The Runge0Lenz Vector (continued)
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1 University of Connecticut Chemistry Education Materials Department of Chemistry 009 The Runge0Lenz Vector continued) Carl W. David University of Connecticut, Follow this and additional works at: Part of the Physical Chemistry Commons Recommended Citation David, Carl W., "The Runge0Lenz Vector continued)" 009). Chemistry Education Materials. 8.
2 The Runge-Lenz Vector continued) C. W. David Department of Chemistry University of Connecticut Storrs, Connecticut Dated: April, 009) I. SYNOPSIS We continue our discussion of the Runge-Lenz vector in a quantum mechanical context. The traditional form of the Runge-Lenz vector is obtained, and the commutation relations between the Runge-Lenz vector, the Hamiltonian, and the Angular Momentum are obtained using Maple. written in slightly different form). Now we seek something of the order of A r, i.e., the symmetric form: form: where ) A r + r A We had II. A r + r A INTRODUCTION Ze ) L + 3 h + r.) so that î ĵ ˆk A r A A A 3 x y z { ) } { L p p ) }) L Ze + ˆr r + r L p p L Ze + ˆr or, expanding ) L p 4Ze r p L ) ) r + ˆr r + r L p r p L ) ) + r ˆr Since r ˆr and its inverse are both zero no partial derivatives here to goof us up!), we have { }} ) { L p 4Ze r p L ) ) ) r + r L p r p ) L }{{}.) where we need to be careful about expanding the triple cross products, since these are not ordinary vectors. We have A B C) A î ĵ ˆk B x B y B z C x C y C z î ĵ ˆk A x A y A z B y C z B z C y B z C x B x C z B x C y B y C x which expands to A B C) î A y B x C y B y C x ) A z B z C x B x C z ) +ĵ A z B y C z B z C y ) A x B x C y B y C x ) +ˆk A x B z C x B x C z ) A y B y C z B z C y ).3) while B C) A î ĵ ˆk B x B y B z C x C y C z A î ĵ ˆk B y C z B z C y B z C x B x C z B x C y B y C x A x A y A z Typeset by REVTEX
3 which expands to B C) A î B z C x B x C z ) A z B x C y B y C x ) A y +ĵ B x C y B y C x ) A x B y C z B z C y ) A z +ˆk B y C z B z C y ) A y B z C x B x C z ) A x.4) We then have since r x x, r y y, etc.) for A B C) with A r, B L and C p is r L p) î y L x p y L y p x ) z L z p x L x p z ) +ĵ z L y p z L z p y ) x L x p y L y p x ) and with A r, B L and C p but with the order reversed, is +ˆk x L z p x L x p z ) y L y p z L z p y ).5) L p) r î L z p x L x p z ) z L x p y L y p x ) y +ĵ L x p y L y p x ) x L y p z L z p y ) z +ˆk L y p z L z p y ) y L z p x L x p z ) x.6) For reference sake, in the next part of this work, we repeat this table from the earlier work: Adding the two, as requested, we have L x, x 0 L y, y 0 L z, z 0 L x, y ı hz -Ly, x L y, x -ı hz L x, y L x, z ı hx - L z, x L x, p y ı hp z L x, p z -ı hp y L x, p x 0 r L p) + L p) r î yl x p y }{{} { yl }}{ y p x zl z p x + zl x p z + L z p x z L x p z z L x p y y }{{} + L y p x y) +ĵ zl y p z }{{} { zl }}{ z p y xl x p y + xl y p x + L x p y x L y p x x L y p z z }{{} + L z p y z) +ˆk xl z p }{{ x } xl x p z yl y p z + yl z p y + L y p z y L z p y y L z p x x }{{} + { L }}{ x p z x).7) or, re-arranging r L p) + L p) r î yl x p y }{{} L xp y y }{{} + yl {}} { y p x {}} { L y p x y zl z p x + L z p x z + zl x p z L x p z z +ĵ zl y p z }{{} {}} { zl z p y + L z p y z L y p z z }{{} xl x p y + L x p y x + xl y p x L y p x x +ˆk xl z p }{{ x } L zp x x }{{} { }}{ {}} xl x p z + { L x p z x) L y p z y + yl y p z + yl z p y L z p y y.8), taking advantage of obvious cancellations: r L p) + L p) r î yl x p y }{{} +zl xp z L x p z z L x p y y }{{}
4 +ĵ zl y p z }{{} +xl yp x L y p x x L y p z z }{{} +ˆk xl z p }{{ x } +yl zp y L z p y y L z p x x }{{}.9) r L p) + L p) r î L x ı h) + ĵ L y ı h) + ˆk L z ı h).) r L p) + L p) r î L x yp y p y y + zp z p z z) +ĵ L y zp z p z z + xp x p x x) +ˆk L z xp x p x x + yp y p y y).0) This means 4Ze r L p) + L ) p) r 4Ze ı h L ).) The other term we need the obverse?) starts with p L which would be p L î ĵ ˆk p x p y p z îp y L z p z L y ) + ĵp z L x p x L z ) + ˆkp x L y p y L x ) L x L y L z Next, we need r p L) î y p x L y p y L x ) z p z L x p x L z ) +ĵ z p y L z p z L y ) x p x L y p y L x ) +ˆk x p z L x p x L z ) y p y L z p z L y ).3) and p L) r î p z L x p x L z ) z p x L y p y L x ) y +ĵ p x L y p y L x ) x p y L z p z L y ) z +ˆk p y L z p z L y ) y p z L x p x L z ) x.4) î y p x L y p y L x ) z p z L x p x L z ) + p z L x p x L z ) z p x L y p y L x ) y +ĵ z p y L z p z L y ) x p x L y p y L x ) + p x L y p y L x ) x p y L z p z L y ) z +ˆk x p z L x p x L z ) y p y L z p z L y ) + p y L z p z L y ) y p z L x p x L z ) x.5) î y p y L x ) z p z L x ) + p z L x ) z p y L x ) y +ĵ z p z L y ) x p x L y ) + p x L y ) x p z L y ) z +ˆk x p x L z ) y p y L z ) + p y L z ) y p x L z ) x.6) î y p y yp }{{} z ) zp y ) z p z yp z zp y ) ) + p z yp z zp y ) ) z p y yp }{{} z ) zp y ) y +ĵ z p z zp }{{} x ) xp z ) x p x zp x xp z ) ) + p x zp x xp z ) ) x p z zp }{{} x ) xp z ) z +ˆk x p x xp y }{{} ) ) ) yp x ) y p y xp y yp x ) + p y xp y yp x ) y p x xp y }{{} ) yp x ) x.7)
5 or, expanding +ĵ +ˆk }{{} + { yp }}{ y zp y zp z yp z + zp z zp y + p z yp z z p z zp y z + p y yp z y }{{} { p }}{ y zp y y }{{} p z xp z z î yp y yp z zp z zp }{{ x } + zp z xp z xp x zp x + xp x xp z + p x zp x x p x xp z x + p z zp x z xp x xp y }{{} + { xp }}{ x yp x yp y xp y + yp y yp x + p y xp y y p y yp x y + p x xp y x }{{} { p }}{ x yp x x.8) or, rearranging or +ĵ +ˆk î yp y yp z + p y yp z y }{{} + { yp }}{ y zp y p y zp y y zp z yp z + p z yp z z + zp z zp y p z zp y z zp z zp x + p z zp x z }{{} + zp z xp z p z xp z z xp x zp x + p x zp x x + xp x xp z p x xp z x xp x xp y + p x xp y x }{{} + { xp }}{ x yp x p x yp x x yp y xp y + p y xp y y + yp y yp x p y yp x y î yp y y + p y y )p z }{{} + yp y p yy)z + zp z + p zz)y + zp z z p z z )p y +ĵ zp z + p z z)zp x }{{} + zp z p zz)x + xp x + p xx)z + xp x x p x x )p z +ˆk xp x x + p x x )p y }{{} + xp x p xx)y yp y + p yy)x + yp y y p y y )p x.9).0) which, using the appropriate commutators, becomes î yp y + p y y)yp z }{{} + yp y p yy)z + zp z + p zz)y + zp z p z z)zp y +ĵ zp z + p z z)zp x }{{} + zp z p zz)x + xp x + p xx)z + xp x p x x)xp z +ˆk xp x + p x x)xp y }{{} + xp x p xx)y yp y + p yy)x + yp y p y y)yp x.) î ı h)p z y }{{} + yp y p yy)z + zp z + p zz)y + ı h)p y z +ĵ ı h)zp x }{{} + zp z p zz)x + xp x + p xx)z + ı h)xp z +ˆk or ı h)xp y }{{} + xp x p xx)y + yp y + yp y)x + ı h)yp.) x î ı h)p z y }{{} + ı hp y )z + ı hp z )y + ı h)p y z
6 or +ĵ ı h)zp x }{{} + ı hp z )x + ı hp x )z + ı h)xp z +ˆk ı h)xp y }{{} + ı hp x )y + ı hp y )x + ı h)yp x.3) r p L) + p L) ) r ı h î yp z zp y + ĵ zp x xp z + ˆk xp y yp x which becomes î L x + ĵ L y + ˆk L z L.4) r p L) + p L) r ı h L.5) We now substitute these two results Equation.5 and Equation.) into Equation.: L ) ) p r + r L p 4Ze p L ) r + r p L) ).6) Substituting, using Equation. r L p) + L ) p) r ı h L.7) and Equation.5 r p L) + p L) ) r ı h L.8) one obtains ) A r + r A which becomes ) A r + r A ) 4Ze ı h L ı h L) ) ) Ze ı h L ) Unfortunately, Pauli s Equation 5 is ) A r + r A h ı III. 3 Ze L m We now assemble all our prior work into a set of commutators, sufficient to attack the Kepler problem. We start with the one already known, i.e., the angular momentum commutator rules. We had L x L y L y L x L x, L z ı hl z and by cyclic permutation, L y L z L z L y L y, L z ı hl x L z L x L x L z L z, L x ı hl y where x y, y z, and z x corresponds to the cyclic permutation we are talking about. There is a sensual formulation of these three rules expressed as a cross product, a highly stylized rendition: L L ı h L where of course the vectors are vector operators, so that the cross product refer to vectors and their components, and therefore, since these components are operators, do not commute and therefore do not cancel. Let s expand the cross product to see what is happening in this condensed notation. We have or L L î ĵ ˆk L x L y L z î L y L z L z L y ) + ĵ L z L x L x L z ) + ˆk L x L y L y L x ) L x L y L z L L î L y L z L z L y ) + ĵ L z L x L x L z ) + ˆk L x L y L y L x ) îl y, L z + ĵl z, L x + ˆkL x, L y Since îl y, L z + ĵl z, L x + ˆkL x, L y îı hl x + ĵı hl y + ˆkı hl z we then have L L ı h L as asserted at the outset.
7 IV. THE RUNGE-LENZ VECTOR COMMUTATORS A. Similar Components Commuting with Angular Momentum Components We had Equation 4.0 ) A ) L p p L Ze + ˆr 4.) this is also Equation 3-3 in Borowitz, loc cit) as the defining equation for the operator form of the Runge- Lenz vector, and we wish now to see how this commutes with L. First, we change over to a better more normal) form. We start with A i, L i ; i which we think of as similar components. First, we need to evaluate î ĵ ˆk L p L x L y L z p x p y p z L p îl y p z L z p y ) ĵl z p x L x p z ) ˆkL x p y L y p x ) 4.) îzp x xp z )p z xp y yp x )p y ) ĵxp y yp x )p x yp z zp y )p z ) 4.3) ˆkyp z zp y )p y zp x xp z )p x ) In the reverse order, we had p L î ĵ ˆk îp y xp y yp x ) p z zp x xp z )) p x p y p z ĵp z yp z zp y ) p x xp y yp x )) 4.4) yp z zp y zp x xp z xp y yp x ˆkp x zp x xp z ) p y yp z zp y )) so, combining these two we have don t worry, we re not making an error here relative to earlier work) îzp x xp z )p z xp y yp x )p y ) L p + p L ĵxp y yp x )p x yp z zp y )p z ) + ˆkyp z zp y )p y zp x xp z )p x ), combining term L p + p L îp y xp y yp x ) p z zp x xp z )) ĵp z yp z zp y ) p x xp y yp x )) ˆkp x zp x xp z ) p y yp z zp y )) 4.5) îzp x xp z )p z xp y yp x )p y + {p y xp y yp x ) p z zp x xp z )} ĵxp y yp x )p x yp z zp y )p z + {p z yp z zp y ) p x xp y yp x )} 4.6) ˆkyp z zp y )p y zp x xp z )p x + {p x zp x xp z ) p y yp z zp y )}, upon expansion, îzp x p z L p + p L xp z xp y + yp x p y + xp y p y yp x p z zp x + xp z ĵxp y p x i yp x yp z + zp y p z + yp z p z zp y p x xp y + yp x 4.7) ˆkyp z p y zp y zp x + xp z p x + zp x p x xp z p y yp z + zp y or L p + p L îzp x p z + yp x p y p y yp x p z zp x ĵxp y p x + zp y p z p z zp y p x xp y 4.8) ˆkyp z p y + xp z p x p x xp z p y yp z or L p + p L îp x zp z p z z) + p x yp y + p y y)) ĵp y xp x p x x) + p y zp z + p z z)) 4.9) ˆkp z yp y p y y) + p z xp x p x x)) L p + p L îı hp x ) ĵı hp y 4.0) ˆkı hp z ) i.e., L p + p L ı h p 4.)
8 so we can rewrite this as p L L p + ı h p 4.) which allows us to use a slightly simpler form in future work for the Runge Lenz vector, i.e., A L p Ze L )) p + ı h p + ˆr 4.3) which results in our final version: A Ze L p ı h p ) + ˆr 4.4) This is the standard form used in most discussions of the Ringe-Lenz vector operator, i.e., not the original form cited and used before. It is a labor of love to obtain the commutators of the Runge-Lenz vector with respect to other quantum- > #Commutator example Zhang > From Hong-Tao Zhang, arxiv.org/ps cache/quantph/pdf/004/00408v.pdf, A Simple Method of Calculating Comtators in Hamiltonian System with Mathematica Software. > restart; > assumer>0); > Comm3D : procf,g)local t, t, t3, t4; > t : difff,q)*diffg,p)-difff,p)*diffg,q); > t : difff,q)*diffg,p)-difff,p)*diffg,q); > t3 : difff,q3)*diffg,p3)-difff,p3)*diffg,q3); > t4 : subs > q*p-q*pl_3,q*p3-q3*pl_,q3*p-q*p3l_,t+t+t3); > t4 : subssqrtq^+q^+q3^)r,t4); > t4 : algsubsq^+q^+q3^r^,t4); > t4 : expandt4); > returni*hbar*t4); > end proc; mechanical operators needed for dealing with the hydrogen atom. The Runge-Lenz vector form used here is from Equation 4.4. What follows is a Maple session showing the commutator relationships needed to proceed. VI. ADDENDUM Given these commutator relations, one can now proceed on to the ladder operator solution to the H-atom s electronic energy levels C. W. David, Am. J. Phys., 34, ). This was the only time that I ever encountered the thrill of discovery first hand. It was exhilarating. It was even cited:blinder, S. M., J. Chem. Educ. 00, 78, 39. V. COMMUTATOR OF A op WITH L i, ETC. Comm3D : procf, g) local t, t, t3, t4 ; t : difff, q ) diffg, p ) difff, p ) diffg, q ) ; t : difff, q ) diffg, p ) difff, p ) diffg, q ) ; t3 : difff, q3 ) diffg, p3 ) difff, p3 ) diffg, q3 ) ; t4 : subsq p q p L 3, q p3 q3 p L, q3 p q p3 L, t + t + t3 ); t4 : subssqrtq + q + q3 ) r, t4 ) ; t4 : algsubsq + q + q3 r, t4 ) ; t4 : expandt4 ) ; return hbar t4 I end proc > ham : procp,p,p3,q,q,q3) > returnp^+p^+p3^)/ - /sqrtq^+q^+q3^)); > end proc; ham : procp, p, p3, q, q, q3 ) return / p + / p + / p3 /sqrtq + q + q3 ) end proc > The angular momentum vector is defined here: > L : q*p3-q3*p: > L : q3*p-q*p3: > L3 : q*p-q*p:
9 > The Runge-Lenz vector is defined here: > RL : L3*p-L*p3-I*hbar*p-q/sqrtq^+q^+q3^): > RL : L*p3-L3*p-I*hbar*p-q/sqrtq^+q^+q3^): > RL3 : L*p-L*p-I*hbar*p3-q3/sqrtq^+q^+q3^): > To show that L x L i hbar L3 > print Comm3DL,L) ); > comm : Comm3DL,L); Comm3DL, L ) comm : hbar L 3 I To show that RL x L equals zero > print Comm3DRL,L) ); > comm : Comm3DRL,L): > comm : simplifycomm); Comm3DRL, L ) comm : 0 > To show that RL x RL - K L3 > comm3 : Comm3DRL,RL): > print Comm3DRL,RL) ); > comm3a : expandsimplifycomm3/i*hbar))); > comm3b : subsq^3q*r^-q^-q3^),comm3a); > comm3b : subsq^3q*r^-q^-q3^),comm3b); > comm3b : subsq3^3q3*r^-q^-q^),comm3b); > comm3c : expandsubsp3^3 p3**en+/r)-p^+p^)),comm3b)); > comm3d : algsubsq*p-q*pl_3,comm3c);#we know that L_ and L_ > are not involved > comm3e : expandsubsp3^ *En- p^+p^)/ - /r)),comm3d)); > January 5, 007 SUCCESS Comm3DRL, RL ) comm3a : q3 q p q q p p 3 q + q p 3 q 3 p + p q p + p3 q p p3 q p p q p q p p q + r r comm3b : q3 q p q q p p 3 q + q p 3 q 3 p + p q p + p3 q p p3 q p p q p q p p q + r r comm3b : q3 q p q q p p 3 q + q p 3 q r q q3 ) p + p q p + p3 q p p3 q p p q p q p p q + r r comm3b : q3 q p q q p p 3 q + q p 3 q r q q3 ) p + p q p + p3 q p p3 q p p q p q p p q + r r comm3c : p 3 q + q p 3 q p + r q p r + p q p + p3 q p p3 q p p p q comm3d : L 3 p L 3 + p3 r ) p L 3 r comm3e : L 3 En > To show that RL x r - K L3 > print L cross r calculation ); > vec : L*Comm3Dq,p)+Comm3Dq3,p3)); > vec : L*Comm3Dq3,p3)+Comm3Dq,p)); > vec3 : L3*Comm3Dq,p)+Comm3Dq,p));#this is Eqn. in h_lada > manuscript. > January 30, 007 SUCCESS
10 L cross r calculation vec : I q p3 q3 p ) hbar vec : I q3 p q p3 ) hbar vec3 : I q p q p ) hbar educ/4 educ op cit
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