Translation versus Momentum and Rotation versus Angular Momentum

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1 APPENDIX F Translation versus Momentum and Rotation versus Angular Momentum In Chapter 2, it was shown that the Hamiltonian Ĥ commutes with any translation p. 68) or rotation p. 69) operator, denoted as Û: [Ĥ, Û] =0. F.1) The Form of the Û Operator Below, it will be demonstrated for κ meaning first a translation vector, and then a rotation angle about an axis in the 3-D space,) that operator Û takes the form Û = exp i ) κ ˆK, F.2) where ˆK stands for a Hermitian operator having the x-, y-, and z- components) acting on functions of points in the 3-D Cartesian space. Translation and Momentum Operators Translation of a function by a vector r is equivalent to the function f in the coordinate system translated in the opposite direction; i.e., f r r) see Fig. 1.3 and p. 68. If the vector r is infinitesimally small, then, in order to establish the relation between f r r) and f r), it is of course sufficient to know the gradient of f neglecting, obviously, the quadratic and higher terms in the Taylor expansion): f r r) = f r) r f = 1 r ) f r). F.3) We will compose a large translation of a function by vector T) from a number of small increments r = N 1 T,whereN is a large natural number. Such a tiny translation will be repeated N times, thus recovering the translation of the function by T. In order for the gradient formula to be exact, one has to ensure that N tends to infinity. Recalling the definition Ideas of Quantum Chemistry, Second Edition Elsevier B.V. All rights reserved. e73

2 e74 Appendix F ) exp ax) = lim N 1 + a N x,wehave ÛT) f r) = f r T) = lim 1 T ) N N N f r) = exp T ) f = exp i ) T ˆp f r), where ˆp = i is the total momentum operator see Chapter 1). Thus, for translations, we have κ T and ˆK ˆp. Rotation and Angular Momentum Operator Imagine a function f r) of positions in the 3-D Cartesian space think, for example, about a probability density distribution centered somewhere in the space). Now, suppose that the function is to be rotated about the z-axis the unit vector showing its direction is e)byanangle α, so we have another function, which we will denote by Ûα; e) f r). What is the relation between f r) and Ûα; 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. 1.1 and p. 89) of the coordinate system: Ûα; e) f r) = f U 1 r) = f U α; e)r), where U is a 3 3 orthogonal matrix. The new coordinates xα), yα),andzα) are expressed by the old coordinates x, y, andz through 1 xα) cos α sin α 0 x r yα) = sin α cos α 0 y. zα) z Therefore, the rotated function Ûα; e) f r) = f xα), yα), zα)). The function can be expanded in the Taylor series about α = 0: ) f Ûα; e) f r) = f xα), yα), zα)) = f x, y, z) + α α α=0 xα) f + = f x, y, z) + α α x + yα) f α [ + = f x, y, z) + α y x x ] f + y y + zα) α ) f z α=0 1 A positive value of the rotation angle means an anticlockwise motion within the xy plane with the x-axis horizontal, y-axis vertical, and z-axis pointing toward us).

3 Translation versus Momentum and Rotation versus Angular Momentum e75 Now, instead of the large rotation angle α, let us consider first an infinitesimally small rotation by angle ε = α/n, wheren is a huge natural number. In such a situation, we retain only the first two terms in the previous equation: α ) Û N ; e f r) = f x, y, z) + α [ y N x x ] f x, y, z) y = 1 + α i N i = 1 + α N [ y x x y ]) f ) f = 1 α N ) i Ĵz f. 1 i [x ˆp y y ˆp x ] If such a rotation is repeated N times, then we recover the rotation of the function by a possibly large) angle α the symbol of limit ensures that ε is infinitesimally small): α )] N Ûα; e) f r) = lim N [Û N ; e f r) = limn 1 α ) i N N Ĵz f r) = exp i α ) Ĵz f = exp i ) αe Ĵ f. ) Thus, for rotations Ûα; e) = exp i αe Ĵ,wehaveκ αe and ˆK Ĵ. This means that in particular, for rotations about the x-, y-, and z-axes with the corresponding unit vectors x, y,andz) we have, respectively, [Ûα; x), Ĵ x ]=0, [Ûα; y), Ĵ y ]=0, [Ûα; z), Ĵ z ]=0. F.4) F.5) F.6) Useful Relation Eq. F.1) means that for any translation or rotation, Û ĤÛ 1 = Ĥ, and taking into account the general form of Eq. F.2), we have for any such transformation a series containing nested commutators valid for any κ): Ĥ = Û ĤÛ 1 = exp i ) ) i κ ˆK Ĥ exp κ ˆK = 1 i ) κ ˆK + Ĥ 1 + i ) κ ˆK + = Ĥ i κ [ˆK, Ĥ] κ2 2 2 [[ ˆK, Ĥ], ˆK]+,

4 e76 Appendix F where each term in + contains [ ˆK, Ĥ]. This means that to satisfy the equation, we must have [ ˆK, Ĥ] =0. F.7) Hamiltonian Commutes with the Total Momentum Operator In particular, this means [ˆp, Ĥ] =0; i.e., [ˆp μ, Ĥ] =0, for μ = x, y, z. Of course, we have also [ˆp μ, ˆp ν ]=0forμ, ν = x, y, z. Since all these four operators mutually commute, the total wave function is simultaneously an eigenfunction of Ĥ and ˆp x, ˆp y, ˆp z ; i.e., the energy and the momentum of the center of mass can both be measured without making any error) in a space-fixed coordinate system see Appendix I available at booksite.elsevier.com/ ). From its definition, the momentum of the center of mass is identical with the total momentum. 2 Hamiltonian, Ĵ 2 and Ĵ z Do Commute Eq. F.7) for rotations means [Ĵ, Ĥ] =0; i.e., in particular, F.8) [Ĵ x, Ĥ] =0, F.9) [Ĵ y, Ĥ] =0, and F.10) [Ĵ z, Ĥ] =0. F.11) The components of the angular momentum operators satisfy the following commutation rules 3 : [Ĵ x, Ĵ y ]=i Ĵ z, [Ĵ y, Ĵ z ]=i Ĵ x, and F.12) [Ĵ z, Ĵ x ]=i Ĵ y. 2 Indeed, the position vector of the center of mass is defined as R CM = i m i r i i m, and after differentiation with i respect to time, i m ) i ṘCM = i m i ṙ i = i p i. The right side represents the momentum of all the particles i.e., the total momentum), whereas the left side is just the momentum of the center of mass. 3 The commutation relations can be obtained by using directly the definitions of the operators involved: Ĵ x = y ˆp z z ˆp y, etc. For instance, [Ĵ x, Ĵ y ] f =[y ˆp z z ˆp y )z ˆp x x ˆp z ) z ˆp x x ˆp z )y ˆp z z ˆp y )] f =[y ˆp z z ˆp x z ˆp x y ˆp z ) y ˆp z x ˆp z x ˆp z y ˆp z ) z ˆp y z ˆp x z ˆp x z ˆp y ) + z ˆp y x ˆp z x ˆp z z ˆp y )] f = y ˆp z z ˆp x z ˆp x y ˆp z ) f yx ˆp z ˆp z yx ˆp z ˆp z ) f z 2 ˆp y ˆp x z 2 ˆp x ˆp y ) f + xz ˆp y ˆp z x ˆp z z ˆp y ) f [ = y ˆp z z ˆp x yz ˆp x ˆp z ) f xz ˆp y ˆp z x ˆp z z ˆp y ) f = i ) 2 y f x x f ] = i Ĵ z f. y

5 Translation versus Momentum and Rotation versus Angular Momentum e77 Eqs. F.9) F.11) are not independent; e.g., from Eqs. F.9) andf.10), Eq. F.11) can be derived. Indeed, [Ĵ z, Ĥ] =Ĵ z Ĥ Ĥ Ĵ z = 1 i [Ĵ x, Ĵ y ]Ĥ 1 i Ĥ[Ĵ x, Ĵ y ]= 1 i [Ĵ x, Ĵ y ]Ĥ 1 i [Ĵ x, Ĵ y ]Ĥ = 0. Also, from Eqs. F.9), F.10), and F.11), it follows that [Ĵ 2, Ĥ] =0, F.13) because from the Pythagorean theorem, Ĵ 2 = Ĵx 2 + Ĵ y 2 + Ĵ z 2. Do Ĵ x, Ĵ y, Ĵ z commute with Ĵ 2? Let us check the commutator [Ĵ z, Ĵ 2 ]: [Ĵ z, Ĵ 2 ]=[Ĵ z, Ĵ 2 x + Ĵ 2 y + Ĵ 2 z ]=[Ĵ z, Ĵ 2 x + Ĵ 2 y ]=Ĵ z Ĵ 2 x Ĵ 2 x Ĵz + Ĵ z Ĵ 2 y Ĵ 2 y Ĵz = i Ĵ y + Ĵ x Ĵ z )Ĵ x Ĵ x i Ĵ y + Ĵ z Ĵ x ) + i Ĵ x + Ĵ y Ĵ z )Ĵ y Ĵ y i Ĵ x + Ĵ z Ĵ y ) = 0. Thus, [Ĵ z, Ĵ 2 ]=0, F.14) and also, by the argument of symmetry the space is isotropic), [Ĵ x, Ĵ 2 ]=0, and F.15) [Ĵ y, Ĵ 2 ]=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 labeled by the quantum numbers, each corresponding to an eigenvalue of the operators in question. We cannot choose as these operators the whole set of Ĥ, Ĵ x, Ĵ y, Ĵ z, Ĵ 2, because as it was shown earlier, Ĵ x, Ĵ y, Ĵ z do not commute among themselves although they do with Ĥ and Ĵ 2 ). The only way is to choose as the set of the operators either Ĥ, Ĵ z, Ĵ 2 or Ĥ, Ĵ x, Ĵ 2 or Ĥ, Ĵ y, Ĵ 2. Traditionally, one chooses as the set of the mutually commuting operators Ĥ, Ĵ z, Ĵ 2 z is known as the quantization axis). Rotation and Translation Operators Do Not Commute Now, we may think to add ˆp x, ˆp y, ˆp z to the above set of the operators. The operators Ĥ, ˆp x, ˆp y, ˆp z, Ĵ 2 and Ĵ z do not represent a set of mutually commuting operators. The reason for this is that [ˆp μ, Ĵ ν ] = 0forμ = ν, which is a result of the fact that in general, rotation and translation operators do not commute, as shown in Fig. F.1.

6 e78 Appendix F a) b) c) d) Fig. F.1. In general, translation ÛT) and rotation Ûα; e) operators do not commute. The example shows what happens to a point belonging to the xy plane. a) If a rotation Ûα; z) by angle α about the z-axis takes place first, and then a translation ÛT) by a vector T restricted to the xy plane) is carried out, and b) shows what happens if the operations are applied in the reverse order. As we can see, the results are different two points 1 have different positions in panels a and b); i.e., the two operators do not commute: ÛT)Ûα; z) = Ûα; z)ût). Expanding ÛT) = exp[ i T x ˆp x + T y ˆp y )] and Ûα; z) = exp i α Ĵ z ) in a Taylor series, and, taking into account that T x, T y,αare arbitrary numbers, leads to the conclusion that [ Ĵ z, ˆp x ] = 0and[ Ĵ z, ˆp y ] = 0. Note that some translations and rotations do commute; e.g., [ Ĵ z, ˆp z ]=[ Ĵ x, ˆp x ]=[ Ĵ y, ˆp y ]=0, because we see by inspection [as shown in panels c) and d)] that any translation by T = 0, 0, T z ) is independent of any rotation about the z-axis, etc.

7 Translation versus Momentum and Rotation versus Angular Momentum e79 Conclusion It is, therefore, impossible to make all the operators Ĥ, ˆp x, ˆp y, ˆp z, Ĵ 2 and Ĵ z commute in a space-fixed coordinate system. What we are able to do, though, is to write down the total wave function pn in the space-fixed coordinate system as a product of the plane wave exp ip CM R CM ), which depends on the center-of-mass variables, and the wave function 0N,which depends on internal coordinates 4 as follows: pn = 0N exp ip CM R CM ), F.17) which is an eigenfunction of the total i.e., center-of-mass) momentum operators: ˆp x = ˆp CM,x, ˆp y = ˆp CM,y, ˆp z = ˆp CM,z. The function 0N is the total wave function written in the centerof-mass coordinate system a special body-fixed coordinate system; see Appendix I available at booksite.elsevier.com/ ), in which the total angular momentum operators Ĵ 2 and Ĵ z are now defined. The three operators Ĥ, Ĵ 2,andĴ z commute in any space-fixed or body-fixed coordinate system including the center-of-mass coordinate system), and therefore, the corresponding physical quantities energy and angular momentum) have exact values. In this particular coordinate system, ˆp = ˆp CM = 0. We may say, therefore, that in the center-of-mass coordinate system, Ĥ, ˆp x, ˆp y, ˆp z, Ĵ 2 and Ĵ z all do commute. 4 See Chapter 2 and see Appendix I available at booksite.elsevier.com/ , where the total Hamiltonian is split into a sum of the center-of-mass and internal coordinate Hamiltonians; N is the quantum number for the spectroscopic states.

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