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A. La Rosa Lecture Notes INTRODUCTION TO QUANTUM MECHANICS PART-III THE HAMILTONIAN OPERATOR and the SCHRODINGER EQUATION ______CHAPTER-11 SCHRODINGER EQUATION in 3D Description of two interacting particles’ motion
11.1 General case of an arbitrary interaction potential 11.2 Case when the potential depends only on the relative position of the particles U U(r1 r2 ) Decoupling the Center of Mass motion and the Relative Motion The CM variable and relative position variable Solution by separation of variables Equation of Motion for the Center of Mass Equation governing the Relative Motion
11.3 Central Potentials U U (r) 11.3A Separation of variables method R(r)Y (,) 11.3B The angular equation The Legendre Eq. 11.3C The radial equation, using the Coulomb Potential
References: "Introduction to Quantum Mechanics" by David Griffiths; Chapter 4
1 CHAPTER-11 SCHRODINGER EQUATION in 3D Description of two interacting particles’ motion
One particle motion In the case in which a particle of mass m moves in 1-D and inside a potential V(x,t) , the Schrodinger Eq. is, 2 2 i V (x,t) , where = (x, t) (1) t 2m x2 When the particle moves in the 3-D space, the equation adopts the form, 2 i 2 V (r,t) , where = ( r , t) (2) t 2m here 2 stands for the Laplacian operator
______In cartesian coordinates: 2 2 2 2 2 2 2 x y z In spherical coordinates: 1 1 1 2 2 r 2 sin θ 2 2 2 2 2 r r r r sin θ r sin
Z (x, y, z, t) r
Y X
2 11.1 General case of an arbitrary interaction potential Two particles motion
m2 m1 r1 =(x1, y1, z1, t)
r2 r1 Z r2 =(x2, y2, z2, t) Y X
The mutual interaction between the two particles is specified through the potential V(r1 ,r2 ,t) and the corresponding Schrodinger Eq. is 2 2 2 2 , (3) i r r V(r1 ,r2 ,t) t 2m1 1 2m1 2 where = ( r1 ,r2 , t)
In the expression above the following notation is being used, 2 2 r ( , , ) ; ; etc 1 x1 y1 z1 r 2 y 2 z 2 1 x1 1 1
11.2 Case when the potential depends only on the relative position of the particles Here we consider that the potential V(r1 ,r2 ,t) depends only on the relative position, V(r1 ,r2 ,t) =U(r1 r2 ,t) (4)
Decoupling the Center of Mass motion and the Relative Motion For this particular case it can be shown that the motion of the two particles can be decoupled into the motion of the center of mass, and the motion relative to each other.
The CM and the relative-position variables Such decoupling can be achieved through the following change of variables:
3 m r m r R 1 1 2 2 , r r1 r2 (5) m1 m2 ( r1 ,r2 , t) = ( R,r, t)
Notice, by calling R (Rx , Ry , Rz ) and r (x, y, z) the definitions in (5) can be re-written more explicitly as, m x m x m y m y m z m z R 1 1 2 2 , R 1 1 2 2 , R 1 1 2 2 x M y M z M
x (x1 x2 ) , y ( y1 y2 ) , y ( z1 z2 ) (6) ( r1 ,r2 , t) = ( R, r, t) where we are using M m1 m2 ______You may want to jump to expression (10), which gives the form in which Eq. (3) is transformed upon the application of the change of variable (5). ( r1 ,r2 , t) = ( R, r, t) implies,
R x m x 1 x1 Rx x1 x x1 Rx M x
2 R R x x m1 x x ( ) ( ) ( ) ( ) x 2 R R x x R x M R x x x x x 1 x x 1 x 1 x 1 1
2 m m 2 1 m1 1 ( ) ( ) ( ) R 2 M x R M R x M 2 x x x x 2 m m 2 ( 1 )2 2 1 ( ) 2 2 Rx M M x Rx x
2 2 m m 2 ( 1 )2 2 1 ( ) 2 2 M M y R 2 y1 Ry y y
4 2 2 m m 2 ( 1 )2 2 1 ( ) 2 2 2 z1 Rz M M z Rz z
2 2 2 2 Since r 2 2 2 , adding the last three results gives, 1 x1 y1 z1
2 m1 2 2 m1 ( ) 2 ( ) r R r 1 M M x Rx y Ry z Rz
m m ( 1 )2 2 2 2 1 M R r M r R where we have used the notation 2 2 2 2 , , ) R , R 2 2 2 and Rx Ry Rz Rx Ry Rz
2 2 2 2 , , ) r , 2 2 2 x y z r x y z Very conveniently, the previous results is expressed as,
1 2 m1 2 1 2 1 2 r 2 R r r R (7) m 1 M m M 1 1
Similarly
Rx x m2 x2 Rx x2 x x2 Rx M x
2 R R x x m2 x x ( ) ( ) ( ) ( ) x 2 R R x x R x M R x x x x x 2 x x 2 x 2 x 2 2
2 m m 2 2 m2 2 ( ) ( ) ( ) R 2 M x R M R x M 2 x x x x 2 m m 2 ( 2 )2 2 2 ( ) 2 2 Rx M M x Rx x
5 2 2 Analogous results for 2 and 2 lead to, y2 z2
1 2 m 2 1 2 1 2 2 r 2 R r r R (8) m 2 M m M 2 2
Replacing (7) and (8) in (3) one obtains, 2 2 m m 2 1 1 2 i 1 2 ( ) U(r,t) 2 R r t 2 M 2 m1 m2
In terms the reduced mass 1 1 1 (9) m1 m2 one obtains the equation
2 2 2 2 i U(r,t) (10) t 2M R 2 r where = ( R,r, t)
Solution by separation of variables We will specialize in the case where the potential U does not depend explicitly on the time t. For such a case, we will be looking for solutions to Eq. (10) in the form, (i / )(E E) t ( R,r, t) ≡ (R)F(r)e CM (11) (i / )(E ) t (i / )E t [ (R)e CM [ F(r)e
Replacing (11) in (10)
2 2 2 2 (E E)(R)F(r) F(r) (R) (R) F(r) U(r)(R)F(r) CM 2M R 2 r Dividing by (R)F(r)
6 2 2 2 2 (ECM E) (R) F(r ) U(r ) 2M (R) R 2 F(r ) r
This expression indicates that the terms on the right, although intrinsically depending on the variables r and R , they add up to a constant value. Selectively we choose, 2 2 ECM (R) , (12) 2M (R) R and 2 2 E F(r) U(r) (13) 2 F(r) r
Equation of Motion for the Center of Mass The first expression leads to, 2 2 (R) E (R) (14) 2M R CM which admit solutions of the form (i / ) 2M E n R CM (15) (R) e where n is an arbitrary constant unit vector. (That means there are many solutions of this type, one for each selected unit vector n .) 2ME CM In terms of the wavevector kCM n , the previous expression is written as, (R) = i (k ) R (15)’ e CM .
In short, if a single vector kCM were used in the separation of variables given in (11), then the motion of the CM would be described by a plane-wave that propagates in the direction of kCM , (i / )(E ) t i [k R - (E / ) t] (R)e CM e CM. CM (16)
Accordingly, the CM is likely to be anywhere in the space but has a very definite orientation of its momentum (that given by kCM ).
7 If a more precise localization of the CM were desired, then we can 2 M E CM select a bunch of wavevectors kCM n within a range k and define a wavepacket i [k R - (E /) t] (R,t) e CM CM . (17) k CM That way the CM will be more localized (because (R,t) is more spatially localized) but, at the same time, there will be a corresponding uncertainty in its momentum.
It is worth mentioning that we have not been looking for a solution R R(t) , as we would have been doing in a classical mechanics approach; rather we have found an amplitude probability (R) .
Equation governing the Relative Motion
Back to Eq. (13). This leads to, 2 2 F(r) U(r)F(r) E F(r) (18) 2 r
In the next section we will specialize for the case for central potentials; that is, when the potential depends only on the magnitude of the particles separation: U(r) U (r)
11.3 Central Potentials U(r) U (r) 2 2 F(r) U(r)F(r) E F(r) (19) 2 r This is an equation for F(r) where the energy parameter E is still to be determined. When working with central potentials, it is more convenient to use spherical coordinates to solve (19). At the beginning of this Chapter the Laplacian operator was given in spherical coordinates. Using that expression in (19) one obtains,
8 2 2 1 2 1 1 r sin θ F(r) 2 2 r r 2 2 2 2 r r sin θ r sin U(r)F(r) E F(r) (20)
11.3A Separation of variables method We will look for solutions of the form F(r) R(r)Y (,) (21) This leads to, 2 1 2 Y (,) r R(r) 2 2 r r r
2 2 1 1 R(r) sin θ Y (,) 2 2 2 2 2 r sin θ r sin
U (r) R(r)Y (,) E R(r)Y (,)
Dividing each term by R(r)Y (,)
2 1 1 2 r R(r) 2 2 R(r) r r r
2 2 1 1 1 sin θ Y (,) 2 Y (,) 2 2 2 2 r sin θ r sin
U (r) E
2 1 2 2 r R(r) r U (r) E + 2 R(r) r r
2 2 1 1 1 sin θ Y (,) 0 2 Y (,) sin θ 2 2 sin
1 2 2 2 r R(r) r U (r) E + 2 R(r) r r
9 2 1 1 1 + sin θ Y (,) 0 Y(,) sin θ 2 2 sin The first two terms depend only on r, while the third term only on the angular variables. This means that the radial part and the angular part are each constant.
2 1 1 1 sin θ Y (,) - Y (,) sin θ 2 2 sin
1 2 2 2 r R(r) r U (r) E 2 R(r) r r Rearranging terms, one obtains 2 1 1 sin θ Y (,) Y (,) Angular Eq. (22) sin θ 2 2 sin and
2 1 2 r U(r) R(r) E R(r) Radial Eq. (23) 2 2 2 r r r r
11.3B The angular Equation To solve the angular equation 2 1 1 sin θ Y (,) Y (,) sin θ 2 2 sin we will look for solutions of the form, Y (,) ( ) () (24)
1 1 2 () sin θ ( ) + ( ) () = ( ) () sin θ sin2 2
2 Dividing by ( ) () and multiplying by sin ()
10 2 1 1 sin θ sin θ ) 2 [ ( + sin () ] + () = 0 ( ) () 2 This means that each of the two term must be constant
1 [ sin θ sin θ () + sin2() ] = m2 (25) ( ) 2 1 2 () = - m (26) () 2 with m2 being a constant value to be determined by the physical interpretation of the second Eq. above. d 2 () + m2 () = 0 2 d where the partial derivative symbol has been dropped, since the function depends only on one variable. i m The solutions of Eq. (26) has the form () =e . But notice that the wavefunction should take the same value whenever is incremented by 2; that is, ( ) = () (physics situation requirement) im( im This implies e = e , or im e = 1 This condition can be satisfied only if m is an integer number. Thus, the dependence of the wavefunction on the variable is given by, im () =e ; m = … , -2, -1, 0, 1, 2, … (27)
11
Z (x, y, z) r
Y X
The other angular Equation (25) has the form,
2 1 d d m sin θ ( ) 0 (28) sin θ d d 2 m sin () where it has been emphasized that the solution depends on the parameters and m. The task is to find a solution over the range 0≤≤. It is convenient introduce the new variable
w = cos (), for 0 ≤ ≤. and the new function ( ) Fm (w) = m
It can be shown that the Eq. for Fm is,
2 d F dF m2 2 m m (29) (1- w ) 2w 2 F 0 d w2 d w 1- w m Case m=0
We will see later that the functions Fm corresponding to m≠0 can be built upon F0 . The latter satisfies the Eq.
d 2 F dF (1- w2 ) 0 2 w 0 F 0 Legendre Eq. (30) 2 w 0 d w d Power series solution c wk F0 (w) = k (31) k 0
12 From (31) one obtains, dF dF 0 k 1 0 k k c w , 2w 2k c w , which gives, d w k d w k k 0 k 0
F k k 2 w 0 F 2k c w + c w = 0 k k w k 0 k 0 k (2k )c w (32) k k 0 Also,
2 2 d F F 0 k 2 2 0 k k(k 1) c w , w k(k 1) c w 2 k 2 k d w k 0 w k 0
2 d F k 2 k 2 0 k(k 1) c w - k(k 1) c w (1- w ) k k d w2 k 0 k 0 k 2 k = k(k 1) c w - k(k 1) c w k k k 2 k 0 Using k - 2 = k’
k k = (k 2)(k 1) c w - k(k 1) c w k 2 k k 0 k 0 where we have replaced k’ for k again (to avoid writing too many variables)
k = [(k 2)(k 1) c k(k 1) c ]w (33) k 2 k k 0 Replacing (32) and (33) in (30), one obtains
k k [(k 2)(k 1) c k(k 1) c ]w + (2k )c w = 0 k 2 k k k 0 k 0
13 k [(k 2)(k 1) c [k(k 1) ] c w = 0 k 2 k k 0 This requires k(k 1) c c recursion relation (34) k2 (k 1)(k 2) k Thus, the solution (31) can be written more explicitly as,
(0)(1) 2 (2)(3) (0)(1) 4 c 1 w w ... F0 (w) = 0 (35) (1)(2) (3)(4) (1)(2)
(1)(2) 3 (3)(4) (1)(2) 5 + c1 1 w w ... (2)(3) (4)(5) (2)(3) c c where 0 and 1 are arbitrary constants. Expression (31) provides two independent solutions. Notice, each subsequent coefficient ck 2 depends on the value of the previous one ck (for example, if one coefficient were equal to zero, then all the other subsequent coefficients will be null as well.)
Divergence at w =1. For large values of k, this series varies as
ck2/ ck k /(k 2) , a behavior which is similar to the divergent series (1/ k) . This indicates that (31) diverges for w=1, thus k 0 constituting an unacceptable solution, since the wavefunction has to have a definite value at =90o. Polynomial solutions. It is still possible to obtain a satisfactory solution out of (31), if were chosen of the form = l(l+1), with l being an integer value, then the coefficient series (31) would c c become a polynomial of degree l (with the corresponding 0 or 1 conveniently chosen equal to zero, depending on whether l is odd or even, respectively.) Summary: The Legendre Eq
14 d 2 F dF (1- w2 ) 0 2 w 0 F 0 Legendre Eq. 2 0 d w d w admits physically acceptable solutions provided that has the form of = l (l +1), with l being an integer. For a given l, the resulting solutions is a polynomial of degree l. l c wk P (w) F0 (w) = k Legendre Polyniomial . k 0 l k (k 1) ( 1 ) where c l l c k2 (k 1)(k 2) k c c Since an arbitrary multiplicative constant, 0 or 1 , accompany to these solutions, the Legendre polynomials are defined such that P (1) =1 l
From expression (35) one obtains, P (w) 1 l = 0: 0 (36) P (w) w l = 1: l
1 2 = 2: P (w) (1 3w ) l 2 2
3 5 3 = 3: P (w) (w w ) l 3 2 3
7 4 3 35 4 = 4: P (w) c (110w 10w ) (110w w ) l 4 0 6 8 3 …
The general case: m≠ 0 We are looking for solutions to Eq. (29)
2 d F dF 2 2 m m m (1- w ) 2w 2 F 0 d w2 d w 1- w m
15 Or that purpose, let’s define the Associated Legendre functions
d I mI PI mI (w) (1- w2 )I mI / 2 P (w) (37) I mI l d w l That is, the polynomials PI mI (w) are obtained by taking derivatives of l the already known Legendre polynomials P (w) given in (36). It turns l out these Associated Legendre functions satisfy Eq. (29), 2 I mI I mI d P dP 2 2 m I mI (38) (1- w ) l 2w l 2 P 0 d w2 d w 1- w l I mI Notice that, since P (w) is a polynomial of degree , P (w) will be l l l identically equal to zero for I mI l. Accordingly, for a given l there will be (2 l + 1) associated Legendre polynomials m = - l , - l +1, … , l -1, l (39)
Using (37) one obtains, 0 = = P (w) P (w) 1 l 0, m 0: 0 0 0 = = P (w) P (w) w l 1, m 0: 1 1
1 2 1/ 2 d 2 1/ 2 m= 1: P (w) (1- w ) P (w) (1- w ) 1 dw 1
1 2 1/ 2 P1 (w) (1 w ) (38)
1 2 1/ 2 P2 (w) 3(1 w ) w
2 2 P2 (w) 3(1 w )
Spherical Harmonics
16 1/ 2 (2l 1)(l m)! im (,) (1)m m (cos)e , Yl, m Pl m 0 4(l m)!
m * Yl, m (,) (1) Yl,m (,), for m 0
http://www.vis.uni-stuttgart.de/~kraus/LiveGraphics3D/java_script/SphericalHarmonics.html
1 Y (,) 0,0 (4 )1/ 2
Absolute value of real part l=0, m=0
Absolute value, colored by complex phase
l=0, m=0 ______
17 1/ 2 1/ 2 3 3 i Y (,) cos Y (,) ∓ sin e 1,0 4 1,1 8
l=1, m=0 l=1, m=1 Absolute value of real part
l=1, m=0 l=1, m=1 Absolute value, colored by complex phase ______
18 l=3, m=1 Absolute value of real part
l=3, m=1 Absolute value, colored by complex phase
11.3B The Radial Equation We have found that in Eq. (23) = l ( l+1 )
2 1 2 l( l 1) r U(r) R(r) E R(r) 2 2 2 r r r r
For the case of a Coulomb field U(r) r Ze2 where , Z is the atomic number (number of 4 0 protons in the nucleus) and e is the elementary charge, the Eq. is,
19 2 2 2 l( l 1) R(r) E R(r) 2 2 r r 2 r r r
When considering the case of a Coulomb field, it is convenient to use 2 3 m , , e m m 2 e e as the units of mass, length and time, respectively. 2 Notice that the unit of energy is m . e 2
For our purpose, instead of me we will use the reduced mass 2 2 instead. Thus, since has units of distance, and has units 2 of energy, the equation above can be expressed in terms of unit-less parameters r’ and E’ instead of the variables r and E, where 2 2 r ≡ r’ , E ≡ E’ . 2
2 R R r R Indeed, using in addition R(r)= R(r’), , … etc., r r' r m r' e one obtains,
2 2 l( l 1) 1 2 E' R(r') 0 2 2 r' r' r' r' r'
Let u(r') R(r’) = r' The purpose of this change of variable is to obtain an equation the resembles the one dimensional case.
R 1 u 1 2 u
r' r' r' r'
20 2 R 2 u 2 3 u 2 r' r' r' r r'
2 2 2 R 1 u 1 u 2 1 u 1 u 2 u 2 u u 2 2 3 2 2 2 3 2 r' r' r' r' r' r' r' r' r' r' r' r' r' Adding the last two expressions
2 2 R 2 R 1 u 2 2 r' r' r' r' r'
1 2u l( l 1) u(r') 1 u(r') 2 2 2 E' 0 r' r' r' r' r' r'
2u l( l 1) 1 u(r') 2 E' u(r') 0 r'2 r'2 r'
2u l( l 1) 1 2 2 E' 2 u(r') 0 r' 2r' r' This Eq. can be expressed as, 2u 2 E' V (r') u(r') 0 r'2 eff where an effective potential has been defined as, l( l 1) 1 V (r') eff 2r'2 r' Looking for bound states Bound states will be obtained with negative values for E’. Let’s define 1 2 k and r' - 2E' k
2 2 2 u v v 2 u 2 v u(r’)=v() , , r' r' k 2 2 r' k
21 2 2 2v 1 l( l 1) 1 2 u(r') 0 2 2 2 r' k -2k 2r'
22