Hydrogen and the Central Force Problem

Home , Constant of motion, Harmonic oscillator

Our treatment of quantum mechanics is getting increasingly more realistic. In the last chapter, we made the leap from 1 to 3 dimensions. In this chapter we solve our ﬁrst problem with a physically realistic potential– the attractive electrostatic (Coulomb) potential between two oppositely charged particles. In the case of atomic hydrogen these particles are an electron and a proton, but the theory can be easily extended to one electron ions such as Li++, muonic atoms where a muon is captured in orbit about a proton, or positronium (an electron in orbit about an antielectron). In this chapter we will ignore the complications of electron spin.

Heretofore we have been discussing the motion of a single particle in an “external” potential of mysterious origin. Because the proton is about 2000 times more massive than the electron and doesn’t move very much, we could think of it as providing an external potential to the “orbiting” electron and still get fairly accurate results. However positronium gives us a great excuse to discuss two particle wave functions which interact in a potential mutually created by both particles.

The Two Particle Reduction

Lets consider two particles in one dimension. The ﬁrst particle has mass m1

and is located at x1 ; the second has mass m2 and is located at x2. The wave function becomes a function of both x1 and x2 : ψ(x1,x2). It is natural to generalize 2 the probabilistic interpretation of the wave function PDF(x1,x2)= |ψ(x1,x2)| , where PDF(x1,x2) gives the probability density of finding particle 1 at position x1 and simultaneously finding particle 2 at position x2. To clarify this last point, the probability of finding particle 2 within a distance ∆ of particle 1 would be written as:

+∞ x1+∆ ∗ P(|x1 − x2| < ∆) = dx1 dx2 ψ (x1,x2)ψ(x1,x2)

−∞ x1−∆

1 Thewavefunctionψ(x1,x2)obeysaSchr¨odinger equation which is built on 2 2 the total energy of the two particle system: E = p1/(2m1)+p1/(2m1)+V (x1,x2).

2 2 −¯h ∂2 −¯h ∂2 2 ψ(x1,x2)+ 2 ψ(x1,x2)+V (x1,x2) ψ(x1,x2)=Eψ(x1,x2)(1) 2m1 ∂x1 2m2 ∂x2

We thus have a partial diﬀerential equation which is traditionally solved using separation of variables. The separation of variables would be trivial if

V (x1,x2)=V1(x1)+V2(x2) since we just have a diﬃQ in x1 added to a diﬃQ in x2. This would be the situation if particle 1 and 2 moved in external potentials but felt no mutual interaction. In the case of positronium, we usually have the opposite problem of zero external potential but the electron and positron (antielectron) attract each other with a potential related to their separation.

The one dimensional analogy would be V (x1 −x2). It is very diﬃcult to separate variables for this potential in the original (x1,x2) coordinates but there is a classic change of variables trick which allows for a separation of variables solutions.

The clue comes from classical mechanics. In many classical situations motion about center of mass is essentially decoupled from motion of the center of mass. An example would be if one were to throw a system consising of two masses coupled with a spring from the surface of the moon. The masses will oscillate following simple harmonic motion about their center of gravity; while the center of mass follows the same parabolic trajectory as a point like particle in the gravitational ﬁeld of the moon. Guided by the classical problem, we tranform the variables of Eq. (1) into the position of the center of mass X and the position of x1 relative to x2:

m1 m2 X = x1 + x2 where M = m1 + m2 M M

x = x1 − x2 (2)

2 We next transform the derivatives using the chain rule.

∂ ∂X ∂ ∂x ∂ ∂ m1 ∂ ∂ = + , = + ∂x1 ∂x1 ∂X ∂x1 ∂x ∂x1 M ∂X ∂x

∂ ∂X ∂ ∂x ∂ ∂ m2 ∂ ∂ = + , = − (3) ∂x2 ∂x2 ∂X ∂x2 ∂x ∂x2 M ∂X ∂x

2 2 The kinetic energy operator Tˇ = p1/(2m1)+p2/(2m2) can be written in terms of the derivatives of Eq. (3)

2 −¯h m1 ∂ ∂ m1 ∂ ∂ Tˇ = + + 2m1 M ∂X ∂x M ∂X ∂x

2 −¯h m2 ∂ ∂ m2 ∂ ∂ + − − (4) 2m2 M ∂X ∂x M ∂X ∂x

We note that the cross terms proportional to ∂2/(∂x∂X) cancel leaving the terms:

− 2 2 − 2 2 ˇ ¯h m1 + m2 ∂ ¯h 1 1 ∂ T = 2 2 + + 2 2 M ∂X 2 m1 m2 ∂x

− 2 2 − 2 2 ˇ ¯h ∂ ¯h ∂ 1 1 1 T = 2 + 2 where = + (5) 2M ∂X 2µ ∂x µ m1 m2

We see that the kinetic energy has a contribution from the center of mass motion

X which enters with a mass M = m1 + m2 and a contribution from the relative −1 −1 −1 motion x = x1 − x2 which enters with an “reduced mass” µ = m1 + m2 .

Lets now assume a potential of the form V (x1,x2)=Vr(x1 − x2)+Vcg(X) which might be applicable, for example, to two masses connected with a spring 2 which experience a potential Vr(x1 − x2)=κ(x1 − x2) /2, as the mass spring system falls in a gravitational ﬁeld Vcg(X)=(m1 gx1 +m2 gx2)=MgX.The

3 Schr¨odinger equation reads:

2 2 −¯h ∂2 −¯h ∂2 + V (x) ψ(x, X)+ + V (X) ψ(x, X)=Eψ(x, X) 2µ ∂x2 r 2M ∂X2 cg r cg (6) We now have a classic separation of variables situation. We factor the wave function as ψ(x, X)=ψr(x) × ψcg(X). Inserting this into Eq. (6) and dividing both sides by ψr(x) × ψcg(X) we get two separate Schr¨odinger equations:

2 −¯h d2 ψ (x)+V (x)ψ (x)=E ψ (x) 2µ dx2 r r r r r

2 −¯h d2 ψ (X)+V (X)ψ (X)=E ψ (X) ,E+ E = E (7) 2M dX2 cg g cg cg cg r cg

To gain insight into Eq. (7) lets consider the speciﬁc example of two masses m1 and m2 undergoing simple harmonic motion, while their center of gravity travels to the right with a total momentumhK ¯ . If the two masses are tied together with a spring with spring constant κ, the oscillation frequency will be −1 −1 −1 ω = κ/µ where the “reduced mass” is µ = m1 + m2 .

The ψcg(X) portion of the wave function which describes the center of gravity is just a free particle traveling wave solution of the Schr¨odinger equation of the form: ψcg(X)=exp(iKX) ; while the ψr(x) portion of the wave function which describes the relative coordinate x1−x2 is constructed out of Hermite polynomials √ 2 and given by ψr(x)=Hn( βx)exp(−βx/2). The complete wave function is:

2 µω ψ(x, X)=N H ( βx) e−βx /2 eiKX where β = (8) n ¯h

The energy of the system, or E in Eq. (6), is just equal to the sum of Er (the harmonic oscillator energy) and Ecg (the kinetic energy of the center of mass

4 which acts as a particle of mass M traveling with a momentumhK ¯ .

2 1 ¯h K2 E = n + ¯hω + (9) 2 2M

A brief summary might be in order. Quite often in reality, we have the problem of describing a situation where two particles interact with each other through a potential which depends on their separation. In such cases, one can affect a separation of variable solution by writing the Schrödinger equation in terms of the relative displacement of two particles and location of their center of mass. We’ve shown that the equation describing the center of mass, becomes totally independent of the equation describing the relative coordinates. The Schrödinger equation for the relative displacement between the two bodies is exactly the same as the equation for a single particle of reduced mass

µ interacting in an external potential V (x)=V (x1 − x2). Hence the fact that we have two particles rather than one creates no additional mathematical complications at all!

The Angular Reduction

Besides the reduction of a two particle system down to a one particle system with a reduced mass µ, the conservation of angular momentum creates additional simpliﬁcations in the central force problem in both classical mechanics and quantum mechanics. A few words about the classical orbit problem (illustrated below) wouldn’t hurt.

V(r) r φ r φ Perigee Apogee

2 L = µ r φ r = µ r φ

5 We have described the elliptical orbit in terms of the polar coordinates r and ˙ 1 φ. The tangential velocity is vφ = r φ; while the radial velocity isr ˙. The energy expression is given by:

1 E = µ r˙2 +(rφ˙)2 + V (r) (10) 2

2 ˙ The angular momentum, given by L = µvφ r = µr φ, is a constant of motion. We can express φ˙ in terms of L and insert it in Eq. (10) to obtain an expression in r only: 2 L 1 2 L φ˙ = ,E= µr˙ + + V (r) (11) µr2 2 2 µr2 Eq. (11) is very analogous to the conservation of energy in a one dimensional 1 2 potential well: E = 2 m x˙ + V (x). To complete the analogy we combine the angular momentum term with the potential actual potential energy to form an “eﬀective potential”. L2 Veﬀ = + V (r) (12) 2 µr2

The L2/(2 µr2) term is often called the “centrifugal potential”. We illustrate radial motion of the orbit for a V = −K/r potential in an eﬀective potential well below: V eff

2 -K L V = + 2 eff r 2µ r

dominated by centrifugal potential

r E Perigee Apogee dominated by Coulomb potential

1 In classical mechanics, one often writesr ˙ to stand for dr/dt.

6 The reduced mass satellite, bound in the well with a negative total energy, oscillates in radius between a closest “perigee” and furthest “apogee” which serve as classical turning points. The centrifugal potential prevents the satellite from

falling into the force center. As r gets small, vφ must get large so as to conserve

L = mrvφ.Whenr gets below the perigee, it becomes energetically impossible to have a large enough vφ to conserve L.

In classical mechanics we used the L constraint and center of mass trans- formation to eﬀectively turn the orbit problem from a problem of 6 coordinates describing both the bodies to just 1 coordinate (the radius between the bodies) via an eﬀective potential.

The angular reduction in quantum mechanics

We have workesd out the Hamiltonian in spherical coordinates in the chapter on Dimensions. − 2 ˇ2 ˇ ¯h 1 ∂ 2 ∂ L H = 2 r + 2 + V (r) . (13) 2µ r ∂r ∂r radial 2µr eﬀ pot

Because of rotation symmetry, we know that the wave functions can be chosen 2 to be eigenstates of both Lˇ and Lz. As disussed in the Dimension chapter, the wave function is of the form:

m ψ = Rn(r) Y (θ, φ) (14)

ˇ2 m Inserting the Eq. (14) wave function into Eq. (13), using the fact that L Y = 2 m m ( +1)¯h Y and dividing both sides of the resulting equation by Y we have a diﬀerential equation for the radial wavefunction only:

2 2 −¯h 1 d dR ¯h ( +1) r2 + R + V (r) R = ER (15) 2µ r2 dr dr 2µr2

Eq. (15) simpliﬁes considerably if we re-cast Eq. (15) in terms of a function u(r)

7 where u(r)=rR(r): 2 2 −¯h d2u ¯h ( +1) u(r) + + V (r) u = Eu where ψ = Y m(θ, φ) (16) 2µ dr2 2µr2 r

We can think of Eq. (16) as the Holy Grail of the radially symmetric two body problem in quantum mechanics! We note that it is of the exact form as the one dimensional Schr¨odinger equation with x substituted for r andwithaneﬀective potential which includes a centrifugal barrier term. In the chapter on Bound State States in One Dimension we built up considerable insight into the solutions of such problems. We do have a novel boundary condition however. Since we do not generally wish the wave function to become singular at the origin and ψ OC u(r)/r we require that u(r) vanish at the origin.

The below ﬁgure shows the usual situation for a binding eﬀective potential well. We illustrate it for the case of a hydrogen atom potential. At small enough radius, the centrifugal potential will generally dominate over the interaction potential as long as =0.

V eff (r)

large l

r E small l

l=0

As increases, the centrifugal potential becomes increasingly more eﬀective at keeping the two interacting bodies apart by moving the classical turning point to larger values. We expect therefore that the PDF(r) for high orbitals will

8 peak further from the origin than small orbitals. In fact this is easy to see formally if r is suﬃciently small such that ( +1)/(2µr2) |V (r)| or |E|.In 2 such a case we can neglect V (r)andE in Eq. (16) and cancel outh ¯ /(2µ):

d2u ( +1) = u (17) dr2 r2

The solution of Eq. (17) is a simple power law of the form u(r)=rp. Substituting this form into Eq. (17) we have:

p(p − 1)rp−2 = ( +1)rp−2 → p = +1or − (18)

2 The only non-singular solution is therefore:

u(r) Near origin u(r)=r+1 → R(r)= = r (19) r

The below sketch compares various R(r)=r in the low r region to try to convince you that the high orbitals are pulled further from the origin.

R(r)

r 0 r1

r2 r 3

Of course, the radial wavefunction cannot be of the form R(r)=r at large r or it would blow up as r →∞. At suﬃciently large r, the centrifugal potential becomes unimportant and the R(r) → 0 in a way that depends on the interaction potential V (r).

2 Technically this should only hold for = 0 but it works for =0aswell

9 The Coulomb Potential

Let us consider the general one ion atom with a nuclear charge of +Ze an electron charge of −e. Wewillusethesymbole to represent the absolute value of the charge of an electron: e =1.6 × 1019 C. The atomic nucleus consists of Z protons each of which has a charge of +e. The attractive potential is given by the Coulomb potential: −Ze2 V (r)= (20) 4π or and the relevant radial equation becomes: − 2 2 2 2 ¯h d u ¯h ( +1)− Ze 2 + 2 u = Eu (21) 2µ dr 2µr 4π or

It would take us a bit far a field to solve Eq. (21) in general. The solutions are of the form of a polynomial in r (called an associated Laguerre polynomial) multiplied by a decaying exponential in r. The exponential fall off of the Coulomb wave function is much slower than the exp (−βr2/2) fall off for the harmonic oscillator. This makes a great deal of sense since the harmonic well becomes much more confining as r increases while the Coulomb potential approaches zero at large r. In the exercises I ask you to solve for the ground state energy and β by assuming a ground state wave function of the form u(r)=r exp (−βr)with =0.

The maximum power of r appearing in the Laguerre polynomial is equal to the principle quantum number n. Interestingly enough the energy of the orbital depends only on n and is exactly the same as predicted by Bohr using the Correspondence Principle, namely:

−Z2 e4µ En = 2 2 (22) 2(4π 0¯h) n

It is quite remarkable that the eigenvalues of Eq. (21) are independent of .At one time this was known as the “accidental” degeneracy. In fact there is a rather subtle symmetry which leads to this degeneracy.

10 A very analogous degeneracy occurs for classical orbits:

κ Degenerate classical orbits with E = - d

High L

Medium L Low L

Circular orbits have the maximum possible orbital momentum for a given orbital energy. Orbits of the same energy but with very small angular momentum ( very little tangential velocity at the apogee) collapse into highly eccentric, thin ellipses. The ﬁgure to the right shows a high L (circular orbit) and low L (elliptical orbit) in relationship to the force center origin (at the focus of the ellipse). As you can see the low L orbit has a much smaller perigee and spends much more time near the origin. This is in exact analogy with the quantum mechanical situation (the r behavior of R(r)asr → 0).

The general solution also shows that the angular momentum quantum number can take on all integral values 0 ≤

11 m = +2 Z m = + 1 l = 2 m = 0 m= -1 m=+2 m = -2 m=+1 N = 3 m = +1 l = 1 m = 0 m=0 m = -1 m=-1 m=-2

l = 0 m = 0 radius= 6

We also show how the = 2 orbitals have 5 possible orientations for L .In general the degeneracy of orbital n is given by n2.

Here is a list of some of the hydrogenic wavefunctions:

3/2 3/2 1 Z −Zr/ao 1 Z Zr −Zr/2ao ψ100 = √ e ,ψ200 = √ 2 − e π ao 4 2π ao ao

3/2 3/2 1 Z Zr −Zr/2ao 1 Z Zr −Zr/2ao ±iφ ψ210 = √ e cos θ, ψ21±1 = √ e sin θe 4 2π ao ao 8 π ao ao

3 2 1 Z / Zr Z2r2 √ − −Zr/3ao ψ300 = 27 18 +2 2 e 81 3π ao ao ao

2 2 The variable ao =4π o ¯h /(µe )=0.0529 nm is known as the Bohr radius. The 2 ratio n ao/Z sets the scale for atomic size. In fact one can easily show from the −1 2 virial theorem that r = Z/(n a0)whereZ is the number of protons in the nucleus which is called the atomic number. Incorporation the atomic number allows us to apply our hydrogen formalism to one electron ions such as Li++ where Z = 3.

Its easy to conﬁrm the expected r behavior as r → 0

12 Hydrogen Wavefunction Plots

It is useful to get a good feeling for the shape of the electron cloud for the electron in various orbitals. Below is a crude sketch of the electon clouds with hopefully evocative names.

z z z z

(2 1 1) (1 0 0) (2 0 0) (2 1 0) (2 1 -1) marble tootsie pop dumbbell donut

The spinning arrow means that the distribution can be spun about the z axis because of its cylindrical symmetry.

Another way of viewing the hydrogen atom wave function is in terms of the radial probability, or the probability density for ﬁnding the electron between r and r + dr. For a spherically symmetric wave function, this probability is given by:

2 ∗ dP 2 ∗ dP =4πr dr ψ 00(r) ψ 00(r)or =4πr ψ 00(r) ψ 00(r) (23) n n dr n n

since a volume element between two spherical shells is dvol = 4πr2 dr.Asketch of the radial probability is shown below for the ﬁrst three spherically symmetric wave functions.

13 4 π r 2 ψ 2

r a (1 0 0) d i a l 49ao ao (2 0 0) p ao r o b (3 0 0)

It is interesting to note that there are n − 1 radial nodes in the ψnoo(r)wave function. These nodes cause the wave PDF to peak up near r → 0. The number of nodes radial nodes decreases with increasing according to # nodes = n−−1.

Electric Dipole Transitions

The study of the hydrogen spectral lines was a major inﬂuence on the devel- opment of quantum physics. Schr¨odinger quantum mechanics as we’ve developed it thus far, very accurately predicts all the energies of the electron orbitals and the location of all of the hydrogen spectral lines. Interestingly enough, however, it does not describe the mechanism for transitions between the various energy levels. In our treatment of quantum mechanics, an electron placed in excited state

(n) has a wave function of the form ψ( x)exp(−iωn t) which produces a time independent PDF. There is nothing in our description which allows the electron in an excited state to drop down to its ground state by emitting a photon and yet typical atomic transitions occur on times scales of ≈ 10−9 → 10−8 seconds.

The solution to this problem is that the quantum mechanics which we have developed thus far has ignored the interaction of the electron with external electromagnetic fields. These fields are present as “zero point” fluctuations in a way analogous to the ground state oscillations of a harmonic oscillator. We can’t begin to do justice to this important topic, which is often called second quan-

14 tization, in a one semester course on quantum mechanics, but a great deal of insight can be obtained using a semiclassical argument based on radiation from an oscillating dipole antenna which we illustrate below:

e 2 ω 4 2 2 S = e a sin θ 32 ε π 2 3 2 ο c r S a θ E

Radiation pattern

One gets electromagnetic radiation from an accelerating electric charge. A static charge has just an E field. A constant moving charge has both an E field and a B field but no Poynting’s vector or electromagnetic intensity since 5 S = E × H =0. Once the charge is accelerating, one can get a non-zero Poynting’s vector which is proportional to the square of the charge × acceleration. A simple way of getting repetitive acceleration is though the use of an oscillating charge executing simple harmonic motion (or its equivalent – the dipole antenna fed with an oscillating current). The above picture shows a charge on a spring executing SHM of the form a sin (ωt) which has an acceleration of −aω2 sin (ωt). We thus anticipate a radiation intensity which is proportional to (a × e)2 × ω4 where the charge times the amplitude is called the dipole moment.

As indicated in the ﬁgure, the intensity falls oﬀ as 1/r2 so there is the same power subtending any sphere around the dipole. The dipole antenna pattern shows that the maximum intensity is directed normal to the dipole. The maxi-

5 It would violate basic relativity principles to get radiation from a charge moving with a constant velocity, since the absence or presence of radiation would be diﬀerent for diﬀerent inertial observers.

15 mum electrical ﬁeld is parallel to the dipole. The time averaged power is given by the expression: 2 4 1 e ω 2 Power = 3 a (24) 3 4π o c

Dividing the power by the energy of a photon Eγ =¯hω wehavetherate(or photons/second):

2 3 3 2 1 e ω 2 1 ω a R = 2 a = α 2 (25) 3 4π o ¯hc c 3 c

2 wherewehaveusedα = e /(4π o¯hc) ≈ 1/137.

But how does an electron in a stationary excited state spontaneously develop an oscillating dipole moment? Imagine that the excited electron is tickled by the the radiation ﬁeld leaving it in a mixed state which is partially excited ψ2 and 6 partially ground ψ1.

−iE2t/h¯ −iE1t/h¯ ψ = ψ2( x) e + γψ1( x) e (26)

We will write the 3 component dipole operator as d ≡ e r = e ( xyz)The expectation value of the dipole operator (d ) for the mixed state ψ is given by d = ψ| e r |ψ which is:

∗ +iωt ∗ −iω t d = ψ2| e r |ψ2 + γ γ ψ1| e r |ψ1 + γe ψ2| e r |ψ1 + γ e ψ1| e r |ψ2

2 ∗ −iω t d = ψ2| e r |ψ2 + |γ| ψ1| e r |ψ1 +2Re γ ψ1| e r |ψ2 e (27) where ω ≡ (E2 −E1)/¯h. We see that this mixed state will have a (real) dipole moment which oscillates sinusoidally with a frequency of ω and a phase determined by the relative phase of γ and and the dipole moment connecting the initial and

6 The dynamics of how spontaneous emission happens is beyond the scope of the course, but books such as Quantum Mechanics by Eugen Merzbacher have complete descriptions.

16 ﬁnal state or ψ1| e r |ψ2. The static dipole moments such as ψ2| e r |ψ2 would not contribute to the radiation if they did exist (but in fact we’ll show shortly that ∗ they vanish). The eﬀective oscillation displacement would be a =2γ ψ1| r |ψ2. Using Eq. (25) and Eq. (27) we anticipate a photon emission rate of:

3 3 2 3 α ω 2 α ω 2 ∗ 4α|γ| ω 2 R = a = 4|γ| ψ1| r |ψ2 ·ψ1| r |ψ2 = |ψ1| r |ψ2| 3 c2 3 c2 3 c2 (28) In order to get an intrinsically real rate which reﬂects all three dipole components 2 we replace |ψ1| r |ψ2| in Eq. (28) by:

∗ ψ1| r |ψ2 ·ψ1| r |ψ2≡

2 2 2 |ψ1| x |ψ2| + |ψ1| y |ψ2| + |ψ1| z |ψ2| (29)

Of course we haven’t a clue without second quantization theory as to the value of γ. Interestingly enough γ = 1 meaning that the rate for an excited state spontaneously decaying to the ground state via an electric dipole transition is just:

4α 3 ∗ R = ω ψ1| r |ψ2 ·ψ1| r |ψ2 (30) 3 c2 It is useful to get Eq. (30) in practical units since we will work some examples in homework. In Eq. (30), |ψ2 represents the initial (higher energy) state which we will now write as |ψi and |ψ1 represents the ﬁnal (lower energy) state which we will now write as |ψf . We are calculating the number of spontaneous decays per second of the form |ψf →γ |ψi. 3 3 4αω | 2 4αc ¯hω | 2 4αc 3 | 2 R = 2 ψf r ψi = ψf r ψi = 3 ∆E ψf r ψi 3c 3 ¯hc 3(¯hc) (31) 3 3 Here ∆E = Ef − Ei. Lets work out the constant 4αc/(3¯h c ) in practical units. 8 4αc 4(1/137)(3 × 10 nm/ns) −1 −2 −3 3 = 3 =0.382 ns nm eV 3(¯hc) 3 ( 197eV nm)

17 Hence putting it all together we have:

−1 −2 −3 3 2 R =0.382 ns nm eV ∆E ψf r |ψi (32)

As an example of a typical electric dipole (sometimes called E1) transition rate lets consider the hydrogen 210 → 100 transition which (as you can easily verify) produces a λ = 122 nm photon. As a guess for 7 |ψ100| r |ψ210| we might choose 0.75 a0 =0.04 nm. The values we need to insert into Eq. (32) are: 1 2 ∆E =13.6 1 − =10.2eV; |ψ100| r |ψ210| ≈ 0.04 nm (33) 22 where we used the Bohr formula to ﬁnd ∆E = E210 −E100. Inserting these values we get

−3 3 2 R =0.382 ns−1 nm−2 eV (10.2eV)(0.04 nm) =0.65 ns−1 =6.5 × 108 s−1 (34) The rate R means we expect 0.65 spontaneous decays per nanosecond from each excited hydrogen atoms in the |210 state. We will relate this to the lifetime of the 210 state in the next section.

Transition rates , lifetimes , and line shapes

The meaning of this transition rate might be somewhat unclear at this point. In the context of a large number N (such as Avagodro’s number) of hydrogen atoms all making 2 → 1 transitions, the total number of photons per second from all N atoms would R × N. In the context of a single hydrogen atom in the n =2 state, the rate R−1 is related to the mean lifetime of the excited state or the average amount of time that the electron spends in the n = 2 orbital. To see this

7 Of course there is no need to guess since you can calculate the actual integral as I ask you to do in the exercises.

18 connection, let us imagine starting at time t = 0 with a collection of N atoms in o the n = 2 state and follow how many remain as a function of time. Let i Ri be the total rate for the atom to leave the n = 2 state and go to a lower orbital. Let N(t) be the number of hydrogen atoms remaining in the n =2orbitaftera given time t. The change in number of atoms which remain at the n =2orbital will be total decay rate i Ri times the time interval ∆t times the number of atoms presently in the excited state N or: dN dN = −Ndt R or = −dt R (35) i N i i i

Integrating both sides of this equation we get: −( Ri) t N(t)=No e i (36)

This is the exponential decay law which describes many decay processes such as radioactive decay.

We can consider applying it to the decay of a single, excited hydrogen atom.

We set No = 1 since we initially have just one excited atom. The probability the atom will still be excited is then:

1 P = exp(−t/τ)whereτ = (37) i Ri

We can think of τ as the “lifetime” of the excited state which is the inverse of the total rate that allows the state to de-excite. In our example, all 210 states decay into 100 state and hence the lifetime of the 210 state is simply τ ≈ 1/R = 1/(0.65 ns−1)=1.5 ns. There is an energy-time Uncertainty Principle which says short lived states are very uncertain in energy. Hence long-lived excited atomic states de-excite with narrow spectral lines ; while short-lived excited atomic states produce very broad spectral lines.

Selection Rules

19 Quite often all three components of an electric dipole matrix element given by the integral in | <ψf | r |ψi > | vanish because of symmetry. In such cases, the 9 transition from the initial to ﬁnal state is forbidden by the electric dipole process. We can formulate a set of selection rules which can be used to determine at a glance whether or not an electric dipole transition is allowed.

A particularly instructive and easy to derive selection rule says that only processes with ∆m ≡ mf − mi ∈ 0, ±1 can have electric dipole (E1) transitions. This is a necessary but insuﬃcient condition which follows from the φ part of the | | <ψnf f mf r ψniimi > which we can write as:

2π 2π iφ −iφ iφ −iφ −imf φ +imiφ −imf φ e +e e −e +imiφ Iφ = dφ e (cosφ sin φ 1) e = dφ e ( 2 2i 1) e o o (38)

In order that all three components vanish for Iφ we must have the condition:

2π 2π 0= dφ e−imf φ e±iφ eimiφ = dφ e−i(mf −±1−mi) o o

2π 2π and 0 = dφ e−imf φ eimiφ = dφ e−i(mf −mi) (39) o o

One can easily show that for any integer N

2π dφ eiNφ = 0 unless N = 0 (40) o

9 However, there always other, suppressed processes which allow the electron to ultimately drop to the ground state including transitions involving magnetic dipoles, electric quadrupoles, and processes where two photons are emitted. There is a suppression hierarchy where each level is suppressed by an additional factor of ≈ α2 ≈ 10−4 in the decay rate.

20 Hence

| | | | − ± <ψnf f mf x ψniimi >=<ψnf f mf y ψniimi >= 0 unless mf mi = 1

| | − <ψnf f mf z ψniimi >= 0 unless mf mi = 0 (41)

A very important set of selection rules for both atomic, nuclear, solid state, and high energy physics processes involve the concept of parity. We have discussed the idea of even (ψ(x)=ψ(−x)) or odd (ψ(x)=−ψ(−x)) one dimensional wave functions earlier in the chapter Bound States in One Dimension for the case of symmetric wells (V (x)=V (−x)). In the case of symmetric potentials, wave functions can be classified according to parity, since parity becomes a symmetry of the Hamiltonian. A central force problem also has a parity symmetry (or inversion symmetry) defined by r →− r. As you can verify using r =(r sin θ cos φrsin θ sin φrcos θ ), one can change r →− r in spherical coordinates by : r → r , φ → φ + π ,andθ → π − θ.Sincether coordinate is unaffected, the parity of the wave function is determined by the parity of the angular wave function , which is the angular wave function of a central force, m means the parity of Y .

0 0 ±1 Yo Y1 Y1 1 3 ∓ 3 ±iφ 4π 4π cos θ 8π e sin θ 0 ±1 ±2 Y2 Y2 Y2 5 2 − ∓ 15 ±iφ 15 ±2iφ 2 16π 3cos θ 1 8π e sin θ cos θ 32π e sin θ

10 You can conﬁrm for a few trial spherical harmonics that the parity of a spherical harmonic is given by parity = (−1) since:

m − − m Y (π θ, φ + π)= ( 1) Y (θ, φ) (42)

10 Note for odd states with even m, the minus sign comes from the Legendre polynomial in θ.Foroddm, the minus sign comes from the exp (imφ) part.

21 | | The ith component of the dipole matrix element. <ψnf f mf ri ψniimi >. ∗ × × will involve a volume element integral over the function Di( r)=ψnf f mf ri

ψniimi . The parity of Di( r) will essentially be the product of the parities of the

intial state, ﬁnal state, and ri itself. Since under parity ri →− ri the parity of a dipole element is just:

f +i+1 Di( r)=(−1) (43) which will be odd (or negative ) if ∆ is even, and even if ∆ is odd.

Symmetric integrals (such as the integrals from −∞ → +∞ which occur in quantum mechanics) over odd functions vanish because of symmetry. This is easy to see in one dimension.

f(x) f(x) Even Odd

x x

For the case of the odd function the integral over the shaded region cancels the clear region. For the case of a two dimensional odd function, the argument is slightly more subtle as illustrated below:

22 For an odd function, such as illustrated on the left, the two clear quadrants will have the opposite sign and cancel as will the two shaded regions. The same sort of thing happens in three dimensions where the space can be divided into 8 quadrants and diagonally opposite quadrants cancel for odd functions.

We thus conclude that E1 transitions will vanish unless the initial and ﬁnal statehavetheoppositeparity(ie ∆ is odd).

The ﬁnal selection rule, which I will discuss here, says that E1 transitions vanish unless ∆ = ±1 which is a consequence of the recurrence formula and orthogonality of the associated Legendre polynomial which describes the θ parts of the spherical harmonics. We illustrate the argument for the case of the z dipole | | | | moment, <ψnf f mf z ψniimi >=<ψnf f mf r cos θ ψniimi >. This recurrence m formula relates cos θ times the θ part of ψniimi to the θ parts of Yi±1 spherical harmonics:

m − m m (2i +1)cosθPi (cos θ)=(i +1 m) Pi+1(cos θ)+(i + m) Pi−1(cos θ) (44)

The orthogonality property of the associated Legendra polynomials then says × × m that integral over ψf cos θ Yi±1 will only survive if the ﬁnal state contains m Yi±1:

+1 m × m d cos θP (cos θ) P (cos θ) = 0 unless = (45) −1

This rule implies the following chart for some of the allowed and forbidden E1 transitions for hydrogen.

23 l=0 l=1 l=2 n=3

E1 allowed n=2 E1 forbidden

n=1

Some allowed and forbidden electric dipole transitions

Photon Spin and the Physics of Selection Rules

The ∆m and ∆ selection rules have a fairly simple physical explanation be- | | sides the detailed mathematical explanation for the vanishing of <ψnf f mf r ψniimi >. The basic idea is that the photon itself carries one unit (¯h) of angular momentum which is directed either along or against its direction of motion. The two spin states of the photon correspond to left and right circular polarized light. If a photon is emitted when an excited hydrogen atom returns to the ground state, there must be a change of one unit of angular momentum in the atom in order to 11 allow for the angular momentum carried by the photon. Classical electromagnetic ﬁelds can carry both linear and angular momentum. Below is a cartoon sketch of the transitions ψ211 → ψ100 and ψ210 → ψ100.

11 This is a bit oversimpliﬁed since, in principle, the photon could have carry orbital momentum which could balance its spin. In fact in E1 transitions, the photon carries no orbital angular momentum with respect to the force center.

24 Transitions from ψ ψ 2 1 m 1 0 0

z E

m=0 y x

m=1

By considering the φ integral it is easy to show that the non-vanishing dipole

matrix element for the ﬁrst transition is <ψ100| x − iy |ψ211 >. This means that the photon is radiated from an x and y dipole antenna which being run 90o out of phase (because of the i coeﬃcient). This will result in circularly polarized light emitted primarily along the ±zˆ axis (perpendicular to the two dipole directions).

The inital Lz = 1 of the atom in the ψ211 is now carried by the circularly polarized photon leaving Lz =0fortheψ100 ﬁnal state. Similarly the transition changed by one and this one unit of angular momentum is carried away by the photon.

We next consider the process ψ210 → ψ100. The non-vanishing dipole matrix element for this transition is <ψ100| z |ψ210 >. This means that the photon is radiated from an z dipole antennae which emits light primarily in the x-y plane. The electrical ﬁeld will tend oscillate along and against the antenna direction which means that this photon will tend to be plane polarized in the z direction. In general, a plane polarized photon consists of equal amplitudes for left and right circularly polarized light. For the ψ210 → ψ100 transition there was no change in Lz in the atom and the emitted photon carries no Lz. There was however a change in the total angular momentum ∆ = 1 which represents the angular momemtum carried by the the photon.

We will show in the next chapter, that a free electron traveling through space carries intrinsic spin as well.

25 Lasers and Masers

As a practical illustration of some of the concepts presented in this chapter, we give a brief, thumbnail sketch of the operaton of the LASER (Light Amplification by Stimulated Emission of Radiation) and MASER (Microwave Amplification by Stimulated Emission of Radiation). Of course, lasers are ubiquitous in science and engineering. In the future, minature solid state lasers driving optical fibers might ultimately replace most wires in electrical circuits. Masers are less common, but still provide the lowest noise method for amplifying microwaves.

One of the primary virues of a laser which is a light amplifier as opposed to just a light source, is that it creates or amplifies a nearly perfectly coherent light beam with a well defined, single , frequency and phase. It is the coherence property which allows its light beam to be nearly perfectly unidirectional and monochromatic.

The operant part of the laser/maser acronym is the term stimulated (as op- pose to spontaneous) emission. Stimulated emission means that excited atom is tickled by an external electromagnetic field, rather than the zero point fluctuations of the radiation field responsible for spontaneous emission. Like a high quality electrical or mechanical resonator, the maximum energy absorption or emission occurs when the incident radiation is very close to a natural resonant frequency, which for an atom is ω =∆E/¯h ,where ∆E is the energy difference between two atomic energy levels.

As you probably know, an electrically or mechanically driven oscillator can either provide or absorb power depending on the phase of the driver relative to the oscillator. The same is true of the atom, which we illustrate for an atomic electron making transitions between the ground state and an excited state.

26 E2 n2 n2 A 2 1 n2 B ρ(ω) s s s 2 1 p t t o i i n m m t a e e b m m s i i o t t r b E1 n1

ρ(ω) n1 B 1 2

According to the classic analysis of Einstein, there are three processes com- prising the dynamics of the two level transition: spontaneous emission, stimulated emission and stimulated absorption. Since stimulated emission or absorption is due to the interaction of atomic electrons with the external radiation ﬁeld, the transition rate per atom is proportional to the density of photons with the resonant frequency or ρ(ω). The rate per atom for spontaneous emission, given by Eq. (32) , is of course independent of the external spectral density ρ(ω). Total rates will be proportional to the rate per atom times the number of atoms at each level. In terms of the famous Einstein A and B coeﬃcients the total rate for

1 → 2 transitions is given by n1 ρ(ω) B1→2 where n1 is the number of atoms at level E1 and B is related to the rate for stimulated emission. The rate for 2 → 1 transitions is fed both by spontaneous emission and stimulated emission and is given by n2 (A2→1 + ρ(ω) B2→1)whereA2→1 describes spontaneous emission.

All three coeﬃcients B1→2 , B2→1 and A2→1 depend on the matrix elements between the initial and ﬁnal state and can be computed using the theory of second quantization.

The name of the game for a laser or maser is to arrange the total rate for stimulated emission to be greater than the rate for stimulated absorption, since the photons produced via stimulated emission will be in phase with those from the external radiation ﬁeld leading to coherant ampliﬁcation. Light produced via

27 spontaneous emission will of course contribute to the ﬁnal radiant output of the device but in a totally uncorrelated (incoherent) manner.

As early as 1917, Einstein was able to relate spontaneous to stimulated emission and demonstrate that the coefficient for stimulated emission equals the rate for stimulated absorption using an ingeneous thermodynamic argument. We will briefly summarize this argument since it gives considerable insight into laser dynamics. Einstein considered the case where the two level are in thermal equilibrium with the external radiation field. If the two level system is in thermal equilibrium , the overall rate for 1 → 2 transitions will equal the rate for 2 → 1 transitions:

n1 ρ(ω) B1→2 = n2 (A2→1 + ρ(ω) B2→1) (46)

The relative n1 and n2 populations of the two orbitals in equilibrium will be given by the temperature dependent Boltzman factor.

n2 = e−∆E/KT (47) n1

Eq. (46) and Eq. (47) can be combined to give a prediction for the equilibrium spectral density ρ(ω):

A2→1/B2→1 ρ(ω)= ¯ ( ) (48) B1→2/B2→1 ehω/ KT − 1

Much earlier, Planck made the ﬁrst use photon quantization arguments to com- pute the spectral density for photons in equilibrium with the walls of a “black 12 cavity” and derived the Planck distribution 2¯hω3 1 ρ(ω)= (49) πc3 ehω/¯ (KT) − 1

Einstein argued that spectral density Eq. (48) describing the spectral density for photons in equilibrium with a collection of atoms must be the same as Planck’s

12 This describes the spectral density of photons found in a hot “black body” such as the ﬁlament of an incandescent bulb. The term “black” means that the surface is as eﬃcient as possible for radiating or absorbing photons.

28 expression Eq.(49) for photons in equilibrium with the walls in a black cavity. One consequence is that the rate for stimulated emission equals the rate for stimulated absorption.

I ﬁnd this very analogous to driving a black car in the summer. During the day the car gets hot quickly since the black surface is a good radiation absorber. At night, the black car cools oﬀ quickly since the black surface is a good radiation emitter.

How can one make an ampliﬁer or strong light source using atomic transitions? The idea is to somhow enhance the total rate for stimulated emission of light over the rate for stimulated absorption: This cannot be done in thermal

equilibrium since in equilibrium n1 ρ(ω) B1→2 = n2 ρ(ω) B2→1 + n2 A2→1 which means that stimulated absorption rate actually exceeds the rate for stimulated emission. Since B2→1 = B1→2 and the spectral density factors cancel , the only way to meet the condition that stimulated emission rate exceeds the stimulated absorption rate to achieve a non-equilibrium situation with n2 >n1 which is called a population inversion. We note that for typical atoms, at reasonable tem- peratures we usually have the reverse situation where the bulk of the atoms are in their ground state n1 >> n2 and absorption will dominate over stimulated emission.

There are many clever ways of achieving the population inversion which in- sures that stimulated emission will dominate over stimulated absorption. A particularly simple method (called optical pumping) was used in the ﬁrst laser, invented in 1960, which is constructed out of a ruby rod. Ruby is a transparent crystal of consisting primarily of Al2O2 with a small percentage of chromium +++ Cr ions which are responsible for the laser properties. A rough schematic of the atomic transitions responsible for the ruby laser:

29 collisional n2 1.96 eV excitation n1

blue n2 metastable 18.7 eV green

1.79 eV

n1 Helium Neon chromium

+++ The red appearance of the ruby crystal is due to the fact that Cr strongly absorbs blue and green light. In a ruby laser a ﬂash tube is used to optically pump electrons up to the green and blue energy levels and thereby de-populate the n1 +++ level. The excited Cr can then de-excite (through somewhat complicated, non-radiative transitions which transfer mechanical energy to the ruby crystal) and feed a metastable state (no E1 transitions) which populates the n2 level and sets up the population inversion. A few of the n2 atoms will spontaneously emit photons which will then be coherently ampliﬁed since stimulated emission will occur at a larger rate than stimulated absorption owing to the population imbalance.

The very common, helium-neon laser, invented in 1967, uses a 4 level optical pumping scheme and can produce a continuous beam. The helium-neon laser consists of a roughly 5 to 1 mix of neon to helium in a gas discharge tube. A gas discharge puts a large number of helium atoms in their ﬁrst excited state. This ﬁrst excited state is nearly degenerate with a particular excited state of neon and therefore is very likely to excite this level through resonant collisional excitation. The transition rates are such that the collisionally excited state drops to a state lying 1.96 eV beneath it, rather than de-exciting to the ground state. Because the n1 state lies so much above the ground state it is virtually unoccupied ini- tally according to the Bolzman factor. The n1 population is continually depleted

30 through spontaneous emission. The n2 population is continually fed from col- lisions with the optically pumped helium. We thus have an automatic, highly efficient population inversion. The lasing action amplifies the familiar 1.96 eV red photons. Clearly lasers are a fairly inefficient way of generating light, but they are highly efficient at concentrating light. One can burn holes though metal plates by concentrating a power equivalent to a 60 Watt light bulb into a very small area.

Masers are similar to lasers but involve transitions of more like 1/1000 of an electron volt rather than 2 eV. One of the ﬁrst masers used the double well ammonia molecular transition which we discussed extensively in the chapter on Bound States in One Dimension.

31 Important Points

1. The hydrogen atom consists two particles which interact via a mutual po-

tential of the form V ( x1 − x2). In close analogy with classical mechanics, it is possible to decouple the two particle Schr¨odinger equation by work-

ing in relative coordinates x = x1 − x2 and center of mass coordinates

X =(m1 x1 + m2 x2)/(m1 + m2). The relative coordinate Schr¨odinger equation is identical to the Schr¨odinger equation for a particle of mass −1 −1 −1 µ = m1 + m2 in an external potential of the form V (x). The re- sultant wave function is the product of a wave function in x times a wave funtion in X .

2. Because of rotational symmetry, the solutions of the central force problem m are eigenstates of angular momentum of the form ψ = R(r) Y . Guided by classical physics, one can write a radial Schr¨odinger Eq. in terms of an “eﬀective” potential which includes the true interaction potential as well as a “centrifugal” potential of the form Lˇ2/(2µr2). 2 2 −¯h d2u ¯h ( +1) u(r) + + V (r) u = Eu where ψ = Y m(θ, φ) 2µ dr2 2µr2 r

At short distances, where the centrifugal potential dominates, one can show that R ∝ r which means large states tend to stay away from the origin more than low states.

3. For the Coulomb potential, relevant to the one electron ion like the hydrogen atom, the energy levels are described by a principle quantum number n in exactly the same as the Bohr model:

1 1 −13.6 eV E = − α2 µc2 = n 2 n2 n2

The radial wave function, on the other hand, depends on both n and . The fact that the energy levels do not depend on is surprising but has

32 an analog for classical, gravitational orbits. The Coulomb potential wave function is described by three integer quantum numbers {n, , m} which obey the relationship

4. We discussed spontaneous electic dipole emission which describes the domi- nent process by which excited atoms return to their ground state. The rate for this process per atom is given by:

4α 3 ∗ R = ω <ψ1| r |ψ2 > · <ψ1| r |ψ2 > 3 c2

−1 −3 −2 3 2 or R =0.382 ns eV nm (∆E) |ψf | r|ψi|

5. The rate for spontaneous emission is related to the lifetime of an excited electron orbital as well as the width of its spectral line according to the Uncertainty Principle.

1 ¯h τ = , Γ= i Ri τ

Γ is the full width at half maximum for the excited state’s uncertain energy.

6. We developed several selection rules which describe when electric dipole operator vanishes because of symmetry. The most general rule is that the parity of the initial state must diﬀer from the parity of the ﬁnal state. The parity is given by (−1).Wealsohavetherules∆ =1and∆m = −1, 0, +1.

Often the components of electric dipole matrix element | <ψf | r |ψi > | are relatively imaginary which implies that their dipole antenna radiate 90o out of phase. This means that circularly polarized light is emitted from the atom.

7. Ultimately the selection rules imply that the photon carries one unit of angular momentum which can be directed either along its axis (left circularly polarized light) or against its axis (right circularly polarized light).

33 8. Finally we discussed the role of stimulated emission in the production or amplification of coherent light in a laser or maser. Stimulated emission occurs when an electron is tickled by an external field and emits a photon which is in phase with the external field. Unfortunately the amplifying effects for stimulated emission are countered by the reverse process called stimulated absorption which occurs at the same rate. In order to achieve laser action, it is necessary to achieve population inversions which means that there are more electrons at a higher energy orbital than at a lower energy orbital. This inversion is often accomplished by optical pumping.