# The math behind the maser

As I skipped the mathematical arguments in my previous post so as to focus on the essential results only, I thought it would be good to complement that post by looking at the math once again, so as to ensure we understand what it is that we’re doing. So let’s do that now. We start with the easy situation: free space.

#### The two-state system in free space

We started with an ammonia molecule in free space, i.e. we assumed there were no external force fields, like a gravitational or an electromagnetic force field. Hence, the picture was as simple as the one below: the nitrogen atom could be ‘up’ or ‘down’ with regard to its spin around its axis of symmetry. It’s important to note that this ‘up’ or ‘down’ direction is defined in regard to the molecule itself, i.e. not in regard to some external reference frame. In other words, the reference frame is that of the molecule itself. For example, if I flip the illustration above – like below – then we’re still talking the same states, i.e. the molecule is still in state 1 in the image on the left-hand side and it’s still in state 2 in the image on the right-hand side. We then modeled the uncertainty about its state by associating two different energy levels with the molecule: E0 + A and E− A. The idea is that the nitrogen atom needs to tunnel through a potential barrier to get to the other side of the plane of the hydrogens, and that requires energy. At the same time, we’ll show the two energy levels are effectively associated with an ‘up’ or ‘down’ direction of the electric dipole moment of the molecule. So that resembles the two spin states of an electron, which we associated with the +ħ/2 and −ħ/2 energies respectively. So if E0 would be zero (we can always take another reference point, remember?), then we’ve got the same thing: two energy levels that are separated by some definite amount: that amount is 2A for the ammonia molecule, and ħ when we’re talking quantum-mechanical spin. I should make a last note here, before I move on: note that these energies only make sense in the presence of some external field, because the + and − signs in the E0 + A and E− A and +ħ/2 and −ħ/2 expressions make sense only with regard to some external direction defining what’s ‘up’ and what’s ‘down’ really. But I am getting ahead of myself here. Let’s go back to free space: no external fields, so what’s ‘up’ or ‘down’ is completely random here. 🙂

Now, we also know an energy level can be associated with a complex-valued wavefunction, or an amplitude as we call it. To be precise, we can associate it with the generic a·e−(i/ħ)·(E·t − px) expression which you know so well by now. Of course,  as the reference frame is that of the molecule itself, its momentum is zero, so the px term in the a·e−(i/ħ)·(E·t − px) expression vanishes and the wavefunction reduces to a·ei·ω·t a·e−(i/ħ)·E·t, with ω = E/ħ. In other words, the energy level determines the temporal frequency, or the temporal variation (as opposed to the spatial frequency or variation), of the amplitude.

We then had to find the amplitudes C1(t) = 〈 1 | ψ 〉 and C2(t) =〈 2 | ψ 〉, so that’s the amplitude to be in state 1 or state 2 respectively. In my post on the Hamiltonian, I explained why the dynamics of a situation like this can be represented by the following set of differential equations: As mentioned, the Cand C2 functions evolve in time, and so we should write them as C= C1(t) and C= C2(t) respectively. In fact, our Hamiltonian coefficients may also evolve in time, which is why it may be very difficult to solve those differential equations! However, as I’ll show below, one usually assumes they are constant, and then one makes informed guesses about them so as to find a solution that makes sense.

Now, I should remind you here of something you surely know: if Cand Care solutions to this set of differential equations, then the superposition principle tells us that any linear combination a·C1 + b·Cwill also be a solution. So we need one or more extra conditions, usually some starting condition, which we can combine with a normalization condition, so we can get some unique solution that makes sense.

The Hij coefficients are referred to as Hamiltonian coefficients and, as shown in the mentioned post, the H11 and H22 coefficients are related to the amplitude of the molecule staying in state 1 and state 2 respectively, while the H12 and H21 coefficients are related to the amplitude of the molecule going from state 1 to state 2 and vice versa. Because of the perfect symmetry of the situation here, it’s easy to see that H11 should equal H22 , and that H12 and H21 should also be equal to each other. Indeed, Nature doesn’t care what we call state 1 or 2 here: as mentioned above, we did not define the ‘up’ and ‘down’ direction with respect to some external direction in space, so the molecule can have any orientation and, hence, switching the i an j indices should not make any difference. So that’s one clue, at least, that we can use to solve those equations: the perfect symmetry of the situation and, hence, the perfect symmetry of the Hamiltonian coefficients—in this case, at least!

The other clue is to think about the solution if we’d not have two states but one state only. In that case, we’d need to solve iħ·[dC1(t)/dt] = H11·C1(t). That’s simple enough, because you’ll remember that the exponential function is its own derivative. To be precise, we write: d(a·eiωt)/dt = a·d(eiωt)/dt = a·iω·eiωt, and please note that can be any complex number: we’re not necessarily talking a real number here! In fact, we’re likely to talk complex coefficients, and we multiply with some other complex number (iω) anyway here! So if we write iħ·[dC1/dt] = H11·C1 as dC1/dt = −(i/ħ)·H11·C1 (remember: i−1 = 1/i = −i), then it’s easy to see that the Ca·e–(i/ħ)·H11·t function is the general solution for this differential equation. Let me write it out for you, just to make sure:

dC1/dt = d[a·e–(i/ħ)H11t]/dt = a·d[e–(i/ħ)H11t]/dt = –a·(i/ħ)·H11·e–(i/ħ)H11t

= –(i/ħ)·H11·a·e–(i/ħ)H11= −(i/ħ)·H11·C1

Of course, that reminds us of our generic wavefunction a·e−(i/ħ)·E0·t wavefunction: we only need to equate H11 with E0 and we’re done! Hence, in a one-state system, the Hamiltonian coefficient is, quite simply, equal to the energy of the system. In fact, that’s a result can be generalized, as we’ll see below, and so that’s why Feynman says the Hamiltonian ought to be called the energy matrix.

In fact, we actually may have two states that are entirely uncoupled, i.e. a system in which there is no dependence of C1 on Cand vice versa. In that case, the two equations reduce to:

iħ·[dC1/dt] = H11·C1 and iħ·[dC2/dt] = H22·C2

These do not form a coupled system and, hence, their solutions are independent:

C1(t) = a·e–(i/ħ)·H11·t and C2(t) = b·e–(i/ħ)·H22·t

The symmetry of the situation suggests we should equate a and b, and then the normalization condition says that the probabilities have to add up to one, so |C1(t)|+ |C2(t)|= 1, so we’ll find that = 1/√2.

OK. That’s simple enough, and this story has become quite long, so we should wrap it up. The two ‘clues’ – about symmetry and about the Hamiltonian coefficients being energy levels – lead Feynman to suggest that the Hamiltonian matrix for this particular case should be equal to: Why? Well… It’s just one of Feynman’s clever guesses, and it yields probability functions that makes sense, i.e. they actually describe something real. That’s all. 🙂 I am only half-joking, because it’s a trial-and-error process indeed and, as I’ll explain in a separate section in this post, one needs to be aware of the various approximations involved when doing this stuff. So let’s be explicit about the reasoning here:

1. We know that H11 = H22 = Eif the two states would be identical. In other words, if we’d have only one state, rather than two – i.e. if H12 and H21 would be zero – then we’d just plug that in. So that’s what Feynman does. So that’s what we do here too! 🙂
2. However, H12 and H21 are not zero, of course, and so assume there’s some amplitude to go from one position to the other by tunneling through the energy barrier and flipping to the other side. Now, we need to assign some value to that amplitude and so we’ll just assume that the energy that’s needed for the nitrogen atom to tunnel through the energy barrier and flip to the other side is equal to A. So we equate H12 and H21 with −A.

Of course, you’ll wonder: why minus A? Why wouldn’t we try H12 = H21 = A? Well… I could say that a particle usually loses potential energy as it moves from one place to another, but… Well… Think about it. Once it’s through, it’s through, isn’t it? And so then the energy is just Eagain. Indeed, if there’s no external field, the + or − sign is quite arbitrary. So what do we choose? The answer is: when considering our molecule in free space, it doesn’t matter. Using +A or −A yields the same probabilities. Indeed, let me give you the amplitudes we get for H11 = H22 = Eand H12 and H21 = −A:

1. C1(t) = 〈 1 | ψ 〉 = (1/2)·e(i/ħ)·(E− A)·t + (1/2)·e(i/ħ)·(E+ A)·t = e(i/ħ)·E0·t·cos[(A/ħ)·t]
2. C2(t) = 〈 2 | ψ 〉 = (1/2)·e(i/ħ)·(E− A)·t – (1/2)·e(i/ħ)·(E+ A)·t = i·e(i/ħ)·E0·t·sin[(A/ħ)·t]

[In case you wonder how we go from those exponentials to a simple sine and cosine factor, remember that the sum of complex conjugates, i.e eiθ eiθ reduces to 2·cosθ, while eiθ − eiθ reduces to 2·i·sinθ.]

Now, it’s easy to see that, if we’d have used +A rather than −A, we would have gotten something very similar:

• C1(t) = 〈 1 | ψ 〉 = (1/2)·e(i/ħ)·(E+ A)·t + (1/2)·e(i/ħ)·(E− A)·t = e(i/ħ)·E0·t·cos[(A/ħ)·t]
• C2(t) = 〈 2 | ψ 〉 = (1/2)·e(i/ħ)·(E+ A)·t – (1/2)·e(i/ħ)·(E− A)·t = −i·e(i/ħ)·E0·t·sin[(A/ħ)·t]

So we get a minus sign in front of our C2(t) function, because cos(α) = cos(–α) but sin(α) = −sin(α). However, the associated probabilities are exactly the same. For both, we get the same P1(t) and P2(t) functions:

• P1(t) = |C1(t)|2 = cos2[(A/ħ)·t]
• P2(t) = |C2(t)|= sin2[(A/ħ)·t]

[Remember: the absolute square of and −is |i|= +√12 = +1 and |i|2 = (−1)2|i|= +1 respectively, so the i and −i in the two C2(t) formulas disappear.]

You’ll remember the graph: Of course, you’ll say: that plus or minus sign in front of C2(t) should matter somehow, doesn’t it? Well… Think about it. Taking the absolute square of some complex number – or some complex function , in this case! – amounts to multiplying it with its complex conjugate. Because the complex conjugate of a product is the product of the complex conjugates, it’s easy to see what happens: the e(i/ħ)·E0·t factor in C1(t) = e(i/ħ)·E0·t·cos[(A/ħ)·t] and C2(t) = ±i·e(i/ħ)·E0·t·sin[(A/ħ)·t] gets multiplied by e+(i/ħ)·E0·t and, hence, doesn’t matter: e(i/ħ)·E0·t·e+(i/ħ)·E0·t = e0 = 1. The cosine factor in C1(t) = e(i/ħ)·E0·t·cos[(A/ħ)·t] is real, and so its complex conjugate is the same. Now, the ±i·sin[(A/ħ)·t] factor in C2(t) = ±i·e(i/ħ)·E0·t·sin[(A/ħ)·t] is a pure imaginary number, and so its complex conjugate is its opposite. For some reason, we’ll find similar solutions for all of the situations we’ll describe below: the factor determining the probability will either be real or, else, a pure imaginary number. Hence, from a math point of view, it really doesn’t matter if we take +A or −A for  or  real factor for those H12 and H21 coefficients. We just need to be consistent in our choice, and I must assume that, in order to be consistent, Feynman likes to think of our nitrogen atom borrowing some energy from the system and, hence, temporarily reducing its energy by an amount that’s equal to −A. If you have a better interpretation, please do let me know! 🙂

OK. We’re done with this section… Except… Well… I have to show you how we got those C1(t) and C1(t) functions, no? Let me copy Feynman here: Note that the ‘trick’ involving the addition and subtraction of the differential equations is a trick we’ll use quite often, so please do have a look at it. As for the value of the a and b coefficients – which, as you can see, we’ve equated to 1 in our solutions for C1(t) and C1(t) – we get those because of the following starting condition: we assume that at t = 0, the molecule will be in state 1. Hence, we assume C1(0) = 1 and C2(0) = 0. In other words: we assume that we start out on that P1(t) curve in that graph with the probability functions above, so the C1(0) = 1 and C2(0) = 0 starting condition is equivalent to P1(0) = 1 and P1(0) = 0. Plugging that in gives us a/2 + b/2 = 1 and a/2 − b/2 = 0, which is possible only if a = b = 1.

Of course, you’ll say: what if we’d choose to start out with state 2, so our starting condition is P1(0) = 0 and P1(0) = 1? Then a = 1 and b = −1, and we get the solution we got when equating H12 and H21 with +A, rather than with −A. So you can think about that symmetry once again: when we’re in free space, then it’s quite arbitrary what we call ‘up’ or ‘down’.

So… Well… That’s all great. I should, perhaps, just add one more note, and that’s on that A/ħ value. We calculated it in the previous post, because we wanted to actually calculate the period of those P1(t) and P2(t) functions. Because we’re talking the square of a cosine and a sine respectively, the period is equal to π, rather than 2π, so we wrote: (A/ħ)·T = π ⇔ T = π·ħ/A. Now, the separation between the two energy levels E+ A and E− A, so that’s 2A, has been measured as being equal, more or less, to 2A ≈ 10−4 eV.

How does one measure that? As mentioned above, I’ll show you, in a moment, that, when applying some external field, the plus and minus sign do matter, and the separation between those two energy levels E+ A and E− A will effectively represent something physical. More in particular, we’ll have transitions from one energy level to another and that corresponds to electromagnetic radiation being emitted or absorbed, and so there’s a relation between the energy and the frequency of that radiation. To be precise, we can write 2A = h·f0. The frequency of the radiation that’s being absorbed or emitted is 23.79 GHz, which corresponds to microwave radiation with a wavelength of λ = c/f0 = 1.26 cm. Hence, 2·A ≈ 25×109 Hz times 4×10−15 eV·s = 10−4 eV, indeed, and, therefore, we can write: T = π·ħ/A ≈ 3.14 × 6.6×10−16 eV·s divided by 0.5×10−4 eV, so that’s 40×10−12 seconds = 40 picoseconds. That’s 40 trillionths of a seconds. So that’s very short, and surely much shorter than the time that’s associated with, say, a freely emitting sodium atom, which is of the order of 3.2×10−8 seconds. You may think that makes sense, because the photon energy is so much lower: a sodium light photon is associated with an energy equal to E = h·f = 500×1012 Hz times 4×10−15  eV·s = 2 eV, so that’s 20,000 times 10−4 eV.

There’s a funny thing, however. An oscillation of a frequency of 500 tera-hertz that lasts 3.2×10−8 seconds is equivalent to 500×1012 Hz times 3.2×10−8 s ≈ 16 million cycles. However, an oscillation of a frequency of 23.97 giga-hertz that only lasts 40×10−12 seconds is equivalent to 23.97×109 Hz times 40×10−12 s ≈ 1000×10−3 = 1 ! One cycle only? We’re surely not talking resonance here!

So… Well… I am just flagging it here. We’ll have to do some more thinking about that later. [I’ve added an addendum that may or may not help us in this regard. :-)]

#### The two-state system in a field

As mentioned above, when there is no external force field, we define the ‘up’ or ‘down’ direction of the nitrogen atom was defined with regard to its its spin around its axis of symmetry, so with regard to the molecule itself. However, when we apply an external electromagnetic field, as shown below, we do have some external reference frame.

Now, the external reference frame – i.e. the physics of the situation, really – may make it more convenient to define the whole system using another set of base states, which we’ll refer to as I and II, rather than 1 and 2. Indeed, you’ve seen the picture below: it shows a state selector, or a filter as we called it. In this case, there’s a filtering according to whether our ammonia molecule is in state I or, alternatively, state II. It’s like a Stern-Gerlach apparatus splitting an electron beam according to the spin state of the electrons, which is ‘up’ or ‘down’ too, but in a totally different way than our ammonia molecule. Indeed, the ‘up’ and ‘down’ spin of an electron has to do with its magnetic moment and its angular momentum. However, there are a lot of similarities here, and so you may want to compare the two situations indeed, i.e. the electron beam in an inhomogeneous magnetic field versus the ammonia beam in an inhomogeneous electric field. Now, when reading Feynman, as he walks us through the relevant Lecture on all of this, you get the impression that it’s the I and II states only that have some kind of physical or geometric interpretation. That’s not the case. Of course, the diagram of the state selector above makes it very obvious that these new I and II base states make very much sense in regard to the orientation of the field, i.e. with regard to external space, rather than with respect to the position of our nitrogen atom vis-á-vis the hydrogens. But… Well… Look at the image below: the direction of the field (which we denote by ε because we’ve been using the E for energy) obviously matters when defining the old ‘up’ and ‘down’ states of our nitrogen atom too!

In other words, our previous | 1 〉 and | 2 〉 base states acquire a new meaning too: it obviously matters whether or not the electric dipole moment of the molecule is in the same or, conversely, in the opposite direction of the field. To be precise, the presence of the electromagnetic field suddenly gives the energy levels that we’d associate with these two states a very different physical interpretation. Indeed, from the illustration above, it’s easy to see that the electric dipole moment of this particular molecule in state 1 is in the opposite direction and, therefore, temporarily ignoring the amplitude to flip over (so we do not think of A for just a brief little moment), the energy that we’d associate with state 1 would be equal to E+ με. Likewise, the energy we’d associate with state 2 is equal to E− με.  Indeed, you’ll remember that the (potential) energy of an electric dipole is equal to the vector dot product of the electric dipole moment μ and the field vector ε, but with a minus sign in front so as to get the sign for the energy righ. So the energy is equal to −μ·ε = −|μ|·|ε|·cosθ, with θ the angle between both vectors. Now, the illustration above makes it clear that state 1 and 2 are defined for θ = π and θ = 0 respectively. [And, yes! Please do note that state 1 is the highest energy level, because it’s associated with the highest potential energy: the electric dipole moment μ of our ammonia molecule will – obviously! – want to align itself with the electric field ε ! Just think of what it would imply to turn the molecule in the field!]

Therefore, using the same hunches as the ones we used in the free space example, Feynman suggests that, when some external electric field is involved, we should use the following Hamiltonian matrix: So we’ll need to solve a similar set of differential equations with this Hamiltonian now. We’ll do that later and, as mentioned above, it will be more convenient to switch to another set of base states, or another ‘representation’ as it’s referred to. But… Well… Let’s not get too much ahead of ourselves: I’ll say something about that before we’ll start solving the thing, but let’s first look at that Hamiltonian once more.

When I say that Feynman uses the same clues here, then… Well.. That’s true and not true. You should note that the diagonal elements in the Hamiltonian above are not the same: E+ με ≠ E+ με. So we’ve lost that symmetry of free space which, from a math point of view, was reflected in those identical H11 = H22 = Ecoefficients.

That should be obvious from what I write above: state 1 and state 2 are no longer those 1 and 2 states we described when looking at the molecule in free space. Indeed, the | 1 〉 and | 2 〉 states are still ‘up’ or ‘down’, but the illustration above also makes it clear we’re defining state 1 and state 2 not only with respect to the molecule’s spin around its own axis of symmetry but also vis-á-vis some direction in space. To be precise, we’re defining state 1 and state 2 here with respect to the direction of the electric field ε. Now that makes a really big difference in terms of interpreting what’s going on.

In fact, the ‘splitting’ of the energy levels because of that amplitude A is now something physical too, i.e. something that goes beyond just modeling the uncertainty involved. In fact, we’ll find it convenient to distinguish two new energy levels, which we’ll write as E= E+ A and EII = E− A respectively. They are, of course, related to those new base states | I 〉 and | II 〉 that we’ll want to use. So the E+ A and E− A energy levels themselves will acquire some physical meaning, and especially the separation between them, i.e. the value of 2A. Indeed, E= E+ A and EII = E− A will effectively represent an ‘upper’ and a ‘lower’ energy level respectively.

But, again, I am getting ahead of myself. Let’s first, as part of working towards a solution for our equations, look at what happens if and when we’d switch to another representation indeed.

#### Switching to another representation

Let me remind you of what I wrote in my post on quantum math in this regard. The actual state of our ammonia molecule – or any quantum-mechanical system really – is always to be described in terms of a set of base states. For example, if we have two possible base states only, we’ll write:

| φ 〉 = | 1 〉 C1 + | 2 〉 C2

You’ll say: why? Our molecule is obviously always in either state 1 or state 2, isn’t it? Well… Yes and no. That’s the mystery of quantum mechanics: it is and it isn’t. As long as we don’t measure it, there is an amplitude for it to be in state 1 and an amplitude for it to be in state 2. So we can only make sense of its state by actually calculating 〈 1 | φ 〉 and 〈 2 | φ 〉 which, unsurprisingly are equal to 〈 1 | φ 〉 = 〈 1 | 1 〉 C1 + 〈 1 | 2 〉 C2  = C1(t) and 〈 2 | φ 〉 = 〈 2 | 1 〉 C1 + 〈 2 | 2 〉 C2  = C2(t) respectively, and so these two functions give us the probabilities P1(t) and  P2(t) respectively. So that’s Schrödinger’s cat really: the cat is dead or alive, but we don’t know until we open the box, and we only have a probability function – so we can say that it’s probably dead or probably alive, depending on the odds – as long as we do not open the box. It’s as simple as that.

Now, the ‘dead’ and ‘alive’ condition are, obviously, the ‘base states’ in Schrödinger’s rather famous example, and we can write them as | DEAD 〉 and | ALIVE 〉 you’d agree it would be difficult to find another representation. For example, it doesn’t make much sense to say that we’ve rotated the two base states over 90 degrees and we now have two new states equal to (1/√2)·| DEAD 〉 – (1/√2)·| ALIVE 〉 and (1/√2)·| DEAD 〉 + (1/√2)·| ALIVE 〉 respectively. There’s no direction in space in regard to which we’re defining those two base states: dead is dead, and alive is alive.

The situation really resembles our ammonia molecule in free space: there’s no external reference against which to define the base states. However, as soon as some external field is involved, we do have a direction in space and, as mentioned above, our base states are now defined with respect to a particular orientation in space. That implies two things. The first is that we should no longer say that our molecule will always be in either state 1 or state 2. There’s no reason for it to be perfectly aligned with or against the field. Its orientation can be anything really, and so its state is likely to be some combination of those two pure base states | 1 〉 and | 2 〉.

The second thing is that we may choose another set of base states, and specify the very same state in terms of the new base states. So, assuming we choose some other set of base states | I 〉 and | II 〉, we can write the very same state | φ 〉 = | 1 〉 C1 + | 2 〉 Cas:

| φ 〉 = | I 〉 CI + | II 〉 CII

It’s really like what you learned about vectors in high school: one can go from one set of base vectors to another by a transformation, such as, for example, a rotation, or a translation. It’s just that, just like in high school, we need some direction in regard to which we define our rotation or our translation.

For state vectors, I showed how a rotation of base states worked in one of my posts on two-state systems. To be specific, we had the following relation between the two representations: The (1/√2) factor is there because of the normalization condition, and the two-by-two matrix equals the transformation matrix for a rotation of a state filtering apparatus about the y-axis, over an angle equal to (minus) 90 degrees, which we wrote as: The y-axis? What y-axis? What state filtering apparatus? Just relax. Think about what you’ve learned already. The orientations are shown below: the S apparatus separates ‘up’ and ‘down’ states along the z-axis, while the T-apparatus does so along an axis that is tilted, about the y-axis, over an angle equal to α, or φ, as it’s written in the table above. Of course, we don’t really introduce an apparatus at this or that angle. We just introduced an electromagnetic field, which re-defined our | 1 〉 and | 2 〉 base states and, therefore, through the rotational transformation matrix, also defines our | I 〉 and | II 〉 base states.

[…] You may have lost me by now, and so then you’ll want to skip to the next section. That’s fine. Just remember that the representations in terms of | I 〉 and | II 〉 base states or in terms of | 1 〉 and | 2 〉 base states are mathematically equivalent. Having said that, if you’re reading this post, and you want to understand it, truly (because you want to truly understand quantum mechanics), then you should try to stick with me here. 🙂 Indeed, there’s a zillion things you could think about right now, but you should stick to the math now. Using that transformation matrix, we can relate the Cand CII coefficients in the | φ 〉 = | I 〉 CI + | II 〉 CII expression to the Cand CII coefficients in the | φ 〉 = | 1 〉 C1 + | 2 〉 C2 expression. Indeed, we wrote:

• C= 〈 I | ψ 〉 = (1/√2)·(C1 − C2)
• CII = 〈 II | ψ 〉 = (1/√2)·(C1 + C2)

That’s exactly the same as writing: OK. […] Waw! You just took a huge leap, because we can now compare the two sets of differential equations: They’re mathematically equivalent, but the mathematical behavior of the functions involved is very different. Indeed, unlike the C1(t) and C2(t) amplitudes, we find that the CI(t) and CII(t) amplitudes are stationary, i.e. the associated probabilities – which we find by taking the absolute square of the amplitudes, as usual – do not vary in time. To be precise, if you write it all out and simplify, you’ll find that the CI(t) and CII(t) amplitudes are equal to:

• CI(t) = 〈 I | ψ 〉 = (1/√2)·(C1 − C2) = (1/√2)·e(i/ħ)·(E0+ A)·t = (1/√2)·e(i/ħ)·EI·t
• CII(t) = 〈 II | ψ 〉 = (1/√2)·(C1 + C2) = (1/√2)·e(i/ħ)·(E0− A)·t = (1/√2)·e(i/ħ)·EII·t

As the absolute square of the exponential is equal to one, the associated probabilities, i.e. |CI(t)|2 and |CII(t)|2, are, quite simply, equal to |1/√2|2 = 1/2. Now, it is very tempting to say that this means that our ammonia molecule has an equal chance to be in state I or state II. In fact, while I may have said something like that in my previous posts, that’s not how one should interpret this. The chance of our molecule being exactly in state I or state II, or in state 1 or state 2 is varying with time, with the probability being ‘dumped’ from one state to the other all of the time.

I mean… The electric dipole moment can point in any direction, really. So saying that our molecule has a 50/50 chance of being in state 1 or state 2 makes no sense. Likewise, saying that our molecule has a 50/50 chance of being in state I or state II makes no sense either. Indeed, the state of our molecule is specified by the | φ 〉 = | I 〉 CI + | II 〉 CII = | 1 〉 C1 + | 2 〉 Cequations, and neither of these two expressions is a stationary state. They mix two frequencies, because they mix two energy levels.

Having said that, we’re talking quantum mechanics here and, therefore, an external inhomogeneous electric field will effectively split the ammonia molecules according to their state. The situation is really like what a Stern-Gerlach apparatus does to a beam of electrons: it will split the beam according to the electron’s spin, which is either ‘up’ or, else, ‘down’, as shown in the graph below: The graph for our ammonia molecule, shown below, is very similar. The vertical axis measures the same: energy. And the horizontal axis measures με, which increases with the strength of the electric field ε. So we see a similar ‘splitting’ of the energy of the molecule in an external electric field. How should we explain this? It is very tempting to think that the presence of an external force field causes the electrons, or the ammonia molecule, to ‘snap into’ one of the two possible states, which are referred to as state I and state II respectively in the illustration of the ammonia state selector below. But… Well… Here we’re entering the murky waters of actually interpreting quantum mechanics, for which (a) we have no time, and (b) we are not qualified. So you should just believe, or take for granted, what’s being shown here: an inhomogeneous electric field will split our ammonia beam according to their state, which we define as I and II respectively, and which are associated with the energy E0+ A and E0− A  respectively. As mentioned above, you should note that these two states are stationary. The Hamiltonian equations which, as they always do, describe the dynamics of this system, imply that the amplitude to go from state I to state II, or vice versa, is zero. To make sure you ‘get’ that, I reproduce the associated Hamiltonian matrix once again: Of course, that will change when we start our analysis of what’s happening in the maser. Indeed, we will have some non-zero HI,II and HII,I amplitudes in the resonant cavity of our ammonia maser, in which we’ll have an oscillating electric field and, as a result, induced transitions from state I to II and vice versa. However, that’s for later. While I’ll quickly insert the full picture diagram below, you should, for the moment, just think about those two stationary states and those two zeroes. 🙂 Capito? If not… Well… Start reading this post again, I’d say. 🙂

#### Intermezzo: on approximations

At this point, I need to say a few things about all of the approximations involved, because it can be quite confusing indeed. So let’s take a closer look at those energy levels and the related Hamiltonian coefficients. In fact, in his LecturesFeynman shows us that we can always have a general solution for the Hamiltonian equations describing a two-state system whenever we have constant Hamiltonian coefficients. That general solution – which, mind you, is derived assuming Hamiltonian coefficients that do not depend on time – can always be written in terms of two stationary base states, i.e. states with a definite energy and, hence, a constant probability. The equations, and the two definite energy levels are:  That yields the following values for the energy levels for the stationary states: Now, that’s very different from the E= E0+ A and EII = E0− A energy levels for those stationary states we had defined in the previous section: those stationary states had no square root, and no μ2ε2, in their energy. In fact, that sort of answers the question: if there’s no external field, then that μ2ε2 factor is zero, and the square root in the expression becomes ±√A= ±A. So then we’re back to our E= E0+ A and EII = E0− A formulas. The whole point, however, is that we will actually have an electric field in that cavity. Moreover, it’s going to be a field that varies in time, which we’ll write: Now, part of the confusion in Feynman’s approach is that he constantly switches between representing the system in terms of the I and II base states and the 1 and 2 base states respectively. For a good understanding, we should compare with our original representation of the dynamics in free space, for which the Hamiltonian was the following one: That matrix can easily be related to the new one we’re going to have to solve, which is equal to: The interpretation is easy if we look at that illustration again: If the direction of the electric dipole moment is opposite to the direction ε, then the associated energy is equal to −μ·ε = −μ·ε = −|μ|·|ε|·cosθ = −μ·ε·cos(π) = +με. Conversely, for state 2, we find −μ·ε·cos(0) = −με for the energy that’s associated with the dipole moment. You can and should think about the physics involved here, because they make sense! Thinking of amplitudes, you should note that the +με and −με terms effectively change the H11 and H22 coefficients, so they change the amplitude to stay in state 1 or state 2 respectively. That, of course, will have an impact on the associated probabilities, and so that’s why we’re talking of induced transitions now.

Having said that, the Hamiltonian matrix above keeps the −A for H12 and H21, so the matrix captures spontaneous transitions too!

Still… You may wonder why Feynman doesn’t use those Eand EII formulas with the square root because that would give us some exact solution, wouldn’t it? The answer to that question is: maybe it would, but would you know how to solve those equations? We’ll have a varying field, remember? So our Hamiltonian H11 and H22 coefficients will no longer be constant, but time-dependent. As you’re going to see, it takes Feynman three pages to solve the whole thing using the +με and −με approximation. So just imagine how complicated it would be using that square root expression! [By the way, do have a look at those asymptotic curves in that illustration showing the splitting of energy levels above, so you see how that approximation looks like.]

So that’s the real answer: we need to simplify somehow, so as to get any solutions at all!

Of course, it’s all quite confusing because, after Feynman first notes that, for strong fields, the A2 in that square root is small as compared to μ2ε2, thereby justifying the use of the simplified E= E0+ με = H11 and EII = E0− με = H22 coefficients, he continues and bluntly uses the very same square root expression to explain how that state selector works, saying that the electric field in the state selector will be rather weak and, hence, that με will be much smaller than A, so one can use the following approximation for the square root in the expressions above:

And then we can calculate the force on the molecules as: So the electric field in the state selector is weak, but the electric field in the cavity is supposed to be strong, and so… Well… That’s it, really. The bottom line is that we’ve a beam of ammonia molecules that are all in state I, and it’s what happens with that beam then, that is being described by our new set of differential equations: #### Solving the equations

As all molecules in our ammonia beam are described in terms of the | I 〉 and | II 〉 base states – as evidenced by the fact that we say all molecules that enter the cavity are state I – we need to switch to that representation. We do that by using that transformation above, so we write:

• C= 〈 I | ψ 〉 = (1/√2)·(C1 − C2)
• CII = 〈 II | ψ 〉 = (1/√2)·(C1 + C2)

Keeping these ‘definitions’ of Cand CII in mind, you should then add the two differential equations, divide the result by the square root of 2, and you should get the following new equation: Please! Do it and verify the result! You want to learn something here, no? 🙂

Likewise, subtracting the two differential equations, we get: Now, the problem is that the Hamiltonian constants here are not constant. To be precise, the electric field ε varies in time. We wrote: So HI,II  and HII,I, which are equal to με, are not constant: we’ve got Hamiltonian coefficients that are a function of time themselves. […] So… Well… We just need to get on with it and try to finally solve this thing. Let me just copy Feynman as he grinds through this: This is only the first step in the process. Feynman just takes two trial functions, which are really similar to the very general Ca·e–(i/ħ)·H11·t function we presented when only one equation was involved, or – if you prefer a set of two equations – those CI(t) = a·e(i/ħ)·EI·t and CI(t) = b·e(i/ħ)·EII·equations above. The difference is that the coefficients in front, i.e. γI and γII are not some (complex) constant, but functions of time themselves. The next step in the derivation is as follows: One needs to do a bit of gymnastics here as well to follow what’s going on, but please do check and you’ll see it works. Feynman derives another set of differential equations here, and they specify these γI = γI(t) and γII = γII(t) functions. These equations are written in terms of the frequency of the field, i.e. ω, and the resonant frequency ω0, which we mentioned above when calculating that 23.79 GHz frequency from the 2A = h·f0 equation. So ω0 is the same molecular resonance frequency but expressed as an angular frequency, so ω0 = f0/2π = ħ/2A. He then proceeds to simplify, using assumptions one should check. He then continues: That gives us what we presented in the previous post: So… Well… What to say? I explained those probability functions in my previous post, indeed. We’ve got two probabilities here:

• P= cos2[(με0/ħ)·t]
• PII = sin2[(με0/ħ)·t]

So that’s just like the P=  cos2[(A/ħ)·t] and P= sin2[(A/ħ)·t] probabilities we found for spontaneous transitions. But so here we are talking induced transitions.

As you can see, the frequency and, hence, the period, depend on the strength, or magnitude, of the electric field, i.e. the εconstant in the ε = 2ε0cos(ω·t) expression. The natural unit for measuring time would be the period once again, which we can easily calculate as (με0/ħ)·T = π ⇔ T = π·ħ/με0.

Now, we had that T = (π·ħ)/(2A) expression above, which allowed us to calculate the period of the spontaneous transition frequency, which we found was like 40 picoseconds, i.e. 40×10−12 seconds. Now, the T = (π·ħ)/(2με0) is very similar, it allows us to calculate the expected, average, or mean time for an induced transition. In fact, if we write Tinduced = (π·ħ)/(2με0) and Tspontaneous = (π·ħ)/(2A), then we can take ratio to find:

Tinduced/Tspontaneous = [(π·ħ)/(2με0)]/[(π·ħ)/(2A)] = A/με0

This A/με0 ratio is greater than one, so Tinduced/Tspontaneous is greater than one, which, in turn, means that the presence of our electric field – which, let me remind you, dances to the beat of the resonant frequency – causes a slower transition than we would have had if the oscillating electric field were not present.

But – Hey! – that’s the wrong comparison! Remember all molecules enter in a stationary state, as they’ve been selected so as to ensure they’re in state I. So there is no such thing as a spontaneous transition frequency here! They’re all polarized, so to speak, and they would remain that way if there was no field in the cavity. So if there was no oscillating electric field, they would never transition. Nothing would happen! Well… In terms of our particular set of base states, of course! Why? Well… Look at the Hamiltonian coefficients HI,II = HII,I = με: these coefficients are zero if ε is zero. So… Well… That says it all.

So that‘s what it’s all about: induced emission and, as I explained in my previous post, because all molecules enter in state I, i.e. the upper energy state, literally, they all ‘dump’ a net amount of energy equal to 2A into the cavity at the occasion of their first transition. The molecules then keep dancing, of course, and so they absorb and emit the same amount as they go through the cavity, but… Well… We’ve got a net contribution here, which is not only enough to maintain the cavity oscillations, but actually also provides a small excess of power that can be drawn from the cavity as microwave radiation of the same frequency.

As Feynman notes, an exact description of what actually happens requires an understanding of the quantum mechanics of the field in the cavity, i.e. quantum field theory, which I haven’t studied yet. But… Well… That’s for later, I guess. 🙂

Post scriptum: The sheer length of this post shows we’re not doing something that’s easy here. Frankly, I feel the whole analysis is still quite obscure, in the sense that – despite looking at this thing again and again – it’s hard to sort of interpret what’s going on, in a physical sense that is. But perhaps one shouldn’t try that. I’ve quoted Feynman’s view on how easy or how difficult it is to ‘understand’ quantum mechanics a couple of times already, so let me do it once more:

“Because atomic behavior is so unlike ordinary experience, it is very difficult to get used to, and it appears peculiar and mysterious to everyone—both to the novice and to the experienced physicist. Even the experts do not understand it the way they would like to, and it is perfectly reasonable that they should not, because all of direct, human experience and human intuition applies to large objects.”

So… Well… I’ll grind through the remaining Lectures now – I am halfway through Volume III now – and then re-visit all of this. Despite Feynman’s warning, I want to understand it the way I like to, even if I don’t quite know what way that is right now. 🙂

Addendum: As for those cycles and periods, I noted a couple of times already that the Planck-Einstein equation E = h·f  can usefully be re-written as E/= h, as it gives a physical interpretation to the value of the Planck constant. In fact, I said h is the energy that’s associated with one cycle, regardless of the frequency of the radiation involved. Indeed, the energy of a photon divided by the number of cycles per second, should give us the energy per cycle, no?

Well… Yes and no. Planck’s constant h and the frequency are both expressed referencing the time unit. However, if we say that a sodium atom emits one photon only as its electron transitions from a higher energy level to a lower one, and if we say that involves a decay time of the order of 3.2×10−8 seconds, then what we’re saying really is that a sodium light photon will ‘pack’ like 16 million cycles, which is what we get when we multiply the number of cycles per second (i.e. the mentioned frequency of 500×1012 Hz) by the decay time (i.e. 3.2×10−8 seconds): (500×1012 Hz)·(3.2×10−8 s) = 16 ×10cycles, indeed. So the energy per cycle is 2.068 eV (i.e. the photon energy) divided by 16×106, so that’s 0.129×10−6 eV. Unsurprisingly, that’s what we get when we we divide h by 3.2×10−8 s: (4.13567×10−15)/(3.2×10−8 s) = 1.29×10−7 eV. We’re just putting some values in to the E/(T) = h/T equation here.

The logic for that 2A = h·f0 is the same. The frequency of the radiation that’s being absorbed or emitted is 23.79 GHz, so the photon energy is (23.97×109 Hz)·(4.13567×10−15 eV·s) ≈ 1×10−4 eV. Now, we calculated the transition period T as T = π·ħ/A ≈ (π·6.626×10−16 eV·s)/(0.5×10−4 eV) ≈ 41.6×10−12 seconds. Now, an oscillation of a frequency of 23.97 giga-hertz that only lasts 41.6×10−12 seconds is an oscillation of one cycle only. The consequence is that, when we continue this style of reasoning, we’d have a photon that packs all of its energy into one cycle!

Let’s think about what this implies in terms of the density in space. The wavelength of our microwave radiation is 1.25×10−2 m, so we’ve got a ‘density’ of 1×10−4 eV/1.25×10−2 m = 0.8×10−2 eV/m = 0.008 eV/m. The wavelength of our sodium light is 0.6×10−6 m, so we get a ‘density’ of 1.29×10−7 eV/0.6×10−6 m = 2.15×10−1 eV/m = 0.215 eV/m. So the energy ‘density’ of our sodium light is 26.875 times that of our microwave radiation. 🙂

Frankly, I am not quite sure if calculations like this make much sense. In fact, when talking about energy densities, I should review my posts on the Poynting vector. However, they may help you think things through. 🙂

# The ammonia maser: transitions in a time-dependent field

Feynman’s analysis of a maser – microwave amplification, by stimulated emission of radiation – combines an awful lot of stuff. Resonance, electromagnetic field theory, and quantum mechanics: it’s all there! Therefore, it’s complicated and, hence, actually very tempting to just skip it when going through his third volume of Lectures. But let’s not do that. What I want to do in this post is not repeat his analysis, but reflect on it and, perhaps, offer some guidance as to how to interpret some of the math.

#### The model: a two-state system

The model is a two-state system, which Feynman illustrates as follows: Don’t shy away now. It’s not so difficult. Try to understand. The nitrogen atom (N) in the ammonia molecule (NH3) can tunnel through the plane of the three hydrogen (H) atoms, so it can be ‘up’ or ‘down’. This ‘up’ or ‘down’ state has nothing to do with the classical or quantum-mechanical notion of spin, which is related to the magnetic moment. Nothing, i.e. nada, niente, rien, nichts! Indeed, it’s much simpler than that. 🙂 The nitrogen atom could be either beneath or, else, above the plane of the hydrogens, as shown above, with ‘beneath’ and ‘above’ being defined in regard to the molecule’s direction of rotation around its axis of symmetry. That’s all. That’s why we prefer simple numbers to denote those two states, instead of the confusing ‘up’ or ‘down’, or ‘↑’ or ‘↓’ symbols. We’ll just call the two states state ‘1’ and state ‘2’ respectively.

Having said that (i.e. having said that you shouldn’t think of spin, which is related to the angular momentum of some (net) electric charge), the NHmolecule does have some electric dipole moment, which is denoted by μ in the illustration and which, depending on the state of the molecule (i.e. the nitrogen atom being above or beneath the plane of the hydrogens), changes the total energy of the molecule by an amount that is equal to +με or −με, with ε some external electric field, as illustrated by the ε arrow on the left-hand side of the diagram. [You may think of that arrow as an electric field vector.] This electric field may vary in time and/or in space, but we’ll not worry about that now. In fact, we should first analyze what happens in the absence of an external field, which is what we’ll do now.

The NHmolecule will spontaneously transition from an ‘up’ to a ‘down’ state, or from ‘1’ to ‘2’—and vice versa, of course! This spontaneous transition is also modeled as an uncertainty in its energy. Indeed, we say that, even in the absence of an external electric field, there will be two energy levels, rather than one only: E+ A and E− A.

We wrote the amplitude to find the molecule in either one of these two states as:

• C1(t) = 〈 1 | ψ 〉 = (1/2)·e(i/ħ)·(E− A)·t + (1/2)·e(i/ħ)·(E+ A)·t = e(i/ħ)·E0·t·cos[(A/ħ)·t]
• C2(t) =〈 2 | ψ 〉 = (1/2)·e(i/ħ)·(E− A)·t – (1/2)·e(i/ħ)·(E+ A)·t = i·e(i/ħ)·E0·t·sin[(A/ħ)·t]

[Remember: the sum of complex conjugates, i.e eiθ eiθ reduces to 2·cosθ, while eiθ − eiθ reduces to 2·i·sinθ.]

That gave us the following probabilities:

• P= |C1|2 = cos2[(A/ħ)·t]
• P= |C2|= sin2[(A/ħ)·t]

[Remember: the absolute square of is |i|= +√12 = +1, so the in the C2(t) formula disappears.]

The graph below shows how these probabilities evolve over time. Note that, because of the square, the period of cos2[(A/ħ)·t] and sin2[(A/ħ)·t] is equal to π, instead of the usual 2π. The interpretation of this is easy enough: if our molecule can be in two states only, and it starts off in one, then the probability that it will remain in that state will gradually decline, while the probability that it flips into the other state will gradually increase. As Feynman puts it: the first state ‘dumps’ probability into the second state as time goes by, and vice versa, so the probability sloshes back and forth between the two states.

The graph above measures time in units of ħ/A but, frankly, the ‘natural’ unit of time would usually be the period, which you can easily calculate as (A/ħ)·T = π ⇔ T = π·ħ/A. In any case, you can go from one unit to another by dividing or multiplying by π. Of course, the period is the reciprocal of the frequency and so we can calculate the molecular transition frequency fas f0 = A/[π·ħ] = 2A/h. [Remember: h = 2π·ħ, so A/[π·ħ] = 2A/h].

Of course, by now we’re used to using angular frequencies, and so we’d rather write: ω= 2π·f= f= 2π·A/[π·ħ] = 2A/ħ. And because it’s always good to have some idea of the actual numbers – as we’re supposed to model something real, after all – I’ll give them to you straight away. The separation between the two energy levels E+ A and E− A has been measured as being equal to 2A = hf0 ≈ 10−4 eV, more or less. 🙂 That’s tiny. To avoid having to convert this to joule, i.e. the SI unit for energy, we can calculate the corresponding frequency using h expressed in eV·s, rather than in J·s. We get: f0 = 2A/h = (1×10−4 eV)/(4×10−15 eV·s) = 25 GHz. Now, we’ve rounded the numbers here: the exact frequency is 23.79 GHz, which corresponds to microwave radiation with a wavelength of λ = c/f0 = 1.26 cm.

How does one measure that? It’s simple: ammonia absorbs light of this frequency. The frequency is also referred to as a resonance frequency, as light of this frequency, i.e. microwave radiation, will also induce transitions from one state to another. In fact, that’s what the stimulated emission of radiation principle is all about. But we’re getting ahead of ourselves here. It’s time to look at what happens if we do apply some external electric field, which is what we’ll do now.

#### Polarization and induced transitions

As mentioned above, an electric field will change the total energy of the molecule by an amount that is equal to +με or −με. Of course, the plus or the minus in front of με depends both on the direction of the electric field ε, as well as on the direction of μ. However, it’s not like our molecule might be in four possible states. No. We assume the direction of the field is given, and then we have two states only, with the following energy levels: Don’t rack your brain over how you get that square root thing. You get it when applying the general solution of a pair of Hamiltonian equations to this particular case. For full details on how to get this general solution, I’ll refer you to Feynman. Of course, we’re talking base states here, which do not always have a physical meaning. However, in this case, they do: a jet of ammonia gas will split in an inhomogeneous electric field, and it will split according to these two states, just like a beam of particles with different spin in a Stern-Gerlach apparatus. A Stern-Gerlach apparatus splits particle beams because of an inhomogeneous magnetic field, however. So here we’re talking an electric field.

It’s important to note that the field should not be homogeneous, for the very same reason as to why the magnetic field in the Stern-Gerlach apparatus should not be homogeneous: it’s because the force on the molecules will be proportional to the derivative of the energy. So if the energy doesn’t vary—so if there is no strong field gradient—then there will be no force. [If you want to get more detail, check the section on the Stern-Gerlach apparatus in my post on spin and angular momentum.] To be precise, if με is much smaller than A, then one can use the following approximation for the square root in the expressions above:

And then we can calculate the force on the molecules as: The bottom line is that our ammonia jet will split into two separate beams: all molecules in state I will be deflected toward the region of lower ε2, and all molecules in state II will be deflected toward the region of larger ε2. [We talk about ε2 rather than ε because of the ε2 gradient in that force formula. However, you could, of course, simplify and write ε2 as ε= 2εε.] So, to make a long story short, we should now understand the left-hand side of the schematic maser diagram below. It’s easy to understand that the ammonia molecules that go into the maser cavity are polarized. To understand the maser, we need to understand how the maser cavity works. It’s a so-called resonant cavity, and we’ve got an electric field in it as well. The field direction happens to be south as we’re looking at it right now, but in an actual maser we’ll have an electric field that varies sinusoidally. Hence, while the direction of the field is always perpendicular to the direction of motion of our ammonia molecules, it switches from south to north and vice versa all of the time. We write ε as:

ε = 2ε0cos(ω·t) = ε0(ei·ω·t ei·ω·t)

Now, you’ve guessed it, of course. If we ensure that ω = ω= 2A/ħ, then we’ve got a maser. In fact, the result is a similar graph: Let’s first explain this graph. We’ve got two probabilities here:

• P= cos2[(με0/ħ)·t]
• PII = sin2[(με0/ħ)·t]

So that’s just like the P=  cos2[(A/ħ)·t] and P= sin2[(A/ħ)·t] probabilities we found for spontaneous transitions. In fact, the formulas for the related amplitudes are also similar to those for C1(t) and C2(t):

• CI(t) = 〈 I | ψ 〉 = e(i/ħ)·EI·t·cos[(με0/ħ)·t], which is equal to:

CI(t) = e(i/ħ)·(E0+A)·t·cos[(με0/ħ)·t] = e(i/ħ)·(E0+A)·t·(1/2)·[ei·(με0/ħ)·t + ei·(με0/ħ)·t] = (1/2)·e(i/ħ)·(E0+A−με0)·t + (1/2)·e(i/ħ)·(E0+A+με0)·t

• CII(t) = 〈 II | ψ 〉 = i·e(i/ħ)·EII·t·sin[(με0/ħ)·t], which is equal to:

CII(t) = e(i/ħ)·(E0−A)·t·i·sin[(με0/ħ)·t] = e(i/ħ)·(E0−A)·t·(1/2)·[ei·(με0/ħ)·t ei·(με0/ħ)·t] = (1/2)·e(i/ħ)·(E0−A−με0)·t – (1/2)·e(i/ħ)·(E0−A+με0)·t

But so here we are talking induced transitions. As you can see, the frequency and, hence, the period, depend on the strength, or magnitude, of the electric field, i.e. the εconstant in the ε = 2ε0cos(ω·t) expression. The natural unit for measuring time would be the period once again, which we can easily calculate as (με0/ħ)·T = π ⇔ T = π·ħ/με0. However, Feynman adds an 1/2 factor so as to ensure it’s the time that corresponds to the time a molecule needs to go through the cavity. Well… That’s what he says, at least. I’ll show he’s actually wrong, but the idea is OK.

First have a look at the diagram of our maser once again. You can see that all molecules come in in state I, but are supposed to leave in state II. Now, Feynman says that’s because the cavity is just long enough so as to more or less ensure that all ammonia molecules switch from state I to state II. Hmm… Let’s have a close look at that. What the functions and the graph are telling us is that, at the point t = 1 (with t being measured in those π·ħ/2με0 units), the probability of being in state I has all been ‘dumped’ into the probability of being in state II!

So… Well… Our molecules had better be in that state then! 🙂 Of course, the idea is that, as they transition from state I to state II, they lose energy. To be precise, according to our expressions for Eand EII above, the difference between the energy levels that are associated with these two states is equal to 2A + μ2ε02/A.

Now, a resonant cavity is a cavity designed to keep electromagnetic waves like the oscillating field that we’re talking about here going with minimal energy loss. Indeed, a microwave cavity – which is what we’re having here – is similar to a resonant circuit, except that it’s much better than any equivalent electric circuit you’d try to build, using inductors and capacitors. ‘Much better’ means it hardly needs energy to keep it going. We express that using the so-called Q-factor (believe it or not: the ‘Q’ stands for quality). The Q factor of a resonant cavity is of the order of 106, as compared to 102 for electric circuits that are designed for the same frequencies. But let’s not get into the technicalities here. Let me quote Feynman as he summarizes the operation of the maser:

“The molecule enters the cavity, [and then] the cavity field—oscillating at exactly the right frequency—induces transitions from the upper to the lower state, and the energy released is fed into the oscillating field. In an operating maser the molecules deliver enough energy to maintain the cavity oscillations—not only providing enough power to make up for the cavity losses but even providing small amounts of excess power that can be drawn from the cavity. Thus, the molecular energy is converted into the energy of an external electromagnetic field.”

As Feynman notes, it is not so simple to explain how exactly the energy of the molecules is being fed into the oscillations of the cavity: it would require to also deal with the quantum mechanics of the field in the cavity, in addition to the quantum mechanics of our molecule. So we won’t get into that nitty-gritty—not here at least. So… Well… That’s it, really.

Of course, you’ll wonder about the orders of magnitude, or minitude, involved. And… Well… That’s where this analysis is somewhat tricky. Let me first say something more about those resonant cavities because, while that’s quite straightforward, you may wonder if they could actually build something like that in the 1950s. 🙂 The condition is that the cavity length must be an integer multiple of the half-wavelength at resonance. We’ve talked about this before. [See, for example, my post on wave modes. More formally, the condition for resonance in a resonator is that the round trip distance, 2·d, is equal to an integral number of the wavelength λ, so we write: 2·d = N·λ, with N = 1, 2, 3, etc. Then, if the velocity of our wave is equal to c, then the resonant frequencies will be equal to f = (N·c)/(2·d).

Does that makes sense? Of course. We’re talking the speed of light, but we’re also talking microwaves. To be specific, we’re talking a frequency of 23.79 GHz and, more importantly, a wavelength that’s equal to λ = c/f0 = 1.26 cm, so for the first normal mode (N = 1), we get 2·d = λ ⇔ d = λ/2 = 63 mm. In short, we’re surely not talking nanotechnology here! In other words, the technological difficulties involved in building the apparatus were not insurmountable. 🙂

But what about the time that’s needed to travel through it? What about that length? Now, that depends on the μεquantity if we are to believe Feynman here. Now, we actually don’t need to know the actual values for μ or ε: we said that the value of the μεproduct is (much) smaller than the value of A. Indeed, the fields that are used in those masers aren’t all that strong, and the electric dipole moment μ is pretty tiny. So let’s say με0 = A/2, which is the upper limit for our approximation of that square root above, so 2με0 = A = 0.5×10−4 eV. [The approximation for that square root expression is only used when y ≤ x/2.]

Let’s now think about the time. It was measured in units equal to T = π·ħ/2με0. So our T here is not the T we defined above, which was the period. Here it’s the period divided by two. First the dimensions: ħ is expressed in eV·s, and με0 is an energy, so we can express it in eV too: 1 eV ≈ 1.6×10−19 J, i.e. 160 zeptojoules. 🙂 π is just a real number, so our T = π·ħ/2μεgives us seconds alright. So we get:

T ≈ (3.14×6.6×10−16 eV·s)/(0.5×10−4 eV) ≈ 40×10−12 seconds

[…] Hmm… That doesn’t look good. Even when traveling at the speed of light – which our ammonia molecule surely doesn’t do! – it would only travel over a distance equal to (3×108 m/s)·(20×10−12 s) = 60×10−4 m = 0.6 cm = 6 mm. The speed of our ammonia molecule is likely to be only a fraction of the speed of light, so we’d have an extremely short cavity then. The time mentioned is also not in line with what Feynman mentions about the ammonia molecule being in the cavity for a ‘reasonable length of time, say for one millisecond.‘ One millisecond is also more in line with the actual dimensions of the cavity which, as you can see from the historical illustration below, is quite long indeed. So what’s going on here? Feynman’s statement that T is “the time that it takes the molecule to go through the cavity” cannot be right. Let’s do some good thinking here. For example, let’s calculate the time that’s needed for a spontaneous state transition and compare with the time we calculated above. From the graph and the formulas above, we know we can calculate that from the (A/ħ)·T = π/2 equation. [Note the added 1/2 factor, because we’re not going through a full probability cycle: we’re going through a half-cycle only.] So that’s equivalent to T = (π·ħ)/(2A). We get:

T ≈ (3.14×6.6×10−16 eV·s)/(1×10−4 eV) ≈ 20×10−12 seconds

The T = π·ħ/2με0 and T = (π·ħ)/(2A) expression make it obvious that the expected, average, or mean time for a spontaneous versus an induced transition depends on A and με respectively. Let’s be systematic now, so we’ll distinguish Tinduced = (π·ħ)/(2με0) from Tspontaneous = (π·ħ)/(2A) respectively. Taking the ratio, we find:

Tinduced/Tspontaneous = [(π·ħ)/(2με0)]/[(π·ħ)/(2A)] = A/με0

However, we know the A/με0 ratio is greater than one, so Tinduced/Tspontaneous is greater than one, which, in turn, means that the presence of our electric field – which, let me remind you, dances to the beat of the resonant frequency – causes a slower transition than we would have had if the oscillating electric field were not present. We may write the equation above as:

Tinduced = [A/με0]·Tspontaneous = [A/με0]·(π·ħ)/(2A) = h/(4με0)

However, that doesn’t tell us anything new. It just says that the transition period (T) is inversely proportional to the strength of the field (as measured by ε0). So a weak field will make for a longer transition period (T), with T → ∞ as ε0 → 0. So it all makes sense, but what do we do with this?

The Tinduced/Tspontaneous = [με0/A]−1 is the most telling. It says that the Tinduced/Tspontaneous is inversely proportional to the με0/A ratio. For example, if the energy με0 is only one fifth of the energy A, then the time for the induced transition will be five times that of a spontaneous transition. To get something like a millisecond, however, we’d need the με0/A ratio to go down to like a billionth or something, which doesn’t make sense.

So what’s the explanation? Is Feynman hiding something from us? He’s obviously aware of these periods because, when discussing the so-called three-state maser, he notes that “The | I 〉 state has a long lifetime, so its population can be increased.” But… Well… That’s just not relevant here. He just made a mistake: the length of the maser has nothing to do with it. The thing is: once the molecule transitions from state I to state II, then that’s basically the end of the story as far as the maser operation is concerned. By transitioning, it dumps that energy 2A + μ2ε02/A into the electric field, and that’s it. That’s energy that came from outside, because the ammonia molecules were selected so as to ensure they were in state I. So all the transitions afterwards don’t really matter: the ammonia molecules involved will absorb energy as they transition, and then give it back as they transition again, and so on and so on. But that’s no extra energy, i.e. no new or outside energy: it’s just energy going back and forth from the field to the molecules and vice versa.

So, in a way, those PI and PII curves become irrelevant. Think of it: the energy that’s related to A and μεis defined with respect to a certain orientation of the molecule as well as with respect to the direction of the electric field before it enters the apparatus, and the induced transition is to happen when the electric field inside of the cavity points south, as shown in the diagram. But then the transition happens, and that’s the end of the story, really. Our molecule is then in state II, and will oscillate between state II and I, and back again, and so on and so on, but it doesn’t mean anything anymore, as these flip-flops do not add any net energy to the system as a whole.

So that’s the crux of the matter, really. Mind you: the energy coming out of the first masers was of the order of one microwatt, i.e. 10−6 joule per second. Not a lot, but it’s something, and so you need to explain it from an ‘energy conservation’ perspective: it’s energy that came in with the molecules as they entered the cavity. So… Well… That’s it.

The obvious question, of course, is: why do we actually need the oscillating field in the cavity? If all molecules come in in the ‘upper’ state, they’ll all dump their energy anyway. Why do we need the field? Well… First, you should note that the whole idea is that our maser keeps going because it uses the energy that the molecules are dumping into its field. The more important thing, however, is that we actually do need the field to induce the transition. That’s obvious from the math. Look at the probability functions once again:

• P= cos2[(με0/ħ)·t]
• PII = sin2[(με0/ħ)·t]

If there would be no electric field, i.e. if ε0 = 0, then P= 1 and PII = 0. So, our ammonia molecules enter in state I and, more importantly, stay in state I forever, so there’s no chance whatsoever to transition to state II. Also note what I wrote above: Tinduced = h/(4με0), and, therefore, we find that T → ∞ as ε0 → 0.

So… Well… That’s it. I know this is not the ‘standard textbook’ explanation of the maser—it surely isn’t Feynman’s! But… Well… Please do let me know what you think about it. What I write above, indicates the analysis is much more complicated than standard textbooks would want it to be.

There’s one more point related to masers that I need to elaborate on, and that’s its use as an ‘atomic’ clock. So let me quickly do that now.

#### The use of a maser as an ‘atomic’ clock

In light of the amazing numbers involved – we talked GHz frequencies, and cycles expressed in picoseconds – we may wonder how it’s possible to ‘tune’ the frequency of the field to the ‘natural’ molecular transition frequency. It will be no surprise to hear that it’s actually not straightforward. It’s got to be right: if the frequency of the field, which we’ll denote by ω, is somewhat ‘off’ – significantly different from the molecular transition frequency ω– then the chance of transitioning from state I to state II shrinks significantly, and actually becomes zero for all practical purposes. That basically means that, if the frequency isn’t right, then the presence of the oscillating field doesn’t matter. In fact, the fact that the frequency has got to be right – with tolerances that, as we will see in a moment, are expressed in billionths – is why a maser can be used as an atomic clock.

The graph below illustrates the principle. If ω = ω0, then the probability that a transition from state I to II will happen is one, so PI→II(ω)/PI→II0) = 1. If it’s slightly off, though, then the ratio decreases quickly, which means that the PI→II probability goes rapidly down to zero. [There’s secondary and tertiary ‘bumps’ because of interference of amplitudes, but they’re insignificant.] As evidenced from the graph, the cut-off point is ω − ω= 2π/T, which we can re-write as 2π·f − 2π·f= 2π/T, which is equivalent to writing: (f − f0)/f0 =   1/(f0T). Now, we know that f= 23.79 GHz, but what’s T in this expression? Well… This time around it actually is the time that our ammonia molecules spend in the resonant cavity, from going in to going out, which Feynman says is of the order of a millisecond—so that’s much more reasonable that those 40 picoseconds we calculated. So 1/(f0T) = 1/[23.79×109·1×−3] ≈ 0.042×10−6 = 42×10−9  , i.e. 42 billionths indeed, which Feynman rounds to “five parts in 108“, i.e. five parts in a hundred million. In short, the frequency must be ‘just right’, so as to get a significant transition probability and, therefore, get some net energy out of our maser, which, of course, will come out of our cavity as microwave radiation of the same frequency. Now that’s how one the first ‘atomic’ clock was built: the maser was the equivalent of a resonant circuit, and one could keep it going with little energy, because it’s so good as a resonant circuit. However, in order to get some net energy out of the system, in the form of microwave radiation of, yes, the ammonia frequency, the applied frequency had to be exactly right. To be precise, the applied frequency ω has to match the ω0 frequency, i.e. the molecular resonance frequency, with a precision expressed in billionths. As mentioned above, the power output is very limited, but it’s real: it comes out through the ‘output waveguide’ in the illustration above or, as the Encyclopædia Brittanica puts it: “Output is obtained by allowing some radiation to escape through a small hole in the resonator.” 🙂

In any case, a maser is not build to produce huge amounts of power. On the contrary, the state selector obviously consumes more power than comes out of the cavity, obviously, so it’s not some generator. Its main use nowadays is as a clock indeed, and so it’s that simple really: if there’s no output, then the ‘clock’ doesn’t work.

It’s an interesting topic, but you can read more about it yourself. I’ll just mention that, while the ammonia maser was effectively used as a timekeeping device, the next-generation of atomic clocks was based on the hydrogen maser, which was introduced in 1960. The principle is the same. Let me quote the Encyclopædia Brittanica on it: “Its output is a radio wave, hose frequency of 1,420,405,751.786 hertz (cycles per second) is reproducible with an accuracy of one part in 30 × 1012. A clock controlled by such a maser would not get out of step more than one second in 100,000 years.”

So… Well… Not bad. 🙂 Of course, one needs another clock to check if one’s clock is still accurate, and so that’s what’s done internationally: national standards agencies in various countries maintain a network of atomic clocks which are intercompared and kept synchronized. So these clocks define a continuous and stable time scale, collectively, which is referred to as the International Atomic Time (TAI, from the French Temps Atomique International).

Well… That’s it for today. I hope you enjoyed it.

Post scriptum:

When I say the ammonia molecule just dumps that energy 2A + μ2ε02/A into the electric field, and that’s “the end of the story”, then I am simplifying, of course. The ammonia molecule still has two energy levels, separated by an energy difference of 2A and, obviously, it keeps its electric dipole moment and so that continues to play as we’ve got an electric field in the cavity. In fact, the ammonia molecule has a high polarizability coefficient, which means it’s highly sensitive to the electric field inside of the cavity. So, yes, the molecules will continue ‘dancing’ to be the beat of the field indeed, and absorbing and releasing energy, in accordance with that 2A and με0 factor, and so the probability curves do remain relevant—of course! However, we talked net energy going into the field, and so that’s where the ‘end of story’ story comes in. I hope I managed to make that clear.

In fact, there’s lots of other complications as well, and Feynman mentions them briefly in his account of things. But let’s keep things simple here. 🙂 Also, if you’d want to know how we get that PI→II(ω)/PI→II0), check it out in Feynman. However, I have to warn you: the math involved is not easy. Not at all, really. The set of differential equations that’s involved is complicated, and it takes a while to understand why Feynman uses the trial functions he uses. So the solution that comes out, i.e. those simple P= cos2[(με0/ħ)·t] and PII = sin2[(με0/ħ)·t] functions, makes sense—but, if you check it out, you’ll see the whole mathematical argument is rather complicated. That’s just how it is, I am afraid. 🙂