Quantum Physics I, Lecture Notes 20 and 21

Lectures 20 and 21: Quantum Mechanics in 3D and Central Potentials B. Zwiebach May 3, 2016 Contents 1 Schrödinger Equation in 3D and Angular Momentum 1 2 The angular momentum operator 3 3 Eigenstates of Angular Momentum 7 4 The Radial Wave Equation 1 10 Schrödinger Equation in 3D and Angular Momentum We have so far considered a number of Hermitian operators: the position operator, the momentum operator, and the energy operator, or the Hamiltonian. These operators are observables and their eigenvalues are the possible results of measuring them on states. We will be discussing here another operator: angular momentum. It is a vector operator, just like momentum. It will lead to three components, each of which is a Hermitian operator, and thus a measurable quantity. The definition of 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. Indeed the 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 of the momentum operator: ipa ˆ e ~ , (1.1) where a is a constant with units of length, making the argument of the exponential unit free. Consider now letting this operator act on a wavefunction ψ(x) e ipa ˆ ~ d ψ(x) = ea dx ψ(x) , (1.2) where we simplified the exponent. Expanding the exponential gives e ipa ˆ ~  d a2 d2 a3 d3 + + + . . . ψ(x) , dx 2! dx2 3! dx3 dψ a2 d2 ψ a3 d3 ψ = ψ(x) + a + + + . . . = ψ(x + a) , dx 2! dx2 3! dx3 ψ(x) =  1+a (1.3) ipa ˆ since we recognize the familiar Taylor expansion. This result means that the operator e ~ moves the wavefunction. In fact it moves it a distance −a, since ψ(x + a) is the displacement of ψ(x) by a distance −a. We say that the momentum operator generates translations. Similarly, we will be able to show that the angular momentum operator generates rotations. Again, this means that suitable exponentials 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 particle rotates around some fixed point. But is can also be spin angular momentum. This is a rather different kind of angular momentum and can be carried by point particles. Much of the mathematics of angular momentum 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 x̂ and p̂ operators are vector operators:   ~ ~ ∂ ∂ ∂ p̂ = (p̂x , p̂y , p̂z ) = ∇ = , , . i i ∂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, the value of the potential depends only on the distance r from the origin. A central potential is spherically symmetric; the surfaces of constant potential are spheres centered at the origin and it is therefore rotationally 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 using spherical coordinates. In spherical coordinates, the Laplacian is ∇2 ψ = (∇ · ∇)ψ =
1 ∂2 1 rψ + 2 2 r r ∂r   1 ∂ ∂ 1 ∂2 sin θ + ψ. sin θ ∂θ ∂θ sin2 θ ∂φ2 Therefore the Schrödinger equation for a particle in a central potential becomes    ~2 1 ∂ 2 1 1 ∂ ∂ 1 ∂2 − r + sin θ + ψ + V (r)ψ = Eψ . 2m r ∂r2 r2 sin θ ∂θ ∂θ sin2 θ ∂φ2 In 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 of the angular momentum operator 1 ∂ ∂ 1 ∂2 L2 sin θ + = − 2 2 2 sin θ ∂θ ∂θ sin θ ∂φ ~ (1.10) ˆ xL ˆx + L ˆyL ˆy + L ˆzL ˆz . L2 = L (1.11) where This will imply that the Schrödinger equation becomes   1 L2 ~2 1 ∂ 2 − r − 2 2 ψ + V (r)ψ = Eψ 2m r ∂r2 r ~ (1.12) or expanding out − ~2 1 ∂ 2 L2 (rψ) + ψ + V (r)ψ = Eψ . 2m r ∂r2 2mr2 (1.13) 2. Eq. (1.7) is the relevant equation for the two-body problem when the potential satisfies V (r1 , r2 ) = V (|r1 − r2 |) , (1.14) namely, if the potential energy is just a function of the distance between the particles. This is true for the electrostatic potential energy between the proton and the electron forming a hydrogen atom. Therefore, we will be able to treat the hydrogen atom as a central potential problem. 2 The angular momentum operator Classically, we are familiar with the angular momentum, defined as the cross product of r and p: L = r × p. We therefore have L = (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 L ˆ ˆ ˆ the operators (Lx , Ly , Lz ): ˆ = (L ˆ x, L ˆy, L ˆz) , L L̂x = ŷp̂z − ẑp̂y , L̂y = ẑp̂x − x̂p̂z , (2.2) L̂z = x̂p̂y − ŷp̂x . In crafting this definition we saw no ordering ambiguities. Each angular momentum operator is the difference of two terms, each term consisting of a product of a coordinate and a momentum. But note that in all cases it is a coordinate and a momentum along different axes, so they commute. Had we ˆ x = p̂z ŷ − p̂y ẑ, it would have not mattered, it is the same as the L ˆ x above. It is simple to written L 3 ˆ x , for example. Recalling that for check that the angular momentum operators are Hermitian. Take L † † † any two operators (AB) = B A we have ˆ x )† = (ŷp̂z − ẑp̂y )† = (ŷp̂z )† − (ẑp̂y )† = p̂†z ŷ † − p̂†y ẑ † . (L (2.3) Since all coordinates and momenta are Hermitian operators, we have ˆ x )† = p̂z ŷ − p̂y ẑ = ŷp̂z − ẑp̂y = L ˆx , (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 have ˆ †x = L ˆx , L ˆ †y = L ˆy , L ˆ †z = L ˆz . 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 of ˆ x with L ˆy: L ˆ x, L ˆ 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 first ˆ x only fails to commute with the first term of L ˆ y . Similarly, the second term of L ˆ x only fails term of L ˆ y . Therefore to commute with the second term of L ˆ x, L ˆ 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 L ˆ x, L ˆ y ] = i~ L ˆz . [L (2.8) The basic commutation relations are completely cyclic, as illustrated in Figure 1. In any commutation relation we can cycle the position operators as in x̂ → ŷ → ẑ → x̂ and the momentum operators as in p̂x → p̂y → p̂z → p̂x and we will obtain another consistent commutation relation. You can also see ˆx → L ˆy → L ˆz → L ˆ x , by looking at (2.2). We therefore claim that we do not that such cycling takes L have to calculate additional angular momentum commutators, and (2.8) leads to ˆ x, L ˆ y ] = i~ L̂z , [L ˆy, L ˆ z ] = i~ L̂x , [L ˆz, L ˆ x ] = i~ L ˆy . [L (2.9) This is the full set of commutators of angular momentum operators. The set is referred to as the ˆ were defined in terms of cooralgebra of angular momentum. Notice that while the operators L dinates and momenta, the final answer for the commutators do not involve coordinates nor momenta: ˆ operators are sometimes referred to commutators of angular momenta give angular momenta! The L 4 as orbital angular momentum, to distinguish them from spin angular momentum operators. The spin angular 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 have ˆ simultaneous eigenstates of position and of momentum. Let us now see what the commutators of L ˆ x and L ˆ y ? As it turns operators tell us. In particular: can we have simultaneous eigenstates of L out, the answer is no, we cannot. We demonstrate this as follows. Let’s assume that there exists a ˆ x and L ˆy, wavefunction φ0 which is simultaneously an eigenstate of L L̂x φ0 = λx φ0 , L̂y φ0 = λy φ0 . (2.11) Letting the first commutator identity of (2.9) act on φ0 we have ˆ xL ˆ y φ0 − L ˆyL ˆ x φ0 ˆ z φ0 = [L ˆ x, L ˆ y ]φ0 = L i~L ˆ x λ y φ0 − L ˆ 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 momentum showing that L algebra we see that they also vanish acting on φ0 and as a result λx and λy must be zero: ˆy, L ˆ z ]φ0 = i~L̂x φ0 = i~ λx φ0 =⇒ λx = 0 , [L | {z } 0 ˆz, L ˆ x ]φ0 = i~Lˆy φ0 = i~ λy φ0 =⇒ λy = 0 . [L } | {z (2.13) 0 ˆ x and L ˆ y has led to L ˆ x φ0 = L ˆ y φ0 = All in all, assuming that φ0 is a simultaneous eigenstate of L L̂z φ0 = 0. The state is annihilated by all angular momentum operators. This trivial situation is not 5 very interesting. We have learned that it is impossible to find states that are nontrivial simultaneous eigenstates 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 we ˆ z as one of the operators we want to measure. Can we now find a second Hermitian operator select L ˆz that commutes with it? The answer is yes. As it turns out, L2 , defined in (1.11) commutes with L and is an interesting choice for a second operator. Indeed, we quickly check ˆ 2 ] = [L ˆz, L ˆz, L ˆ xL ˆ x ] + [L ˆz, L ˆyL ˆy] [L ˆz, L ˆ x ]L ˆx + L ˆ x [L ˆz, L ˆ x ] + [L ˆz, L ˆ y ]L ˆy + L ˆ y [L ˆz, L ˆy] = [L ˆyL ˆ x + i~L ˆ xL ˆ y − i~L ˆ xL ˆ y − i~L ˆ xL ˆy = i~L (2.14) = 0. ˆ z and L ˆ 2 . We will do this shortly. So we should be able to find simultaneous eigenstates of both L 2 ˆ The operator L is Casimir operator, which means that it commutes with all angular momentum ˆ z , it commutes also with L ˆ x and L ˆy. operators. Just like it commutes with L To 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): p x = r sin θ cos φ , r = x2 + y 2 + z 2 ,
(2.15) y = r sin θ sin φ , θ = cos−1 zr ,
−1 y z = r cos θ , φ = tan x . 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. Both ˆ z is simple in spherical rotations about the x and y axes change θ and φ. We can therefore hope that L ˆ z = x̂p̂y − ŷp̂x we have coordinates. Using the definition L   ˆz = ~ x ∂ − y ∂ . (2.16) L ∂x i ∂y Notice that this is related to ∂ ∂φ since, by the chain rule 0 ∂ ∂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 identify ˆz = ~ ∂ . L i ∂φ (2.18) ˆ z generates This is a very simple and useful representation. It confirms the interpretation that L ˆ z is like a momentum along rotations about the z axis, as it has to do with changes of φ. Note that L the “circle” defined by the φ coordinate (φ = φ + 2π). The other angular momentum operators are a bit more complicated. A longer calculation shows what we suggested earlier, that   L̂2 1 ∂ ∂ 1 ∂2 − 2 = sin θ + . (2.19) ~ sin θ ∂θ ∂θ sin2 θ ∂φ2 6 3 Eigenstates of Angular Momentum ˆ z and L2 commute. We now aim to construct We demonstrated before that the Hermitian operators L the simultaneous eigenfunctions of these operators. They will be functions of θ and φ and we will call them ψ`m (θ, φ). The conditions that they be eigenfunctions are m∈R L̂z ψ`m = ~m ψ`m , 2 2 L̂ 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 ~ in ˆ 2 we have an ~2 . ˆ z eigenvalue because angular momentum has units of ~. For the eigenvalue of L the L 2 ˆ Note that we have written the eigenvalue of L as `(` + 1) and for ` real this is always greater than or equal to −1/4. In fact `(` + 1) ranges from zero to infinity as ` ranges from zero to infinity. We can ˆ 2 can’t be negative. For this we first claim that show that the eigenvalues of L
ˆ 2ψ ≥ 0 , ψ, L (3.2) ˆ 2 eigenvalue λ we immediately see that the above and taking ψ
to be a normalized eigenfunction with L gives ψ, λψ = λ ≥ 0, as desired. To prove the above equation we simply expand and use Hermiticity ˆ 2ψ ψ, L
= =


ˆ 2x ψ + ψ, L ˆ 2x ψ + ψ, L ˆ 2x ψ ψ, L


ˆ x ψ, L ˆ xψ + L ˆ y ψ, L ˆyψ + L ˆ z ψ, L ˆzψ ≥ 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 ∂φ This determines the φ dependence of the solution and we write ψ`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 . We will require that ψ`m be uniquely defined as a function of the angles and this requires that1 ψ`m (θ, φ + 2π) = ψ`m (θ, φ) . (3.6) There is no similar condition for θ. The above condition requires that eim(φ+2π) = eimφ → e2πim = 1 . (3.7) This equation implies that m must be an integer: m ∈ Z. (3.8) 1 One may have tried to require that after φ increases by 2π the wavefunction changes sign, but this does not lead to a 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 , gives using our expression (2.19) for L     1 ∂ ∂ 1 ∂2 2 2 sin θ −~ + (3.9) 2 ∂φ2 ψ`m = ~ `(` + 1)ψ`m . sin θ ∂θ ∂θ sin θ We 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 ∂φ2 Using ψ`m = eimφ P`m (θ) we can evaluate the action of arrive at the differential equation sin θ ∂2 ∂φ2 (3.10) on ψ`m and then cancel the overall eimφ to dP m  d sin θ ` − m2 P`m = −`(` + 1)P`m sin2 θ , dθ dθ (3.11) or, equivalently,
dP m  d sin θ ` + `(` + 1) sin2 θ − m2 P`m = 0 . (3.12) dθ dθ We now want to make it clear that we can view P`m as a function of cos θ by writing the differential equation in terms of x = cos θ. Indeed, this gives sin θ d dx d d = = − sin θ dθ dθ dx dx → sin θ d d = −(1 − x2 ) . dθ dx (3.13) The differential equation becomes (1 − x2 )  dP m i  dh (1 − x2 ) ` + `(` + 1)(1 − x2 ) − m2 P`m (x) = 0 , dx dx (3.14) and dividing by 1 − x2 we get the final form: dP m i h m2 i m dh P (x) = 0 . (1 − x2 ) ` + `(` + 1) − dx dx 1 − x2 ` (3.15) The P`m (x) are called the associated Legendre functions. They are not polynomials. All we know at this point is that m is an integer. We will discover soon that ` is a non-negative integer and that for a 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 satisfy dh dP` i (1 − x2 ) + `(` + 1)P` (x) = 0 . dx dx (3.16) This is the Legendre differential equation. We try finding a series solution by writing P` (x) = ∞ X k=0 8 ak xk , (3.17) assuming that P` (x) is regular at x = 0, as it better be. Plugging this into the DE yields we find that the 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 have The 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 ` 2 P` (x) = ` x −1 . (3.21) 2 `! dx The Legendre polynomials have a nice generating function ∞ X P` (x)s` = √ 1 . 1 − 2xs + s2 P1 (x) = x , P2 (x) = `=j (3.22) A few examples are P0 (x) = 1 , 1 2
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). The differential 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  d dx |m| P` (x) . (3.24) Since P` is a polynomial of degree `, the above gives a non-zero answer only for |m| ≤ `. We thus have solutions for − ` ≤ m ≤ `. (3.25) It is possible to prove that no other solutions exist. One can think of the ψ`m eigenfunctions as first determined 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 are s 2` + 1 (` − m)! Y`,m (θ, φ) ≡ (−1)m eimφ P`m (cos θ). (3.26) 4π (` + m)! 9 For m < 0, we use Y`,m (θ, φ) = (−1)m [Y`,−m (θ, φ)]∗ . (3.27) We thus have L̂z Y`m = ~m Y`m , (3.28) L̂2 Y`m = ~2 `(` + 1) Y`m . The first few spherical harmonics are 1 Y0,0 (θ, φ) = √ 4π r (3.29) r 3 ±iφ 3 x ± iy Y1,±1 (θ, φ) = ∓ e sin θ = ∓ 8π 8π r r r 3 3 z Y1,0 (θ, φ) = cos θ = . 4π 4π 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 to make them have unit normalization. The spherical harmonics form an orthonormal set with respect to integration over the solid angle. This integration can be written in many forms: Z Z 2π Z π Z 2π Z 1 dΩ · · · = dφ dθ sin θ · · · = dφ d(cos θ) · · · . (3.32) 0 θ=0 0 −1 The statement that the spherical harmonics form an orthonormal set with respect to this integration means that Z dΩ Y`∗0 ,m0 (θ, φ)Y`,m (θ, φ) = δ`,`0 δm,m0 . (3.33) 4 The Radial Wave Equation Let us now write an ansatz for the solution of the Schrödinger equation. For this we take the product of a purely radial function RE` (r) and a spherical harmonic ψ(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 will see the equation for R does not depend on m. We can now insert this into the Schrödinger equation (1.13) ˆ2 ~2 1 ∂ 2 L − (rR Y ) + RE` Y`m + V (r)RE` Y`m = ERE` Y`m . (4.2) E` `m 2m r ∂r2 2mr2 ˆ 2 eigenstates we can simplify the equation to give Since the spherical harmonics are L − ~2 1 d2 (rRE` ) ~2 `(` + 1) Y + RE` Y`m + V (r)RE` Y`m = ERE` Y`m . `m 2m r dr2 2mr2 (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 dr2 2mr2 10 (4.4) It is now convenient to define uE` (r) ≡ rRE` (r) . (4.5) This allows us to rewrite the entire DE as − ~2 d2 uE`  ~2 `(` + 1)  V (r) + uE` = EuE` . + 2m dr2 2mr2 (4.6) This is called the radial equation. It looks like the familiar time-independent Schrödinger equation in one dimension, but with an effective potential Veff (r) = V (r) + ~2 `(` + 1) , 2mr2 (4.7) that features the original potential supplemented by a centrifugal term, a repulsive potential proportional to the angular momentum squared. Because of this term, the radial equation is slightly different for each value of `. As anticipated, the quantum number m does not appear in the differential equation. The same radial solution uE` (r) must be used for all allowed values of m. Recall our decomposition of the wavefunction: ψ(r, θ, φ) = RE` (r)Y`,m (θ, φ) = uE` (r) Y`,m (θ, φ) . r The normalization condition requires Z Z |uE ` |2 ∗ 3 2 1 = d x |ψ| = r2 drdΩ Y`,m Y`,m . r2 The angular integral gives one, the explicit factors of r cancel and we get Z ∞ dr |uE` |2 = 1. (4.8) (4.9) (4.10) 0 Indeed uE` (r) plays the role of a one-dimensional wavefunction for a particle moving in the effective potential along r. Since only r > 0 is allowed, we must consider the possible behavior of the wavefunction for r = 0. We can learn about the behavior of the radial solution at the origin under the reasonable assumption that the centrifugal barrier dominates the potential as r → 0. In this case the most singular terms of the radial differential equation must cancel each other out, leaving less singular terms that we can ignore in this leading order calculation. So we set: − ~2 d2 uE` ~2 `(` + 1) + uE` = 0 , 2m dr2 2mr2 as r → 0 . (4.11) or equivalently `(` + 1) d2 uE` = uE` . (4.12) 2 dr r2 The solutions of this can be taken to be uE` = rs with s a constant to be determined. We then find s(s − 1) = `(` + 1) → 11 s = ` + 1, s = −` , (4.13) thus leading to two possible behaviors near r = 0: uE` ∼ r`+1 , or uE` ∼ 1 . r` (4.14) For ` > 0 the second behavior is not consistent with normalization, the wavefunction diverges at r → 0 far too quickly. For ` = 0 the second behavior, leading to R ∼ 1/r, is in fact not a solution of the Schrödinger equation. Therefore we have established that for all ` ≥ 0 we must have uE` ∼ c r`+1 , as r → 0 . (4.15) Note that uE` vanishes at r = 0. Even for ` = 0, we have u ∼ r and u vanishes at r = 0. Effectively there is an infinite wall at r = 0 consistent with the impossibility of extending r to negative values. Recall that the full radial dependence of the wavefunction is obtained by dividing uE` by r, so that RE` ∼ c r` . (4.16) This allows for a constant non-zero wavefunction at the origin only for ` = 0. Only for ` = 0 a particle can be at the origin. For ` 6= 0 the angular momentum “barrier” prevents the particle from reaching the origin. Sarah Geller and Andrew Turner transcribed Zwiebach’s handwritten notes to create the first LaTeX version of this document. 12