : (2) With these boundary conditions we can formally extend the domain of de¯nition of our wavefunction from the interval x 2 [0;L] to the whole number axis, with the requirement that the function be periodic with the period L: Ã(x + L) = Ã(x) : (3) The solution to the problem (1)-(2)is eiknx Ãn(x) = p ; (4) L where 2¼n k = ; n = 0; §1; §2;:::; (5) n L and, correspondingly, µ ¶ 2 ¼¹hn 2 E = ; n = 0; §1; §2;:::: (6) n m L Being the eigenstates of the Hamiltonian operator, the states (4) are also the eigenstate of the operator of momentum: @ p^ à (x) = ¡i à (x) = p à (x) ; (7) n @x n n n 2¼¹hn p = ¹hk = : (8) n n L Note also that p2 E = n : (9) n 2m Similarly to the case of non-penetrable walls, we get Gibbs distribution in the form 1 1 X X 2 ¡En=T ¡γn Z1D = e = e ; (10) n=¡1 n=¡1 but now the summation is from n = ¡1 and 2¼2¹h2 γ = (11) mT L2 1 (di®ers by a numeric factor). Once again we are interested in the classical limit|that is γ À 1|and replace summation with integration: Z Z X1 1 L 1 (:::) ! dn (:::) ! dp (:::) : (12) n=¡1 ¡1 2¼¹h ¡1 As a result we get the same partition function as in the case of non-penetrable walls: Z 1 µ ¶1=2 L ¡p2=2mT mT Z1D = dp e = L 2 : (13) 2¼¹h ¡1 2¼¹h To generalize the treatment to 3D case, we note that in the rectangular geometry we can decompose the wave function into the product of three one-dimensional functions: Ãn(x; y; z) = Ãnx (x) Ãny (y) Ãnz (z) ; n = (nx; ny; nz) ; (14) and arrive at three independent one-dimensional problems with periodic boundary conditions. We thus get à ! 2¼¹hnx 2¼¹hny 2¼¹hnz pn = ; ; ; (15) Lx Ly Lz and, correspondingly, p2 E = E + E + E = n ; (16) n nx ny nz 2m The partition function can be written as Z V 2 Z = dp dp dp e¡p =2mT : (17) (2¼¹h)3 x y z Eq. (17) contains Plank's constant only as a pre-factor. This is due to the fact that in the limit γ ¿ 1, the statistics is the classical one, and the only role ofh ¹ here is to ¯x the units of measuring Z, which are not ¯xed in classical statistics. The distribution (17) is nothing else than the Maxwell distribution of a classical particle over momenta/velocities (v = p=m): ¡p2=2mT dW (p) / e dpx dpy dpz : (18) In fact, it is easy to understand why γ ¿ 1 implies classical statistics. Up to a numeric factor, this condition means ¹h2 ¿ 1 : (19) mT L2 Consider the quantity ¹h ¸T = p (20) mT called thermal de Broglie wavelength of a particle. It is the de Broglie wavelength of a particle which energy is of order T . That is it is just a typical de Broglie wavelength corresponding to a given temperature at a given particle mass. Now we see that the inequality (19) is the requirement that typical de Broglie wavelength be much smaller than the system size: ¸T ¿ L: (21) Under this condition, instead of working with genuine eigenstate wavefunctions one can introduce localized wave packets of the size much larger than ¸T , but much smaller than L. On one hand, 2 these wavepackets are almost the eignstates of the Hamiltonian, and, on the other hand, they behave like classical particles. Below we explicitly construct such packets and use them to derive classical Maxwell-Boltzmann distribution from quantum statistics. Maxwell-Boltzmann Distribution As we demonstrated above, for a particle in a box of the size L, classical-mechanical Maxwell dis- tribution follows from the Quantum Statistics in the limit of ¸T ¿ L. What changes if we add an external potential?|When does Quantum Statistics become equivalent to the classical one (that is to Maxwell-Boltzmann distribution)? It turns out that the criterion ¸T ¿ L works in the inhomogeneous case as well, if by L we understand a typical size of the distribution, which now is essentially a function of the external potential and temperature. To arrive at this result, we use a trick of reducing the inhomogeneous problem to the previously solved homogeneous one. We notice that in a homogeneous case the boundary conditions are not important for the ¯nal answers, provided the condition L À ¸T is met. This allows us to break up the bulk of a homogeneous system into cubic cells of the size ¢L À ¸T and, instead of considering global genuine eigenstates of the Hamiltonian, introduce the following wave packets. The wavefunction of our wave packet is equal to zero in all cells, but one. Within the cell where it is non-zero, the wavefunction is nothing else than the solution of the SchrÄodingerequation with periodic boundary conditions on the surface of the cube. Due to the condition ¢L À ¸T our wave packets e®ectively behave as the energy eigenstates. This can be checked by calculating the partition function, which is nothing else than a partition function of one cell times the number of cells. As is readily seen, this partition function coincides with Eq. (17). The size of the cell ¢L drops out from the ¯nal answer. Now we introduce an external potential and utilize our freedom of choosing the size of the cell, provided ¢L À ¸T . For de¯niteness, below we consider the 1D case. A generalization to higher dimensions is straightforward. Let us label each cell by discrete coordinate x0 corresponding to the cell's center. Consider SchrÄodingerequation for the eigenfunctions of the given cell, ¹h2 ¡ Ã00 + U(x)à = Eà ; (22) 2m where m is the particle mass, U(x) is the external potential, x 2 [x0 ¡ ¢L=2; x0 + ¢L=2], periodic boundary conditions are assumed: Ã(x) = Ã(x + L). If ¢L is small enough, we can neglect variation of the potential in the second term and write ¹h2 ¡ Ã00 + U(x )à = Eà ; (23) 2m 0 which is equivalent to ¹h2 ¡ Ã00 = [E ¡ U(x )]Ã; (24) 2m 0 that is the external potential leads only to the global energy shift and does not a®ect the form of the wavefunctions. We thus get the solution eiknx Ãn;x (x) = p ; (25) 0 ¢L 2¼n k = ; n = 0; §1; §2;:::; (26) n ¢L 3 p2 E = n + U(x ) ; (27) n;x0 2m 0 2¼¹hn p = ¹hk = : (28) n n ¢L Now we need to establish the criterion that allows us to do the replacement U(x) ! U(x0). Identically rewriting the original equation as ¹h2 ¡ Ã00 + [U(x) ¡ U(x )]à = [E ¡ U(x )] Ã; (29) 2m 0 0 we see that it is necessary and su±cient to require that the second term be negligibly small as compared to the ¯rst one: ¯ ¯ ¯ 2 ¯ ¯ ¹h ¯ j [U(x) ¡ U(x )] à j ¿ ¯ Ã00¯ : (30) 0 ¯2m ¯ From calculus we know that 0 U(x) ¡ U(x0) = U (x¤)(x ¡ x0) ; (31) where x¤ is some point within the interval [x0; x]. Then, taking into account the estimate ¯ ¯ ¯ 2 ¯ 2 ¯ ¹h ¯ ¹h ¯ Ã00¯ » jÃj ; (32) ¯2m ¯ 2m(¢L)2 and also remembering that jx ¡ x0j < ¢L, we arrive at the condition ¯ ¯ ¹h2 ¯U 0(x)¯ ¿ ; (33) m(¢L)3 that, generally speaking, should be met for any x inside our cell. This condition is compatible with the requirement ¢L À ¸T if and only if ¯ ¯ 2 ¯ 0 ¯ ¹h U (x) ¿ 3 (34) m¸T for any x within the characteristic region of the particle distribution. In terms of the particle mass and temperature, this condition reads ¯ ¯ m1=2 T 3=2 ¯U 0(x)¯ ¿ : (35) ¹h Assuming that condition (35) is met, we use the energy levels (27) to ¯nd the partition function: X X X ¡E =T ¡U(x )=T ¡p2 =2mT Z = e n;x0 = e 0 e n : (36) n;x0 x0 n The sum over n is the same as in the homogeneous case. Since we have the condition ¢L À ¸T , we replace it with an integral: Z 1 Z 1 X ¢L 2 dp 2 X Z = e¡U(x0)=T dp e¡p =2mT = e¡p =2mT ¢L e¡U(x0)=T : (37) 2¼¹h ¡1 ¡1 2¼¹h x0 x0 Finally, we take into account that our potential changes very little at the distance ¢L and replace the summation over x0 with integration: X Z 1 ¢L e¡U(x0)=T ! dx e¡U(x)=T : (38) ¡1 x0 4 We arrive at the Maxwell-Boltzmann distribution: Z 1 Z 1 Z dp 2 dp dx 2 Z = e¡p =2mT dx e¡U(x)=T = e¡[p =2m+U(x)]=T : (39) ¡1 2¼¹h ¡1 2¼¹h We can write the distribution (39) in the di®erential form by introducing the probability density W (x; p) for the coordinate x and momentum p: 2 dW (x; p) / e¡[p =2m+U(x)]=T dp dx : (40) Note that Plank's constant does not totally disappear from the answer for the partition function.