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966 G. VECTOR AND SCALAR POTENTIALS Fig. G.1. How do we understand the arbitrariness of the vector potential A? Figs. (a), (b), (c) repre- sent schematically three physically equivalent vector potentials A. Fig. (a) shows a section in the plane z =0(axisz protrudes towards the reader from the xy plane) of the vector field A = 1 2 (H × r) with H = (00H) and H>0 We see that vectors A become longer and longer, when we leave the origin (where A =0), they “rotate” counter-clockwise. Such A therefore determines H directed perpendic- ularly to the page and oriented towards the reader. By the way, note that any shift of the potential obtained should give the same magnetic field perpendicular to the drawing, Fig. (b). This is what we get (Fig. (b)) after adding, according to eq. (G.11), the gradient of function f =ax +by +c to potential A because A +∇f =A +(ia +jb) =A −R =A ,whereR =−(ia +jb) =const The transformation is only one of the possibilities. If we took an arbitrary smooth function f(xy), e.g., with many maxima, minima and saddle points (as in the mountains), we would deform Fig. (b) by expanding or shrinking it like a pancake. In this way we might obtain the situation shown in Fig. (c). All these situations a,b,c are physically indistinguishable (on condition that the scalar potential φ is changed appropriately). G. VECTOR AND SCALAR POTENTIALS 967 +2(∇) − iq ¯ hc exp − iq ¯ hc χ ∇χ +V exp − iq ¯ hc χ =− ¯ h 2 2m exp − iq ¯ hc χ + − iq ¯ hc − iq ¯ hc exp − iq ¯ hc χ (∇χ) 2 +exp − iq ¯ hc χ χ − ¯ h 2 2m 2(∇) − iq ¯ hc exp − iq ¯ hc χ ∇χ +V exp − iq ¯ hc χ Dividing the Schrödinger equation by exp(− iq ¯ hc χ) we obtain − ¯ h 2 2m + − iq ¯ hc − iq ¯ hc (∇χ) 2 +χ +2(∇) − iq ¯ hc ∇χ +V =E(r) Let us define a vector field A(r) using function χ(r) A(r) =∇χ(r) (G.14) Hence, we have − ¯ h 2 2m + − iq ¯ hc − iq ¯ hc A 2 +∇A +2(∇) − iq ¯ hc A +V=E(r) and introducing the momentum operator ˆ p =−i ¯ h∇ we obtain 1 2m ˆ p 2 + q c 2 A 2 − q c ˆ pA −2( ˆ p) q c A +V=E(r) or finally 1 2m ˆ p− q c A 2 +V=E (G.15) which is the equation corresponding to the particle moving in electromagnetic field with vector potential A, see p. 654. Indeed, the last equation can be transformed in the following way 1 2m ˆ p 2 + q c 2 A 2 − q c ˆ p(A) − q c A ˆ p +V=E which after using the equality 3 ˆ p(A) = ˆ pA + A ˆ p gives the expected result [eq. (G.15)]. 3 Remember that ˆ p is proportional to the first derivative operator. 968 G. VECTOR AND SCALAR POTENTIALS In conclusion, if a particle moves in a vector potential field A from r 0 to r,then its wave function changes the phase by δ δ =− q ¯ hc r r 0 A(r) dr or, putting it in a different way: if the wave function undergoes a phase change, it means that the particle moves in the vector potential of an electromagnetic field. The incredible Aharonov–Bohm effect In a small area (say, in the centre of the Grand Place in Brussels, where we like to locate the origin of the coordinate system) there is a magnetic field flux corre- sponding to field intensity H directed along the z axis (perpendicular to the market place surface). Now let us imagine a particle of electric charge q enclosed in a 3D box (say, a cube) of small dimensions located at a very long distance from the ori- gin, and therefore from the magnetic flux, say, in Lisbon. Therefore, the magnetic field in the box is equal to zero. Now we decide to travel with the box: Lisbon, Cairo, Ankara, StPetersburg, Stockholm, Paris, and back to Lisbon. Did the wave function of the particle in the box change during the journey? Let us see. The magnetic field H is related to the vector potential A through the relation ∇×A =H. This means that the particle was subject to a huge vector po- tential field (see Fig. G.1) all the time, although the magnetic field was practically zero. Since the box is back to Lisbon, the phase acquired by the particle in the box 4 is an integral over the closed trajectory (loop) δ =− q ¯ hc A(r) dr However, from the Stokes equation, we can replace the integral by an integral over a surface enclosed by the loop δ =− q ¯ hc A(r) dr =− q ¯ hc ∇×A(r) dS This may be written as δ =− q ¯ hc H dS =− q ¯ hc where is the magnetic flux (of the magnetic field H) intersecting the loop sur- face, which contains, in particular, the famous market place of Brussels. Thus, de- spite the fact that the particle could not feel the magnetic field H (because it was zero inthe box), its wave function underwent a change of phase, which is detectable experimentally (in interference experiments). Does the pair of potentials A and φ contain the same information as E and H? The Aharonov–Bohm effect (see also p. 780) suggests that A and φ are more important. 4 Anon-zeroδ requires a more general A than that satisfying eq. (G.14). H. OPTIMAL WAVE FUNCTION FOR A HYDROGEN-LIKE ATOM In several contexts we encounter the problem of the mean value of the Hamil- tonian for a hydrogen-like atom (the a.u. are used throughout) ˆ H =− 1 2 − Z r with the normalized function (r θ φ;c) = c 3 π exp(−cr) where r θ φ are the spherical coordinates of the electron (the position of the nu- cleus is fixed at the origin). Calculation of the mean value of the Hamiltonian, i.e. the mean value of the energy ε() = ˆ H requires calculation of the mean value of the kinetic energy: ¯ T = − 1 2 and the mean value of the potential energy (Coulombic attraction of an electron by a nucleus of charge Z) ¯ V =−Z 1 r Therefore, ε = ¯ T + ¯ V First, the Laplacian = ∂ 2 ∂x 2 + ∂ 2 ∂y 2 + ∂ 2 ∂z 2 may be expressed in spherical coordi- nates = 1 r 2 ∂ ∂r r 2 ∂ ∂r + 1 r 2 sinθ ∂ ∂θ sinθ ∂ ∂θ + 1 r 2 sin 2 θ ∂ 2 ∂φ 2 (H.1) 969 970 H. OPTIMAL WAVE FUNCTION FOR A HYDROGEN-LIKE ATOM and in view of the fact that is spherically symmetric (it only depends on r) − 1 2 =− 1 2 1 r 2 ∂ ∂r r 2 ∂ ∂r =− 1 2 c 3 π (−c) ∞ 0 r 2 2 r −c exp(−2cr)dr π 0 sinθdθ 2π 0 dφ = 1 2 c 4 4 ∞ 0 2r −cr 2 exp(−2cr)dr = 2c 4 2 ∞ 0 r exp(−2cr)dr −c ∞ 0 r 2 exp(−2cr)dr = 4c 4 (2c) −2 −2c 5 2(2c) −3 =c 2 − 1 2 c 2 = 1 2 c 2 where we have used (this formula is often exploited throughout the book) ∞ 0 r n exp(−βr)dr =n!β −(n+1) (H.2) Similarly the second integral gives −Z 1 r =−Z c 3 π ∞ 0 r exp(−2cr)dr π 0 sinθdθ 2π 0 dφ =−4Zc 3 (2c) −2 =−Zc Therefore, finally ε = 1 2 c 2 −Zc (H.3) We may want to use the variational method for finding the ground-state wave function. In this method we minimize the mean value of the Hamiltonian with respect to parameters in the variational function . We may treat c as such a para- meter. Hence, minimizing ε we force ∂ε ∂c =0, and therefore c opt =Z Note that in this particular case: • Such a value of c gives the exact ground-state of the hydrogen-like atom from the variational function. • The ground-state energy computed with c opt =Z gives ε = 1 2 Z 2 −ZZ =− 1 2 Z 2 , which is the exact ground-state energy. • The quantity − ¯ V ¯ T = Zc 1 2 c 2 =2 Z c For c =c opt =Z we have what is called the virialvirial theorem theorem − ¯ V ¯ T =2 (H.4) I. SPACE- AND BODY-FIXED COORDINATE SYSTEMS Space-fixed coordinate system (SFCS) A planetoid (or molecule) moves through empty space, we observe it from our (inertial 1 ) space ship. To carry out observations of the planetoid (molecule), we have to install some equipment in our space ship and to fix a Cartesian coordinate system on it. This will enable us to describe the planetoid whatever happens to it. This is the Space-Fixed Coordinate System (SFCS), its orientation with respect to distant stars does not change in time. If the molecule does not interact with anything, then with respect to the SFCS (see Chapter 2) • its total energy remains invariant (because of the homogeneity of time), • its total momentum remains invariant (because of the homogeneity of space), • its total angular momentum vector remains invariant (because of the isotropy of space). An observer on another space ship (also inertial) will see the same phenomena in exactly the same way, 2 the energy, momentum and angular momentum will also be invariant, but in general they will be different from what was measured in the first space ship. Let us introduce the vectors r i = (x i y i z i ) into the SFCS showing (from the origin of the coordinate system) the particles, from which our molecule is com- posed (i.e. the electrons and the nuclei), i =1 2N. Then, using the SFCS, we write the Hamiltonian of the system, the operators of the mechanical quantities we are interested in, we calculate all the wave functions we need, compare with spectra measured in the SFCS, etc. Body-fixed coordinate system (BFCS) One day, however, we may feel that we do not like the SFCS, because to describe the molecule we use too many variables. Of course, this is not a sin, but only a 1 No rotation. We will convince ourselves that our SFCS is inertial by measuring how a point-like mass moves (assumed to be non-interacting with the rest of the space ship). If it moves along a straight line with a constant velocity, the SFCS is inertial. In a non-inertial coordinate system the description of the physical phenomena in the molecule will look different. 2 In the non-relativistic approximation. The Doppler effect, with the change in electromagnetic wave frequency due to the motion (even uniform) of the emitting object is seen in the experiment. The effect is of a relativistic character, i.e. vanishes, if we assume an infinite velocity of light. 971 972 I. SPACE- AND BODY-FIXED COORDINATE SYSTEMS waste of our time. Indeed, since in all inertial systems we have the same physics, we can separate the motion of the centre of mass 3 (the total mass M = i m i ). The centre of mass with position R CM = i m i r i M moves with a constant velocity along a straight line in the SFCS, which can easily be taken into account after the solution is obtained, and in most cases it is irrelevant. This is why we decide to introduce the Cartesian coordinates (X CM Y CM Z CM ) = R CM inthehopethatinfuturewewillbeabletogetridofthem.Nowweneed to introduce a coordinate system (of the missing 3N −3 variables) located on the molecule, called the body-fixed coordinate system (BFCS). How to define this? Well, it should be a coordinate system that will define any configuration of the particles in the molecule unambiguously. There are a lot of such coordinate systems. Here you have some of the possibilities for the BFCS (in all of them their axes are parallel to the corresponding axes of the SFCS 4 ). We may choose one of the following sets 5 of position vectors: • R CM then,welocateintheBFCSon any of the particles (say, the one indicated by vector r 1 ), and the BFCS positions of the other particles are shown by: r i = r i −r 1 for i =2 3N. • R CM the vector R =r 2 −r 1 indicating particle 2 from particle 1, and the remain- ing particles are shown by the vectors which begin in the centre of the section linking particles 1 and 2: r i =r i − (r 1 +r 2 ) 2 for i =3 4N. • R CM , and all the vectors showing the particles (except particle 1): r i =r i −R CM for i =23N. the position vector of the particle 1 can be calculated from the coordinates already given. Centre-of-mass separation After writing the Hamiltonian ˆ H in the SFCS, and introducing any of the above choices of coordinate system, we obtain ˆ H = ˆ H CM + ˆ H,where ˆ H CM =− ¯ h 2 2M CM with CM = ∂ 2 ∂X 2 CM + ∂ 2 ∂Y 2 CM + ∂ 2 ∂Z 2 CM ,and ˆ H that does not contain X CM , Y CM , Z CM . 3 The exact separation of the centre-of-mass motion in SFCS, as well as (not shown in this Appen- dix) the exact separation of rotation of the molecule have been shown in the paper by R.T. Pack, J.O. Hirschfelder, J. Chem. Phys. 49 (1968) 4009 for the first time. 4 Only after introducing the axes of the coordinate system associated with the particles, and not with the SFCS, separation of rotation is possible. 5 There are other possible choices. I. SPACE- AND BODY-FIXED COORDINATE SYSTEMS 973 At any of the choices the operator ˆ H is identical, but the mathematical formula for ˆ H will be different, because different coordinates are used. Thus, the total Hamiltonian in the SFCS is ˆ H = ˆ H CM (X CM Y CM Z CM ) + ˆ H(r) where r symbolizes 6 all the other variables. The key result is that the two operators on the right do depend on different variables. The goal of the above changes to the coordinate system was to show that the Schrödinger equation written in the SFCS, i.e. ˆ H = E,splitsintotwo Schrödinger equations (“separation of variables”): • ˆ H CM ψ CM = E CM ψ CM describing the motion of a free “particle” of mass M and coordinates X CM Y CM Z CM (the “centre-of-mass motion”), with ψ CM = exp(ip CM ·R CM ),wherep CM stands for the total momentum of the system; • ˆ Hψ =Eψ,where E =E +E CM (R CM r) = ψ CM (R CM ) ·ψ(r) The proof is simple. Let us check that the product wave function satisfies the Schrödinger equation. The left-hand side is: ˆ H ψ CM (R CM ) ·ψ(r) = ˆ H CM ψ CM (R CM ) ·ψ(r) + ˆ H ψ CM (R CM ) ·ψ(r) = ψ(r) · ˆ H CM ψ CM (R CM ) +ψ CM (R CM ) · ˆ Hψ(r) = ψ(r) ·E CM ψ CM (R CM ) +ψ CM (R CM ) ·Eψ(r) = (E +E CM ) ψ CM (R CM ) ·ψ(r) andthisequalstherightsideE. Example 1. Centre-of-mass separation for the first choice of the coordinates. We use the first choice of coordinates for the system of two particles. In the SFCS ˆ H =− ¯ h 2 2m 1 1 − ¯ h 2 2m 2 2 +V The new coordinates are: X CM = i m i x i M Y CM = i m i y i M Z CM = i m i z i M x =x 2 −x 1 y=y 2 −y 1 z=z 2 −z 1 6 For the sake of brevity. 974 I. SPACE- AND BODY-FIXED COORDINATE SYSTEMS Then, 7 ∂ ∂x 1 = ∂X CM ∂x 1 ∂ ∂X CM + ∂Y CM ∂x 1 ∂ ∂Y CM + ∂Z CM ∂x 1 ∂ ∂Z CM + ∂x ∂x 1 ∂ ∂x + ∂y ∂x 1 ∂ ∂y + ∂z ∂x 1 ∂ ∂z = m 1 M ∂ ∂X CM +0 +0 − ∂ ∂x +0 +0 = m 1 M ∂ ∂X CM − ∂ ∂x and similarly for y 1 and z 1 Further, ∂ ∂x 2 = ∂X CM ∂x 2 ∂ ∂X CM + ∂Y CM ∂x 2 ∂ ∂Y CM + ∂Z CM ∂x 2 ∂ ∂Z CM + ∂x ∂x 2 ∂ ∂x + ∂y ∂x 2 ∂ ∂y + ∂z ∂x 2 ∂ ∂z = m 2 M ∂ ∂X CM +0 +0 + ∂ ∂x +0 +0 = m 2 M ∂ ∂X CM + ∂ ∂x and similarly for y 2 and z 2 Hence, the kinetic energy operator (after constructing the proper Laplacians from the operators above) ˆ T =− ¯ h 2 2m 1 1 − ¯ h 2 2m 2 2 =− ¯ h 2 2m 1 m 1 M 2 ∂ 2 ∂X 2 CM + ∂ 2 ∂x 2 −2 m 1 M ∂ 2 ∂X CM ∂x +(similarly for y and z) − ¯ h 2 2m 2 m 2 M 2 ∂ 2 ∂X 2 CM + ∂ 2 ∂x 2 +2 m 2 M ∂ 2 ∂X CM ∂x +(similarly for y and z) =− ¯ h 2 2M CM − ¯ h 2 2μ where the reduced mass μ of the two particles: 1 μ = 1 m 1 + 1 m 2 and = ∂ 2 ∂x 2 + ∂ 2 ∂y 2 + ∂ 2 ∂z 2 Our derivation is over, and the operator ˆ H has been found It turns out to be 8 (note, that the new coordinates also have to be introduced in the potential energy V )oftheform 7 According to the mathematical analysis we have to write the contributions of all the differential operators ∂ ∂u of the new coordinates u multiplied by their “coupling constants” ∂u ∂x 1 with the coordinate x 1 . 8 The kinetic energy operator has a quite interesting form. Particle 1 rests right at the origin of the BFCS (x = 0, y = 0, z = 0), and therefore its kinetic energy operator is absent in ˆ H.Thereisthe kinetic energy of particle 2, but its mass is equal to μ,nottom 2 The coordinates x yz (measured from the origin of the BFCS) correspond to particle 2. For example, for the hydrogen-like atom, if someone takes the nucleus as particle 1, and the electron as particle 2, then x yz show the electron I. SPACE- AND BODY-FIXED COORDINATE SYSTEMS 975 ˆ H =− ¯ h 2 2μ +V Example 2. Centre-of-mass separation for the third choice of coordinates. Let us take the same two particles again, but this time use the third choice of coordinate sys- tem. X CM = i m i x i M Y CM = i m i y i M Z CM = i m i z i M x =x 2 −X CM y=y 2 −Y CM z=z 2 −Y CM Then, ∂ ∂x 1 = ∂X CM ∂x 1 ∂ ∂X CM + ∂Y CM ∂x 1 ∂ ∂Y CM + ∂Z CM ∂x 1 ∂ ∂Z CM + ∂x ∂x 1 ∂ ∂x + ∂y ∂x 1 ∂ ∂y + ∂z ∂x 1 ∂ ∂z = m 1 M ∂ ∂X CM +0 +0 − m 1 M ∂ ∂x +0 +0 = m 1 M ∂ ∂X CM − ∂ ∂x and similarly for y 1 and z 1 Further, ∂ ∂x 2 = ∂X CM ∂x 2 ∂ ∂X CM + ∂Y CM ∂x 2 ∂ ∂Y CM + ∂Z CM ∂x 2 ∂ ∂Z CM + ∂x ∂x 2 ∂ ∂x + ∂y ∂x 2 ∂ ∂y + ∂z ∂x 2 ∂ ∂z = m 2 M ∂ ∂X CM +0 +0 + 1 − m 2 M ∂ ∂x +0 +0 = m 2 M ∂ ∂X CM + 1 − m 2 M ∂ ∂x = m 2 M ∂ ∂X CM + m 1 M ∂ ∂x and similarly for y 2 and z 2 Thus, the kinetic energy operator takes the form (after inserting the squares of the corresponding operators) from the Cartesian coordinate system BFCS located on the nucleus. The potential energy operator V =− Ze 2 (x 2 −x 1 ) 2 +(y 2 −y 1 ) 2 +(z 2 −z 1 ) 2 =− Ze 2 x 2 +y 2 +z 2 corresponds to the Coulombic interaction of the electron of charge −e and the nucleus of charge Ze After the separation of the centre of mass, we are left with equation ˆ Hψ = Eψ The electron of mass μ is described by the wave function ψ. In the ground state ψ = 1 √ π e − x 2 +y 2 +z 2 This a the description of the hydrogen-like atom according to an observer sitting at the nucleus. If another observer puts his armchair (with the axes of the BFCS carved on it) at the electron,then he would see the hydrogen-like atom “according to the electron”. Since in V there are squares of x yz, and in the kinetic energy operator there are the second derivatives with respect to x y z,we would obtain the same wave function as before: ψ = 1 √ π e − x 2 +y 2 +z 2 , where the particle moving with respect to the electron is the nucleus, but with mass equal to μ,i.e.thesameasbefore.Bytheway,this μ is almost equal to the mass of the electron. Thus, the two descriptions mean the same. . (because of the homogeneity of time), • its total momentum remains invariant (because of the homogeneity of space), • its total angular momentum vector remains invariant (because of the isotropy of space). An. Coulombic interaction of the electron of charge −e and the nucleus of charge Ze After the separation of the centre of mass, we are left with equation ˆ Hψ = Eψ The electron of mass μ is described. the spherical coordinates of the electron (the position of the nu- cleus is fixed at the origin). Calculation of the mean value of the Hamiltonian, i.e. the mean value of the energy ε() = ˆ H requires
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