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956 F. TRANSLATION vs MOMENTUM and ROTATION vs ANGULAR MOMENTUM where p =−ih∇ is the total momentum operator (see Chapter 1). Thus, for trans-lations we have κ ≡T and K ≡p: Rotation and angular momentum operator Imagine a function f(r) of positions in 3D Cartesian space (think, e.g., about a probability density distribution centred somewhere in space). Now suppose the function is to be rotated about the z axis (the unit vector showing its direction is e) by an angle α, so we have another function, let us denote it by U(α;e)f(r): What is the relation between f(r) and U(α;e)f(r)? This is what we want to establish. This relation corresponds to the opposite rotation (i.e. by the angle −α, see Fig. 2.1 and p. 58) of the coordinate system: U(α;e)f(r)=f¡U−1r¢=f¡U(−α;e)r¢; where U is a 3 × 3 orthogonal matrix. The new coordinates x(α);y(α);z(α) are expressed by the old coordinates x;y;z through1 ⎛x(α)⎞ ⎛ cosα r0 ≡⎝y(α)⎠=⎝−sinα z(α) 0 sinα 0⎞⎛x⎞ cosα 0⎠⎝y⎠: 0 1 z Therefore the rotated function U(α;e)f(r) = f(x(α);y(α);z(α)): The func-tion can be expanded in the Taylor series about α=0: U(α;e)f(r) = f¡x(α);y(α);z(α)¢=f(x;y;z)+αµ∂f ¶ +··· α=0 ∂x(α) ∂f ∂y(α) ∂f ∂z(α) ∂f · ∂α ∂x ¸ ∂α ∂y ∂α ∂z α=0 = f(x;y;z)+α y∂x −x∂y f +··· Now instead of the large rotation angle α, let us consider first an infinitesimally small rotation by angle ε = α/N, where N is a huge natural number. In such a situation we retain only the first two terms in the previous equation: µ ¶ · ¸ ˆ N;e f(r) = f(x;y;z)+ N y∂x −x∂y f(x;y;z) · ¸¶ = 1+ N ih y∂x −x∂y f = 1+ N ih xpy −ypx f = 1− α i ˆ f: 1A positive value of the rotation angle means an anticlockwise motion within the xy plane (x axis horizontal, y vertical, z axis pointing to us). 2 The Hamiltonian commutes with the total momentum operator 957 If such a rotation is repeated N times, we recover the rotation of the function by a (possibly large) angle α (the limit assures that ε is infinitesimally small): · µ ¶¸N µ ¶N U(α;e)f(r) = lim U ;e f(r)= lim 1− ˆ f(r) N→∞ ¶ µ N→∞¶ = exp −i ¯ Jz f =exp −¯ αe·J f: Thus for rotations U(α;e)=exp(− i αe·J), and, therefore, we have κ ≡αe and K ≡J. This means that, in particular for rotations about the x;y;z axes (with the cor-responding unit vectors x;y;z) we have, respectively £U(α;x);Jx¤ = 0; (F.4) £U(α;y);Jy¤ = 0; (F.5) £U(α;z);Jz¤ = 0: (F.6) Useful relation The relation (F.1) means that for any translation or rotation UHU−1 =H and taking into account the general form of eq. (F.2) we have for any such trans- formation a series containing nested commutators (valid for any κ) µ ¶ µ ¶ H = UHU−1 =exp −hκ·K Hexp hκ·K = 1− ¯ κ·K+··· H 1+ ¯ κ·K+··· = H − i κ·£K;H¤− 2h2 ££K;H¤;K¤+···; where each term in “+···” contains [K;H]:This means that to satisfy the equation we necessarily have K;H =0: (F.7) 2 THE HAMILTONIAN COMMUTES WITH THE TOTAL MOMENTUM OPERATOR In particular this means [p;H]=0, i.e. £pμ;H¤=0 (F.8) 958 F. TRANSLATION vs MOMENTUM and ROTATION vs ANGULAR MOMENTUM for μ=x;y;z. Of course, we also have [pμ;pν]=0 for μ;ν =x;y;z. Since all these four operators mutually commute, the total wave function is si- multaneously an eigenfunction of H and px;py;pz, i.e. the energy and the mo-mentum of the centre of mass can both be measured (without making any error) in a space-fixed coordinate system (see Appendix I). From the definition, the mo-mentum of the centre of mass is identical to the total momentum.2 3 THE HAMILTONIAN, ˆ2 AND ˆ DO COMMUTE Eq. (F.7) for rotations means [J;H]=0, i.e. in particular £Jx;H¤ = 0; (F.9) £ ˆ ;H¤ = 0; (F.10) £ ˆ ;H¤ = 0: (F.11) The components of the angular momentum operators satisfy the following com-mutation rules: 3 £Jx;Jy¤ = ihJz; £Jy;Jz¤ = ihJx; (F.12) £Jz;Jx¤ = ihJy: 2Indeed the position vector of the centre of mass is defined as RCM = Pmiri , and after differ- entiation with respect to time ( i mi)RCM = i mi˙i = i pi. The right-hand side represents the momentum of all the particles (i.e. the total momentum), whereas the left is simply the momentum of the centre of mass. 3The commutation relations can be obtained by using the definitions of the operators involved di-rectly: Jx =ypz −zpy; etc. For example, £Jx;Jy¤f = £¡ypz −zpy¢¡zpx −xpz¢−¡zpx −xpz¢¡ypz −zpy¢¤f = £¡ypzzpx −zpxypz¢−¡ypzxpz −xpzypz¢ −¡zpyzpx −zpxzpy¢+¡zpyxpz −xpzzpy¢¤f = ¡ypzzpx −zpxypz¢f −¡yxpzpz −yxpzpz¢ −¡z2 ˆy ˆx −z2 ˆx ˆy¢+¡xzpy ˆz −xpzzpy¢ = ¡ypzzpx −yzpx ˆz¢f −0−0+¡xzpy ˆz −xpzzpy¢f · ¸ = (−ih)2 y∂x −x∂y =ihJzf: 3 The Hamiltonian, J2 and Jz do commute 959 Eqs. (F.9)–(F.11) are not independent, e.g., eq. (F.11) can be derived from eqs. (F.9) and (F.10). Indeed, £Jz;H¤ = JzH −HJz = ih£Jx;Jy¤H − ih ˆ£Jx;Jy¤ = ih£Jx;Jy¤H − ih£Jx;Jy¤H = 0: Also, from eqs. (F.9), (F.10) and (F.11) it also follows that £J2;H¤=0; (F.13) because from Pythagoras’ theorem J2 =J2 +J2 +J2: Do Jx;Jy;Jz commute with J2? Let us check the commutator [Jz;J2]: £Jz;J2¤ = £Jz;J2 +J2 +L2¤ = £Jz;Jx +Jy¤ = Jz ˆ2 −J2Jz +Jz ˆ2 −J2Jz = ¡ihJy +JxJz¢Jx −Jx¡−ihJy +JzJx¢+¡−ihJx +JyJz¢Jy −Jy¡ihJx +JzJy¢ = 0: Thus, £ ˆ ;J2¤ = 0; (F.14) and also by the argument of symmetry (the space is isotropic) £Jx;J2¤ = 0; (F.15) £Jy;J2¤ = 0: (F.16) Now we need to determine the set of the operators that all mutually commute. Only then can all the physical quantities, to which the operators correspond, have definite values when measured. Also the wave function can be an eigenfunction of all of these operators and it can be labelled by quantum numbers, each corre-sponding to an eigenvalue of the operators in question. We cannot choose, for these operators, the whole set of H;Jx;Jy;Jz;J2; because, as was shown above, Jx;Jy;Jz do not commute among themselves (although they do with H and J2). 960 F. TRANSLATION vs MOMENTUM and ROTATION vs ANGULAR MOMENTUM The only way is to choose as the set of operators either H;Jz;J2 or H;Jx;J2 or H;Jy;J2: Traditionally, we choose H;Jz;J2 as the set of mutually com-muting operators (z is known as the quantization axis). 4 ROTATION AND TRANSLATION OPERATORS DO NOT COMMUTE Now we may think about adding px;py;pz, to the above set of operators. The operators H;px;py;pz;J2 and Jz do not represent a set of mutually commuting operators. The reason for this is that [pμ;Jν]=0 for μ=ν, which is a consequence of the fact that, in general, rotation and translation operators do not commute as shown in Fig. F.1. 5 CONCLUSION It is, therefore, impossible to make all the operators H;px;py;pz;J2 and Jz com-mute in a space fixed coordinate system. What we are able to do, though, is to write the total wave function 9pN in the space fixed coordinate system as a product of the plane wave exp(ipCM · RCM) depending on the centre-of-mass variables and on the wave function 90N depending on internal coordinates 9pN =90N exp(ipCM ·RCM); (F.17) which is an eigenfunction of the total (i.e. centre-of-mass) momentum operators: ˆx =pCM;x; py =pCM;y; pz =pCM;z: The function 90N is the total wave function written in the centre-of-mass coor-dinate system (a special body-fixed coordinate system, see Appendix I), in which the total angular momentum operators J2 and Jz are now defined. The three op-erators H;J2 and Jz commute in any space-fixed or body-fixed coordinate system (including the centre-of-mass coordinate system), and therefore the correspond-ing physical quantities (energy and angular momentum) have exact values. In this particular coordinate system: p =pCM =0: We may say, therefore, that in the centre-of-mass coordinate system H;px;py;pz;J2 and Jz all do com-mute. 4See Chapter 2 and Appendix I, where the total Hamiltonian is split into the sum of the centre-of-mass and internal coordinate Hamiltonians; N is the quantum number for the spectroscopic states. ... - tailieumienphi.vn
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