Quantum Physics I, Lecture Notes 20-21

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Lectures 20 and 21: Quantum Mechanics in 3D and Central PotentialsB. ZwiebachMay 3, 2016Contents1 Schrödinger Equation in 3D and Angular Momentum12 The angular momentum operator33 Eigenstates of Angular Momentum74 The Radial Wave Equation110Schrödinger Equation in 3D and Angular MomentumWe have so far considered a number of Hermitian operators: the position operator, the momentumoperator, and the energy operator, or the Hamiltonian. These operators are observables and theireigenvalues are the possible results of measuring them on states. We will be discussing here anotheroperator: angular momentum. It is a vector operator, just like momentum. It will lead to threecomponents, each of which is a Hermitian operator, and thus a measurable quantity. The definitionof the angular momentum operator, as you will see, arises from the classical mechanics counterpart.The properties of the operator, however, will be rather new and surprising.You may have noticed that the momentum operator has something to do with translations. Indeedthe momentum operator is a derivative in coordinate space and derivatives are related to translations.The precise way in which this happens is through exponentiation. Consider a suitable exponential ofthe momentum operator:ipaˆe ,(1.1)where a is a constant with units of length, making the argument of the exponential unit free. Considernow letting this operator act on a wavefunction ψ(x)eipaˆ dψ(x) ea dx ψ(x) ,(1.2)where we simplified the exponent. Expanding the exponential giveseipaˆ da2 d2a3 d3 .ψ(x) ,dx2! dx23! dx3dψ a2 d2 ψ a3 d3 ψ ψ(x) a . . . ψ(x a) ,dx2! dx23! dx3ψ(x) 1 a(1.3)ipaˆsince we recognize the familiar Taylor expansion. This result means that the operator e movesthe wavefunction. In fact it moves it a distance a, since ψ(x a) is the displacement of ψ(x) by adistance a. We say that the momentum operator generates translations. Similarly, we will be ableto show that the angular momentum operator generates rotations. Again, this means that suitableexponentials of the angular momentum operator acting on wavefunctions will rotate them in space.1

Angular momentum can be of the orbital type, this is the familiar case that occurs when a particlerotates around some fixed point. But is can also be spin angular momentum. This is a rather differentkind of angular momentum and can be carried by point particles. Much of the mathematics of angularmomentum is valid both for orbital and spin angular momentum.Let us begin our analysis of angular momentum by recalling that in three dimensions the usual x̂and p̂ operators are vector operators: p̂ (p̂x , p̂y , p̂z ) , ,.ii x y z(1.4)x̂ (x,ˆ ŷ, ẑ) .The commutation relations are as follows:[ x,ˆ p̂x ] i ,[ ŷ, p̂y ] i ,(1.5)[ ẑ, p̂z ] i .All other commutators are involving the three coordinates and the three momenta are zero!Consider a particle represented by a three-dimensional wavefunction ψ(x, y, z) moving in a threedimensional potential V (r). The Schrödinger equation takes the form 2 2 ψ(r) V (r)ψ(r) Eψ(r) .2m(1.6)We have a central potential if V (r) V (r). A central potential has no angular dependence, thevalue of the potential depends only on the distance r from the origin. A central potential is sphericallysymmetric; the surfaces of constant potential are spheres centered at the origin and it is thereforerotationally invariant. The equation above for a central potential is 2 2 ψ(r) V (r)ψ(r) Eψ(r) .2m(1.7)This equation will be the main subject of our study. Note that the wavefunction is a full function of r,it will only be rotational invariant for the simplest kinds of solutions. Given the rotational symmetry of the potential we are led to express the Schroödinger equation and energy eigenfunctions usingspherical coordinates.In spherical coordinates, the Laplacian is 2 ψ ( · )ψ 1 21rψ 22rr r 1 1 2sin θ ψ.sin θ θ θ sin2 θ φ2Therefore the Schrödinger equation for a particle in a central potential becomes 2 1 211 1 2 r sinθ ψ V (r)ψ Eψ .2m r r2r2 sin θ θ θsin2 θ φ2In our work that follows we will aim to establish two facts:2(1.8)(1.9)

1. The angular dependent piece of the 2 operator can be identified as the magnitude squared ofthe angular momentum operator1 1 2L2sin θ 222sin θ θ θ sin θ φ (1.10)ˆ xLˆx LˆyLˆy LˆzLˆz .L2 L(1.11)whereThis will imply that the Schrödinger equation becomes 1 L2 2 1 2 r 2 2 ψ V (r)ψ Eψ2m r r2r (1.12)or expanding out 2 1 2L2(rψ) ψ V (r)ψ Eψ .2m r r22mr2(1.13)2. Eq. (1.7) is the relevant equation for the two-body problem when the potential satisfiesV (r1 , r2 ) V ( r1 r2 ) ,(1.14)namely, if the potential energy is just a function of the distance between the particles. Thisis true for the electrostatic potential energy between the proton and the electron forming ahydrogen atom. Therefore, we will be able to treat the hydrogen atom as a central potentialproblem.2The angular momentum operatorClassically, we are familiar with the angular momentum, defined as the cross product of r and p:L r p. We therefore haveL (Lx , Ly , Lz ) r p ,Lx ypz zpy ,Ly zpx xpz ,(2.1)Lz xpy ypx .ˆ and its components,We use the above relations to define the quantum angular momentum operator Lˆˆˆthe operators (Lx , Ly , Lz ):ˆ (Lˆ x, Lˆy, Lˆz) ,LL̂x ŷp̂z ẑp̂y ,L̂y ẑp̂x x̂p̂z ,(2.2)L̂z x̂p̂y ŷp̂x .In crafting this definition we saw no ordering ambiguities. Each angular momentum operator is thedifference of two terms, each term consisting of a product of a coordinate and a momentum. But notethat in all cases it is a coordinate and a momentum along different axes, so they commute. Had weˆ x p̂z ŷ p̂y ẑ, it would have not mattered, it is the same as the Lˆ x above. It is simple towritten L3

ˆ x , for example. Recalling that forcheck that the angular momentum operators are Hermitian. Take L†††any two operators (AB) B A we haveˆ x )† (ŷp̂z ẑp̂y )† (ŷp̂z )† (ẑp̂y )† p̂†z ŷ † p̂†y ẑ † .(L(2.3)Since all coordinates and momenta are Hermitian operators, we haveˆ x )† p̂z ŷ p̂y ẑ ŷp̂z ẑp̂y Lˆx ,(L(2.4)where we moved the momenta to the right of the coordinates by virtue of vanishing commutators.The other two angular momentum operators are also Hermitian, so we haveˆ †x Lˆx ,Lˆ †y Lˆy ,Lˆ †z Lˆz .L(2.5)All the angular momentum operators are observables.Given a set of Hermitian operators, it is natural to ask what are their commutators. This computation enables us to see if we can measure them simultaneously. Let us compute the commutator ofˆ x with Lˆy:Lˆ x, Lˆ y ] [ ŷp̂z ẑp̂y , ẑp̂x x̂p̂z ][L(2.6)We now see that these terms fail to commute only because ẑ and p̂z fail to commute. In fact the firstˆ x only fails to commute with the first term of Lˆ y . Similarly, the second term of Lˆ x only failsterm of Lˆ y . Thereforeto commute with the second term of Lˆ x, Lˆ y ] [ ŷp̂z , ẑp̂x ] [ẑp̂y , x̂p̂z ][L [ ŷp̂z , ẑ]p̂x x̂[ẑp̂y , p̂z ] ŷ [ p̂z , ẑ] p̂x x̂ [ẑ, p̂z ] p̂y(2.7) ŷ( i )p̂x x̂(i )p̂y i (x̂p̂y ŷp̂x ) .ˆ z and therefore,We now recognize that the operator on the final right hand side is Lˆ x, Lˆ y ] i Lˆz .[L(2.8)The basic commutation relations are completely cyclic, as illustrated in Figure 1. In any commutationrelation we can cycle the position operators as in x̂ ŷ ẑ x̂ and the momentum operators asin p̂x p̂y p̂z p̂x and we will obtain another consistent commutation relation. You can also seeˆx Lˆy Lˆz Lˆ x , by looking at (2.2). We therefore claim that we do notthat such cycling takes Lhave to calculate additional angular momentum commutators, and (2.8) leads toˆ x, Lˆ y ] i L̂z ,[Lˆy, Lˆ z ] i L̂x ,[Lˆz, Lˆ x ] i Lˆy .[L(2.9)This is the full set of commutators of angular momentum operators. The set is referred to as theˆ were defined in terms of cooralgebra of angular momentum. Notice that while the operators Ldinates and momenta, the final answer for the commutators do not involve coordinates nor momenta:ˆ operators are sometimes referred tocommutators of angular momenta give angular momenta! The L4

as orbital angular momentum, to distinguish them from spin angular momentum operators. The spinangular momentum operators Sˆx , Sˆy , and Sˆz cannot be written in terms of coordinates and momenta.They are more abstract entities, in fact their simplest representation is as two-by-two matrices! Still,being angular momenta they satisfy exactly the same algebra as their orbital cousins. We have[ Sˆx , Sˆy ] i Sˆz ,[ Sˆy , Sˆz ] i Ŝx ,[ Sˆz , Sˆx ] i Sˆy .(2.10)Figure 1: The commutation relations for angular momentum satisfy cyclicity).We have seen that the commutator [x,ˆ p̂] i is associated with the fact that we cannot haveˆsimultaneous eigenstates of position and of momentum. Let us now see what the commutators of Lˆ x and Lˆ y ? As it turnsoperators tell us. In particular: can we have simultaneous eigenstates of Lout, the answer is no, we cannot. We demonstrate this as follows. Let’s assume that there exists aˆ x and Lˆy,wavefunction φ0 which is simultaneously an eigenstate of LL̂x φ0 λx φ0 ,L̂y φ0 λy φ0 .(2.11)Letting the first commutator identity of (2.9) act on φ0 we haveˆ xLˆ y φ0 LˆyLˆ x φ0ˆ z φ0 [Lˆ x, Lˆ y ]φ0 Li Lˆ x λ y φ0 Lˆ y λ x φ0 L(2.12) (λx λy λy λx )φ0 0 ,ˆ z φ0 0. But this is not all, looking at the other commutators in the angular momentumshowing that Lalgebra we see that they also vanish acting on φ0 and as a result λx and λy must be zero:ˆy, Lˆ z ]φ0 i L̂x φ0 i λx φ0 λx 0 ,[L {z }0ˆz, Lˆ x ]φ0 i Lˆy φ0 i λy φ0 λy 0 .[L} {z(2.13)0ˆ x and Lˆ y has led to Lˆ x φ0 Lˆ y φ0 All in all, assuming that φ0 is a simultaneous eigenstate of LL̂z φ0 0. The state is annihilated by all angular momentum operators. This trivial situation is not5

very interesting. We have learned that it is impossible to find states that are nontrivial simultaneouseigenstates of any two of the angular momentum operators.For commuting Hermitian operators, there is no problem finding simultaneous eigenstates. In fact,commuting Hermitian operators always have a complete set of simultaneous eigenstates. Suppose weˆ z as one of the operators we want to measure. Can we now find a second Hermitian operatorselect Lˆzthat commutes with it? The answer is yes. As it turns out, L2 , defined in (1.11) commutes with Land is an interesting choice for a second operator. Indeed, we quickly checkˆ 2 ] [Lˆz, Lˆz, Lˆ xLˆ x ] [Lˆz, LˆyLˆy][Lˆz, Lˆ x ]Lˆx Lˆ x [Lˆz, Lˆ x ] [Lˆz, Lˆ y ]Lˆy Lˆ y [Lˆz, Lˆy] [LˆyLˆ x i Lˆ xLˆ y i Lˆ xLˆ y i Lˆ xLˆy i L(2.14) 0.ˆ z and Lˆ 2 . We will do this shortly.So we should be able to find simultaneous eigenstates of both L2ˆThe operator L is Casimir operator, which means that it commutes with all angular momentumˆ z , it commutes also with Lˆ x and Lˆy.operators. Just like it commutes with LTo understand the angular momentum operators a little better, let’s write them in spherical coordinates. For this we need the relation between (r, θ, φ) and the cartesian coordinates (x, y, z):px r sin θ cos φ ,r x2 y 2 z 2 , (2.15)y r sin θ sin φ ,θ cos 1 zr , 1 yz r cos θ ,φ tanx .We have hinted at the fact that angular momentum operators generate rotations. In spherical coordinates rotations about the z axis are the simplest: they change φ but leave θ invariant. Bothˆ z is simple in sphericalrotations about the x and y axes change θ and φ. We can therefore hope that Lˆ z x̂p̂y ŷp̂x we havecoordinates. Using the definition L ˆz x y .(2.16)L xi yNotice that this is related to φsince, by the chain rule0 y x z x y, φ φ y φ x φ z y x(2.17)where we used (2.15) to evaluate the partial derivatives. Using the last two equations we can identifyˆz .Li φ(2.18)ˆ z generatesThis is a very simple and useful representation. It confirms the interpretation that Lˆ z is like a momentum alongrotations about the z axis, as it has to do with changes of φ. Note that Lthe “circle” defined by the φ coordinate (φ φ 2π). The other angular momentum operators are abit more complicated. A longer calculation shows what we suggested earlier, that L̂21 1 2 2 sin θ .(2.19) sin θ θ θsin2 θ φ26

3Eigenstates of Angular Momentumˆ z and L2 commute. We now aim to constructWe demonstrated before that the Hermitian operators Lthe simultaneous eigenfunctions of these operators. They will be functions of θ and φ and we will callthem ψ m (θ, φ). The conditions that they be eigenfunctions arem RL̂z ψ m m ψ m ,22L̂ ψ m ( 1) ψ m , R.(3.1)As befits Hermitian operators, the eigenvalues are real. Both m and l are unit free; there is an inˆ 2 we have an 2 .ˆ z eigenvalue because angular momentum has units of . For the eigenvalue of Lthe L2ˆNote that we have written the eigenvalue of L as ( 1) and for real this is always greater than orequal to 1/4. In fact ( 1) ranges from zero to infinity as ranges from zero to infinity. We canˆ 2 can’t be negative. For this we first claim thatshow that the eigenvalues of L ˆ 2ψ 0 ,ψ, L(3.2)ˆ 2 eigenvalue λ we immediately see that the aboveand taking ψ to be a normalized eigenfunction with Lgives ψ, λψ λ 0, as desired. To prove the above equation we simply expand and use Hermiticityˆ 2ψψ, L ˆ 2x ψ ψ, Lˆ 2x ψ ψ, Lˆ 2x ψψ, L ˆ x ψ, Lˆ xψ Lˆ y ψ, Lˆyψ Lˆ z ψ, Lˆzψ 0 ,L(3.3)because each of the three summands is greater than or equal to zero.Let us now solve the first eigenvalue equation in (3.1) using the coordinate representation (2.18)ˆ z operator:for the L ψ m ψ m mψ m imψ m .(3.4)i φ φThis determines the φ dependence of the solution and we writeψ m (θ, φ) eimφ P m (θ) ,(3.5)where the function P m (θ) captures the still undetermined θ dependence of the eigenfunction ψ m . Wewill require that ψ m be uniquely defined as a function of the angles and this requires that1ψ m (θ, φ 2π) ψ m (θ, φ) .(3.6)There is no similar condition for θ. The above condition requires thateim(φ 2π) eimφ e2πim 1 .(3.7)This equation implies that m must be an integer:m Z.(3.8)1One may have tried to require that after φ increases by 2π the wavefunction changes sign, but this does not lead toa consistent set of ψ m ’s.7

This completes our analysis of the first eigenvalue equation. The second eigenvalue equation in (3.1),ˆ 2 , givesusing our expression (2.19) for L 1 1 222sin θ (3.9)2 φ2 ψ m ( 1)ψ m .sin θ θ θsin θWe multiply through by sin2 θ and cancel the 2 to get sin θ 2 sin θ ψ m ( 1) sin2 θ ψ m . θ θ φ2Using ψ m eimφ P m (θ) we can evaluate the action ofarrive at the differential equationsin θ 2 φ2(3.10)on ψ m and then cancel the overall eimφ todP m d sin θ m2 P m ( 1)P m sin2 θ ,dθdθ(3.11)or, equivalently, dP m d sin θ ( 1) sin2 θ m2 P m 0 .(3.12)dθdθWe now want to make it clear that we can view P m as a function of cos θ by writing the differentialequation in terms of x cos θ. Indeed, this givessin θddx dd sin θdθdθ dxdx sin θdd (1 x2 ) .dθdx(3.13)The differential equation becomes(1 x2 ) dP m i dh(1 x2 ) ( 1)(1 x2 ) m2 P m (x) 0 ,dxdx(3.14)and dividing by 1 x2 we get the final form:dP m i hm2 i mdhP (x) 0 .(1 x2 ) ( 1) dxdx1 x2 (3.15)The P m (x) are called the associated Legendre functions. They are not polynomials. All we know atthis point is that m is an integer. We will discover soon that is a non-negative integer and that fora given value of there is a range of possible values of m.To find out about we consider the above equation for m 0. In that case we write P (x) P 0 (x)and the P (x) must satisfydhdP i(1 x2 ) ( 1)P (x) 0 .dxdx(3.16)This is the Legendre differential equation. We try finding a series solution by writingP (x) Xk 08ak xk ,(3.17)

assuming that P (x) is regular at x 0, as it better be. Plugging this into the DE yields we find thatthe vanishing of the coefficient of xk requires:(k 1)(k 2)ak 2 [ ( 1) k(k 1)]ak 0 .(3.18)ak 2 ( 1) k(k 1) .ak(k 1)(k 2)(3.19)Equivalently, we haveThe large k behavior of the coefficients is such that unless the series terminates P diverges at x 1(since x cos θ this corresponds to θ 0, π). In order for the series to terminate, we must have ( 1) k(k 1) for some integer k 0. We can simply pick k so that ak 2 0, making Pk (x)a degree k polynomial. We have thus learned that the possible values of are 0, 1, 2, 3, . . . .(3.20)This is quantization! Just like the m values are quantized, so are the values. The Legendre polynomials P (x) are given by the Rodriguez formula: 1 d 2P (x) x 1 .(3.21)2 ! dxThe Legendre polynomials have a nice generating function XP (x)s 1.1 2xs s2P1 (x) x ,P2 (x) j(3.22)A few examples areP0 (x) 1 ,12 3x2 1 .(3.23)P (x) is a degree polynomial of definite parity.Having solved the m 0 equation we now have to discuss the general equation for P m (x). Thedifferential equation involves m2 and not m, so we can take the solutions for m and m to be the same.One can show that taking m derivatives of the Legendre polynomials gives a solution for P m (x):P m (x) (1 x2 ) m /2 ddx m P (x) .(3.24)Since P is a polynomial of degree , the above gives a non-zero answer only for m . We thus havesolutions for m .(3.25)It is possible to prove that no other solutions exist. One can think of the ψ m eigenfunctions as firstdetermined by the integer and, for a fixed , there are 2 1 choices of m: , 1, . . . , .Our ψ m eigenfunctions, with suitable normalization, are called the spherical harmonics Y m (θ, φ).The properly normalized spherical harmonics for m 0 ares2 1 ( m)!Y ,m (θ, φ) ( 1)m eimφ P m (cos θ).(3.26)4π ( m)!9

For m 0, we useY ,m (θ, φ) ( 1)m [Y , m (θ, φ)] .(3.27)We thus haveL̂z Y m m Y m ,(3.28)L̂2 Y m 2 ( 1) Y m .The first few spherical harmonics are1Y0,0 (θ, φ) 4πr(3.29)r3 iφ3 x iyY1, 1 (θ, φ) esin θ 8π8π rrr33 zY1,0 (θ, φ) cos θ .4π4π r2(3.30)(3.31)Being eigenstates of Hermitian operators with different eigenvalues, spherical harmonics with different and m subscripts are automatically orthogonal. The complicated normalization factor is needed tomake them have unit normalization. The spherical harmonics form an orthonormal set with respectto integration over the solid angle. This integration can be written in many forms:ZZ 2πZ πZ 2πZ 1dΩ · · · dφdθ sin θ · · · dφd(cos θ) · · · .(3.32)0θ 00 1The statement that the spherical harmonics form an orthonormal set with respect to this integrationmeans thatZdΩ Y 0 ,m0 (θ, φ)Y ,m (θ, φ) δ , 0 δm,m0 .(3.33)4The Radial Wave EquationLet us now write an ansatz for the solution of the Schrödinger equation. For this we take the productof a purely radial function RE (r) and a spherical harmonicψ(r, θ, φ) RE (r)Y ,m (θ, φ).(4.1)We have put subscripts E and for the radial function. We did not include m, because, as we willsee the equation for R does not depend on m. We can now insert this into the Schrödinger equation(1.13)ˆ2 2 1 2L (rRY) RE Y m V (r)RE Y m ERE Y m .(4.2)E m2m r r22mr2ˆ 2 eigenstates we can simplify the equation to giveSince the spherical harmonics are L 2 1 d2 (rRE ) 2 ( 1)Y RE Y m V (r)RE Y m ERE Y m . m2m rdr22mr2(4.3)Canceling the common spherical harmonic and multiplying by r we get a purely radial equation 2 d2 (rRE ) 2 ( 1) (rRE ) V (r)(rRE ) E (rRE ) ,2m dr22mr210(4.4)

It is now convenient to defineuE (r) rRE (r) .(4.5)This allows us to rewrite the entire DE as 2 d2 uE 2 ( 1) V(r) uE EuE . 2m dr22mr2(4.6)This is called the radial equation. It looks like the familiar time-indep

Lectures 20 and 21: Quantum Mechanics in 3D and Central Potentials B. Zwiebach May 3, 2016 Contents 1 Schr odinger Equation in 3D and Angular Momentum 1 2 The angular momentum operator 3 3 Eigenstates of Angular Momentum 7 4 The Radial Wave Equation 10 1 Schr odinger Equation in 3D and Angular Momentum

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