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Quantum Physics I, Lecture Notes 20-21

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Quantum Physics I, Lecture Notes 20-21 Lectures 20 and 21: Quantum Mechanics in 3D and Central Potentials B. Zwiebach May 3, 2016 Contents 1 Schr¨odingerEquation 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¨odingerEquation 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^ 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 (x) ipa^ a d e ~ (x) = e dx (x) ; (1.2) where we simplified the exponent. Expanding the exponential gives 2 2 3 3 ipa^ d a d a d e ~ (x) = 1 + a +2 +3 + : : : (x) ; dx 2! dx 3! dx (1.3) d a2 d2 a3 d3 = (x) + a + + + ::: = (x + a) ; dx 2! dx2 3! dx3 ipa^ 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 (x + a) is the displacement of (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 x^ and p^ operators are vector operators: ~ ~ @@@ p^ = (p^x; p^y; p^z) = r = ;;: i i @x @y @z (1.4) x^ = (x;^ y^; z^) : The commutation relations are as follows: [ x;^ p^x ] = i~ ; [ y^; p^y ] = i~ ; (1.5) [ z^; p^z ] = i~ : All other commutators are involving the three coordinates and the three momenta are zero! Consider a particle represented by a three-dimensional wavefunction (x; y; z) moving in a three- dimensional potential V (r). The Schr¨odingerequation takes the form 2 − ~ r2 (r) + V (r) (r) = E (r) : (1.6) 2m 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 − ~ r2 (r) + V (r) (r) = E (r) : (1.7) 2m 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 symme- try of the potential we are led to express the Schro¨odingerequation and energy eigenfunctions using spherical coordinates. In spherical coordinates, the Laplacian is 1 @2 1 1 @@ 1 @2 r2 = (r · r) = r + sin θ + : (1.8) r @r2 r2 sin θ @θ @θ sin2 θ @φ2 Therefore the Schr¨odingerequation for a particle in a central potential becomes 2 1 @2 1 1 @ @ 1 @2 − ~ r + sin θ + + V (r) = E : (1.9) 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. The angular dependent piece of the r2 operator can be identified as the magnitude squared of the angular momentum operator 1 @@ 1 @2 L2 sin θ + = − (1.10) sin θ @θ @θ sin2 θ @φ2 ~2 where 2 L = L^xL ^x + L ^yL ^y + L ^zL ^z : (1.11) This will imply that the Schr¨odingerequation becomes 2 1 @2 1 L2 − ~ r− + V (r) = E (1.12) 2m r @r2 r2 ~2 or expanding out 2 1 @2 L2 − ~ (r ) + + V (r) = E : (1.13) 2m r @r2 2mr2 2. Eq. (1.7) is the relevant equation for the two-body problem when the potential satisfies V (r1; r2) = V (jr1 − r2j) ; (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 ; (2.1) Ly = zpx − xpz ; Lz = xpy − ypx : We use the above relations to define the quantum angular momentum operator L^ and its components, the operators (L^x;L ^y;L ^z): L^ = (L^x;L ^y;L ^z) ; L^x = y^p^z − z^p^y ; (2.2) L^y = z^p^x − x^p^z ; L^z = x^p^y − y^p^x : 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 written L^x = p^zy^ − p^yz^, it would have not mattered, it is the same as the L^x above. It is simple to 3 check that the angular momentum operators are Hermitian. Take L^x, for example. Recalling that for any two operators (AB)y = ByAy we have ^ y y y y y y y y (Lx) = (y^p^z − z^p^y) = (y^p^z) − (z^p^y) = p^zy^ − p^yz^ : (2.3) Since all coordinates and momenta are Hermitian operators, we have y (L^x) = p^zy^ − p^yz^ = y^p^z − z^p^y = L^x ; (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 ^y ^ ^y ^ ^y ^ Lx = Lx ;Ly = Ly ;Lz = Lz : (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 compu- tation enables us to see if we can measure them simultaneously. Let us compute the commutator of L^x with L ^y: [L^x;L ^y] = [ y^p^z − z^p^y; z^p^x − x^p^z ] (2.6) We now see that these terms fail to commute only because z^ and p^z fail to commute. In fact the first term of L^x only fails to commute with the first term of L^y. Similarly, the second term of L^x only fails to commute with the second term of L^y. Therefore [L^x;L ^y] = [ y^p^z; z^p^x ] + [z^p^y; x^p^z ] = [ y^p^z; z^]p^x + x^[z^p^y; p^z ] = y^[ p^z; z^] p^x + x^ [z^; p^z] p^y (2.7) = y^(−i~)p^x + x^(i~)p^y = i~(x^p^y − y^p^x) : We now recognize that the operator on the final right hand side is L^z and therefore, ^ ^ ^ [ Lx;Ly ] = i~ Lz : (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 x^ ! y^ ! z^ ! x^ and the momentum operators as in p^x ! p^y ! p^z ! p^x and we will obtain another consistent commutation relation. You can also see that such cycling takes L^x ! L^y ! L ^z ! L ^x, by looking at (2.2). We therefore claim that we do not have to calculate additional angular momentum commutators, and (2.8) leads to ^ ^ ^ [ Lx;Ly ] = i~ Lz ; ^ ^ ^ [ Ly;Lz ] = i~ Lx ; (2.9) ^ ^ ^ [ Lz;Lx ] = i~ Ly : This is the full set of commutators of angular momentum operators. The set is referred to as the algebra of angular momentum. Notice that while the operators L^ were defined in terms of coor- dinates and momenta, the final answer for the commutators do not involve coordinates nor momenta: commutators of angular momenta give angular momenta! The L^ operators are sometimes referred to 4 as orbital angular momentum, to distinguish them from spin angular momentum operators. The spin angular momentum operators S^x;S ^y; and S ^z cannot be written in terms of coordinates and momenta.
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