H7C Section Material

Physics H7C Section Notes

Kevin Grosvenor

Last Edited: April 24, 2011

b

Contents

1 Special Relativity 11.1 Relativistic Train 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Relativistic Train 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Colliding Photons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.4 Length Contraction in Arbitrary Direction . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 The Propagation of Light 92.1 Rayleigh Scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.2 Apparent Depth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

3 Geometrical Optics 113.1 Refraction at a Spherical Interface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113.2 Hecht P.6.8 p.278 (Modified) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123.3 Silvered Thin Lens . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

4 Interference 154.1 Laser Wavelength Measurement via Metal Ruler . . . . . . . . . . . . . . . . . . . . . . 154.2 Continuous Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

5 Polarization 195.1 Hecht P.8.32 p.381 (Augmented) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195.2 Hecht P.8.41 p.382 (Augmented) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

6 Midterm Review 236.1 Michelson Interferometer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236.2 Pion Photoproduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256.3 Microscopes and Embryos . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266.4 Three-Slit Interference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276.5 Polarizers Away! . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

7 Blackbody Radiation 317.1 Planck Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327.2 Planck Spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337.3 Stefan-Boltzmann Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

8 Quantum Operator Expectation Values 378.1 Infinite Square Well . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

i

ii CONTENTS

9 Operators and States 419.1 Harmonic Oscillator Bra-Kets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

10 Final Review 4510.1 Wavefunction Shapes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4510.2 Harmonic Oscillator Coherent States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4510.3 Finite Square Well Bound States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4610.4 A Theorem about Bound States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4610.5 Spin-Spin Interaction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4610.6 Fine and Hyperfine Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

Chapter 1Special Relativity

1.1 Relativistic Train 1

A relativistic train moves at velocity 0.8c in the +x direction. Observer A is at the back of thetrain. Observer B is at the front of the train. The conductor, C, is at the middle of the train. Thestationmaster, S, stands on the platform as the train passes. The conductor, C, and the stationmaster,S, agree to set their clocks to t = 0 when they are directly next to each other and to define this positionas x = 0. A, B and C all have identical synchronized clocks. The conductor, C, measures the train tobe two lightseconds (ls) long.

(a) At t = 0, C turns on a bulb. What time will A and B read on their clocks when they see the light?

(b) What does S measure for the length of the train?

(c) What time(s) does S observe on his clock when he sees the light reach A and B.

(d) What do your answers to (a) and (c) tell you about simultaneity? Draw two spacetime diagramsfor this process: one for the train reference frame and one for the platform reference frame.

SOLUTION:

(a) A and B are at rest relative to C and thus measure the same length for the train. In theirperspective, they are each 1 ls away from C. From their perspective, it takes the light one secondto go from C to them. Thus, A and B will both read 1 sec.

(b) The train will appear contracted to S by a factor of γ = 5/3. Thus, its length is 2/(5/3) = 6/5 = 1.2ls for the stationmaster.

(c) From the point of view of S, observer A is moving towards the light ray and observer B is movingaway. Thus, from the point of view of S, the relative speed of the light ray and A is 1.8c whereasit is 0.2c for the light ray and B. The distance that needs to be covered for each is 0.6 ls (half ofpart (b)). Thus, the time as viewed by S for A is (0.6 ls)/(1.8c) = 1/3 = 0.33 sec and the time forB is (0.6 ls)/(0.2c) = 3 sec.

You might want to think of this in the follow way instead. We can write the trajectory of A asxA = −0.6 ls + 0.8ct and the trajectory of B as xB = 0.6 ls + 0.8ct. Meanwhile, the light ray goingtowards A has the trajectory x`1 = −ct and the one heading for B is x`2 = ct. The time whenxA = x`1 is t = 0.33 sec and the time when xb = x`2 is t = 3 sec.

1

2 CHAPTER 1. SPECIAL RELATIVITY

(d) In the frame of reference of the train (and A, B and C), the light reaches the back and the frontof the train at the same time. But, from the point of view of S, it reaches the back in 0.33 sec andreaches the front in 3 sec. Thus, events that appear simultaneous in one frame will generically notappear simultaneous in a different frame.

You might be wondering what S observes on the clocks of A and B when the light reaches eachone (as opposed to what he observes on his own clock at these two events). On the answer to thisquestion, the two reference frames cannot possibly disagree. That is, A and B see their clocksread 1 sec, when the light reaches them. S must also see 1 sec on A’s and B’s clock when thelight reaches them. But, we already know that S sees the light reach A long before he sees itreach B. Hence, we must conclude that, while S observes A’s and B’s clocks ticking at the samerate, he observes that A’s clock is systematically ahead of B’s clock by a fixed amount, namely3 − 0.33 = 2.67 sec. Even though A’s and B’s clocks are synchronized in their reference frame,they are not synchronized in S’s reference frame!

Below (left) is the spacetime diagram from the train perspective and below (right) in the platformperspective. The vertical direction is time with each mark corresponding to 0.5 sec. The horizontaldirection is the x-direction with each mark corresponding to 0.5 ls. The dotted lines are thetrajectories of each observer and the solid directed line shows the light signal propagating fromC to A and B. In the train perspective, S is moving with velocity −0.8c x. Thus, the slope ofthe trajectory of S in this frame is 1/(−0.8) = −5/4. Similarly, the slopes of the A, B and Ctrajectories in the platform frame is +5/4. Meanwhile, all light signals have slopes of ±1 in bothframes. Note that A and B are 1 ls away from C in the train frame, whereas they are 0.6 ls awayfrom C in the platform frame. The squiggly dots show the moment when the light signal reachesA and B. In the train frame, these occur at the same time whereas in the platform frame, theseoccur at drastically different times.

1.2 Relativistic Train 2

A train and a tunnel both have proper lengths L. The train moves towards the tunnel at speed v. Abomb is located at the front of the train and is designed to explode when the front of the train passesthe far end of the tunnel. A deactivation sensor is located at the back of the train. When the back ofthe train passes the near end of the tunnel, the sensor tells the bomb to disarm itself. Does the bombexplode?

SOLUTION:

Yes, the bomb explodes. Let us first consider the train reference frame, in which the answer is obvious.In this frame, the train has length L and the tunnel has length L/γ < L and is heading towards the

1.2. RELATIVISTIC TRAIN 2 3

train at speed v. Therefore, it is clear that the back end of the tunnel will pass the front of the trainbefore the front end of the tunnel reaches the back of the train.

In the tunnel frame, the tunnel has length L and the train has length L/γ < L. Therefore, theback of the train reaches the near end of the tunnel before the front of the train reaches the back ofthe tunnel. You might be tempted to say that the bomb is then deactivated before it can explode.However, you have to keep in mind that the deactivator at the back of the train needs to send a signalto the bomb at the front of the train saying that it has reached the front end of the tunnel and thatthe bomb should therefore disarm itself. At best, that signal can travel at the speed of light. It willtake time for that signal to reach the bomb at the front of the train. If that time is longer than thetime it takes for the front of the train to reach the back of the tunnel, then it will be too late and thebomb will explode.

Let the front of the tunnel correspond to x = 0 and let t = 0 be when the back of the train passesthe front of the tunnel. Henceforth, the signal sent by the deactivator travels forward at the speed oflight, its worldline described by xs = ct (the s subscript stands for “signal”). At t = 0, the front of thetrain is at x = L/γ, since that is the length of the train in the tunnel reference frame. The trajectoryof the front of the train is xb = L

γ + vt (the b subscript stands for “bomb”). Which one reaches x = L

(the back of the tunnel) first? Well, the time it takes for the signal is ts = L/c whereas for the bombtakes tb =

(L− L

γ

)/v = ts

β

(1− 1

γ

), where β ≡ v/c. We claim that tb < ts and so the bomb explodes.

To prove this, start with the inequality β < 1, which just says that the train must be moving atless than the speed of light. Multiply by 2β and add 1 to both sides to get 1 + 2β2 < 1 + 2β. Now,subtract 2β + β2 from both sides to get 1− 2β + β2 < 1− β2. Rewrite the left hand side as (1− β)2,

then take the positive square root of both sides to get 1− β <√

1− β2 = 1γ . This final inequality can

be rearranged to read 1β

(1− 1

γ

)< 1. But, the left hand side is just tb/ts, and so tb < ts.

Below are spacetime diagrams in both reference frames in the case β = 4/5. Note that we have sett = t′ = 0 when the front of the train lines up with the front of the tunnel. But, note that we have notset x = x′ = 0 to be the position of this event. The spatial origins of the frames are different: x′ = 0for the center of the train and x = 0 for the center of the tunnel. In the train frame, the explosionhappens before the deactivation signal is sent. In the tunnel frame, those two events occur in theopposite order. However, in both reference frames, the bomb explodes; it certainly cannot be the casethat the train explodes in one frame whereas it does not in the other! Assuming that the explosion“signal” (i.e. the fires, etc.) travel at the speed of light, the red shaded regions represent the region ofthe train that is engulfed in fire, or at least the region that is aware of the fact that the explosion hasoccurred.


1.3 Colliding Photons

[Goldstein, Poole & Safko 7.22 ] A photon of energy E′ collides at angle θ with a photon of energy E.Determine the minimum value of E′ permitting the formation of a pair of particles of mass m.

SOLUTION:

Set c = 1 for the moment. Let the four-momenta of the photons be

p = E(1, 1, 0, 0), p′ = E′(1, cos θ, sin θ, 0).

This is written in a coordinate system where x points in the direction of propagation of the photonwith energy E. In these coordinates, which we will call x′ and y′, the total 4-momentum is

p′t = p+ p′ = (E + E′, E + E′ cos θ,E′ sin θ, 0).

These coordinates turn out to be inconvenient. Define the x and y coordinates to be such that x is inthe direction of the total 3-momentum (the last three components of the total 4-momentum). Thesecoordinates will simply be rotated with respect to the x′ and y′ coordinates. Thankfully, we do notreally need to know what this rotation is, since we don’t care about what p and p′ look like in thesecoordinates. What we know is that, if pt is the total 4-momentum in these unprimed coordinates,and p′t and pt are the 3-momentum parts of p′t and pt, then |p′t| = |pt|, since rotations do not changemagnitudes. Conveniently, in these coordinates, the y and z components of pt vanish, so we can justneglect them altogether and just write the t and x components:

pt =

(E + E′

|p′t|

)=

(E + E′√

E2 + E′2 + 2EE′ cos θ

)= (E + E′)

(1√

1− 2EE′(1−cos θ)(E+E′)2︸︷︷︸A

).

Henceforth, we may just drop the y and z entries since they are identically zero. Now, boost in the+x direction by some amount β:

γ

(1 −β−β 1

)(E + E′)

(1A

)= γ(E + E′)

(1− βA−β +A

).

To make the boosted frame the center of momentum frame (COM), we need the x component of theboosted total momentum to vanish, which implies β = A. The corresponding γ factor is

γ = (1− β2)−1/2 =

(1− 1 +

2EE′(1− cos θ)

(E + E′)2

)−1/2

=E + E′√

2EE′(1− cos θ).

Thus, the total energy in the COM is

Ecom = γ(E + E′)(1− βA) =E + E′

γ=√

2EE′(1− cos θ),

where use was made of β = A and 1− β2 = γ−2.

1.4. LENGTH CONTRACTION IN ARBITRARY DIRECTION 5

Actually, there is a much simpler way to get to this result for Ecom. The combination AµAµ is

Lorentz inveriant for any covariant vector, Aµ. Recall that AµAµ ≡ A2

0 − A21 − A2

2 − A23. For p′t this

gives(p′t)µ(p′t)

µ = (E + E′)2 − (E + E′ cos θ)2 − (E′ sin θ)2 = 2EE′(1− cos θ).

The total 4-momentum in the COM is pt,com = (Ecom, 0, 0, 0) with invariant

(pt,com)µ(pt,com)µ = E2com.

The invariant for p′t must be equal to the invariant for pt,com thereby yielding the previously calculatedresult for Ecom.

We must have Ecom ≥ 2m in order to form a pair of particles of mass m, thus yielding

E′ ≥ 2m2

E(1− cos θ).

Now, we must reintroduce c. In order for the RHS to have units of energy, we must multiply by c4:

E′ ≥ 2m2c4

E(1− cos θ).

1.4 Length Contraction in Arbitrary Direction

Consider a cube of side-length 1 (in some appropriate units). Let S be the rest frame of the cube inwhich the sides of the cube are parallel to the coordinate axes, one corner is at the origin, and the cubeoccupies the positive octant (i.e. the coordinates of any point in the cube are non-negative). Let S′

be a frame whose axes are parallel with those of S, but with respect to which the cube is moving withvelocity βcx′. Let S′′ be a frame that is rotated in the xy-plane relative to S′ in the counter-clockwisedirection by an angle θ. Determine the volume of the cube in the frames S′ and S′′.

[Note: check out Wikipedia for the volume of a Parallelepiped.]

SOLUTION:

The volume in S′ is 1/γ, where γ =[1− β2

]−1/2. Hopefully, this much is familiar from problem set 1.

Incorrect solution for S”: First, let us try to follow the naıve procedure that I had proposedin section: break up the velocity into its x′′ and y′′ components in frame S′′ and contract eachdirection accordingly. So, βx′′ = β cos θ and βy′′ = β sin θ, with corresponding gamma factors, γx′′ =[1− β2

x′′

]−1/2and γy′′ =

[1− β2

y′′

]−1/2. After some algebra, we can solve for these gamma factors in

terms of the gamma factor of S′ with respect to S, namely γ =[1− β2

]−1/2:

1

γx′′=[cos2 θ

γ2+ sin2 θ

]1/2,

1

γy′′=[ sin2 θ

γ2+ cos2 θ

]1/2.

Therefore, we might expect the volume to decrease by

1

γx′′

1

γy′′=

1

γ

[1 + β4γ2 cos2 θ sin2 θ

]1/2,

where I have performed a fair amount of algebra to simplify the answer to this form.Notice that this is not the same as the volume we found in S′, which is 1/γ. However, the difference

is of order β4, which is at least very small except in the limit β → 1.


Correct solution: Since S′′ is related to S′ only by a spatial rotation, and spatial rotations DO NOTchange spatial distances, the volume in S′ ought to be the same as the volume in S′′ exactly, not onlyapproximately! Let us rigorously trace the steps to get the volume in S′, then it should be clear howto extend the procedure to S′′.

Let the spacetime coordinates of the cube corners in S be

p1 = (t, 0, 0, 0), p5 = (t, 1, 1, 0),

p2 = (t, 1, 0, 0), p6 = (t, 1, 0, 1),

p3 = (t, 0, 1, 0), p7 = (t, 0, 1, 1),

p4 = (t, 0, 0, 1), p8 = (t, 1, 1, 1).

Note that the spatial coordinates are t-independent since S is the cube’s rest frame. Also, so that wecan keep track of volume, p2 can be considered the width, p3 the length, and p4 the height, where piis the spatial vector corresponding to pi (i.e. the last three components). Therefore, the volume in Sis 1, as expected:

(p2 × p3) · p4 =[(1, 0, 0)× (0, 1, 0)

]· (0, 0, 1) = (0, 0, 1) · (0, 0, 1) = 1.

We must boost these coordinates into the S′ frame, which is moving relative to S with a velocity−βc x. The appropriate boost transformation is

Λ =

γ βγ 0 0βγ γ 0 00 0 1 00 0 0 1

.

The coordinates of the corners in S′ are p′i = Λpi:

p′1 = (γt, βγt, 0, 0), p′5 = (γ(t+ β), γ(βt+ 1), 1, 0),

p′2 = (γ(t+ β), γ(βt+ 1), 0, 0), p′6 = (γ(t+ β), γ(βt+ 1), 0, 1),

p′3 = (γt, βγt, 1, 0), p′7 = (γt, βγt, 1, 1),

p′4 = (γt, βγt, 0, 1), p′8 = (γ(t+ β), γ(βt+ 1), 1, 1).

This form is not too useful. We want to write each 4-vector in terms of t′, where t′ is set equal to thezeroth component of each 4-vector separately. For example, for p′1, set t′ = γt, then x′1 = βγt = βt′. Asanother example, for p′2, set t′ = γ(t+β), then, after some algebra, one finds x′2 = γ(βt+ 1) = βt′+ 1

γ :

p′1 = (t′, βt′, 0, 0), p′5 = (t′, βt′ + 1γ , 1, 0),

p′2 = (t′, βt′ + 1γ , 0, 0), p′6 = (t′, βt′ + 1

γ , 0, 1),

p′3 = (t′, βt′, 1, 0), p′7 = (t′, βt′, 1, 1),

p′4 = (t′, βt′, 0, 1), p′8 = (t′, βt′ + 1γ , 1, 1).

Now, let us see the positions of the cube corners in S′ at some fixed time in S′, say t′ = 0:

p′1 = (0, 0, 0), p′5 = ( 1γ , 1, 0),

p′2 = ( 1γ , 0, 0), p′6 = ( 1

γ , 0, 1),

p′3 = (0, 1, 0), p′7 = (0, 1, 1),

p′4 = (0, 0, 1), p′8 = ( 1γ , 1, 1).

The volume in S′ is 1/γ, as expected:

(p′2 × p′3) · p′4 =[( 1γ , 0, 0)× (0, 1, 0)

]· (0, 0, 1) = (0, 0, 1

γ ) · (0, 0, 1) = 1γ .

1.4. LENGTH CONTRACTION IN ARBITRARY DIRECTION 7

Now, the appropriate rotation matrix that takes S′ into S′′ is

R =

1 0 0 00 cos θ sin θ 00 − sin θ cos θ 00 0 0 1

Then, p′′i = Rp′i are the spacetime coordinates of the cube corners in S′′. Note that R does not changethe zeroth (time) component and so t′′ = t′. You can go ahead and do the algebra to calculate thevolume, but the result is obvious: spatial rotations cannot change lengths and thus the volume willstill be 1/γ.

[Note: I invite you to try and see what happens if you first rotate using R from S to another frame,S′′′, then boost by −βx′′ in the horizontal direction, and then boost by +βy′′ in the vertical direction.My initial calculations are encouraging in the sense that I get approximately the same result as the“incorrect solution”. But, it’s such a big mess of a calculation that I’m just going to stop and notwrite any more. You may want to ruminate on the distinction between the frame one gets with thisset of transformations versus the frame one gets from first boosting from S to S′ and then rotatingfrom S′ to S′′.]


Chapter 2The Propagation of Light

2.1 Rayleigh Scattering

[Hecht P.4.2 (modified and augmented)] A white floodlight beam crosses a large volume containinga tenuous molecular gas mixture of mostly oxygen and nitrogen. Compare the relative amount ofscattering occurring for the red (∼ 750 nm) component with that of the blue (∼ 450 nm) component.What does this have to say about the color of sky and the sunset?

SOLUTION:

This problem is about Rayleigh scattering, whose intensity goes like ∼ 1/λ4. Thus,

red scattering

blue scattering=

(1/750 nm)4

(1/450 nm)4= 0.13. (2.1)

When sunlight passes through the atmosphere, the blue component is scattered more than the red.What we see as the color of the sky is just scattered sunlight, which will be dominated by blue. Whatwe see in the sunset is mostly direct sunlight, the blue component of which has mostly been scatteredby the time it travels through so much atmosphere, leaving mostly red.

2.2 Apparent Depth

[Hecht P.4.25 ] A coin is resting on the bottom of a tank of water (nw = 1.33) 1.00 m deep. Ontop of the water floats a layer of benzene (nb = 1.50), which is 20.0 cm thick. Looking down nearlyperpendicularly, how far beneath the topmost surface does the coin appear?

SOLUTION:

Consider the following diagram

9

10 CHAPTER 2. THE PROPAGATION OF LIGHT

There are a couple of things to note about this diagram: in order to actually show the angles, we havemade them quite large, when, in actual fact, they’re supposed to be very small. Secondly, in orderto show the separate triangles (black, red, and blue), we have displaced them relative to each other,slightly. The light ray (lines with arrows) is black in the water, red in benzene, and blue in air, butthis is unrelated to their actual color in those media! The incident and transmitted angles relative thenormal at each surface in water, benzene, and air are denoted θw, θb, and θa. By simple geometry,these also happen to be the angles at the bottom corners of the appropriate triangles (θw for the blacktriangle, θb for the red triangles, and θa for the blue triangle). Finally, the apparent depth relative tothe topmost surface (benzene-air interface) is h′′.

The point of looking down nearly perpendicularly is to make the incident and refracted anglessmall, and thus to justify the small angle approximation sin θ ≈ tan θ ≈ θ. In this case, Snell’s lawreads niθi = ntθt, or θi

θt= nt

ni. Thus, θwθb = nb

nwand θb

θa= na

nb.

Comparing the black triangle with the smaller red triangle, we have the two equations

tan θw =x

h, tan θb =

x

h′.

Dividing these two equations and using the small angle approximation yields

h′

h=

tan θwtan θb

≈ θwθb≈ nbnw

.

Comparing the larger red triangle with the blue triangle, we have the two equations

tan θb =y

h′ + t, tan θa =

y

h′′.

Dividing these two equations and using the small angle approximation yields

h′′

h′ + t=

tan θbtan θa

≈ θbθa≈ nanb.

Thus, the apparent depth is

h′′ =nanb

(h′ + t) =nanb

[ nbnw

h+ t]

=1

1.50

[1.50

1.33(1.00 m) + 0.20 m

]= 0.883 m = 88.3 cm.

Chapter 3Geometrical Optics

3.1 Refraction at a Spherical Interface

[Hecht P.5.5 ] Making use of the diagram below, Snell’s Law, and small angle approximations for allthe relevant angles, derive the equation n1

so+ n2

si= n2−n1

R .

SOLUTION:

θ1 is the supplement of ∠ABG = 180− α− ϕ. Therefore, θ1 = α+ ϕ. Similarly, ϕ is the supplementof ∠BGE = 180− θ2 − β. Therefore, ϕ = θ2 + β, or θ2 = ϕ− β.

CD = R −√R2 − h2 = R

[1−

√1− (h/R)2

]≈ R

[1− 1 + 1

2 (h/R)2]≈ 0, where we have dropped

the quadratic term, (h/R)2, since small angles α, β and ϕ imply small h/R.The small angle approximations for α, β and ϕ now read

α ≈ tanα ≈ h

so, β ≈ tanβ ≈ h

si, ϕ ≈ sinϕ =

h

R.

Therefore, the corresponding small angle approximations for θ1 and θ2 read

sin θ1 = sin(α+ ϕ) ≈ α+ ϕ ≈ h

so+h

R, sin θ2 = sin(ϕ− β) ≈ ϕ− β ≈ h

R− h

si.

Therefore, Snell’s Law reads

n1 sin θ1 = n2 sin θ2 =⇒ n1h

so+n1h

R=n2h

R− n2h

si,

from which the desired relation follows directly.From here on, the derivation of the lensmaker formula proceeds as in Hecht resulting in Hecht’s

equation (5.14) with (5.15) as the thin lens limit.

11

12 CHAPTER 3. GEOMETRICAL OPTICS

3.2 Hecht P.6.8 p.278 (Modified)

Compute the magnification that results when the image of a flower 4.0 m from the center of a solid,clear-plastic sphere with a 0.20-m radius (and a refractive index of 1.4) is cast on a nearby wall. De-scribe the image in detail.

SOLUTION:

Denote the object and image distances with respect to the vertices with overlines. Those withoutoverlines will be taken with respect to the principal points. Note that so1 = 3.8 m since it is withrespect to the left vertex and not the center! Let nm = 1 be the index of the medium and let n` = 1.4be the index of the lens. Let R1 = 0.2 m and R2 = −0.2 m be the radii of the front and back surfaces,respectively. Let d = 0.4 m be the thickness of the lens.

Method 1: For the first surface,

nmso1

+n`si1

=n` − nmR1

.

Everything is known here except si1 whose solution yields si1 = 0.806 m, which is with respect to the

left vertex. The transverse magnification is MT1 = −nmsi1n`so1= − 1(0.806)

1.4(3.8) = −0.151.

Now, for the second surface:n`

d− si1+nmsi2

=nm − n`R2

.

Again, everything is known except for si2 whose solution yields si2 = 0.184 m, which is now with

respect to the right vertex. The transverse magnification is MT2 = − n`si2nmso2

= 1.4(0.184)1(0.4−0.806) = 0.633.

The total transverse magnification is MT = MT1MT2 = −0.0959. The image height is about a tenthof the original flower height and it is up-side-down. Let us assume the paraxial limit and neglect allaberrations.

Method 2: Let us first calculate the focal length with respect to the principal points. Definen`m ≡ n`/nm = 1.4 as the relative index of refraction.

1

f= (n`m − 1)

( 1

R1− 1

R2+

(n`m − 1)d

n`R1R2

).

Plugging in the numbers gives f = 0.35 m. Next, we calculate the principal points:

h1 = − (n`m − 1)df

n`R2= 0.2 m, h2 = − (n`m − 1)df

n`R1= −0.2 m.

This means that H1 is 0.2 m to the right of V1 and H2 is 0.2 m to the left of V2, which puts them bothright at the center. We could just remember the fact that for a perfect sphere, the principal pointsconverge at the optical center. Thus, the distances with respect to the principal points are actuallywith respect to the optical center, which simplifies things! In particular, so = 4 m and we have

1

so+

1

si=

1

f=⇒ si = 0.384 m.

This agrees with Method 1 since si = 0.384 = 0.184 + 0.2 = si2 − h2. The transverse magnificationalso agrees:

MT = − f

xo= − f

so − f= − 0.35

4− 0.35= −0.0959.

3.3. SILVERED THIN LENS 13

3.3 Silvered Thin Lens

Suppose you have a double convex lens made of glass (n = 1.5) such that the magnitude of the radiusof curvature of both sides is R = 42 m. One of the sides is painted with a layer of silver and thus actslike a spherical mirror. Assume the thin lens approximation and thus, that the lens and the mirrorare essentially at the exact same position. An object lies to the left of the unsilvered face of the lensa distance so = 84 m away.

(a) What is the focal length of the lens by itself? What about the mirror by itself?

(b) How many stages are there in this optical system. For each stage, state whether the light is incidentfrom the left or from the right.

(c) For each stage, calculate the image distance and linear magnification using the image of one stage asthe object for the following stage. Make a diagram for each stage separately. Caution: rememberthat using the image of one stage as the object for another does not mean using the image distanceof one stage as the object distance of another.

(d) Where is the final image located and what is the total linear magnification? Describe the image(i.e. is it real or virtual, upright or inverted, bigger or smaller?)

SOLUTION:

(a) Since the lens is double convex, the center of curvature of the left surface is on the right andthe right surface is on the left. Thus, R1 = R and R2 = −R, by the sign conventions, for lightoriginating from the object on the left. From the Lensmaker equation,

1f`

= (n− 1)(

1R1− 1

R2

)= (1.5− 1)

(1R −

1−R)

= 1R =⇒ f` = R = 42 m.

For the mirror, the focal length is just half of the radius of curvature. Since this is a concavemirror, it is converging and thus has a positive focal length:

fm = R/2 = 21 m.

(b) There are three stages:

(1) The lens. Light goes from left to right through the lens;

(2) The mirror. Light goes from left to right, reflects off of the mirror, and goes right to left;

(3) The lens again. Light goes from right to left through the lens.

(c) Let soj and sij be the object and image distances for stage j = 1, 2, 3 and so so1 ≡ so = 84 m = 2Ris the initial object distance and si3 ≡ si is the final image distance.

(1) The image distance just for the lens alone (stage 1) is

1si1

= 1f`− 1

so1= 1

R −1

2R = 12R =⇒ si1 = 2R = 84 m.

The linear magnification of stage 1 is

M1 = − si1so1

= − 84 m84 m = −1.

14 CHAPTER 3. GEOMETRICAL OPTICS

(2) Since si1 is positive, image 1 lies to the right of the lens. This is the same as object 2. Thus,object 2 lies to the right of the mirror whereas the light is actually hitting the mirror from theleft! This is a virtual object with negative object distance: so2 = −si1 = −2R = −84 m. So,

1si2

= 1fm− 1

so2= 2

R −1−2R = 5

2R =⇒ si2 = 25R = 16.8 m.


M2 = − si2so2

= − 16.8 m−84 m = 1

5 .

(3) Since si2 is positive, image 2 lies to the left of the mirror. This is the same as object 3. Thus,object 3 lies to the left of the lens whereas the light is actually entering the lens from the right!This is a virtual object with negative object distance: so3 = −si2 = − 2

5R = −16.8 m. So,

1si3

= 1f`− 1

so3= 1

R −5−2R = 7

2R =⇒ si3 = 27R = 12 m.


M3 = − si3so3

= − 12 m−16.8 m = 5

7 .

Below, virtual objects are drawn in gray. Some rules are tricky: for example, in stage 2, wecan’t actually draw the light ray that starts from the object parallel to the axis and reflectsoff of the mirror heading towards the focal point. This is because the object is virtual andbehind the mirror - there is no actual light there! In fact, the light has to go left to right. Sowe have to draw the light ray that looks like its headed towards the object parallel to the axis,and then this ray reflects towards the focal point.

(d) The final image is 12 m to the left of the lens. The total linear magnification is M = M1M2M3 =−1/7. The image is real, inverted and smaller by a factor of 7.

Chapter 4Interference

4.1 Laser Wavelength Measurement via Metal Ruler

Devise and explain a method for measuring the wavelength of a laser pointer chiefly using a finelygraded metal ruler (e.g. with 1/64 inch markings or smaller).

SOLUTION:

Consider reflecting the laser off of the ruler at a shallow angle. If there were no notches on the surfaceof the ruler, then each point on the ruler where the light hits becomes a source for outwardly spreadingspherical waves. The superposition of these waves produces wavefronts that travel in the direction ofspecular reflection. That is, on a far-away screen, we get constructive interference only around thepoint of specular reflection, as expected.

Imagine we make wide notices with narrow reflective bands in between. Then, consider the followingdiagram showing two adjacent light beams headed towards a far-away screen (e.g. a wall) havingreflected off of two adjacent reflective bands.

The optical path length difference is d(cosα− cosβ), which must be set equal to mλ for some integerm for constructive interference. For a fixed α (angle at which we shine the laser on the ruler), thisgives discrete values for β where bright spots occur (i.e. we get a diffraction pattern).

The claim is that we see the same thing if instead we have narrow non-reflective notches with widerreflective bands. Can you think of why? Hint: superposition. This goes under the name of Babinet’sprinciple, by the way.

Consider the following setup

15

16 CHAPTER 4. INTERFERENCE

This gives us an expression for the wavelength

λ =d

m

[ 1√1 + (s0/L)2

− 1√1 + (sm/L)2

].

If you make α very small and use only low orders, then we can assume that sm/L << 1:

λ ≈ d(s2m − s2

0)

2mL2.

As an example, when I did this experiment at home, I used the marks on the ruler that were d = 0.5mm apart and the distance to the wall was L = 105 cm. I found s0 = 10.5 cm and s1 = 11.9 cm.Assuming small angles, this gives

λ =(5× 10−4 m)[(11.9 cm)2 − (10.5 cm)2]

2× 1× (105 cm)2≈ 711 nm.

That’s not bad! The wavelength should be around 635 nm. Before we rejoice, however, we should notethat a millimeter difference in any sm makes a huge difference in the final answer. For example, if Ichange s1 to 11.8 cm, I get λ = 657 nm! So, unless I can measure sm and L with very high precision,the uncertainties are likely to swamp the final measurement of λ anyway.

Note: In section, I claimed that if the laser light reflects off of a smooth metalic surface, then we onlyget specular reflection. This is true only if the region on the surface that is illuminated is much widerthan the wavelength of the light. Well, our laser beam has a width of a few millimeters, which willobviously do since its wavelength is on the order of 10−4 mm! As calculated above, the extra opticalpath length travelled by one beam relative to another that hits the surface a distance x to the left of itis x(cosα− cosβ). So, the phase shift is φ = 2π xλ (cosα− cosβ). Let a be the width of the illuminatedregion and let x run from −a/2 to a/2 with the “zero phase” corresponding to x = 0. The intensity isproportional to

I ∝∣∣∣∫ a/2

−a/2e2πi xλ (cosα−cos β) dx

∣∣∣2 ∝ cos2A

A2,

where A = π aλ (cosα− cosβ).In the limit a/λ << 1, the intensity becomes a delta function:

Ia/λ→0−−−−→ δ(A).

Thus, the intensity vanishes everywhere except when A = 0, or when cosα = cosβ, or α = β, whichis the condition for specular reflection!

4.2 Continuous Case

A single slit has length a and width b. Determine the interference pattern that it produces on a farawayscreen assuming monochromatic, coherent light is normally incident upon it.

SOLUTION:

Consider the following setup of coordinates:

4.2. CONTINUOUS CASE 17

We have separate but parallel coordinates on the slit screen (X and Y ) and the observation screen(x and y), a large distance L away. I have chosen the very center of the slit to be the origin ofcoordinates on the slit screen and the point on the observation screen directly in line with that tobe the corresponding origin on the observation screen. Let the length of the slit be along horizontaldirection, so that −a2 ≤ X ≤

a2 and − b

2 ≤ Y ≤b2 .

Consider a little rectangle located in the slit at the coordinates (X,Y ) and with length dX andwidth (or height?) dY . The contribution, dE, of this little bit of area to the electric field at the pointon the observation screen with coordinates (x, y) should be proportional to its area, dX dY :

dE =E dX dY

rei(kr−ωt),

where E is some constant having to do with the source about which we need not care.We calculate r using the Pythagorean theorem:

r =√L2 + (x−X)2 + (y − Y )2 = L

[1 +

(xL

)2+(yL

)2 − 2xXL2 − 2yY

L2 +(XL

)2+(YL

)2]1/2≈ L

[1 + 1

2

(xL

)2+ 1

2

(yL

)2 − xXL2 − yY

L2

]= R− xX

L −yYL ,

where we have definedR ≡ L+ x2+y2

2L .

R is a constant as far as position on the slit screen is concerned (i.e. it does not change with X orY ). Therefore, it does not contribute to the diffraction pattern. We don’t care about the differencesin the path lengths, r, when it comes to the amplitude of the electric field contribution of the little bitof area. This is because tiny differences in amplitude make tiny overall differences to the total electricfield. But, we must keep this tiny differences in the complex exponential because even tiny values ofX and Y can produce dramatic phase shifts. Thus,

dE ≈ ER ei(kR−ωt) e−ikxX/L e−ikyY/L dX dY.

Denote ERei(kR−ωt) by A. We must integrate this:

E =

∫slit

dE = A

∫ a/2

−a/2e−ikxX/L dX

∫ b/2

−b/2e−ikyY/L dY

= AL

−ikxe−ikxX/L

∣∣∣a/2−a/2

L

−ikye−ikyY/L

∣∣∣b/2−b/2

=AL2

k2xy

[e−ikax/2L − eikax/2L−i

][e−ikby/2L − eikby/2L−i

]=

4AL2

k2xysin(kax

2L

)sin(kby

2L

).

18 CHAPTER 4. INTERFERENCE

This gives us E as a function of the coordinate (x, y) on the observation screen. Let us take the limitas (x, y)→ (0, 0). For this, we can use the small angle approximation, sin θ ≈ θ:

E(0, 0) = lim(x,y)→(0,0)

4AL2

k2xy· kax

2L· kby

2L= abA.

Therefore, we can eliminate A and relate the electric field at an arbitrary point (x, y) on the observationscreen to the electric field at the origin of the observation screen, (0, 0):

E(x, y) =4L2E(0, 0)

k2abxysin(kax

2L

)sin(kby

2L

).

Define the following expressions,

α ≡ ka

2L=πa

λL, β ≡ kb

2L=πb

λL.

Then, we can write

E(x, y) = E(0, 0)sin(αx)

αx

sin(βy)

βy= E(0, 0) sinc(αx) sinc(βy).

Finally, the intensity is proportional to the square of this:

I(x, y) = I(0, 0) sinc2(αx) sinc2(βy).

You should recognize this as simply the product of the single slit diffraction patterns for slits of widtha in the x-direction and b in the y-direction.

We could use this result almost directly to calculate the diffraction pattern produced by a cross-shapedaperture, which you considered in a previous problem set. The cross can be thought of as the sum oftwo of our single slits considered above, one with its length along the x-direction and one along they-direction. However, we would be counting their intersection twice! So, we must subtract the resultof a square slit of sidelength b:

E(x, y) = E(0, 0)[sinc(αx) sinc(βy) + sinc(βx) sinc(αy)− sinc(βx)sinc(βy)

].

Again, the intensity is just proportional to the square of this, which doesn’t really simplify; there’sno point in multiplying it all out. Below is a contourplot of the intensity as a function of (x, y). Thediagram on the left (right) is for a cross with arms whose length is three (nine) times their width.

Chapter 5Polarization

5.1 Hecht P.8.32 p.381 (Augmented)

A beam of natural light is incident on an air-glass interface (nit = 1.5) at 40◦. Compute the degree ofpolarization of the reflected light.

SOLUTION:

From Hecht Eqns. 8.26 and 8.27 p.351, the parallel and perpendicular reflectances are

R|| =tan2(θi − θt)tan2(θi + θt)

, R⊥ =sin2(θi − θt)sin2(θi + θt)

.

In this case, θt = sin−1 nint

sin θi = 25.37◦ and so

R|| = 0.014, R⊥ = 0.077.

If the total incident intensity is I, then the incident intensity that is polarized in the parallel direction isI/2, and the same for perpendicular: Ii|| = Ii⊥ = I/2. Then, the reflected intensities in the respectivepolarizations is

Ir|| = R||Ii|| =R||2 I, Ir⊥ = R⊥Ii⊥ = R⊥

2 I.

The difference between the two is the part that is completely polarized and the remainder is the partthat is completely unpolarized:

Ip =(R⊥ −R||

2

)I, In = Ir,tot − Ip =

(R⊥ +R||

2

)I −

(R⊥ −R||2

)I = R||I.

Thus, the degree of polarization is

V =Ip

Ip + In=

IpIr,tot

=R⊥ −R||R⊥ +R||

= 0.687.

The reflected light is 68.7% polarized in the direction perpendicular to the plane of incidence.

19

20 CHAPTER 5. POLARIZATION

5.2 Hecht P.8.41 p.382 (Augmented)

Take two ideal Polaroids (the first with its axis horizontal and the second, vertical) and insert betweenthem a stack of 10 half-wave plates, the first with its fast axis rotated π/40 rad from the horizontal,and each subsequent one rotated π/40 rad from the previous one. Determine the ratio of the emergingto incident irradiance.

SOLUTION:

Method 1: If I0 is the incident intensity, then after the first polarizer, the intensity is I0/2 andpolarized in the x direction. Its angle relative to the optical axis of the first HWP is −π/40, whichbecomes π/40 after the first HWP. Remember that a HWP turns a polarization angle α into −α (orπ−α or any whole number of π’s −α). But, π/40 relative to the optical axis of the first HWP is π/20relative to the x-axis (i.e. the transmission axis of the first polarizer. That happens to be the angleof the optical axis of the second HWP relative to the x-axis. Thus, after the first HWP, the light ispolarized along the optical axis of the second HWP. Therefore, the second HWP does not change itspolarization. The third HWP will rotate the polarization to point along the axis of the fourth HWP,and so on. Therefore, as long as we have an even number of HWP’s, the final polarization will be alongthe final HWP. In this case, that would be at an angle of π/4 relative to the x-axis. The HWP’s changethe polarization but not the intensity. Passing it through the final polarization cuts the intensity downby a factor of cos2(π/4) = 1/2. Thus, the final intensity is I0/4.

Method 2: This is quite a bit more involved, but it’s also more algorithmic and would be superiorin dealing with, for example, the case when the HWP’s are not all evenly rotated. Let N = 10 be thenumber of half-wave plates. First, we would like to relate the fast and slow axes of the nth half-waveplate to the original axes of the first Polaroid. We’ll call the optical axis the fast axis. Consider thediagram below.

It should be clear that

fn = cos(nπ4N

)x + sin

(nπ4N

)y, sn = − sin

(nπ4N

)x + cos

(nπ4N

)y.

In the sequel, we will abbreviate cos(nπ4N

)by cn and the sine by sn.

Denote the vector xx + yy by(xy

)and the vector fnfn + snsn by

(fnsn

)n. Note that the subscript n

on the column vector reminds us that it is written in terms of the {fn, sn} basis and not the original{x, y} basis. We have(

fnsn

)n

= fnfn + snsn = fn(cn x + sn y) + sn(−sn x + cn y) =

(cn sn−sn cn

)(fnsn

).

The inverse relation would read (xy

)=

(cn −snsn cn

)(xy

)n

.

Let’s unpack these equations. The first one says that if I know the components of a vector in the xy-basis, namely fn for the x component and sn for the y component, then the corresponding componentsin the fnsn-basis is given by the rotation matrix

(cn sn−sn cn

)multiplying the vector in the xy-basis. The

5.2. HECHT P.8.41 P.382 (AUGMENTED) 21

second equation is just the inverse: if you know the components in the fnsn-basis, namely x for the fncomponent and y for the sn component, then the equation will give you the corresponding componentsin the xy-basis.

Let’s just check one example to make sure this makes sense. In the xy-basis, the horizontal vectorpointing to the right is

(10

). In the fnsn-basis, just by looking at the diagram, you should see that it

should be(

cn−sn)n. Well, this is indeed what we get using the transformation rule:(

cn sn−sn cn

)(10

)=

(cn−sn

)n

.

Now, the matrix representing a HWP with its optical axis along the x-direction is

Φ ≡(eiπ 00 1

)=

(−1 0

0 1

).

It is important to note that this matrix is written with respect to the xy-basis. Also, we have madethe choice, as was done in the lab, of saying that the electric field along the optical axis is flipped whilethe perpendicular component is unchanged.

What matrix would represent one of our HWP’s whose optical axis is rotated by nπ/4N relativeto the positive x-axis? Well, let’s call it

(A BC D

). This means that if you pass light with polarization

given by(fnsn

)(in the xy-basis!), then what comes out is(

A BC D

)(fnsn

).

This is in the xy-basis. To convert it to the fnsn-basis, we multiply by the appropriate rotation:(cn sn−sn cn

)(A BC D

)(fnsn

).

However, we could have written the initial vector in the fnsn-basis before it passed through the HWP.Then, we are already in the fnsn-basis and we just have to write the matrix for the HWP in this basis,which is just

(−1 00 1

), since the optical axis of the HWP points along fn. Therefore, the result above

should be the same as (−1 0

0 1

)(cn sn−sn cn

)(fnsn

).

This implies that (cn sn−sn cn

)(A BC D

)=

(−1 0

0 1

)(cn sn−sn cn

).

Therefore, the matrix representing the rotated HWP in the xy-basis is related to its matrix in thefnsn-basis by a similarity transformation, each step of which is just a rotation:(

A BC D

)=

(cn −snsn cn

)(−1 0

0 1

)(cn sn−sn cn

)≡ RnΦR−1

n .

Note the relation R−1n+1Rn = R−1

1 , which just says that a forward rotation by nπ/4N followed by abackward rotation by (n + 1)π/4N is equivalent to a backward rotated by −π/4N . Hopefully that’sobvious.

The matrix representing the N successive HWP’s is

(RNΦR−1N )(RN−1ΦR−1

N−1) · · · (R1ΦR−11 ) = RNΦ(R−1

N RN−1)Φ · · · (R−12 R1)ΦR−1

1

= RNΦR−11 Φ · · ·R−1

1 ΦR−11

= RN (ΦR−11 )N .

22 CHAPTER 5. POLARIZATION

Now, we calculate:

ΦR−11 =

(−1 0

0 1

)(c s−s c

)=

(−c −ss c

).

Conveniently, we find

(ΦR−11 )2 =

(cos2 + sin2 cos sin− sin cos

sin cos− cos sin sin2 + cos2

)=

(1 00 1

).

Therefore, as long as N is even, the matrix for the N HWP’s is

RN (ΦR−11 )N = RN =

(cos(π/4) − sin(π/4)sin(π/4) cos(π/4)

)=

1√2

(1 −11 1

),

Now, the light incident on this collection of HWP’s has a polarization vector(

10

)(i.e. horizontal

direction). After the HWP’s, it has a polarization vector given by

1√2

(1 −11 1

)(10

)=

1√2

(11

),

which is indeed in the direction π/4 = 45◦ from the positive x-axis. As in method 1, the final polarizerlets cos2(π/4) = 1/2 of this intensity through. So the first polarizer cuts the intensity down to I0/2,the ten HWP’s rotate the polarization by 45◦, and the final polarizer cuts the intensity down to I0/4.

Chapter 6Midterm Review

6.1 Michelson Interferometer

Suppose the ether theory were true (i.e. that the speed of light is only c in the rest frame of theether) and that we lived in a Galilean-relativistic world, rather than an Einsteinian-relativistic one.Suppose that the Michelson interferometer were moving relative to the rest frame of the ether with avelocity v = (vx, vy, vz), where the plane of the interferometer is the xy-plane with one arm along thex direction and one along the y direction. Let ` be the length of both arms. We do not know whatv is, but suppose we assume that each of its components is characterized by a Gaussian probabilitydistribution centered at zero with a width, v0 << c. This means that we are assuming that it is veryunlikely for us to be moving relative to the ether with a speed anywhere near c or greater; in fact,our most likely speed is 0. We are also assuming that it is equally likely for us to be moving in anydirection (i.e. the distribution is isotropic). So, let the probability distribution for vx be

P (vx) = 1√π v0

e−(vx/v0)2 ,

with the exact same distribution for vy and vz. Note that we have already ensured that this probabilitydistribution is normalized:

∫∞−∞ P (vx) dvx = 1.

Let N be the number of fringes through which the interferometer cycles in the course of turningclockwise by 90◦. Determine the root-mean-squared value of N as a function of `, v0, and λ, which isthe wavelength of the laser used in the experiment.

You may need to use the following integrals:∫ ∞−∞

e−x2

dx =√π,

∫ ∞−∞

x2 e−x2

dx =

√π

2,

∫ ∞−∞

x4 e−x2

dx =3√π

4.

SOLUTION:

Relative to the rest frame of the interferometer, the ether is moving with a velocity, −v = −(vx, vy, vz).This is what French calls the “ether wind”. Let arm 1 be the one along the x direction and arm 2 bethe one along the y direction. Let u±1 = (u±1 , 0, 0) be the velocity of the light beam in the rest frameof the interferometer along arm 1 with the + sign corresponding to going “up” the arm (i.e. in the +xdirection) and the − sign corresponding to going “down” the arm (i.e. in the −x direction). Similarly,define u±2 = (0, u±2 , 0) for the velocity up and down arm 2. Let v±1 = c(ξ±1 , ψ

±1 , ζ

±1 ), and similarly for

v±2 , be the corresponding velocities in the ether rest frame. Galilean relativity relates these via

u±1 = v±1 − v,

23

24 CHAPTER 6. MIDTERM REVIEW

and similarly for the velocities along arm 2. This equation reads

(u±1 , 0, 0) = (cξ±1 − vx, cψ±1 − vy, cζ

±1 − vz),

which implies that

ψ±1 =vyc , ζ±1 = vz

c .

The speed of the beam in the rest frame of the ether is supposed to be c, which gives the relation

c = |v±1 | = c

√(ξ±1 )2 + (ψ±1 )2 + (ζ±1 )2 =⇒ ξ±1 = ±

√1− (ψ±1 )2 − (ζ±1 )2 = ±

√1− v2y+v2z

c2 .

We have chosen signs appropriately, so that ξ+1 is positive (i.e. up the arm is the positive direction

and down the arm, the negative). Therefore,

u±1 = cξ±1 − vx = ±c√

1− v2y+v2zc2 − vx.

The exact same analysis yields

u±2 = cψ±2 − vy = ±c√

1− v2x+v2zc2 − vy.

The time that it takes for the beam to go up and down arm 1 is

t1 =`

|u+1 |

+`

|u−1 |=

`

c

√1− v2y+v2z

c2 − vx+

`

c

√1− v2y+v2z

c2 + vx

.

Define βx ≡ vx/c, and similarly for the other components, define v2 = v2x + v2

y + v2z and β ≡ v/c.

Finally, define γ = 1/√

1− β2, as usual. Then, after some algebra, we can simplifty t1 to read

t1 =(γ2√

1− β2y − β2

z

)2`c .

Let us check the limit as βx,y,z → 0 (i.e. as we approach the ether rest frame). We should expect toget t1 = 2`

c , which is indeed what we get if we plug βx,y,z = 0 into the formula above.The exact same analysis for the second arm yields

t2 =(γ2√

1− β2x − β2

z

)2`c .

The difference between these times is

∆ ≡ t1 − t2 = γ2(√

1− β2y − β2

z −√

1− β2x − β2

z

)2`c .

Let us now make the small βx,y,z approximation, since we know that the probability for any of theseto be anywhere near or greater than 1 is very small. That is, β0 ≡ v0/c << 1, and the Gaussianprobability distribution vanishes very quickly above this value. We find

∆ ≈ (1 + β2)[�1− 1

2β2y −��

12β

2z −

(�1− 1

2β2x −��

12β

2z

)]2`c ≈ (β2

x − β2y) `c .

Note that we first approximated γ2 = 11−β2 ≈ 1 + β2, and then dropped the β2 anyway because it is

multiplied by terms quadratic in βx and βy.Rotating the interferometer by 90◦ clockwise is equivalent to βx → −βy and βy → βx. The time

difference between the two arms in this case is

∆′ ≡ t′1 − t′2 ≈ (β2y − β2

x) `c .

6.2. PION PHOTOPRODUCTION 25

The phase of a beam increases by 2πctλ after travelling for time t. Each 2π worth of phase is one fringe.

Thus, the number of fringes through which the inteference pattern cycles during a 90◦ rotation of theinterferometer is

N = cλ (∆−∆′) ≈ (β2

x − β2y) 2`

λ .

It shouldn’t surprise you that if you take the average of N , you will get 0, because the effect of avelocity v is cancelled by the effect of the exact opposite velocity, since the probability distribution isisotropic. The RMS value is

N2rms = N2 =

∫∫∫ ∞−∞

dβx dβy dβz P (βx)P (βy)P (βz)N2

=(2`

λ

)2( 1√π β0

)3∫∫ ∞−∞

dβx dβy e−(βx/β0)2e−(βy/β0)2 (β2

x − β2y)2

∫ ∞−∞

dβz e−(βz/β0)2

=(2`

λ

)2[

1√π β0

(∫ ∞−∞

β4xe−(βx/β0)2 dβx +

∫ ∞−∞

β4ye−(βy/β0)2 dβy

)− 2

πβ20

∫ ∞−∞

β2xe−(βx/β0)2 dβx

∫ ∞−∞

β2ye−(βy/β0)2 dβy

].

Define A ≡ βx/β0 and B ≡ βy/β0, then

N2rms =

(2`

λ

)2[

β50√π β0

(∫ ∞−∞

A4e−A2

dA+

∫ ∞−∞

B4e−B2

dB)

− 2β60

πβ20

∫ ∞−∞

A2e−A2

dA

∫ ∞−∞

B2e−B2

dB

]=(2`

λ

)2[β4

0

��√π

(3��√π

4+

3��√π

4

)− 2β4

0

�π·��√π

2·��√π

2

]=(2`

λ

)2

β40 .

Therefore, the RMS is

Nrms =2`β2

0

λ=

2(v0/c)2

λ/`.

Note: This is the same result as French Eqn. (2-14) p.56 if we suppose that the interferometer ismoving relative to the ether only in the x direction with a speed v = v0.

6.2 Pion Photoproduction

Consider the pion photoproduction reaction γ + p→ p+ π0, where the rest energy is 938 MeV/c2 forthe proton and 135 MeV/c2 for the neutral pion.

(a) If the initial proton is at rest in the laboratory, find the laboratory threshold gamma-ray energyfor this reaction to go.

(b) Consider a head-on collision between a proton and a photon from the isotropic 3K cosmic blackbodyradiation, which has an average photon energy of 10−3 eV. Find the minimum proton energy thatwill allow this pion photoproduction reaction to go.

(c) Speculate briefly on the implications of your result for part (b) for the energy spectrum of cosmic-ray protons.

SOLUTION:


(a) The initial value of E2−(pc)2 in the lab frame is (Eγ +mpc2)2−

(Eγc c)2

= 2Eγmpc2 +(mpc

2)2. Atthreshold, in the center of mass frame, the resulting proton and pion are at rest. Thus, the finalvalue of E2 − (pc)2 is (mp +mπ)2c4 = (mpc

2)2 + 2mpmπc4 + (mπc

2)2. Since these are invariant,set these two equal and solve for Eγ :

Eγ = mπc2(1 + mπ

2mp

)= 145 MeV.

(b) Take the energy of the photon in the lab frame to be Eγ = 10−3 eV. Again, equating the initialand final values of E2 − (pc)2 yields

(γmpc2 + Eγ)2 −

(Eγc − γmpβc

)2c2 = (mp +mπ)2c4.

Solving for β gives

γ(1 + β) =

√1 + β

1− β=mπc

2

Eγ

(1 +

mπ

2mp

)= 1.447× 1011.

Since γ(1 + β) is so large, β must be very close to 1. Let us solve approximately for γ by settingβ = 1 in the equation above. This gives γ = 7.235× 1010 and so the proton energy is

Ep = γmpc2 = 7.235× 1010 × 0.938 MeV = 6.787× 1010 GeV .

(c) This implies that the energy spectrum of cosmic ray protons with E > 6.79 × 1010 GeV will bedepleted to some degree due to interaction with the cosmic blackbody radiation.

6.3 Microscopes and Embryos

The egg of a very tiny creature has a radius of R = 110 mm and an index of refraction negg = 3

2 . A tinyembryo sits in its center. This egg sits on the surface of a microscope slide submerged in a layer ofwater 3

10 mm thick. Above this sits a compound microscope made with identical converging objectiveand eyepiece thin lenses with focal length f = 2 cm. The objective lens sits a distance of 2.65 cmabove the surface of the water and the distance between the objective and eyepiece is 10 cm. Locatethe final image of the embryo and calculate the angular magnification of the microscope by itself. Use10 cm as your near point.

SOLUTION:

At the egg-water interface, we have

negg

R+nwater

si1=nwater − negg

−R.

6.4. THREE-SLIT INTERFERENCE 27

Plugging in the numbers gives

si1 = −R = −0.1 mm,

which places it exactly where the embryo started out in the first place! You can show that this is trueno matter the medium in which the egg is immersed nor the index of refraction of the egg itself!

At the water-air interface, we have

nwater

2R+nair

si2=nair − nwater

∞= 0.

Plugging in the numbers gives

si2 = −3R

2= −0.15 mm.

Note that this provides a different method for calculating apparent depth.The distance between this intermediate image and the objective lens is so3 = (2.65 + 0.015) cm =

2.665 cm. The lens makes an image at

1

si3=

1

f− 1

so3=⇒ si3 = 8.01504 cm.

The distance between this intermediate image and the eyepiece lens is so4 = (10 − 8.01504) cm =1.98496 cm. The lens makes an image at

1

si4=

1

f− 1

so4=⇒ si4 = −264 cm.

That is, the final image is produced 2.64 m behind the eyepiece (i.e. it is a virtual image).The linear magnification of the objective by itself is

mob = − si3so3≈ −3.

The angular magnification of the eyepiece by itself is

Mep =dnpso4≈ dnp

f= 5.

Therefore, the overall angular magnification of the microscope is

Mm = mobMep = −15.

6.4 Three-Slit Interference

[Hecht P.9.18 ] Imagine that you have an opaque screen with three horizontal very narrow parallel slitsin it. The second slit is a center-to-center ditance a beneath the first, and the third is a distance 5a/2beneath the first.

(a) Write a complex exponential expression in terms of δ = ka sin θ for the amplitude of the electricfield at some point P at an elevation θ on a distant screen.

(b) Prove that I(θ) = I(0)3 + 2I(0)

9

(cos δ + cos 3δ

2 + cos 5δ2

).

SOLUTION:


(a) Let y be the distance of P from the point on the detection screen directly facing the midpointbetween the top and bottom slits. We take the latter point to be the point from which θ ismeasured.

Let ri be the straight-line distance from the center of slit i to the point P . Let L be the distancebetween the slit and detection screens. Then, making the approximation a/L << 1,

r1 =[L2 +

(y − 5a

4

)2]1/2 ≈ L[1 + 12

(y−5a/4L

)2] ≈ L+ y2

2L −5ay4L .

Define R = L+ y2

2L , which is actually common to all ri and thus does not contribute to interference.Note that sin θ ≈ θ ≈ tan θ = y/L. Thus, we may write ay/L ≈ a sin θ = δ/k. The samecalculation can be done for r2 and r3. In terms of δ,

r1 = R− 5δ4k , r2 = R− δ

4k , r3 = R+ 5δ4k .

The electric field at P due to ray 1 is Er1ei(kr1−ωt) ≈ E

Rei(kR−ωt)e−i(5δ/4). Note that we have

neglected the small difference in r1 relative to R that is involved in the amplitude, but NOT whenit is involved in the phase (i.e. in the complex exponential). Define A ≡ E

Rei(kR−ωt), which is

common to the electric field contributions of the beams. The total electric field is

E(θ) = A(e−i5δ/4 + e−iδ/4 + ei5δ/4

).

Plugging in θ = 0 gives δ = 0 and E(0) = 3A. Therefore, we may write

E(θ) = 13E(0)

(e−i5δ/4 + e−iδ/4 + ei5δ/4

).

(b) Since the irradiance is proportional to |E|2, we have

I(θ) = 19I(0)

(e−i5δ/4 + e−iδ/4 + ei5δ/4

)(ei5δ/4 + eiδ/4 + e−i5δ/4

)= 1

9I(0)(1 + e−iδ + e−i5δ/2 + eiδ + 1 + e−i3δ/2 + ei5δ/2 + ei3δ/2 + 1

)= 1

9I(0)(3 + 2 cos δ + 2 cos 3δ

2 + 2 cos 5δ2

)= I(0)

3 + 2I(0)9

(cos δ + cos 3δ

2 + cos 5δ2

).

6.5 Polarizers Away!

Two identical HN-46 polarizers, at rest relative to each other, with area 10 cm2 (shaped as disks,perhaps) and mass 1 g, are parallel to each other, separated by 1 cm and floating around in space faraway from any gravitational fields. Their polarization axes make an angle 30◦ relative to each other.A beam of natural light, with cross-sectional area larger than the polarizers’ and with irradiance 488.6W/cm2, is normally incident on the first polarizer with the other polarizer directly behind. How muchtime will it take for the two polarizers to make contact with each other?

SOLUTION:

Let the polarizers be HN-100x, so x = 0.46 in this case. Let A = 10 cm2 be the area of the polarizers,m = 1 g their mass, d = 1 cm their separation, θ = 30◦ their relative angle and I = 488.6 W/cm2 theirradiance of the light beam. Let us trace what happens to the beam. 100% of the light is incidenton the first polarizer. Of this, 8% is reflected away, 46% is absorbed and 46% is transmitted overto the second polarizer. At this point, 8% of the remaining 46% is reflected back towards the firstpolarizer, and, were θ to equal 0 (i.e. the polarization axes were to be parallel), the remaining 92% ofthe 46% would all be transmitted. But, since θ 6= 0, instead, sin2 θ = 1

4 is absorbed and cos2 θ = 34 is

6.5. POLARIZERS AWAY! 29

transmitted. The light that was reflected off of the second polarizer can partially reflect and transmitthrough the first polarizer, but it won’t be absorbed because it is now polarized along the direction ofthe first polarizer. We can draw this schematically as follows

Each successive absorption at polarizer 2 gets a factor of 2x sin2 θ, which has been abbreviated to 2xs2

in the diagram, each successive transmission at polarizer 2 gets a factor of 2x cos2 θ, each successivereflection (on either polarizer) picks up a factor of y ≡ 1− 2x, and each successive transmission backthrough polarizer 1 gets a factor of 2x.

However, recall that reflections count twice as much as absorptions as far as radiation pressure isconcerned. For the first polarizer, the initial reflection and absorption count as pressure to the rightwhile the successive reflected waves going from 1 to 2 contribute leftward pressure. The total pressureto the right in units of I/c is

P1 = 2y + x− 2xy2 − 2xy4 − · · · = x+ 2y − 2xy2

1− y2,

where we have used the geometric series formula 1 + y2 + y4 + · · · = 11−y2 , for |y| < 1. Plugging in

y = 1− 2x and simplifying gives

P1 =3− 6x+ 2x2

2(1− x)= 0.614.

As is to be expected, this is only slightly smaller than 0.62 = 0.46 + 2(0.08), which is the result if weneglect all subsequent reflections between the two polarizers.

The pressure on polarizer 2 just due to absorption is

P abs2 = (2x2 sin2 θ)(1 + y2 + y4 + · · · ) =

x sin2 θ

2(1− x)= 0.1065.

The pressure on polarizer 2 just due to reflection is

P ref2 = 2xy(1 + y2 + y4 + · · · ) =

1− 2x

2(1− x)= 0.0741.

Therefore, the total pressure on polarizer 2 is

P2 = 0.181.

First, let’s get some units right:

W

cm2=

J/s

cm2=

Nm

cm2s= 102 N

cm · s= 107 dyne

cm · s,

where dyne = g · cm/s2.The force on polarizer j is Fj = Pj

IcA:

F1 = 0.614× 4.886× 109 dyne/cm · s3× 1010 cm/s

× 10 cm2 = 1 dyne,

F2 =0.181

0.614F1 = 0.295 dyne.


The accelerations, aj , are the same, with dynes replaced by cm/s2, since m = 1 g.Set z = 0 and z = d to be the initial positions of the polarizers. Their positions as functions of

time are simply given by kinematics:

z1(t) = 12a1t

2, z2(t) = d+ 12a2t

2.

The time of coincidence (z1 = z2) is

t =

√2d

a1 − a2=

√2× 1 cm

(1− 0.295) cm/s2 = 1.68 sec.

Chapter 7Blackbody Radiation

One lesson we learn from the photoelectric effect is that light may be thought of as being built outof particles called photons, even though it behaves like a wave in most familiar situations. Somehow,very many photons conspire to produce wave-like behavior. This concept of photons is what startsus down the road towards blackbody radiation, although historically the ideas of Planck that followpreceded, and in fact inspired, Einstein’s explanation of the photoelectric effect.

One thoroughly embarrassing problem that remained before Planck came on the scene is called theultraviolet catastrophe. Consider a thermally insulated box of sidelength L containing electromagneticfields. The boundary conditions at the inner edges of the box imply that the electric and magneticfields have to vanish there. Thus, the fields must form standing waves. In one dimension, these look likethe fundamental and harmonic vibrations on a string that is fixed at both ends. In three dimensions,we just multiply three of these standing waves (they can all have different wavelengths), one in the xdirection, one in the y direction, and one in the z direction. That gives us three degrees of freedom,namely the wavelengths of the standing wave in each direction. We can actually denote the differentmodes by integers: the fundamental mode fits half of its wavelength within L, the first harmonic fits afull wavelength within L, the second harmonic fits 3/2 of a wavelength in L, etc. Thus, the nth modefits n/2 of a wavelength in L, or, in other words λn = 2L/n is the wavelength of the nth mode. Notethat the fundamental mode corresponds to n = 1, and so on. Thus, a single standing wave mode inthe cube can be distinguished by three positive integers (nx, ny, nz), one for each direction. However,since this is an electromagnetic wave, each standing wave component (one in each direction) has twoorthogonal choices for the direction of the electric field (i.e. the polarization). This gives us a total ofsix degrees of freedom for each mode (nx, ny, nz). The equipartition theorem states that each degreeof freedom will contribute an average of kT/2 worth of energy, where k is Boltzmann’s constant and Tis the temperature of the system. That means that each standing wave mode contributes 3kT worth ofenergy on average. But, there are infinitely many possible modes: the three integers can get arbitrarilylarge, or, in other words, the wavelengths can get arbitrarily small. That would make the total energyinfinite! Schroeder describes just how embarrassing this conclusion is: if it were correct, you wouldexpect to be blasted with an infinite amount of radiation every time you open the oven door to checkthe cookies!

The classical assumption is that each mode, denoted by the three integers, can have any non-negative energy, E. From H7B, we know that the probability for a mode to have energy E is propor-tional to the Boltzmann factor: P (E) ∝ e−E/kT . Calculating the average energy per mode as we didfor an ideal gas in H7B for such a continuous spectrum produces the equipartition theorem.

Planck’s neat idea was that electromagnetic energy is not continuously distributed, but is quantizedin integer units of hν, where ν is the frequency of radiation and h is Planck’s constant. He proposedthat light was absorbed and emitted by matter in quanta called photons. So, a single mode withfrequency ν can have an energy of 0, or hν, or 2hν, etc. But it cannot have an energy between these

31

32 CHAPTER 7. BLACKBODY RADIATION

values, like hν/2, since that would correspond to half a photon!

7.1 Planck Distribution

I know of two main paths to Planck’s solution for the UV catastrophe. One makes use of the classicalidea of enumerating modes of electromagnetic fields within a cube and then jumping to the quantumconcept of distributing photons among these modes. The second is to adopt a quantum mechanicalpicture right from the start and enumerate the possible states of photons within some arbitrary enclosedvolume.

There are pros and cons to both methods. The first method makes it easy to compare Planck’s fixto the fully classical approach and clearly highlights exactly where the classical picture breaks down.On the other hand, I have always been dissatisfied with its mixture of classical and quantum concepts:on the one hand describing modes of classical electromagnetic fields and on the other hand populatingthese modes with quanta (photons) while never really clarifying how the former arise from the latter.

The second method starts from the photon immediately, without recourse to classical fields. It alsopossesses the virtue of not requiring a specific shape like a cube for the enclosure. On the other hand,fully describing the photon states will be a bit more challenging. Since the first method is describedin Serway and most other statistical mechanics textbooks, I have elected to take the second path.

First, we must determine which variables we must consider in order to fully describe the state ofphotons inside some arbitrary enclosure. There are four variables: (1) the number of photons, (2)their helicities, (3) their positions, and (4) their momenta. Since we are in three dimensions, the lasttwo actually constitute six variables altogether: x, y, z, px, py, and pz. The combination of theselast six variables taken as an abstract six-dimensional “space” is called “phase space”. By the way,helicity describes the intrinsic angular momentum of photons and can come in two flavors: left andright handed.

Suppose we fix the helicity, ε, the position, x, and the momentum, p, of a one-photon state. Whenthe system sits at temperature T , what is the average number of photons in this particular state? Thatis, how heavily occupied is this particular one-photon state (OPS)? Hopefully, you remember from 7Bthat the probability, Ps, of a state, s, with energy, Es, to be occupied is proportional to the Boltzmannfactor: Ps ∝ e−Es/kT . We must require that the sum of the probabilities of occupation over all of thestates be 1. Thus,

Ps =1

Ze−Es/kT where Z =

∑s

e−Es/kT , (7.1)

is called the partition function.In the case at hand, the energy of the OPS is pc, where p = |p|. The OPS may be occupied by any

number, n, of photons, with total energy npc. Thus, the partition function for a particular OPS withhelicity, position and momentum, (ε,x,p) is

Z(ε,x,p) =

∞∑n=0

e−npc/kT (7.2a)

=1

1− e−pc/kT. (7.2b)

The average number of photons in this OPS is

〈np〉 =∑

states

(photon #

in state

)(probability

of state

)=

∞∑n=0

ne−npc/kT

Zp=

1

Zp

∞∑n=0

ne−npc/kT , (7.3)

where, I will refrain from writing (ε,x,p) over and over again. It happens that Z and n actually onlydepend on p, so we will write Zp and 〈np〉 for brevity.

7.2. PLANCK SPECTRUM 33

Now, consider differentiating Eqn. (7.2a) with respect to temperature, T :

dZpdT

=

∞∑n=0

( npckT 2

)e−npc/kT =

pc

kT 2

∞∑n=0

ne−npc/kT . (7.4)

On the other hand, differentiating Eqn. (7.2b) yields

dZpdT

= − 1

(1− e−pc/kT )2(−e−pc/kT )

pc

kT 2=

pc

kT 2Z2pe−pc/kT . (7.5)

Setting Eqn. (7.4) and Eqn. (7.5) equal yields

∞∑n=0

ne−npc/kT = Z2pe−pc/kT . (7.6)

Plugging this into Eqn. (7.3) yields

〈np〉 =1

ZpZ2pe−pc/kT =

e−pc/kT

1− e−pc/kT=

1

epc/kT − 1. (7.7)

This is called the Planck distribution. It measures how many photons will occupy a particular OPSwith momentum magnitude p when the system sits at temperature T .

7.2 Planck Spectrum

We would like to calculate the average electromagnetic energy (actually, energy density) inside theenclosure at temperature T . To do so, we will multiply the average number, 〈np〉, of photons in thespecific OPS, (ε,x,p), with the energy, pc, of this OPS, and then “sum” over all possible OPS’s.

Conceptually, this is easy enough. But, how exactly do we enumerate the possible OPS’s? Helicityis easy: left and right handedness does not affect energy, and so this just gives a factor of 2 to theenumeration of OPS’s. What about phase space? Well, these are continuous variables taking anypossible value. Aren’t the states infinitely dense? The answer is no and the reason is the uncertaintyprinciple.

The uncertainty principle says that there is an intrinsic lower limit to how well one can determine ordefine the position and momentum of a particle. This means that points in phase space that are so closetogether as to be within the uncertainty principle of each other must be considered as representing theexact same state. In other words, there is a minimum volume of phase space and all points within thisvolume of some phase space point are considered to be indistinguishable from each other and representone and the same state. It turns out that the correct quantization is that there is one quantum statefor every h3 worth of phase space volume. Therefore, to “sum” over all states in phase space simplymeans to integrate over phase space assuming that there is only one state per h3 phase space volume.Thus, we can now calculate the average energy:

U ≡ 〈E〉 =∑OPS

〈np〉 pc

=∑hel.

1

h3

∫d3x

∫d3p

pc

epc/kT − 1

= 2× 1

h3× V × 4π

∫ ∞0

dp p2 pc

epc/kT − 1. (7.8)

Since the integrand is independent of helicity and spatial coordinates, the helicity sum gives the factorof 2 and the spatial integral gives the factor of the volume, V , of the enclosure. Also, since the integrand


is independent of the angular variables (i.e. direction) of the momentum, that integral just gives afactor of 4π. Note that we converted the momentum integral into spherical momentum coordinates.

We can divide by the volume to get the energy density. We would also like to change the integrationvariable to the wavelength of the OPS, rather than the momentum, where the two are related via deBroglie’s equation p = h/λ. When p = 0, we have λ = ∞, and vice versa. Thus, the integral over λgoes from ∞ to 0. Also, dp = − h

λ2 dλ. Thus,

U

V=

8πc

h3

∫ 0

∞

(− h

λ2dλ)(h

λ

)3 1

ehc/λkT − 1=

∫ ∞0

8πhc

λ5(ehc/λkT − 1) dλ. (7.9)

What if we punch a small hole through the wall of the enclosure? Since there is electromagnetic energydensity inside, it will escape. The energy density multiplied by the rate at which the energy is moving(speed of light), should give the intensity of the radiation. However, most of the photons are notheading towards the hole and thus do not escape. It turns out that the intensity radiated through thehold is only one-fourth of the energy density multiplied by the speed of light:

I =1

4

U

Vc =

∫ ∞0

2πhc2

λ5(ehc/λkT − 1)dλ. (7.10)

I believe that the factor of one-fourth is described in your textbook. This immediately gives us thePlanck spectrum:

dI

dλ=

2πhc2

λ5(ehc/λkT − 1). (7.11)

Setting d2I/dλ2 = 0, we find that dI/dλ peaks at

λpeak ≈hc

4.97 kT. (7.12)

This is Wien’s displacement law. This is essentially how a radiation thermometer works. You pointthis machine at something and it tells you the temperature of that object without direct contact, whichis extremely important for things that are far away (like the Sun) and/or things that are too hot orcold for contact thermometers to work properly. The machine analyzes the spectrum of radiation thatit receives, finds the wavelength at which the spectrum peaks and uses Wien’s displacement law todetermine the temperature, T .

Let us go back to Eqn. (7.8) for U/V and change variables to u = pc/kT :

U

V=

8π(kT )4

(hc)3

∫ ∞0

u3 du

eu − 1. (7.13)

It is possible to calculate the integral above. This is done at the end of this chapter. The result isπ4/15. Thus,

U

V=

8π5(kT )4

15(hc)3. (7.14)

Therefore, the total irradiance emitted by a blackbody at temperature T is

I =cU

4V=

2π5k4T 4

15h3c2= σT 4 where σ =

2π5k4

15h3c2. (7.15)

This is Stefan’s law and the constant σ is the Stefan-Boltzmann constant.

7.3. STEFAN-BOLTZMANN INTEGRAL 35

7.3 Stefan-Boltzmann Integral

We would like to calculate the integral

J ≡∫ ∞

0

u3 du

eu − 1.

First rewrite the integrand by multiplying above and below by e−u. Then, for u > 0, we have e−u < 1and so we may use the appropriate geometric series as seen below:

u3

eu − 1=

u3e−u

1− e−u= u3e−u

∞∑n=0

e−nu = u3∞∑n=1

e−nu.

We would like to integrate each term separately. Before doing so, a simple calculation gives∫ ∞0

e−nu du =1

n.

Then, taking three derivatives with respect to n pulls down three factors of −u, and thus

− d3

dn3

∫ ∞0

e−nu du =

∫ ∞0

u3 e−nu du.

On the other hand,

− d3

dn3

∫ ∞0

e−nu du = − d3

dn3

1

n=

d2

dn2

1

n2= − d

dn

2

n3=

6

n4.

Therefore, the integral of each term in the expansion is∫ ∞0

u3 e−nu du =6

n4,

and the total integral is

J =

∞∑n=1

∫ ∞0

u3 e−nu du =

∞∑n=1

6

n4≡ 6 ζ(4), (7.16)

where ζ(s) ≡∑∞n=1

1ns is the Riemann zeta function.

Hecht Eqn. 7.49 p.306 gives the Fourier series for the square wave:

f(x) =

{1 for mλ < x < (m+ 1

2 )λ,

−1 for (m+ 12 )λ < x < mλ,

=4

π

∑odd n

1

nsin(2πnx

λ

).

where m can take on any integer value. Note we have used the relation k = 2π/λ to change HechtEqn. 7.49 to the above form.

In a moment, we will be integrating both expressions for f(x) multiple times. We just need to becareful about integration constants. We want to set the integration constants for the integral of eachterm in the Fourier series to zero. This simplifies things because the integral itself remains periodicwith the same period and remains centered around the x-axis. This uniquely fixes the constants ofintegration for the first definition of f(x).

For example, in the region −λ/2 < x < 0, we have f(x) = −1. You might therefore think thatthe integral is simply F 1(x) = −x in this region. Similarly, you might think that F 1(x) = x in theregion 0 < x < λ/2. This completes one period and the whole thing is repeated over and over again.But that would make F 1(x) always ≥ 0. Namely, it would not be centered around the x-axis and thesecond integral, F 2(x), would deviate further and further away from the x-axis as |x| → ∞. Therefore,


to make F 1(x) centered around the x-axis, we must set F 1(x) = −x− λ4 in the region −λ/2 < x < 0

and F 1(x) = x− λ4 in the region 0 < x < λ/2, with all other regions just translated copies of these.

The second integral is a littler easier because F 2(x) = 12x

2 − λ4x + C in the region 0 < x < λ/2,

where C is some constant, but we also need F 2(0) = 0, which implies C = 0. Hence, F 2(x) = 12x(x− λ

2 )in this region.

Perhaps the pattern is clear now: for odd s, we require F s(0) = −F s(λ/2), while for even s, we

require F s(0) = F s(λ/2) = 0. The third integral must therefore be F 3(x) = 16x

3 − λ8x

2 + λ3

26·3 in theregion 0 < x < λ/2. Setting this equal to the term-by-term integral of the Fourier series yields

1

6x3 − λ

8x2 +

λ3

26 · 3= F 3(x) =

4

π

∑odd n

1

n

( λ

2πn

)3

cos(2πnx

λ

)=

λ3

2π4

∑odd n

1

n4cos(2πnx

λ

).

Evaluating both sides at x = 0 yields

λ3

26 · 3=

λ3

2π4

∑odd n

1

n4=⇒

∑odd n

1

n4=

π4

25 · 3.

Now, we can separate the sum in the zeta function into one over odds and evens. The sum over theevens is ∑

even n

1

n4=

1

24+

1

44+

1

64+

1

84+ · · · = 1

24

[ 1

14+

1

24+

1

34+

1

44+ · · ·

]=

1

24ζ(4).

Hence,

ζ(4) =∑

odd n

1

n4+∑

even n

1

n4=

π4

25 · 3+

1

24ζ(4).

Solving for ζ(4) yields

ζ(4) =π4

2 · 32 · 5=π4

90.

Plugging this into Eqn. (7.16) gives us the desired integral,

J ≡∫ ∞

0

u3 du

eu − 1= 6 ζ(4) =

π4

15. (7.17)

Chapter 8Quantum Operator Expectation Values

8.1 Infinite Square Well

Calculate the expectation values of x, x2, p and p2 in the nth bound state of the infinite square wellof width L. Statistically speaking, the uncertainty in position is its standard deviation, defined by

∆x =

√⟨(x− 〈x〉

)2⟩,

and similarly for momentum. Calculate ∆x and ∆p and show that they satisfy the uncertainty rela-tion. How does this alter your estimate for the force exerted on the confining walls by a particle inthis bound state from problem set 8 question 2?

SOLUTION:

Let |n〉 denote the appropriate abstract state in the abstract Hilbert space for this potential. Let |x〉represent the eigenstates of the position operator, which are orthogonal and complete. In other words,

〈y|x〉 = δ(x− y),

∫ ∞−∞|x〉〈x| = 1,

where δ(x− y) is the Dirac delta function and 1 is the identity operator. The position and momentumoperators act on these states via

x |x〉 = |x〉 x, p |x〉 = −i~ |x〉 ∂∂x .

Notice that |x〉 can be moved to the left of the derivative because the state itself does not change as afunction of position.

The wavefunction of the state |n〉 expressed in position space is ψn(x) = 〈x|n〉. We already knowwhat this wavefunction looks like:

ψn(x) =

√2

Lsin(nπxL

).

37

38 CHAPTER 8. QUANTUM OPERATOR EXPECTATION VALUES

Thus, the expectation value of x is

〈x〉 = 〈n|x|n〉 = 〈n| x∫ ∞−∞|x〉〈x| dx |n〉

=

∫ ∞−∞〈n|x|x〉〈x|n〉 dx

=

∫ ∞−∞〈n|x〉x〈x|n〉 dx

=

∫ ∞∞

ψ∗n(x)xψn(x) dx

=2

L

∫ L

0

sin(nπxL

)x sin

(nπxL

)dx

=2

L

( Lnπ

)2∫ nπ

0

ξ sin2 ξ dξ.

In the first line, we inserted the identity operatore, 1, in the form given by the completeness relationof the position eigenstates. Note that we changed the integration variable to ξ ≡ nπx/L.

To calculate the integral, write sin2 ξ as 12

(1 − cos 2ξ

). The integral of ξ cos 2ξ can be calculated

by integration by parts and one finds that it vanishes. Therefore, we get 〈x〉 = L/2, which you mighthave guessed anyway: the average position of the particle is the center of the well.

Similarly, the expectation value of x2 is⟨x2⟩

=2

L

( Lnπ

)3∫ nπ

0

ξ2 sin2 ξ dξ.

The integral can be calculated the same way, but requries two integration by parts. Since I messedthis up initially, I will write down the steps.⟨

x2⟩

=2L2

(nπ)3

∫ nπ

0

ξ2 1− cos 2ξ

2dξ

=2L2

(nπ)3

[1

6

∫ nπ

0

3ξ2 dξ − 1

4

∫ nπ

0

ξ2 · 2 cos 2ξ dξ]

=2L2

(nπ)3

[1

6ξ3∣∣∣nπ0−��

��1

4ξ2 sin 2ξ

∣∣∣nπ0

+1

4

∫ nπ

0

2ξ sin 2ξ dξ]

=2L2

(nπ)3

[ (nπ)3

6− 1

4ξ cos 2ξ

∣∣∣nπ0

+��

��1

4

∫ nπ

0

cos 2ξ dξ]

=L2

3− L2

2(nπ)2.

The expectation value of p is

〈p〉 = 〈n|p|n〉 = 〈n| p∫ ∞−∞|x〉〈x| dx |n〉

=

∫ ∞−∞〈n|p|x〉〈x|n〉 dx

= −i~∫ ∞−∞

ψn(x)d

dxψn(x) dx

= −i~ 2

L

nπ

L

∫ L

0

sin(nπxL

)cos(nπxL

)dx

= −2i~L

∫ nπ

0

sin ξ cos ξ dξ.

8.1. INFINITE SQUARE WELL 39

The integral can be calculated in numerous ways (e.g. using 2 sin ξ cos ξ = sin 2ξ). One finds that theintegral vanishes and so 〈p〉 = 0. This is not surprising since the particle moves left and right withequal probability by symmetry.

We can read off the expectation value of p2 from the kinetic energy, which is the total energy inthis case since the potential is zero in the well:

En ≡ 〈H〉 =n2π2~2

2mL2=⇒

⟨p2⟩

= 2m 〈H〉 =n2π2~2

L2.

Before we calculate the uncertainties, let us simplify its expression:

(∆x)2 =⟨x2 − 2 〈x〉x+ 〈x〉2

⟩=⟨x2⟩− 2 〈x〉2 + 〈x〉2 =

⟨x2⟩− 〈x〉2 .

Note, of course, that 〈〈x〉〉 = 〈x〉.Now, we can calculate the uncertainties in position and momentum:

∆x =

√L2

3− L2

2(nπ)2−(L

2

)2

=L

2√

3

√1− 6

(nπ)2, ∆p =

√n2π2~2

L2− 0 =

nπ~L

.

The product of the two satisfies the uncertainty relation:

∆x∆p =nπ√

3

√1− 6

(nπ)2

~2

=

√(nπ)2 − 6

3

~2>

~2.

I was concerned that this might not satisfy the uncertainty principle for the ground state (n = 1), butthose fears were unwarranted; it turns out that the prefactor of ~/2 is about 1.14 when n = 1. Phew!It works out after all.

The wavelength of the nth energy eigenstate is λ = 2L/n. Just check this for a couple low valuesof n; it should be pretty obvious if you draw the states. De Broglie’s relation gives the momentummagnitude:

|p| = h

λ=

2π~2L/n

=nπ~L

.

Note that this doesn’t take into account the direction of the momentum, which is the thing that made〈p〉 vanish. So, hopefully you don’t get confused as to why |p| 6= 0 while 〈p〉 = 0. The speed of theparticle is

|v| = |p|m

=nπ~mL

.

Thus, the time between collisions against one of the walls is the time it takes for the particle to goback and forth once across the well:

∆t =2L

|v|=

2L

nπ~/mL=

2mL2

nπ~.

Finally, our estimate for the force is

F =∆p

∆t=n2π2~2

2mL3,

which is smaller than F = −dEn/dL by a factor of 2. That is a huge improvement from the previousestimate which was smaller by a factor of 16π2 ≈ 160.

40 CHAPTER 8. QUANTUM OPERATOR EXPECTATION VALUES

Chapter 9Operators and States

The bra-ket notation may seem like simply a rewriting of stuff we already know from the languageof wavefunctions, albeit in more compact form. For example, with the ket |n〉 representing the samequantum harmonic oscillator state as the wavefunction ψn(x), the expectation value of the positionoperator in this state may be written as

〈n|x|n〉 =

∫ ∞−∞

ψ∗n(x)xψn(x) dx.

In the language of wavefunctions, the “state” of a system is represented by the wavefunction. Anoperator will almost always be a function of position and momentum, and all you have to do is replacemomentum is −i~ ∂

∂x . To take the expectation value of operators, you sandwich it between the complexconjugate of the wavefunction on the left and the wavefunction on the right and integrate over position.There is a somewhat physical interpretation underlying this perspective, which is that ψ∗ψ = |ψ|2 issupposed to be the probability density function.

In the bra-ket language, the “state” of a system is represented by a vector in an abstract complexvector space, V, (a Hilbert space, to be precise). So, we must be able to add two vectors and the resultis supposed to represent the superposition of the two states that are represented by each vector. Wemust be able to multiply a vector by an arbitrary complex number. We must also have an inner productdefined between vectors so that we actually mean something when we say one state is orthogonal toanother. An inner product is nothing more than a map from two copies of the vector space to thenon-negative real numbers: it takes two vectors and spits out a non-negative real number. If it spitsout 0, then the two vectors are orthogonal. Often, the inner product is denoted by (·, ·), where each ·is slot in which one inserts a vector in the vector space. So, (·, ·) : V × V → R≥0.

In the case of the harmonic oscillator, the most convenient choice of basis for the vector space isthe basis of energy eigenstates denoted by |n〉 for n = 0, 1, . . .. Note that this is an abstract vectorrepresenting the nth eigenstate of the harmonic oscillator. It is NOT the wavefunction representingthe same state. The wavefunction and the ket (or vector) are two different representations.

In this picture, an operator, O, is nothing more than a map from the vector space to itself:O : V → V. Having chosen a basis, one is free to represent that map by its matrix relative to the basis,just as in linear algebra. In principle, this is no different than stating that once you have chosen thestandard Cartesian basis for R2 (Euclidean plane), then a rotation of the plane by an angle φ in thecounter-clockwise direction is represented by the matrix(

(x, Rx) (x, Ry)(y, Rx) (y, Ry)

)=

(cos θ − sin θsin θ cos θ

),

where R is the rotation operator, e.g. Rx = cos θ x + sin θ y. The inner product is the standard dotproduct in two dimensions, e.g. (x, x) = x · x = 1, (x, y) = x · y = 0, etc.

41

42 CHAPTER 9. OPERATORS AND STATES

Actually, there is some difference. For one thing, the vector (Hilbert) spaces are often infinitedimensional. Sometimes, they are not even discrete, as in the case of position and momentum eigen-states. But, let’s not worry too much about this. Anyway, it’s not difficult to think about infinitedimensional matrices, as long as there is a discrete basis. For example, we could tack on as manyextra dimensions as we want to R2. The matrix representing the rotation described above would bean infinite-dimensional matrix that has the same top-left 2× 2 components as the one above, then 1’salong the remaining diagonals and 0’s everywhere else.

Using the analogy of matrices to represent operators, you can think of kets as column vectors. Anoperator turns a ket into another ket just like a square matrix turns a column vector into anothercolumn vector. We can dream up a different vector space, called the dual vector space, V∗. Note thatthe star here does not, a priori, have anything to do with complex conjugation. In the most abstractsense, the elements of V∗ are defined to be linear maps from V into R. In our particular case, werestrict to maps into R≥0.

To make this slightly more concrete, consider the example of the harmonic oscillator. A basis forV is furnished by the kets {|0〉 , |1〉 , |2〉 , . . .}. Elements of V∗ are linear maps V → R≥0. One suchmap is the map that sends |0〉 to 1 and everything else to 0. Another is the map that sends |1〉 to1 and everything else to 0. In fact, these types of maps furnish a basis for V∗ since the map thatsends |0〉 to a, |1〉 to b and so on is just a times the first map mentioned plus b times the second mapmentioned, etc. We denote the first map by the bra, 〈0|, the second map by 〈1|, etc. There is anobvious isomorphism between V and V∗, which sends |0〉 to 〈0|, etc. If in addition to this, an arbitrarycomplex constant multiplying a ket becomes complex conjugated under this isomorphism, then thismap is called the adjoint. In other words, an arbitrary linear combination, a |0〉+ b |1〉+ · · · is mappedvia the adjoint map into a∗ 〈0|+ b∗ 〈1|+ · · · . Then, we can define a bilinear map (·, ·) : V∗ ×V → R≥0

as follows: (〈m| , |n〉) is equal to the number to which 〈m| maps |n〉, which happens to be 1 if m = nand 0 if m 6= n. To save ink, we write 〈m|n〉 ≡ (〈m| , |n〉).

In terms of the inner product in V, all that the adjoint of an operator means is that it must satisfythe relation (|m〉 , O† |n〉) = (O |m〉 , |n〉).

Big deal! We’ve managed to double the number of ways of describing or notating the states of oursystem. Well, it turns out that thinking of abstract vectors and operators instead of wavefunctionsand differential operators is often more fruitful than it might seem, as exemplified by the followingquestion.

9.1 Harmonic Oscillator Bra-Kets

Consider the one-dimensional harmonic oscillator with frequency ω. Define

a =1√

2m~ω(mωx+ ip).

Use the results of problem set 9 question 4 to show that

a |n〉 =√n |n− 1〉 , a† |n〉 =

√n+ 1 |n+ 1〉 .

Let |ψ〉 be a superposition of the ground and first excited states. Calculate the time-dependent ex-pectation value of the position operator in this state, which is the average position as a function of time.

Note: This small a is related to the big A in the problem set via a = A/√~.

SOLUTION:

We have the commutation relation

[H, a] = −~ωa.

9.1. HARMONIC OSCILLATOR BRA-KETS 43

Since [H, a] ≡ Ha− aH, we may write

Ha = a(H − ~ω1

),

where 1 is just the identity operator: a = a1.Act both sides of this equation on |n〉 and bracket out a |n〉:

H(a |n〉) = a(H − ~ω1

)|n〉 = a

[~ω(n+ 1

2

)− ~ω

]|n〉 = ~ω

[(n− 1) + 1

2

](a |n〉).

This shows that a |n〉 is an eigenstate of the Hamiltonian, which means that it must be proportional toone of the |m〉’s. The eigenvalue, or energy, happens to coincide with the energy of |n− 1〉. Therefore,a |n〉 = c(n) |n− 1〉, where c is just a number, which may depend on n.

Now, take the inner product of a |n〉 with itself:

(a |n〉 , a |n〉) = 〈n|a†a|n〉 = 〈n| 1~ω H −

12 |n〉 = 〈n| 1

~ω~ω(n+ 1

2

)− 1

2 |n〉 = n〈n|n〉 = n.

But, this is also equal to

(c ˆn− 1, c ˆn− 1) = |c|2〈n− 1|n− 1〉 = |c|2.

Since we don’t distinguish between two states that are related to each other be a complex phase, wemay assume c to be real. Thus, c =

√n, which is the result we wanted.

The same procedure will yield a† |n〉 =√n+ 1 |n+ 1〉, but you may have to use the commutation

relation [a, a†] = 1.

To get a sense of how much simpler using abstract states and operators is, let us first try to calculate〈x〉 using the wavefunctions. The wavefunction of our superposition state is

ψ(x, t) =(mωπ~

)1/4

e−mωx2/2~

[Ae−iωt/2 +B

(2mω

~

)1/2

xe−3iωt/2].

Here, A and B are arbitrary constants, which we will take to be real, for convenience. Then,

〈x〉 (t) =

∫ ∞−∞

ψ∗(x, t)xψ(x, t) dx

=(mωπ~

)1/2{��

��

�A2

∫ ∞−∞

xe−mωx2/~ dx−AB

(2mω

~

)1/2

(eiωt + e−iωt)

∫ ∞−∞

x2e−mωx2/~ dx

+((((

(((((((

(((B2(2mω

~

)∫ ∞−∞

x3e−mωx2/~ dx

}=

2√

2√πAB(mω

~

)cosωt

( ~mω

)3/2∫ ∞−∞

ξ2e−ξ2

dξ

=2√πAB( 2~mω

)1/2

cosωt

√π

2

= AB( 2~mω

)1/2

cosωt.

Note that an odd function in x multiplied by the Gaussian is an odd function whose integral from −∞to +∞ vanishes. Now, first of all, I had to look up what the wavefunctions to begin with. Secondly, Ihad to know the integral of ξ2e−ξ

2

. Ugh!

Now, let us use kets and operators instead. We have

|ψ〉 = Ae−iωt/2 |0〉+Be−3iωt/2 |1〉 .

44 CHAPTER 9. OPERATORS AND STATES

We want to calculate

〈x〉 = 〈ψ|x|ψ〉 = A2〈0|x|0〉+AB(eiωt〈1|x|0〉+ e−iωt〈0|x|1〉) +B2〈1|x|1〉

But, x can be written in terms of the raising and lowering operators:

x =( ~

2mω

)1/2

(a+ a†).

So a piece of x lowers |n〉 and a piece raises |n〉. Therefore, in order for 〈m|x|n〉 to be nonzero, wemust have either m = n− 1 or m = n+ 1. Also,

〈n− 1|a|n〉 =√n, 〈n+ 1|a†|n〉 =

√n+ 1.

Hence, we find

〈x〉 = AB( ~

2mω

)1/2

(eiωt + e−iωt) = AB( 2~mω

)1/2

cosωt.

Not once did I actually need to know what the exact wavefunctions were. Nor did I ever have tocalculate a single integral!

Chapter 10Final Review

10.1 Wavefunction Shapes

Below is picture of an infinite potential well with a non-flat bottom. Explain your answers to thefollowing questions.

(a) For some arbitrary allowed energy, E, rank positions A, B and C by the classical kinetic energyof the particle at these positions from largest to smallest.

(b) Repeat for de Broglie wavelength.

(c) Repeat for the amount of time a classical particle spends traversing an interval of width δx at eachposition.

(d) Repeat for the spacings between the zeros of the wavefunction in the regions near each point.Assume that the energy level is sufficiently high that the wavefunction oscillates many timesbetween the two walls.

(e) Repeat for the amplitude of the wavefunction in the region near each point.

(f) Sketch plausible wavefunctions for the n = 4 and n = 8 energy levels.

10.2 Harmonic Oscillator Coherent States

Eigenvectors of the lowering operator, a = 1√2m~ω (mωx + ip), are called Coherent states. So, these

satisfy a |α〉 = α |α〉 for some α ∈ C.

(a) Calculate 〈x〉,⟨x2⟩, 〈p〉 and

⟨p2⟩

in the state |α〉.

(b) Calculate ∆x and ∆p and show that the uncertainty is saturated.

45

46 CHAPTER 10. FINAL REVIEW

(c) Expand |α〉 in terms of the basis of energy eigenfunctions, |n〉 for n = 0, 1, 2, . . .. Determine thecoefficients of the expansion as functions of α. Remember to properly normalize the state. Youmay need the Taylor expansion, ex =

∑∞n=0

xn

n! .

10.3 Finite Square Well Bound States

Let V (x) = −V0 for |x| ≤ L/2 and V (x) = 0 otherwise. By symmetry, the bound states are eithereven or odd functions. Let E < 0 be the energy of a bound state. Define the constants

η ≡L√

2m|E|2~

, ξ ≡L√

2m(V0 − |E|)2~

.

(a) Show that η2 + ξ2 = mL2V0/2~2.

(b) Show that η = ξ tan ξ for even bound states and ξ = −ξ cot ξ for odd bound states.

(c) Draw a plot of η vs. ξ. Count the number of bound states as a function of V0, L and m.

10.4 A Theorem about Bound States

Let V (1)(x) and V (2)(x) be two potentials such that V (1)(x) ≤ V (2)(x) for all x. Suppose E(1)1 <

E(1)2 < · · · and E

(2)1 < E

(2)2 < · · · are the bound state energies of the two potentials. Note: they need

not have the same number of bound states; they don’t even have to have any, necessarily. A theorem

claims that E(1)n ≤ E(2)

n .Use this theorem to give a lower and upper bound to the number of bound states for the potential

V (x) = 12mω

2x2 for |x| ≤ 8√~/mω and V (x) = 32~ω otherwise.

10.5 Spin-Spin Interaction

Two particles with spins s1 = 32 and s2 = 1

2 interact with each other via a spin-spin interaction whose

Hamiltonian is H = α~ S1 · S2, where α is some constant. At t = 0, the system is in the following

eigenstate of S21 , S1z, S

22 , S2z: ∣∣ 3

2 ,12 ; 1

2 ,12

⟩.

Find the state of the system at times t > 0. What is the probability of finding the system in the state∣∣ 32 ,

32 ; 1

2 ,−12

⟩as a function of t?

10.6 Fine and Hyperfine Structure

(a) What is the zeroth order (Bohr model) value for the n = 2 energy for Hydrogen? Counting thepossible spin states of the proton, what is the degeneracy of this state (i.e. how many differentstates have this same energy)?

(b) Describe qualitatively the effect of fine structure on this energy level. What does it do to thedegeneracy? Give an estimate (order of magnitude) of how much the energy levels change. Labelthe states as n“`”j , where “`” = s, p for ` = 0, 1 respectively and give the degeneracy of each.

(c) Do the same thing for hyperfine structure.

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