I want to come back to something I mentioned in a previous post: when looking at that formula for those U_{ij} amplitudes—which I’ll jot down once more:

U_{ij}(t + Δt, t) = δ_{ij} + ΔU_{ij}(t + Δt, t) = δ_{ij} + K_{ij}(t)·Δt ⇔ U_{ij}(t + Δt, t) = δ_{ij} − (*i*/ħ)·H_{ij}(t)·Δt

—I noted that it resembles the general y(t + Δt) = y(t) + Δy = y(t) + (dy/dt)·Δt formula. So we can look at our K_{ij}(t) function as being equal to the time derivative of the U_{ij}(t + Δt, t) function. I want to re-visit that here, as it triggers a whole range of questions, which may or may not help to understand quantum math somewhat more *intuitively*. Let’s quickly sum up what we’ve learned so far: it’s basically all about quantum-mechanical stuff that does *not *move in space. Hence, the x in our wavefunction ψ(x, t) is some fixed point in space and, therefore, our elementary wavefunction—which we wrote as:

ψ(x, t) = *a·e*^{−i·θ }= *a·e*^{−i·(ω·t − k∙x)} = *a·e*^{−i·[(E/ħ)·t − (p/ħ)∙x]}

—reduces to ψ(t) = *a·e*^{−i·ω·t} = *a·e*^{−i·[(E/ħ)·t}.

Unlike what you might think, we’re *not *equating x with zero here. No. It’s the p = m·*v* factor that becomes zero, because our reference frame is that of the system that we’re looking at, so its velocity is zero: it doesn’t move in *our *reference frame. That immediately answers an obvious question: does our wavefunction look any different when choosing another reference frame? The answer is obviously: yes! It surely matters if the system moves or not, and it also matters how fast it moves, because it changes the energy and momentum values from E and p to some E’ and p’. However, we’ll not consider such complications here: that’s the realm of *relativistic *quantum mechanics. Let’s start with the simplest of situations.

**A simple two-state system**

One of the simplest examples of a quantum-mechanical system that does not move in space, is the textbook example of the ammonia molecule. The picture was as simple as the one below: an ammonia molecule consists of one nitrogen atom and three hydrogen atoms, and the nitrogen atom could be ‘up’ or ‘down’ with regard to the motion of the NH_{3 }molecule around its axis of symmetry, as shown below.

It’s important to note that this ‘up’ or ‘down’ direction is, once again, defined with respect to the reference frame of the system itself. The motion of the molecule around its axis of symmetry is referred to as its *spin*—a term that’s used in a variety of contexts and, therefore, is annoyingly ambiguous. When we use the term ‘spin’ (up or down) to describe an electron state, for example, we’d associate it with the direction of its *magnetic *moment. Such *magnetic *moment arises from the fact that, for all practical purposes, we can think of an electron as a spinning *electric *charge. Now, while our ammonia molecule is electrically neutral, as a whole, the two states are actually associated with opposite *electric *dipole moments, as illustrated below. Hence, when we’d apply an electric field (denoted as ε) below, the two states are effectively associated with different energy levels, which we wrote as E_{0} ± εμ.

But we’re getting ahead of ourselves here. Let’s revert to the system in free space, i.e. without an electromagnetic force field—or, what amounts to saying the same, without *potential*. Now, the ammonia molecule is a quantum-mechanical system, and so there is some* amplitude* for the nitrogen atom to *tunnel* through the plane of hydrogens. I told you before that this is the key to understanding quantum mechanics really: there *is *an energy barrier there and, classically, the nitrogen atom should *not *sneak across. But it does. It’s like it can *borrow *some energy – which we denote by A – to penetrate the energy barrier.

In quantum mechanics, the dynamics of this system are modeled using a set of two differential equations. These differential equations are really the equivalent of Newton’s classical Law of Motion (I am referring to the F = m·(d*v*/dt) = m·*a *equation here) in quantum mechanics, so I’ll have to explain them—which is *not *so easy as explaining Newton’s Law, because we’re talking complex-valued functions, but… Well… Let me first insert the *solution *of that set of differential equations:

This graph shows how the probability of the nitrogen atom (or the ammonia molecule itself) being in state 1 (i.e. ‘up’) *or*, else, in state 2 (i.e. ‘down’), varies sinusoidally in time. Let me also give you the equations for the *amplitudes *to be in state 1 or 2 respectively:

- C
_{1}(t) = 〈 1 | ψ 〉 = (1/2)·*e*^{−(i/ħ)·(E0 − A)·t }+ (1/2)·*e*^{−(i/ħ)·(E0 + A)·t }=*e*^{−(i/ħ)·E0·t}·cos[(A/ħ)·t] - C
_{2}(t) = 〈 2 | ψ 〉 = (1/2)·*e*^{−(i/ħ)·(E0 − A)·t }– (1/2)·*e*^{−(i/ħ)·(E0 + A)·t }=*i*·*e*^{−(i/ħ)·E0·t}·sin[(A/ħ)·t]

So the P_{1}(t) and P_{2}(t) *probabilities *above are just the *absolute *square of these C_{1}(t) and C_{2}(t) functions. So as to help you understand what’s going on here, let me quickly insert the following technical remarks:

- 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
*e*^{iθ }+*e*^{−iθ }reduces to 2·cosθ, while*e*^{iθ }−*e*^{−iθ }reduces to 2·*i*·sinθ. - As for how to take the absolute square… Well… I shouldn’t be explaining that here, but you should be able to work that out remembering that (i) |a·b·c|
^{2}= |a|^{2}·|b|^{2}·|c|^{2}; (ii) |*e*^{iθ}|^{2}= |e^{−iθ}|^{2 }= 1^{2}= 1 (for any value of θ); and (iii) |*i*|^{2}= 1. - As for the periodicity of both probability functions, note that the period of the
*squared*sine and cosine functions is equal to π. Hence, the argument of our sine and cosine function will be equal to 0, π, 2π, 3π etcetera if (A/ħ)·t = 0, π, 2π, 3π etcetera, i.e. if t = 0·ħ/A, π·ħ/A, 2π·ħ/A, 3π·ħ/A etcetera. So that’s why we measure time in units of ħ/A above.

The graph above is actually tricky to interpret, as it assumes that we *know *in what state the molecule starts out with at t = 0. This assumption is tricky because we usually do *not *know that: we have to make some observation which, curiously enough, will always yield one of the two states—nothing in-between. Or, else, we can use a state selector—an inhomogeneous electric field which will separate the ammonia molecules according to their state. It’s a weird thing really, and it summarizes all of the ‘craziness’ of quantum-mechanics: as long as we don’t measure anything – by applying that force field – our molecule is in some kind of abstract state, which mixes the two *base states*. But when we do make the measurement, always along some specific direction (which we usually take to be the *z*-direction in our reference frame), we’ll always find the molecule is either ‘up’ or, else, ‘down’. We never *measure *it as something in-between. Personally, I like to think the measurement apparatus – I am talking the electric field here – causes the nitrogen atom to sort of ‘snap into place’. However, physicists use more precise language here: they would say that the electric field does result in the two positions having very different energy levels (E_{0} + εμ and E_{0} – εμ, to be precise) and that, as a result, *the amplitude for the nitrogen atom to flip back and forth has little effect*. Now how do we *model *that?

**The Hamiltonian equations**

I shouldn’t be using the term above, as it usually refers to a set of differential equations describing *classical* systems. However, I’ll also use it for the quantum-mechanical analog, which amounts to the following for our simple two-state example above:

Don’t panic. We’ll explain. The equations above are all the same but use different formats: the first block writes them as a *set *of equations, while the second uses the *matrix *notation, which involves the use of that rather infamous *Hamiltonian matrix*, which we denote by H = [H_{ij}]. Now, we’ve postponed a lot of technical stuff, so… Well… We can’t avoid it any longer. Let’s look at those Hamiltonian coefficients H_{ij} first. Where do they come from?

You’ll remember we thought of time as some kind of apparatus, with particles entering in some *initial *state φ and coming out in some *final *state χ. Both are to be described in terms of our base states. To be precise, we associated the (complex) coefficients C_{1} and C_{2} with |φ〉 and D_{1} and D_{2} with |χ〉. However, the χ state is a *final *state, so we have to write it as 〈χ| = |χ〉† (read: *chi dagger*). The dagger symbol tells us we need to take the conjugate transpose of |χ〉, so the column vector becomes a row vector, and its coefficients are the *complex conjugate *of D_{1} and D_{2}, which we denote as D_{1}* and D_{2}*. We combined this with Dirac’s bra-ket notation for the *amplitude *to go from one *base* state to another, *as a function in time* (or a function *of* time, I should say):

U_{ij}(t + Δt, t) = 〈i|U(t + Δt, t)|j〉

This allowed us to write the following matrix equation:

To see what it means, you should write it all out:

〈χ|U(t + Δt, t)|φ〉 = D_{1}*·(U_{11}(t + Δt, t)·C_{1} + U_{12}(t + Δt, t)·C_{2}) + D_{2}*·(U_{21}(t + Δt, t)·C_{1} + U_{22}(t + Δt, t)·C_{2})

= D_{1}*·U_{11}(t + Δt, t)·C_{1 }+ D_{1}*·U_{12}(t + Δt, t)·C_{2 }+ D_{2}*·U_{21}(t + Δt, t)·C_{1 }+ D_{2}*·U_{22}(t + Δt, t)·C_{2}

It’s a horrendous expression, but it’s a complex-valued amplitude or, quite simply, a complex number. So this is *not *nonsensical. We can now take the next step, and that’s to go from those U_{ij} amplitudes to the H_{ij} amplitudes of the Hamiltonian matrix. The key is to consider the following: if Δt goes to zero, nothing happens, so we write: U_{ij} = 〈i|U|j〉 → 〈i|j〉 = δ_{ij} for Δt → 0, with δ_{ij} = 1 if i = j, and δ_{ij} = 0 if i ≠ j. We then assume that, for small t, those U_{ij} amplitudes should *differ *from δ_{ij} (i.e. from 1 or 0) by amounts* *that are *proportional to *Δt. So we write:

U_{ij}(t + Δt, t) = δ_{ij} + ΔU_{ij}(t + Δt, t) = δ_{ij} + K_{ij}(t)·Δt

We then equated those K_{ij}(t) factors with − (*i*/ħ)·H_{ij}(t), and we were done: U_{ij}(t + Δt, t) = δ_{ij} − (*i*/ħ)·H_{ij}(t)·Δt. […] Well… I show you how we get those differential equations in a moment. Let’s pause here for a while to see what’s going on really. You’ll probably remember how one can mathematically ‘construct’ the complex exponential *e*^{iθ }by using the *linear* approximation *e ^{i}*

^{ε}= 1 +

*i*ε near θ = 0 and for infinitesimally small values of ε. In case you forgot, we basically used the

*definition*of the derivative of the

*real*exponential

*e*

^{ε }for ε going to zero:

So we’ve got something similar here for U_{11}(t + Δt, t) = 1 − *i*·[H_{11}(t)/ħ]·Δt and U_{22}(t + Δt, t) = 1 − *i*·[H_{22}(t)/ħ]·Δt. Just replace the ε in *e ^{i}*

^{ε}= 1 +

*i*ε by ε = − (E

_{0}/ħ)·Δt. Indeed, we know that H

_{11}= H

_{22}= E

_{0}, and E

_{0}/ħ is, of course, just the energy

*measured in (reduced) Planck units*, i.e. in its

*natural*unit. Hence, if our ammonia molecule is in one of the two base states, we start at θ = 0 and then we just start moving on the unit circle,

*clockwise*, because of the

*minus*sign in

*e*

^{−iθ}. Let’s write it out:

U_{11}(t + Δt, t) = 1 − *i*·[H_{11}(t)/ħ]·Δt = 1 − *i*·[E_{0}/ħ]·Δt and

U_{22}(t + Δt, t) = 1 − *i*·[H_{22}(t)/ħ]·Δt = 1 − *i*·[E_{0}/ħ]·Δt

But what about U_{12 }and U_{21}? Is there a similar interpretation? Let’s write those equations down and think about them:

U_{12}(t + Δt, t) = 0 − *i*·[H_{12}(t)/ħ]·Δt = 0 + *i*·[A/ħ]·Δt and

U_{21}(t + Δt, t) = 0 − *i*·[H_{21}(t)/ħ]·Δt = 0 + *i*·[A/ħ]·Δt

We can visualize this as follows:

Let’s remind ourselves of the definition of the derivative of a function by looking at the illustration below:The *f*(x_{0}) *value* in this illustration corresponds to the U_{ij}(t, t), obviously. So *now *things make somewhat more sense: U_{11}(t, t) = U_{11}(t, t) = 1, obviously, and U_{12}(t, t) = U_{21}(t, t) = 0. We then *add* the ΔU_{ij}(t + Δt, t) to U_{ij}(t, t). Hence, we can, and probably *should*, think of those K_{ij}(t) coefficients as the *derivative *of the U_{ij}(t, t) functions with respect to time. So we can write something like this:

These derivatives are *pure *imaginary numbers. That does *not *mean that the U_{ij}(t + Δt, t) functions are purely imaginary: U_{11}(t + Δt, t) and U_{22}(t + Δt, t) can be approximated by 1 − *i*·[E_{0}/ħ]·Δt for small Δt, so they do have a *real *part. In contrast, U_{12}(t + Δt, t) and U_{21}(t + Δt, t) are, effectively, purely imaginary (for small Δt, that is).

I can’t help thinking these formulas reflect a deep and beautiful geometry, but its meaning escapes me so far. 😦 When everything is said and done, none of the reflections above makes things somewhat more intuitive: these wavefunctions remain as mysterious as ever.

I keep staring at those P_{1}(t) and P_{2}(t) functions, and the C_{1}(t) and C_{2}(t) functions that ‘generate’ them, so to speak. They’re not independent, obviously. In fact, they’re exactly the same, except for a phase *difference*, which corresponds to the phase difference between the sine and cosine. So it’s all *one *reality, really: all can be described in one single functional form, so to speak. I hope things become more obvious as I move forward.

**Post scriptum: **I promised I’d show you how to get those differential equations but… Well… I’ve done that in other posts, so I’ll refer you to one of those. Sorry for not repeating myself. 🙂