R 3 ψ 2 d 3 x = 1. Observables are constructed by tripling everything we did before. The position operator ˆr is*

Quantum Mechanics in 3-d Relevant sections in text: Chapter 4 Quantum mechanics in three dimensions Our study of a particle moving in one dimension has been instructive, but it is now time to consider more realistic models of nature by working in three dimensions. It is reasonably straightforward to generalize our model of a particle moving in one dimension to a particle moving in three dimensions. In terms of our postulates, the generalization is as follows. The Hilbert space is the space of square-integrable, complex valued functions of three variables: ψ ψ(x, y, z) ψ(r). The scalar product is φ ψ = φ (r)ψ(r) d 3 x φ (x, y, z)ψ(x, y, z) dx dy dz. R 3 The meaning of the wave function ψ(r) is that ψ(r) 2 is the probability density for position measurements to find the particle at the position r = xi + yj + zk. The probability P (R) for finding the particle in some region R is P (R) = ψ(r) 2 d 3 x. R Normalization of the wave function now means R 3 ψ 2 d 3 x = 1. Observables are constructed by tripling everything we did before. The position operator ˆr is* (ˆrψ)(r) = rψ(r), that is, (ˆxψ)(r) = xψ, (ŷψ)(r) = yψ(r), (ẑψ)(r) = zψ(r). Momentum operators are (ˆpψ)(r) = h i ψ, * Note that we are using the hatˆto denote linear operators not unit vectors! 1

that is, (ˆp x ψ)(r) = h i ψ x, (ˆp yψ)(r) = h i ψ y, (ˆp zψ)(r) = h i ψ z. Let us note that the different components of the position commute, as do the different components of momentum (exercise). Moreover, each canonical pair of coordinates and momenta have the usual canonical commutation relation: [ˆx, ˆp x ] = i hˆ1, [ŷ, ˆp y ] = i hˆ1, [ẑ, ˆp z ] = i hˆ1. Consequently, we can determine all three components of the position of the particle to arbitrary statistical accuracy and likewise for the momentum. Corresponding components of the position and momentum are incompatible. But non-corresponding components are compatible. So, for example, while there is a non-trivial uncertainty relation for x and p x it is possible to determine both x and p y with arbitrary statistical accuracy. so that The energy operator, i.e., the Hamiltonian, is given by Ĥ = 1 2m ˆp2 + ˆV (ˆr, t), h2 (Ĥψ)(r) = 2m 2 ψ(r) + V (r, t)ψ(r), where V (r, t) V (x, y, z, t) is the potential energy function and 2 is the Laplacian. A significant new feature of particle motion three dimensions is that we can now build an operator representative of the angular momentum. Recall that classically the angular momentum is given by L = r p. The quantum operator representative of this triplet of observables is that is, (ˆLψ)(r) = r ˆL = ˆr ˆp, ) ( h i ψ(r) = h i r ψ. More on angular momentum later. A good exercise for you at this point is to verify that the ordering of the position and momentum operators does not matter in the angular momentum. So, for example, we could just as well define the operator as ˆL = ˆp ˆr. The Schrödinger equation in three dimensions The Schrödinger equation, ĤΨ = i h Ψ t 2

in 3-d is a partial differential equation for Ψ = Ψ(r, t): h2 2m 2 Ψ + V (r, t)ψ = i h Ψ t. For time independent potentials, V = V (r), separation of the time variable works as before. With Ψ(r, t) = ψ(r)e ī h Et, we get the time independent Schrödinger equation: Ĥψ = Eψ. As usual, the spectrum of Ĥ may be discrete, continuous, or both. Normalizable solutions of the TISE corresponding to eigenvectors of Ĥ will have discrete eigenvalues E n, and corresponding stationary states Ψ n (r, t). The corresponding stationary states Ψ n evolve in time via the (physically irrelevant) phase factor e ī h Ent. The general solution to the SE is given by a superposition Ψ(r, t) = n c n ψ n (r)e ī h E nt. Solving the TISE for central forces Let us turn to a simple but important class of interactions in which the potential energy function for the particle depends only upon the distance from a given point. Placing our origin at this point we have V = V ( r ) V (r). These potentials correspond to central forces, since the classical force field is given by (exercise) F(r) = V (r) = V (r) r r. Three good examples of central force potentials are the isotropic oscillator V (r) = 1 2 mω2 r 2, the Coulomb or Kepler potential and the Yukawa potential V (r) = k r, V (r) = αe µr. 3

The isotropic harmonic oscillator gets its name from the fact that, for any displacement in a radial direction, the restoring force is proportional to the displacement. In particular, isotropic indicates that the restoring force is independent of (radial) direction of displacement. The Coulomb potential is of course quite familiar to you and is the basis for the simplest model of an atom. The Yukawa potential models the strong, short-ranged attractive force between nucleons. Without this force nuclei would not exist. The nice thing about central potentials is that they allow a straightforward solution of the TISE using spherical coordinates (r, θ, φ) and separation of variables. Let us now investigate this. To begin, the Laplacian in spherical polar coordinates is 2 ψ = 1 ( r 2 r 2 ψ ) ( 1 + r r r 2 sin θ ψ ) 1 + sin θ θ θ r 2 sin 2 θ We use this in the TISE h2 2µ 2 ψ(r, θ, φ) + V (r)ψ(r, θ, φ) = Eψ(r, θ, φ). ( 2 ) ψ φ 2. Note that we use µ to denote the mass of the particle; this will avoid a clash of conventions later. It is not obvious, but this partial differential equation can be solved by separation of variables. So, let us try ψ(r, θ, φ) = R(r)Y (θ, φ). We substitute this into the TISE, and (i) divide both sides by ψ, (ii) multiply both sides by 2µr2 h 2 to find (exercise) 1 (r 2 R ) 2µr 2 ( 1 [V E] = sin θ Y ) R h 2 Y sin θ θ θ 1 Y sin 2 θ 2 Y φ 2. We see the familiar separation of variables result: the left-hand side is a function of r, the right hand side is independent of r. Therefore, each side must equal a constant (exercise), which we denote for later convenience by l(l + 1): 1 (r 2 R ) 2µr 2 ( 1 [V E] = l(l + 1) = sin θ Y ) + R h 2 Y sin θ θ θ 1 Y sin 2 θ 2 Y φ 2. The strategy now is to solve these two equations separately. Although I won t quite prove it, the definition of L as self-adjoint operators will require l to be a non-negative integer. For each choice of that integer there is a solution of the radial equation. Note that it is the radial equation that will be used to find the allowed energies (although the angular equation s influence will be felt through the integer values of l). Of course, to solve the radial equation we need to know what is the potential energy function. 4

Let us begin with the angular equation. Note that it is universal in the sense that it is the same for all choices of the central force potential energy function, and for all possible energies E. You probably have encountered this equation before. Note that if we consider the TISE with E = V = 0, then we are solving the Laplace equation. In this case the radial equation is different, but the angular equation is unchanged. Thus the angular part of the solution to the TISE for central forces is the same as the angular part of solutions to the Laplace equation for an electrostatic potential in a charge free region. When we solve the angular equation we are solving 1 sin θ ( sin θ Y ) 1 θ θ sin 2 θ 2 Y = l(l + 1)Y. φ2 This is the eigenvalue equation for the operator which is the angular part of Laplacian. We shall see that the angular part of the Laplacian represents the observable 1 h 2 L 2, so h l(l + 1) will be the possible values of the magnitude of angular momentum vector. The angular part of the Laplace eigenvalue problem can also be solved by separation of variables. The result is Y (θ, φ) = AP m l (cos θ)e imφ, where l is a non-negative integer, m is any integer l m l, and P m l is the associated Legendre function Pl m (x) = 1 ( ) 2 l l! (1 x2 ) m /2 d m +l (x 2 1) l. dx The associated Legendre functions look a little messy, but they are not so bad. For example (exercises) P 0 0 = 1, P 0 1 (x) = x = P 1(cos θ) = cos θ, P1 1 (x) = P 1 1 (x) = (1 x2 ) 1/2 = P1 1 and so on. See your text for more examples. (cos θ) = P 1 1 (cos θ) = sin θ, It can be shown that the associated Legendre functions Pl 0 (x) form a complete set of orthogonal polynomials (like the Hermite polynomials). This means in particular that these functions form a basis for the vector space of square-integrable functions on the interval [ 1, 1]. These polynomial functions Pl 0 (x) are usually called the Legendre functions. 5

The integer values of l and m are required so that the wave function is suitably nonsingular, single-valued, and define a domain for the angular momentum operators such that they are self-adjoint. The single-valued requirement is ψ(r, θ, 0) = ψ(r, θ, 2π). This single-valuedness requirement might not appear to be necessary a priori (like it would be for electrostatics) since we only require that ψ be square-integrable. However, in order for ψ to be in the domain of self-adjoint operators H and L this restriction is in fact needed. It is also for this reason that l should be an integer. We normalize the solutions of the TISE via 2π π 1 = dr R(r) 2 r 2 dφ dθ Y (θ, φ) 2 sin θ. 0 0 0 It is conventional to normalize the radial and angular parts separately: If we define Y m l = ɛ 2π π 1 = dr R(r) 2 r 2 dr, 1 = dφ dθ Y (θ, φ) 2 sin θ. 0 0 0 (2l + 1)(l m )! e imφ P m 4π(l + m )! l (cos θ), ɛ = ( 1) m, m 0, ɛ = 1, m 0. then the angular functions are normalized. In fact, they form the orthonormal set of functions on the unit sphere called spherical harmonics. See table 4.2 in your text for a list of several spherical harmonics. The orthonormality relation for spherical harmonics is 2π π (Yl m 0 0 ) Yl m sin θ dθ dφ = δ ll δ mm. Exercise: Show that (Yl m ) = ( 1) m Yl m. The radial equation We now have a look at radial equation, (r 2 R ) 2µr 2 h 2 (V E)R = l(l + 1)R. 6

The detailed nature of the solutions will depend upon the form of the potential energy and on the boundary conditions, but we can make some progress without committing to either of these. Make a change of variables u = rr, and you get (exercise) ( ) h2 2µ u + V + h2 l(l + 1) 2µr 2 u = Eu. This is mathematically the same as the 1-d TISE for a particle moving on the positive real line in a potential V eff = V + h2 l(l + 1) 2µr 2. Indeed, our normalization convention for R becomes u 2 dr = 1. 0 Without loss of generality you can assume that u (and hence R) is a real function (exercise). You can think of u as defining the radial position probability distribution for a particle moving in a central force field. The effective potential V eff governing the radial part of the wave function includes the central force potential and the centrifugal potential h2 l(l+1). You can check that the 2µr 2 centrifugal potential corresponds (classically) to a repulsive force. This kind of effective potential also arises in classical mechanics. It is worth spending a moment to recall this. In classical mechanics, a particle moving in a central force field always moves in a plane orthogonal to the conserved angular momentum vector L. If you introduce polar coordinates (r, φ) in the plane, the equations of motion are µ d2 r dt 2 = d ) (V (r) + L2 dr 2µr 2, and d dφ (µr2 dt dt ) = 0. The latter equation is conservation of the magnitude of angular momentum (exercise), where L = µr 2 dφ dt. The former equation governs the radial motion. You see that the effective potential energy governing the radial motion involves a centrifugal term representing the effect of the angular 7

motion on the radial motion. This classical picture suggests that we might identify h 2 l(l+1) with the squared-magnitude of the angular momentum in the quantum description. This is correct, and will be shown later. To proceed further, we need to specify the central force potential energy function. Let us briefly consider a couple of elementary examples, and then have a more detailed look at the Coulomb potential. Example: The free particle in 3-d A free particle certainly has a central force (rather trivially), since V (r) = 0. The radial equation is h2 2µ u + h2 l(l + 1) 2µr 2 u = Eu. This differential equation is a form of the spherical Bessel equation. The solutions, labeled by l, are r times the spherical Bessel functions and spherical Neumann functions of order l. Only the spherical Bessel functions are non-singular as r 0, so we only use them. They are denoted by j l (x). In terms of these special functions, the solution to the radial equation is u = rj l (kr), where can take any non-negative value. k = 1 h 2µE The spherical Bessel functions are oscillatory in functions with decreasing amplitude as the argument (x = kr) increases. They all vanish at r = 0 except the function defined by l = 0 (see below for more on this function). For details on the behavior of the spherical Bessel functions see the text. We will look at the simplest of them in a moment. Keeping in mind that u = rr, we have then the solution to the TISE of the form ψ(r, θ, φ) = Aj l (kr)yl m (θ, φ). Note that these solutions are determined by (i) a choice of E (through k), (ii) a choice of l (non-negative integer) and (iii) a choice of m (an integer such that l m l). Thus there is quite a bit of degeneracy, that is, there are many solutions (idealized stationary states) with the same energy. This corresponds to the fact in classical mechanics that a free particle can move in a variety of ways (same speed, different directions) and still have the same energy (exercise). We saw (in a homework problem) that for normalizable solutions of the TISE in 1-d there can be no degeneracy. Here this result doesn t apply for two reasons. First, the TISE was solved in 3-d. Still, one might argue, the radial equation was equivalent to a TISE in 1-d. We escape that objection by pointing out that (1) the degeneracy does not appear via the radial equation, for each fixed l and E there is only 8

one solution to that equation, (2) the solutions in this example cannot, in any case, be normalizable (since, e.g., the spectrum of H is continuous). We will content ourselves with examining the simplest of the solutions to the TISE. Suppose we choose l = 0. Then the radial TISE is (exercise) h2 2µ u = Eu. This is the familiar harmonic oscillator equation, with solutions u = A cos(kr) + B sin(kr). Since R = u r, we will only get a non-singular solution to the TISE if A = 0. The spherical Bessel function of order 0 is in fact j 0 (kr) = sin(kr), kr in agreement with our direct solution when l = 0. When l = 0 we must have m = 0 (exercise), in which case the angular functions are just a constant so the corresponding solution of the TISE with energy E is of the form ψ k,l=0=m = A sin(kr). kr Thus the probability distribution is spherically symmetric, and oscillatory with amplitude decreasing with increasing radius. The frequency of the oscillation increases with increasing energy. You can check that this solution is not normalizable, as expected (exercise). It is interesting to think about what the particle is actually doing in this family of stationary states, especially in comparison to the classical mechanics behavior. What do we know? Well, we have a free particle with some statistically definite energy and vanishing angular momentum (relative to the origin). In classical mechanics with that information, since L = r p, the particle is either at rest (in which case the energy vanishes) or its momentum is parallel to r. Just given the energy and vanishing of angular momentum, any radial motion at constant speed could occur. Of course, we know that in classical mechanics the particle traces out a definite path from one of the possibilities. We could in principle use additional measurements in position and momentum to prepare the particle in a state where it follows a single radial trajectory with the given energy. In the quantum state you can imagine similar possibilities for motion given the energy and angular momentum. But you should keep in mind that the state is already completely determined by the specification of energy and vanishing angular momentum, so the particle cannot really be said to take this or that radial path.* Indeed, you have to keep in mind the uncertainty * Notice that the wave function is spherically symmetric, so perhaps it is not surprising to learn that all directions of motion are equally likely. 9

relation between position and momentum which prohibits the classical path of motion to be determined with statistical certainty. Indeed, in this (idealized) stationary state the angular momentum is determined with certainty, which implies maximum uncertainty in angular position. The angular momentum and linear momentum are also incompatible, leading to maximum uncertainty in the direction of motion. So, while the particle is in some sense undergoing radial motion, the position and direction of motion are not determined! One can also prove that in this state (with proper care about its non-normalizability) the radial motion is equally likely to be toward the origin as away from it. One way to check this is to compute the integrand for, say, the expectation value for the x component of momentum. You can check that this integrand is an odd function of x. Consequently any symmetric integral over an arbitrarily large (but finite) region of this integrand will vanish. One can conlude that the expectation value of momentum is zero, even though the magnitude of the momentum need not be zero. This is consistent with equal probability for motion toward or away from the origin. Finally, I remind you that this idealized state is an eigenfunction of energy and hence determines an idealized stationary state. In contrast to the classical motion, no observable quantities for the particle will change in time in this state. The probability current vanishes for this wave function. Particle in a spherical box Let us now consider the stationary states of a particle confined to a region r < a. The potential energy function is that of an infinite spherical well. We can view the free particle of the last section as a limiting case a. We set ψ(r, θ, φ) = 0 when r > a. Inside the well, the radial equation is exactly as before. The only difference now is that we must take account of the boundary condition that R(a) = 0. In general, this leads to a transcendental equation involving the spherical Bessel functions: j l (ka) = 0. The zeros of the spherical Bessel functions are known; they form a discrete set. Thus k is determined by these discrete values. In this way we get a discrete set of allowed energies, as expected. Let us revisit the l = 0 case. We found that R(r) = A sin(kr) kr solves the radial equation (for r < a). To impose the boundary condition is now easy (exercise) k = nπ, n = 1, 2,.... a 10

(Exercise: why do we exclude n = 0, 1, 2,...?). The allowed energies are Quantum Mechanics in 3-d E n,l=0 = n2 π 2 h 2 2µa 2, which are identical to the 1-d particle in a box! This is really not so surprising. By ignoring angular motion effects (l = 0), we have essentially reduced the problem to a 1-d problem (exercise). The normalized stationary states that have this energy are (exercise) ψ n,l=0 = 1 1 2πa r sin(nπr a ). Keep in mind that these states and their energies correspond to choosing l = 0. For each l 0 there will also be a sequence of allowed energies; these energies will depend upon the choice of l and another integer labeling the zeros of the spherical Bessel function. This integer is usually called the principal quantum number, n. When l 0, it is also possible to have stationary states with m 0. Because m does not enter the radial equation, it will play no role in determining the energy. This is quite general: associated with each l there will be 2l+1 states, corresponding to m = l, l+1,..., l 1, l (exercise), all having the same energy. One says that a stationary state with a given value of l is (2l + 1)-fold degenerate. This degeneracy reflects the rotational invariance of the system. Hydrogen atom The simplest successful model of a hydrogen atom is the quantum mechanical system associated with the central force potential V (r) = e2 4πɛ 0 1 r. Here e is the charge on the electron, r is the value of the observable interpreted as distance from proton to electron, and ɛ 0 is the permittivity of the vacuum, familiar from eletrostatics. The Coulomb potential being used here is everywhere negative. It diverges as r 0, and it vanishes as r. The behavior as r 0 suggests, at least classically, that the system is not stable since the energy is unbounded from below. The behavior as r means that positive energy, unbound (scattering) states can exist. This is good, since our model should be able to describe ionized states and/or scattering states. We shall restrict our attention to the bound states, so as to understand the famous success of quantum mechanics in predicting the spectrum of hydrogen. The TISE for the hydrogen atom separates into the angular equations, which we solve using spherical harmonics, and the radial equation: [ ] h2 2µ u + e2 1 4πɛ 0 r + h2 l(l + 1) 2µ r 2 u = Eu. 11

The effective potential energy for the radial equation includes a centrifugal barrier when l 0. Recall that a similar term arises in classical mechanics and guarantees that bound states are kept away from r = 0. When l = 0 the classical particle may reach r = 0. We shall see how these features fare in the quantum domain. In the radial equation above, the mass µ can be interpreted as the reduced mass µ = m em p m e + m p, built from the electron mass m e and the proton mass m p. Because m p >> m e we have µ m e (exercise). We seek solutions to the radial equation that are well behaved everywhere and that vanish as r, so that the solutions define normalizable functions, i.e., bound states. As usual, we must also determine the allowed values of the energy E; these energy eigenvalues will form a discrete spectrum since we are looking for normalizable solutions. The text provides all the gory details; we shall be content with understanding the results. Energy spectrum To begin, the allowed energies are labeled by a positive definite integer n = 1, 2,.... This integer is called the principal quantum number. In terms of it the energy eigenvalues are { ( µ e 2 ) 2 } 1 E n = 2 h 2 4πɛ 0 n 2. We can write this as (exercise) where E n = e2 8πɛ 0 n 2 a, a = 4 h2 πɛ 0 µe 2 0.53 10 11 m. The quantity a, which has dimensions of length, is the Bohr radius. It gets its name because, in the Bohr model of the atom, this is the radius a classical electron must maintain to be in the ground state. Indeed, you can see that when n = 1 the atom is in its lowest allowed energy state and the resulting energy is precisely that of a classical particle in a circular orbit at the Bohr radius (nice exercise). More generally, the energies for any n are the same you would get for a classical particle at the radius na. This is why the Bohr model was successful for the hydrogen atom. The Bohr model is now known to be physically incorrect as a model of nature since (a) it fails to describe any other atom, and (b) it does not provide a satisfactory means for describing any phenomenon besides the energy levels of hydrogen. The energy, expressed in terms of the principal quantum 12

number and the Bohr radius is, nevertheless, often called the Bohr formula for the hydrogen energies. I emphasize that this formula was obtained by Bohr by using a voodoo blend of classical mechanics and preliminary ideas from the not yet formed quantum theory. It is unfortunate that modern textbooks propagate this intermediate step (Bohr theory) on the road to quantum theory as if it were a viable model of nature*. The ground state, when n = 1, has energy E 1 = 13.6eV. The energy is negative, positive energies correspond to unbound (scattering states). This ground state energy, therefore, is the energy needed to ionize the atom, in good agreement with experiment. For this reason, E 1 is often called the binding energy of the atom in its ground state. Excited states have, of course, greater energy and hence less binding energy. Stationary states The stationary states are, of course, labeled by n. But they also depend upon l and m. One can anticipate a relation between l and n since l appears in the radial equation. What one finds when solving the radial equation and imposing boundary conditions is that, for each value of n, one can have any value of l between 0 and n 1, inclusive: l = 0, 1, 2,... n 1. Keep in mind also that for each l there are the 2l + 1 allowed values of m. Thus, the stationary states are labeled as follows. Pick any n > 0 this determines the energy level ; pick any integer l such that 0 l n 1; pick any integer m such that l m l. As we shall see, the choices of l and m correspond to specifying the angular momentum of the stationary state. With this algorithm in mind, the orthonormal basis of stationary states is, in all of its glory: ( ) ψ nlm (r, θ, φ) = N nlm e r 2r l na L 2l+1 na n l 1 (2r/na)Y l m (θ, φ). Here N nlm is a normalization constant (see the text for a formula) and L s r is the associated Laguerre polynomial, defined by ) L s r(x) = ( 1) (e s ds dx s x dr+s dx r+s (e x x r+s ). Note that this special function appears in ψ nlm with x = 2r na. * In my mind, teaching the Bohr model as fact is analogous to using the Ptolemaic model of the universe (geocentric, epicycles and all that) as a viable model of the universe just because it appeared on the road from the theory in which heavenly bodies were gods to the Copernican theory of the solar system. 13

The radial dependence of the stationary states is that of a polynomial in r times a decaying exponential e r na in r. Here are a few of the associated Laguerre polynomials: L 0 0 (x) = 1, L0 1 (x) = 1 x, L2 2 (x) = 12x2 96x + 144; See your text for some more. Such functions polynomials times decaying exponentials are normalizable because of the decaying exponential. Note that the exponential decay of the wave function is characterized by n times the Bohr radius, so we can say that measurements of the size of the atom (via observables that characterize the position probability distribution) will yield values on the order of magnitude of na. One amusing feature of the stationary states stems from the fact that the states with l = 0 have wave functions that do not vanish at the origin (exercise), since R(0) 0. For example, in its ground state we have ψ n=1,l=0,m=0 = 1 πa 3 e r/a. When l 0 it is not hard to see that R(0) = 0 (exercise). Does this means that the atom has (in states with l = 0) a non-zero probability for finding the electron occupying the same space as the proton? Let s see. The probability P (ɛ) for finding a ground state electron in, say, a spherical region of radius ɛ about the origin is 2π π ɛ [ 1 ( P (ɛ) = dφ dθ dr 0 0 0 πa 3 e r/a 2 r 2 ɛ ) 2 ( ] ɛ sin(θ) = 1 + 2 2 1 e a a) 2ɛ/a As ɛ 0 this probability vanishes. On the other hand, there is a real sense in which the nucleus (here, a proton) has a finite size (on the order of Fermis). From the preceding computation we can expect that the probability for finding the electron at the surface of the proton is non-zero, and by continuity we expect that probability to remain nonzero inside the proton. Explicit computations with a finite-sized charge density confirm this. Thus, figuratively speaking, it is possible that the electron is inside the proton! Again we see the tension between the quantum meaning of the word particle and the classical, macroscopic meaning. Because the nucleus has a finite size, the overlap between the electron wave function and the nuclear charge density indicates that we could improve our description of the atom by modifying the potential energy function (1/r outside the proton, r 2 inside exercise). This results in a small change in the l = 0 energy spectrum (thus partially lifting some degeneracy (see below)). For a hydrogen atom this effect is very small, on the order of 10 9 ev. For hydrogenic atoms with larger atomic number, the effect is more pronounced. This effect has been measured experimentally. More on the spectrum of hydrogen There is a good bit of degeneracy in the energies of hydrogen. Recall that an eigenvalue is called p-fold degenerate if there are p linearly independent eigenvectors with that 14

eigenvalue. Physically, this means that the outcome of a measurement corresponding to that eigenvalue can occur with probability one via a p-dimensional set of states. For the hydrogen atom, the energy is only determined by n. For a given n, there are n possible l values (exercise). For each l there are 2l+1 possible m values. All together, for each energy one has (exercise) n 2 different (i.e., linearly independent) energy eigenfunctions obtained by varying l and m in their allowed ranges. The freedom to vary m reflects the rotational symmetry of the problem: by rotating the atom we can obtain new states with the same energy. The freedom to vary l is a special feature of the Coulomb potential and reflects a hidden symmetry in the Coulomb (and Kepler) dynamical motion. Classically, this hidden symmetry is responsible for the fact that the motion of a particle in a r 1 central 2 force field has bounded orbits that are closed (in fact, ellipses).* One can probe the energy levels of hydrogen experimentally by adding energy to a hydrogen atom and looking at the electromagnetic radiation that comes out when the atom decays. For example, one can pass current through a container containing hydrogen gas. This has the effect of transferring energy to and from to the atom. The atom absorbs energy from the perturbation and moves to an excited state. The perturbation will then cause the atom to emit energy (via one or more photons) as it makes a transition to a state of lower energy. This process is called stimulated emission. This picture of stimulated emission is understood by looking at the time dependent Schrödinger equation. We will develop techniques for understanding stimulated emission via the SE next semester. (Why the atom decays at all in the absence of a perturbation (spontaneous emission) is another story... wait until next semester.) The following ideas will be given a full quantum mechanical treatment next semester. The bottom line is as follows. The atom will absorb any energy equal to the difference between any two allowed energies. The effect is to raise that atom into a higher energy state. After a certain amount of time, the atom will decay to a lower energy state by emission of a photon. The energy of this photon is the energy difference between the stationary states before and after the emission of the photon. Typically we measure the energy of the photons by using a spectrometer, which discerns the wavelength λ of the photon. The relation between the energy level difference 1 1 E = ( n 2 final n 2 )( 13.6 ev ), initial and the photon wavelength is (c is the speed of light) E = 2πc h λ. * You may have encountered the Laplace-Runge-Lenz vector in the study of the 2-body inverse-square central force problem in classical mechanics. This conserved vector, like all conservation laws, comes from a symmetry and it is this symmetry which is responsible for the degeneracy of energies in hydrogen associated to varying l for a given n. 15

Transitions from excited states to the ground state (n final = 1) give off the most energy; the emitted light is ultraviolet. Transitions to the first excited state yield visible light. Transitions down to the second excited state give off infrared radiation, etc. The wavelengths predicted by quantum mechanics are in very good agreement with experiment. There are differences, however, since there are other physical effects not accounted for in this simplest of models of the hydrogen atom. Angular momentum So far we have focused on the stationary states of a particle moving in a central force field. In particular, we have focused on the energies of the hydrogen atom. Another very important observable that one can study for a particle moving in three dimensions is its angular momentum. We shall spend a fair amount of time talking about this observable. Recall that we defined the operator representative of angular momentum via ˆL = ˆr ˆp. What this means is that, on a suitably well-behaved wave function ψ, the linear operation is ˆLψ = h i r ψ. More explicitly, L x = h i (y z z y ), L y = h i (z x x z ), L z = h i (x y y x ), Note that once you know the first of these formulas the other two follow by cyclic permutations (exercise). Probably the most important aspects of angular momentum are (1) it is conserved by central forces (more generally, we believe that the total angular momentum of a closed system is always conserved); (2) the 3 components of L are incompatible, that is, they do not commute. A direct computation (see the text) shows that [L x, L y ] = i hl z, [L y, L z ] = i hl x, [L z, L x ] = i hl y. 16

These are the very important angular momentum commutation relations, or angular momentum algebra. Note again that any two formulas follow from a third by cyclic permutations. If you have ever studied the Lie algebra of infinitesimal rotations, you will notice a similarity with the angular momentum algebra. This is no accident, the linear transformation defined by the angular momentum operators can be viewed as the transformation of a state under an infinitesimal rotation. Another way to compute the angular momentum algebra is outlined in one of your homework problems. The idea is to find out how the commutators of components of L with components of r and p look, and then use the fact about commutators (nice exercise) [A, BC] = [A, B]C + B[A, C]. (The order of the operators is crucial for the formula to work.) Because angular momentum is conserved for central potentials, the angular momentum is a useful observable for describing states and their dynamical evolution in this setting. The incompatibility of the components of L means that angular momentum is going to have some surprising properties relative to its classical behavior (recall that position and momentum are also incompatible, hence the uncertainty principle, etc.) In particular, we will find that the possible values of any component of angular momentum do not form a continuum but rather a discrete set. The spectra of the angular momentum operators (L x, L y, L z ) are discrete. One says that angular momentum is quantized. Because of the incompatibility of (L x, L y, L z ), it is generally impossible to have a particle in a state in which two or more components of L are known with certainty, unless the angular momentum is in fact zero. To see this, simply apply our general form of the uncertainty principle to, say, L x and L y to find (exercise) σ Lx σ Ly h 2 L z. As long as the state is such that L z 0, the variances σ Lx and σ Ly cannot vanish, as they should if one knows either of these observables with probability unity. Indeed, suppose that the state ψ is an eigenvector of L x and L y, then it is easy to see that (exercise) [L x, L y ] ψ = 0. But, by the angular momentum commutation relations, this implies L z ψ = 0. So, in fact, the state must be a state of definite value for all three components. By using the other two commutation relations you can show that (exercise) L x ψ = 0 = L y ψ, 17

so that the angular momentum is zero with probability unity in such a state. Quantum Mechanics in 3-d Each of the components (L x, L y, L z ) has its basis of eigenvectors, where the value of that observable is known with probability one. But because the three angular momentum operators do not commute, these basis are not the same. If the particle is in a state in which the angular momentum component, say, L x is known to be (non-zero) with certainty, then in general there will be statistical uncertainty in the values of L y and L z. When working in a basis of eigenvectors of angular momentum the component which is determined with certainty is conventionally chosen to be L z. There is nothing special about the z-axis, since we can call any direction in space the z direction. Classically we are used to thinking of this observable as a vector. Quantum mechanically, there are some vectorial aspects to angular momentum (e.g., it has three components), but it is impossible to say with certainty which way the vector points. As it turns out, the length of the angular momentum vector and any one component (e.g., L z ) can be determined with statistical certainty. This is because the operator representing the length-squared, L 2 = L 2 x + L 2 y + L 2 z, commutes with any component of L. For example (exercise), [L z, L 2 ] = [L z, L 2 x + L 2 y + L 2 z] = [L z, L x ]L x + L x [L z, L x ] + [L z, L y ]L y + L y [L z, L y ] = i h ( L y L x + L x L y L x L y L y L x ) = 0. Recall that commuting Hermitian operators always admit a simultaneous basis of eignvectors. Thus it is possible to find a basis of vectors that are eigenvectors of both one component (conventionally taken to be L z ) and L 2. The states defined by these angular momentum eigenvectors are states in which the value of L z and L 2 are known with probability unity. Our task now is to determine these eigenvectors. We will see that they correspond to the spherical harmonics. Angular momentum eigenvalues and eigenfunctions It is possible to systematically deduce the eigenvalues of L 2 and L z just using the angular momentum commutator algebra. This is logically the same game we played when finding the spectrum of the harmonic oscillator Hamiltonian using raising and lowering ( ladder ) operators. See the text for details. The result is that the operator L 2 has eigenvalues h 2 l(l + 1), where l = 0, 1, 2,.... Given the value l, any given component, say L z, has eigenvalues m h, where m = l, l + 1,..., l 1, l. This is true for L x and L y, too. The catch is that the eigenfunctions for each component are not, in general, the 18

same. We can pick any one component, find its eigenfunctions, and arrange it so that these eigenfunctions are also the eigenfunction of L 2. It turns out that the simultaneous eigenfunctions of L z and L 2 are given by any function of r times the spherical harmonics Y m l. Let us see a little bit more how this works. If we write the angular momentum operators L and L 2 in terms of spherical polar coordinates we find (via a nice, but long exercise) L x = h i L y = h i ( sin φ ) cos φ cot θ. θ φ ( cos φ ) sin φ cot θ. θ φ L z = h i You can see why one likes to use the component L z ; the spherical polar coordinates are adapted to the z and L z takes a very simple form. You really should try to at least verify the formula for L z. φ. It is easy to see that eigenfunctions of L z are any function of r and θ times e imφ : ψ m (r, θ, φ) = F (r, θ)e imφ, m = 0, ±1, ±2,..., Here m is any integer. The restriction to integer values is so that the eigenfunction is a well-defined function in space. You can see that there is a lot of degeneracy here. Since F is arbitrary, there are many, many functions that have the same eigenvalue m h for L z. We can narrow down the possibilities by demanding that ψ m also be an eigenfunction of L 2. The operator L 2 is a little more complicated to write out in spherical polar coordinates. Up to a factor of r 2, it turns out to be the angular part of the Laplacian: L 2 ψ = h 2 [ 1 sin θ ( sin θ ψ ) + 1 θ θ sin 2 θ ( 2 )] ψ φ 2. This differential operator is the restriction of the Laplace operator to a (unit) sphere (exercise). To find the eigenfunctions is a story we have already touched upon when we solved the TISE for central forces via separation of variables (exercise). Indeed, we recover our results from that analysis. In particular, we have that (i) the eigenvalues of L 2 are l(l + 1) h 2, where l = 0, 1,...; (ii) for a given choice of l, the eigenvalues of L z are restricted by m = l, l + 1,..., l 1, l; (iii) the eigenfunctions ψ l,m (r, θ, φ) for a given choice of l and m are built from the spherical harmonics Y m l : ψ l,m (r, θ, φ) = f(r)yl m (θ, φ). 19

We see that there is still some degeneracy owing to the appearance of the arbitrary function f(r). Thus, many different states correspond to the same values of L 2 and L z. These states all differ by the choice of f(r). To uniquely specify the states one must find another observable which commutes with L z and L 2 and demand that ψ l,m also be an eigenfunction of that operator with a specific eigenvalue. For example, in the central force problems it turns out (see homework) that [H, L] = [ p2 + V (r), L] = 0. 2m This fact is equivalent to the statement that the Hamiltonian operator is rotationally invariant. As you will see in a homework problem, the fact that H commutes with L means that angular momentum is conserved (probabilities for outcomes of momentum measurements will not change in time.) This is important in the present discussion because it means that it is possible to further specialize the eigenfunctions of angular momentum to also be energy eigenfunctions. How is this done? Well, you have already done it! The stationary states in a central force problem are radial functions times the spherical harmonics, as above. The radial functions are determined by the radial equation, which also determines the allowed energies. Thus f(r) can be determined by a choice of energy. Indeed, the stationary states of the hydrogen atom (more generally, any central force system) are simultaneously (i) energy eigenfunctions, (ii) eigenfunctions of a component (L z ) of angular momentum and (iii) eigenfunctions of the total angular momentum L 2. We now see the physical significance of l and m. Infinitesimal translations and rotations There is a nice (if a little sophisticated) way to think of momentum-like quantities in classical and quantum mechanics, namely, as generators of transformations. In fact, linear momentum generates translations while angular momentum generates rotations. Let us briefly explore this. We begin with translations. Let ψ(r) describe the state of a system. How does the value of ψ change as we translate from r to the point r + a? Simple: ψ(r) ψ(r + a). Now, we can built any translation by a sequence of many small translations. Suppose that the magnitude a is very small (infinitesimal), then a standard result from calculus is that ψ(r + a) ψ(r) + a ψ(r) = ψ(r) + ī h a (ˆpψ)(r). We say that the momentum is the infinitesimal generator of translations, since a finite translation can be built by many infinitesimal translations associated to p as above. In 20

general, the infinitesimal change in a wave function associated with a translation along the infinitesimal vector a is given by ī h a pψ(r). A similar result is available for angular momentum and infinitesimal rotations. Every rotation is a rotation by some angle about some axis. For a given rotation, we can choose our z axis to be along the axis of rotation. In these coordinates a rotation by an angle φ about z is (exercise) x x = x cos φ y sin φ, y y = y cos φ + x sin φ, z z. Again, we can think of any rotation as the net result of a sequence of many small rotations. Suppose φ << 1, then to first order in φ (exercise) x x φy, y y + φx. Consider a function ψ(r), how does it change under an infinitesimal rotation? We have (exercise) ψ(x, y, z ψ(x, y, z) ψ(x, y, z) ) ψ(x, y, z) φy + φx. x y We can write this as Better yet, we can write ψ(x, y, z ) ψ(x, y, z) + ī h φ(ˆl z ψ)(x, y, z). ψ(x, y, z ) ψ(x, y, z) + ī h φn (ˆLψ)(x, y, z), where n is a unit vector along the rotation axis. We say that angular momentum is the infinitesimal generator of rotations since n L gives the infinitesimal change in the wave function due to a rotation about the axis along n. A finite rotation can be built by many infinitesimal rotations, generated by L. Physically, you can think of the translations and rotations as transformations of the physical system (here a particle). Thus the operators that represent momentum and angular momentum give the infinitesimal change in the state of the system as one performs the indicated transformation (translation and/or rotation) of the system. Exercise: What operator is the infinitesimal generator of time translation of a given system? 21

Spin I am assuming you have had an introduction to the phenomenology of intrinsic spin in an earlier class in modern physics. Here I just want to introduce you to the quantum mechanical basic formalism and a couple of nice results. Spin is an intrinsic property of an elementary particle, like its mass and charge. Spin is a form of angular momentum conservation of angular momentum requires the spin to be taken into account. Charged particles with spin, e.g., the electron, have an intrinsic magnetic moment which is proportional to the spin vector, but neutral particles can have spin, e.g., the neutrino. Unlike mass and charge, which only take a single value for a given particle, the spin can be in various states. More precisely, the magnitude of the spin vector is fixed for a given type of particle unlike the orbital angular momentum whose magnitude can vary while any given component of that spin vector can take various discrete values. Here I will focus on spin-1/2, which pertains to the most significant matter in the universe, e.g., particles such as electrons, protons, neutrons, positrons, neutrinos, quarks, etc. The observable of interest is the spin vector, which represents angular momentum carried intrinsically by the particle. Like orbital angular momentum it can be represented by 3 Hermitian operators, S = (S x, S y, S z ) satisfying the angular momentum commutation relations: [S x, S y ] = i hs z, [S y, S z ] = i hs x, [S z, S x ] = i hs y. As mentioned when we studied the orbital angular momentum, these relationships can be used to deduce the possible eigenvalues and eigenvectors for any given component. It is not too hard to do, but we don t have time for it. Let me just state the main results. First of all, while each component will have a basis of eigenvectors, the bases will be different for different components. The magnitude squared of the spin, is compatible with any component: S 2 = S 2 x + S 2 y + S 2 z [S 2, S x ] = [S 2, S y ] = [S 2, S z ] = 0. Consequently, we can find a basis of eigenvectors of S 2 and any component conventionally we work with eigenvectors of S z. It can be shown from the commutator algebra that the eigenvalues of S 2 are given by s(s + 1) h 2, s = 0, 1 2, 1, 3 2, 2,... where it is understood that for a given type of particle the s value is fixed. Conventionally, it is the s-value which is used to characterize the spin. The spin 1/2 particles are those 22

with s = 1/2. It is pedagogically unfortunate that the particles with spin s actually have their total angular momentum taking the value s(s + 1) h. The eigenvalues of any one component, say S z, are given by m s h, m s = s, s + 1,... s 1, s. This is all very much like orbital angular momentum, of course. But there are two key differences to keep in mind. First, the quantum number s controlling the total spin value is not subject to change with the change of state of the particle; it is fixed by the type of particle (as is the mass and electric charge). The electron may have different l values in an atom, but its s value is always 1/2. This means that every state has the value of S 2 determined with certainty, and the operator representing S 2 must take the form Ŝ 2 = s(s + 1) h 2ˆ1, where s is fixed and 1 is the identity operator. Second, the orbital angular momentum operators require that only integer values of l (and m l ) can occur. This stems from the fact that L = r p with the momentum acting via the gradient on wave functions. The integer values of l are needed so the wave functions are sufficiently well-defined for L to act on them. The spin operators are not constructed as r p and do not have that restriction; half integral values of s (and m s ) may occur for elementary particles. Halfintegral spin particles of a given type (e.g., electrons) obey Fermi-Dirac statistics and are called fermions. Integral spin particles of a given type (e.g., photons) obey Bose-Einstein statistics and are called bosons. Even though the spin (s) is fixed for a given type of particle, if s 0 the value of m s can vary so that the particle can have more than one spin state. Of course the particle can also have a variety of states of motion, but we will ignore that for now and just consider the spin states as if that is all the particle can ever do. This is for pedagogical reasons, but there are physical situations where the spin state is pretty much all one is interested in. For example it is the spin state of the electron and proton in interstellar hydrogen which explains the famous 21 centimeter hydrogen emission line, and it is the spin state of electrons in a lattice which is used to explain ferromagnetic phase transitions in crystalline solids. Spin 1/2 Let s now specialize to the consideration of spin 1/2, i.e., particles with s = 1/2. This means we are studying the spin properties of particles like the electron. The magnitude of the spin for an electron is therefore 3 h/2. The m s values are ± 1 2, so that the component of the spin vector along any direction takes only two values, ± h/2. Therefore the basis of 23

eigenvectors of any of the spin components will have only 2 elements and the space of spin states is two-dimensional.* Because the space of states of a spin 1/2 particle is 2-dimensional, we can fruitfully use a matrix representation of vectors and operators. The vectors representing states are then column vectors with two components: ( ) ψ1 ψ =. ψ 2 The scalar product between two vectors is the usual one from matrix algebra, but keep in mind there is a complex conjugation: ( ) φ ψ = (φ 1 φ 2 ) ψ1 = φ ψ 1 ψ 1 + φ 2 ψ 2. 2 The normalization condition for ψ is then S x = h 2 1 = ψ ψ = ψ 1 2 + ψ 2 2. The spin operators can be constructed as h/2 times the Pauli spin matrices: ( ( ( ) 0 1 0 i 1 0. 1 0 i 0 0 1 ), S y = h 2 ), S z = h 2 It is straightforward, if a bit tedious (unless you use a computer) to check that these matrices do satisfy the angular momentum commutation relations. It is also straightfoward to check that S 2 = Sx 2 + Sy 2 + Sz 2 = 3 ( ) 1 0 4 h2, 0 1 as it should. Notice that the basis is a basis of eigenvectors of S z : S z, + ( ) ( ) 1 0, S 0 z, 1 S z S z, ± = ± h 2 S z, ±. It is straightforward to show that the other spin components have the following column vectors representing their eigenvectors: S x, ± = 1 ( ) 1, S 2 ±1 y, ± = 1 ( ) 1. 2 ±i * Thus we have an example of a 2 state system. This does not mean there are only two states there are infinitely many vectors in a 2-d vector space. What it does mean is that any observable can have (at most) 2 eigenvalues and 2 eigenvectors, so there are 2 possible outcomes of a measurement and two corresponding states. 24