The quantum-mechanical view of chemical binding

In my post on the hydrogen atom, I explained its stability using the following graph out of Feynman’s Lectures. It shows an equilibrium state for the Hmolecule with an energy level that’s about 5 eV (ΔE/E≈ −0.375 ⇔ ΔE ≈ −0.375×13.6 eV = 5.1 eV) lower than the energy of two separate hydrogen atoms (2H).

raph3The lower energy level is denoted by EII and refers to a state, which we also denoted as state II, that’s associated with some kind of molecular orbital for both electrons, resulting in more (shared) space where the two electrons can have a low potential energy, as Feynman puts it, so “the electron can spread out—lowering its kinetic energy—without increasing its potential energy.” The electrons have opposite spin. The have to have opposite spin because our formula for state II would violate the Fermi exclusion principle if they would not have opposite spin. Indeed, if the two electrons would not have opposite spin, the formula for our CII amplitude, would be violating the rule that, when identical fermions are involved, and we’re adding amplitudes, then we should do so with a negative sign for the exchanged case. So our CII = 〈II|ψ〉 = (1/√2)[〈1|ψ〉 + 〈2|ψ〉] = (1/√2)[〈2|ψ〉 + 〈1|ψ〉] would be problematic: when we switch the electrons, we should get a minus sign.

We do get that minus sign for state I:

〈I|ψ〉 = (1/√2)[〈1|ψ〉 − 〈2|ψ〉] = −(1/√2)[〈2|ψ〉 − 〈1|ψ〉]

To make a long story short, state II is the equilibrium state, and so that’s an Hmolecule with two electrons with opposite spins that share a molecular orbital, rather than moving around in some atomic orbital.

The question is: can we generalize this analysis? I mean… We’ve spent a lot of time so as to make sure we understand this one particular case. What’s the use of such analysis if we can’t generalize? We shouldn’t be doing nitty-gritty all of the time, isn’t it?

You’re right. The thing is: we can easily generalize. We’ve learned to play with those Hamiltonian matrices now, and so let’s do the ‘same-same but different’ with other systems. Let’s replace one of the two protons in the two-protons-one-electron model by a much heavier ion—say, lithium. [The example is not random, of course: lithium is very easily ionized, which is why it’s used in batteries.]

We need to think of the Hamiltonian again, right? We’re now in a situation in which the Hamiltonian coefficients H11 and H22 are likely to be different. We’ve lost the symmetry: if the electron is near the lithium ion, then we can’t assume the system has the same energy as when it’s near the hydrogen nucleus (in case you forgot: that’s what the proton is, really). Because we’ve lost the symmetry, we no longer have these ‘easy’ Hamiltonians:

equi

We need to look at the original formulas for Eand E2 once again. Let me write them down:

energies

Of course, H12 and H21 will still be equal to A, and so… Well… Let me simplify my life and copy Feynman:

Feynman

There’s several things here. First, note that approximation to the square root:

square root sum of squares

We’re only allowed to do that if y is much smaller than x, with = 1 and = 2A/(H11 − H22). In fact, the condition is usually written as 0 ≤ y/x ≤ 1/2, so we take the A/(H11 − H22) ratio as (much) less than one, indeed. So the second term in the energy difference E− EII = (H11 − H22) + 2A·A/(H11 − H22) is surely smaller than 2A. But there’s the first term, of course: H11 − H22. However, that’s there anyway, and so we should actually be looking at the additional separation, so that’s where the A comes in, and so that’s the second term: 2A·A/(H11 − H22) which, as mentioned, is smaller by the factor A/(H11 − H22), which is less than one. So Feynman’s conclusion is correct: “The binding of unsymmetric diatomic molecules is generally very weak.

However, that’s not the case when binding two ions by two electrons, which is referred to as a two-electron binding, which is the most common valence bond. Let me simplify my life once more and quote once again:

Feynman 2

What he’s saying is that H11 and H22 are one and the same once again, and equal to E0, because both ions can take one electron, so there’s no difference between state 1 and state 2 in that regard. So the energy difference is 2A once more and we’ve got good covalent binding. [Note that the term ‘covalent’ just refers to sharing electrons, so their value is shared, so to speak.]

Now, this result is, of course, subject to the hypothesis that the electron is more or less equally attracted to both ions, which may or may not be the case. If it’s not the case, we’ll have what’s referred to as ‘ionic’ binding. Again, I’ll let Feynman explain it, as it’s pretty straightforward and so it’s no use to try to write another summary of this:

Feynman3

So… That’s it, really. As Feynman puts it, by way of conclusion: “You can now begin to see how it is that many of the facts of chemistry can be most clearly understood in terms of a quantum mechanical description.”

Most clearly? Well… I guess that, at the very least, we’re “beginning to see” something here, aren’t we? 🙂

An introduction to virtual particles

In one of my posts on the rules of quantum math, I introduced the propagator function, which gives us the amplitude for a particle to go from one place to another. It looks like this:

propagator

The rand r2 vectors are, obviously, position vectors describing (1) where the particle is right now, so the initial state is written as |r1〉, and (2) where it might go, so the final state is |r2〉. Now we can combine this with the analysis in my previous post to think about what might happen when an electron sort of ‘jumps’ from one state to another. It’s a rather funny analysis, but it will give you some feel of what these so-called ‘virtual’ particles might represent.

Let’s first look at the shape of that function. The e(i/ħ)·(pr12function in the numerator is now familiar to you. Note the r12 in the argument, i.e. the vector pointing from r1 to r2. The pr12 dot product equals |p|∙|r12|·cosθ = p∙r12·cosθ, with θ the angle between p and r12. If the angle is the same, then cosθ is equal to 1. If the angle is π/2, then it’s 0, and the function reduces to 1/r12. So the angle θ, through the cosθ factor, sort of scales the spatial frequency. Let me try to give you some idea of how this looks like by assuming the angle between p and r12 is the same, so we’re looking at the space in the direction of the momentum only and |p|∙|r12|·cosθ = p∙r12. Now, we can look at the p/ħ factor as a scaling factor, and measure the distance x in units defined by that scale, so we write: x = p∙r12/ħ. The whole function, including the denominator, then reduces to (ħ/p)·eix/x = (ħ/p)·cos(x)/x + i·(ħ/p)·sin(x)/x, and we just need to square this to get the probability. All of the graphs are drawn hereunder: I’ll let you analyze them. [Note that the graphs do not include the ħ/p factor, which you may look at as yet another scaling factor.] You’ll see – I hope! – that it all makes perfect sense: the probability quickly drops off with distance, both in the positive as well as in the negative x-direction, while going to infinity when very near, i.e. for very small x. [Note that the absolute square, using cos(x)/x and sin(x)/x yields the same graph as squaring 1/x—obviously!]

graph

Now, this propagator function is not dependent on time: it’s only the momentum that enters the argument. Of course, we assume p to be some positive real number. Of course?

This is where Feynman starts an interesting conversation. In the previous post, we studied a model in which we had two protons, and one electron jumping from one to another, as shown below.

hydrogen

This model told us the equilibrium state is a stable ionized hydrogen molecule (so that’s an H2+ molecule), with an interproton distance that’s equal to 1 Ångstrom – so that’s like twice the size of a hydrogen atom (which we simply write as H) – and an energy that’s 2.72 eV less than the energy of a hydrogen atom and a proton (so that’s not an H2+ molecule but a system consisting of a separate hydrogen atom and a proton). The why and how of that equilibrium state is illustrated below. [For more details, see my previous post.]

raph2

Now, the model implies there is a sort of attractive force pulling the two protons together even when the protons are at larger distances than 1 Å. One can see that from the graph indeed. Now, we would not associate any molecular orbital with those distances, as the system is, quite simply, not a molecule but a separate hydrogen atom and a proton. Nevertheless, the amplitude A is non-zero, and so we have an electron jumping back and forth.

We know how that works from our post on tunneling: particles can cross an energy barrier and tunnel through. One of the weird things we had to consider when a particle crosses such potential barrier, is that the momentum factor p in its wavefunction was some pure imaginary number, which we wrote as p = i·p’. We then re-wrote that wavefunction as a·e−iθ = a·e−i[(E/ħ)∙t − (i·p’/ħ)x] = a·e−i(E/ħ)∙t·ei2·p’·x/ħ = a·e−i(E/ħ)∙t·e−p’·x/ħ. The e−p’·x/ħ factor in this formula is a real-valued exponential function, that sort of ‘kills’ our wavefunction as we move across the potential barrier, which is what is illustrated below: if the distance is too large, then the amplitude for tunneling goes to zero.

potential barrier

From a mathematical point of view, the analysis of our electron jumping back and forth is very similar. However, there are differences too. We can’t really analyze this in terms of a potential barrier in space. The barrier is the potential energy of the electron itself: it’s happy when it’s bound, because its energy then contributes to a reduction of the total energy of the hydrogen atomic system that is equal to the ionization energy, or the Rydberg energy as it’s called, which is equal to not less than 13.6 eV (which, as mentioned, is pretty big at the atomic level). Well… We can take that propagator function (1/re(i/ħ)·p∙r (note the argument has no minus sign: it can be quite tricky!), and just fill in the value for the momentum of the electron.

Huh? What momentum? It’s got no momentum to spare. On the contrary, it wants to stay with the proton, so it has no energy whatsoever to escape. Well… Not in quantum mechanics. In quantum mechanics it can use all its potential energy and convert it into kinetic energy, so it can get away from its proton and convert the energy that’s being released into kinetic energy.

But there is no release of energy! The energy is negative!

Exactly! You’re right. So we boldly write: K.E. = m·v2/2 = p2/(2m) = −13.6 eV, and, because we’re working with complex numbers, we can take a square root of negative number, using the definition of the imaginary unit: i = √(−1), so we get a purely imaginary value for the momentum p, which we write as:

p = ±i·√(2m·EH)

The sign of p is chosen so it makes sense: our electron should go in one direction only. It’s going to be the plus sign. [If you’d take the negative root, you’d get a nonsensical propagator function.] To make a long story short, our propagator function becomes:

(1/re(i/ħ)·i·√(2m·EH)∙r = (1/re(i/ħ)·i·√(2m·EH)∙r = (1/rei2/ħ·√(2m·EH)∙r = (1/r)·e−√(2m·EH)/ħ∙r

Of course, from a mathematical point of view, that’s the same function as e−p’·x/ħ: it’s a real-valued exponential function that quickly dies. But it’s an amplitude alright, and it’s just like an amplitude for tunneling indeed: if the distance is too large, then the amplitude goes to zero. The final cherry on the cake, of course, is to write:

A ∼ (1/r)·e−√(2m·EH)/ħ∙r

Well… No. It gets better. This amplitude is an amplitude for an electron bond between the two protons which, as we know, lowers the energy of the system. By how much? Well… By A itself. Now we know that work or energy is an integral or antiderivative of force over distance, so force is the derivative of energy with respect to the distance. So we can just take the derivative of the expression above to get the force. I’ll leave that you as an exercise: don’t forget to use the product rule! 🙂

So are we done? No. First, we didn’t talk about virtual particles yet! Let me do that now. However, first note that we should add one more effect in our two-proton-one-electron system: the coulomb field (ε) caused by the bare proton will cause the hydrogen molecule to take on an induced electric dipole moment (μ), so we should integrate that in our energy equation. Feynman shows how, but I won’t bother you with that here. Let’s talk about those virtual particles. What are they?

Well… There’s various definitions, but Feynman’s definition is this one:

“There is an exchange of a virtual electron when–as here–the electron has to jump across a space where it would have a negative energy. More specifically, a ‘virtual exchange’ means that the phenomenon involves a quantum-mechanical interference between an exchanged state and a non-exchanged state.”

You’ll say: what’s virtual about it? The electron does go from one place to another, doesn’t it? Well… Yes and no. We can’t observe it while it’s supposed to be doing that. Our analysis just tells us it seems to be useful to distinguish two different states and analyze all in terms of those differential equations. Who knows what’s really going on? What’s actual and what’s virtual? We just have some ‘model’ here: a model for the interaction between a hydrogen atom and a proton. It explains the attraction between them in terms of a sort of continuous exchange of an electron, but is it real?

The point is: in physics, it’s assumed that the coulomb interaction, i.e. all of electrostatics really, comes from the exchange of virtual photons: one electron, or proton, emits a photon, and then another absorbs it in the reverse of the same reaction. Furthermore, it is assumed that the amplitude for doing so is like that formula we found for the amplitude to exchange a virtual electron, except that the rest mass of a photon is zero, and so the formula reduces to 1/r. Such simple relationship makes sense, of course, because that’s how the electrostatic potential varies in space!

That, in essence, is all what there is to the quantum-mechanical theory of electromagnetism, which Feynman refers to as the ‘particle point of view’.

So… Yes. It’s that simple. Yes! For a change! 🙂

Post scriptum: Feynman’s Lecture on virtual particles is actually focused on a model for the nuclear forces. Most of it is devoted to a discussion of the virtual ‘pion’, or π-meson, which was then, when Feynman wrote his Lectures, supposed to mediate the force between two nucleons. However, this theory is clearly outdated: nuclear forces are described by quantum chromodynamics. So I’ll just skip the Yukawa theory here. It’s actually kinda strange his theory, which he proposed in 1935, was the theory for nuclear forces for such a long time. Hence, it’s surely all very interesting from a historical point of view.

The hydrogen molecule as a two-state system

My posts on the state transitions of an ammonia molecule weren’t easy, were they? So let’s try another two-state system. The illustration below shows an ionized hydrogen molecule in two possible states which, as usual, we’ll denote as |1〉 and |2〉. An ionized hydrogen molecule is an H2 molecule which lost an electron, so it’s two protons with one electron only, so we denote it as H2+. The difference between the two states is obvious: the electron is either with the first proton or with the second.

hydrogen

It’s an example taken from Feynman’s Lecture on two-state systems. The illustration itself raises a lot of questions, of course. The most obvious question is: how do we know which proton is which? We’re talking identical particles, right? Right. We should think of the proton spins! However, protons are fermions and, hence, they can’t be in the same state, so they must have opposite spins. Of course, now you’ll say: they’re not in the same state because they’re at different locations. Well… Now you’ve answered your own question. 🙂 However you want to look at this, the point is: we can distinguish both protons. Having said that, the reflections above raise other questions: what reference frame are we using? The answer is: it’s the reference frame of the system. We can mirror or rotate this image however we want – as I am doing below – but state |1〉 is state |1〉, and state |2〉 is state |2〉.

flip

The other obvious question is more difficult. If you’ve read anything at all about quantum mechanics, you’ll ask: what about the in-between states? The electron is actually being shared by the two protons, isn’t it? That’s what chemical bonds are all about, no? Molecular orbitals rather than atomic orbitals, right? Right. That’s actually what this post is all about. We know that, in quantum mechanics, the actual state – or what we think is the actual state – is always expressed as some linear combination of so-called base states. We wrote:

|ψ〉 = |1〉C|2〉C= |1〉〈1|ψ〉 + |2 〉〈2|ψ 〉

In terms of representing what’s actually going on, we only have these probability functions: they say that, if we would take a measurement, the probability of finding the electron near the first or the second proton varies as shown below:

graph

If the |1〉 and |2〉 states were actually representing two dual physical realities, the actual state of our H2molecule would be represented by some square or some pulse wave, as illustrated below. [We should be calling it a square function but that term has been reserved for a function like y = x2.]

Dutycycle

Of course, the symmetry of the situation implies that the average pulse duration τ would be one-half of the (average) period T, so we’d be talking a square wavefunction indeed. The two wavefunctions both qualify as probability density functions: the system is always in one state or the other, and the probabilities add up to one. But you’ll agree we prefer the smooth squared sine and cosine functions. To be precise, these smooth functions are:

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

So now we only need to explain A here (you know ħ already). But… Well… Why would we actually prefer those smooth functions? An irregular pulse function would seem to be doing a better job when it comes to modeling reality, doesn’t it? The electron should be either here, or there. Isn’t it?

Well… No. At least that’s why am slowly starting to understand. These pure base states |1〉 and |2〉 are real and not real at the same time. They’re real, because it’s what we’ll get when we verify, or measure, the state, so our measurement will tell us that it’s here or there. There’s no in-between. [I still need to study weak measurement theory.] But then they are not real, because our molecule will never ever be in those two states, except for those ephemeral moments when (A/ħ)·t = n·π (n = 0, 1, 2,…). So we’re really modeling uncertainty here and, while I am still exploring what that actually means, you should think of the electron as being everywhere really, but with an unequal density in space—sort of. 🙂

Now, we’ve learned we can describe the state of a system in terms of an alternative set of base states. We wrote: |ψ〉 = |I〉C|II〉CII = |I〉〈I|ψ〉 + |II〉〈II|ψ〉, with the CI, II and C1, 2 coefficients being related to each other in exactly the same way as the associated base states, i.e. through a transformation matrix, which we summarized as:

transformation

To be specific, the two sets of base states we’ve been working with so far were related as follows:

transformation

So we’d write: |ψ〉 = |I〉C|II〉CII = |I〉〈I|ψ〉 + |II〉〈II|ψ〉 = |1〉C|2〉C= |1〉〈1|ψ〉 + |2 〉〈2|ψ 〉, and the CI, II and C1, 2 coefficients would be related in exactly the same way as the base states:

Eq 4

[In case you’d want to review how that works, see my post on the Hamiltonian and base states.] Now, we cautioned that it’s difficult to try to interpret such base transformations – often referred to as a change in the representation or a different projection – geometrically. Indeed, we acknowledged that (base) states were very much like (base) vectors – from a mathematical point of view, that is – but, at the same time, we said that they were ‘objects’, really: elements in some Hilbert space, which means you can do the operations we’re doing here, i.e. adding and multiplying. Something like |I〉CI doesn’t mean all that much: Cis a complex number – and so we can work with numbers, of course, because we can visualize them – but |I〉 is a ‘base state’, and so what’s the meaning of that, and what’s the meaning of the |I〉CI or CI|I〉 product? I could babble about that, but it’s no use: a base state is a base state. It’s some state of the system that makes sense to us. In fact, it may be some state that does not make sense to us—in terms of the physics of the situation, that is – but then there will always be some mathematical sense to it because of that transformation matrix, which establishes a one-to-one relationship between all sets of base states.

You’ll say: why don’t you try to give it some kind of geometrical or whatever meaning? OK. Let’s try. State |1〉 is obviously like minus state |2〉 in space, so let’s see what happens when we equate |1〉 to 1 on the real axis, and |2〉 to −1. Geometrically, that corresponds to the (1, 0) and (−1, 0) points on the unit circle. So let’s multiply those points with (1/√2, −1/√2) and (1/√2, 1/√2) respectively. What do we get? Well… What product should we take? The dot product, the cross product, or the ordinary complex-number product? The dot product gives us a number, so we don’t want that. [If we’re going to represent base states by vectors, we want all states to be vectors.] A cross product will give us a vector that’s orthogonal to both vectors, so it’s a vector in ‘outer space’, so to say. We don’t want that, I must assume, and so we’re left with the complex-number product, which projects our  (1, 0) and (−1, 0) vectors into the (1/√2, −1/√2)·(1, 0) = (1/√2−i/√2)·(1+0·i) = √2−i/√2 = (1/√2, −i/√2) and (1/√2, 1/√2)·(−1, 0) = (1/√2+i/√2)·(−1+0·i) = −√2−i/√2 = (−1/√2, −i/√2) respectively.

transformation 2

What does this say? Nothing. Stuff like this only causes confusion. We had two base states that were ‘180 degrees’ apart, and now our new base states are only ’90 degrees’ apart. If we’d ‘transform’ the two new base states once more, they collapse into each other: (1/√2, −1/√2)·(1/√2, −1/√2) = (1/√2−i/√2)2 = −= (0, −1) = (1/√2, 1/√2)·(−1/√2, −1/√2) = −i. This is nonsense, of course. It’s got nothing to do with the angle we picked for our original set of base states: we could have separated our original set of base states by 90 degrees, or 45 degrees. It doesn’t matter. It’s the transformation itself: multiplying by (+1/√2, −1/√2) amounts to a clockwise rotation by 45 degrees, while multiplying by (+1/√2, +1/√2) amounts to the same, but counter-clockwise. So… Well… We should not try to think of our base vectors in any geometric way, because it just doesn’t make any sense. So Let’s not waste time on this: the ‘base states’ are a bit of a mystery, in the sense that they just are what they are: we can’t ‘reduce’ them any further, and trying to interpret them geometrically leads to contradictions, as evidenced by what I tried to do above. Base states are ‘vectors’ in a so-called Hilbert space, and… Well… That’s not your standard vector space. [If you think you can make more sense of it, please do let me know!]

Onwards!

Let’s take our transformation again:

  • |I〉 = (1/√2)|1〉 − (1/√2)|2〉 = (1/√2)[|1〉 − |2〉]
  • |II〉 = (1/√2)|1〉 + (1/√2)|2〉 = (1/√2)[|1〉 + |2〉]

Again, trying to geometrically interpret what it means to add or subtract two base states is not what you should be trying to do. In a way, the two expressions above only make sense when combining them with a final state, so when writing:

  • 〈ψ|I〉 = (1/√2)〈ψ|1〉 − (1/√2)〈ψ|2〉 = (1/√2)[〈ψ|1〉 − 〈ψ|2〉]
  • 〈ψ|II〉 = (1/√2)〈ψ|1〉 + (1/√2)〈ψ|2〉 = (1/√2)[〈ψ|1〉 + 〈ψ|2〉]

Taking the complex conjugate of this gives us the amplitudes of the system to be in state I or state II:

  • 〈I|ψ〉 = 〈ψ|I〉* = (1/√2)[〈ψ|1〉* − 〈ψ|2〉*] = (1/√2)[〈1|ψ〉 − 〈2|ψ〉]
  • 〈II|ψ〉 = 〈ψ|II〉* = (1/√2)[〈ψ|1〉* + 〈ψ|2〉*] = (1/√2)[〈1|ψ〉 + 〈2|ψ〉]

That still doesn’t tell us much, because we’d need to know the 〈1|ψ〉 and 〈2|ψ〉 functions, i.e. the amplitudes of the system to be in state 1 and state 2 respectively. What we do know, however, is that the 〈1|ψ〉 and 〈2|ψ〉 functions will have some rather special amplitudes. We wrote:

  • C= 〈 I | ψ 〉 =  e−(i/ħ)·EI·t
  • CII = 〈 II | ψ 〉 = e−(i/ħ)·EII·t

These are amplitudes of so-called stationary states: the associated probabilities – i.e. the absolute square of these functions – do not vary in time: |e−(i/ħ)·EI·t|2 = |e−(i/ħ)·EII·t|2 = 1. For our ionized hydrogen molecule, it means that, if it would happen to be in state I, it will stay in state I, and the same goes for state II. We write:

〈 I | I 〉 = 〈 II | II 〉 = 1 and 〈 I | II 〉 = 〈 II | I 〉 = 0

That’s actually just the so-called ‘orthogonality’ condition for base states, which we wrote as 〈i|j〉 = 〈j|i〉 = δij, but, in light of the fact that we can’t interpret them geometrically, we shouldn’t be calling it like that. The point is: we had those differential equations describing a system like this. If the amplitude to go from state 1 to state 2 was equal to some real- or complex-valued constant A, then we could write those equations either in terms of Cand C2, or in terms of Cand CII:

set of equations

So the two sets of equations are equivalent. However, what we want to do here is look at it in terms of Cand CII. Let’s first analyze those two energy levels E= E+ A and EII = E− A. Feynman graphs them as follows:

raph1raph2

Let me explain. In the first graph, we have E= E+ A and EII = E− A, and they are depicted as being symmetric, with A depending on the distance between the two protons. As for E0, that’s the energy of a hydrogen atom, i.e. a proton with a bound electron, and a separate proton. So it’s the energy of a system consisting of a hydrogen atom and a proton, which is obviously not the same as that of an ionized hydrogen molecule. The concept of a molecule assumes the protons are closely together. We assume E= 0 if the interproton distance is relatively large but, of course, as the protons come closer, we shouldn’t forget the repulsive electrostatic force between the two protons, which is represented by the dashed line in the first graph. Indeed, unlike the electron and the proton, the two protons will want to push apart, rather than pull together, so the potential energy of the system increases as the interproton distance decreases. So Eis not constant either: it also depends on the interproton distance. But let’s forget about Efor a while. Let’s look at the two curves for A now.

A is not varying in time, but its value does depend on the distance between the two protons. We’ll use this in a moment to calculate the approximate size of the hydrogen nucleus in a calculation that closely resembles Feynman’s calculation of the size of a hydrogen atom. That A should be some function of the interproton distance makes sense: the transition probability, and therefore A, will exponentially decrease with distance. There are a few things to reflect on here:

1. In the mentioned calculation of the size of a hydrogen atom, which is based on the Uncertainty Principle, Feynman shows that the energy of the system decreases when an electron is bound to the proton. The reasoning is that, if the potential energy of the electron is zero when it is not bound, then its potential energy will be negative when bound. Think of it: the electron and the proton attract each other, so it requires force to separate them, and force over a distance is energy. From our course in electromagnetics, we know that the potential energy, when bound, should be equal to −e2/a0, with ethe squared charge of the electron divided by 4πε0, and a0 the so-called Bohr radius of the atom. Of course, the electron also has kinetic energy. It can’t just sit on top of the proton because that would violate the Uncertainty Principle: we’d know where it was. Combining the two, Feynman calculates both a0 as well as the so-called Rydberg energy, i.e. the total energy of the bound electron, which is equal to −13.6 eV. So, yes, the bound state has less energy, so the electron will want to be bound, i.e. it will want to be close to one of the two protons.

2. Now, while that’s not what’s depicted above, it’s clear the magnitude of A will be related to that Rydberg energy which − please note − is quite high. Just compare it with the A for the ammonia molecule, which we calculated in our post on the maser: we found an A of about 0.5×10−4 eV there, so that’s like 270,000 times less! Nevertheless, the possibility is there, and what happens when the electron flips over amounts to tunneling: it penetrates and crosses a potential barrier. We did a post on that, and so you may want to look at how that works. One of the weird things we had to consider when a particle crosses such potential barrier, is that the momentum factor p in its wavefunction was some pure imaginary number, which we wrote as p = i·p’. We then re-wrote that wavefunction as a·e−iθ = a·e−i[(E/ħ)∙t − (i·p’/ħ)x] = a·e−i(E/ħ)∙t·ei2·p’·x/ħ = a·e−i(E/ħ)∙t·e−p’·x/ħ. Now, it’s easy to see that the e−p’·x/ħ factor in this formula is a real-valued exponential function, with the same shape as the general e−x function, which I depict below.

graph

This e−p’·x/ħ basically ‘kills’ our wavefunction as we move in the positive x-direction, across the potential barrier, which is what is illustrated below: if the distance is too large, then the amplitude for tunneling goes to zero.

potential barrier

So that’s what depicted in those graphs of E= E+ A and EII = E− A: A goes to zero when the interproton distance becomes too large. We also recognize the exponential shape for A in those graphs, which can also be derived from the same tunneling story.

Now we can calculate EA and E− A taking into account that both terms vary with the interproton distance as explained, and so that gives us the final curves on the right-hand side, which tell us that the equilibrium configuration of the ionized hydrogen molecule is state II, i.e. the lowest energy state, and the interproton distance there is approximately one Ångstrom, i.e. 1×10−10 m. [You can compare this with the Bohr radius, which we calculated as a0 = 0.528×10−10 m, so that all makes sense.] Also note the energy scale: ΔE is the excess energy over a proton plus a hydrogen atom, so that’s the energy when the two protons are far apart. Because it’s the excess energy, we have a zero point. That zero point is, obviously, the energy of a hydrogen atom and a proton. [Read this carefully, and please refer back to what I wrote above. The energy of a system consisting of a hydrogen atom and a proton is not the same as that of an ionized hydrogen molecule: the concept of a molecule assumes the protons are closely together.] We then re-scale by dividing by the Rydberg energy E= 13.6 eV. So ΔE/E≈ −0.2 ⇔ ΔE ≈ −0.2×13.6 = –2.72 eV. That basically says that the energy of our ionized hydrogen molecule is 2.72 eV lower than the energy of a hydrogen atom and a proton.

Why is it lower? We need to think about our model of the hydrogen atom once more: the energy of the electron was minimized by striking a balance between (1) being close to the proton and, therefore, having a low potential energy (or a low coulomb energy, as Feynman calls it) and (2) being further away from the proton and, therefore, lowering its kinetic energy according to the Uncertainty Principle ΔxΔp ≥ ħ/2, which Feynman boldly re-wrote as p = ħ/a0. Now, a molecular orbital, i.e. the electron being around two protons, results in “more space where the electron can have a low potential energy”, as Feynman puts it, so “the electron can spread out—lowering its kinetic energy—without increasing its potential energy.”

The whole discussion here actually amounts to an explanation for the mechanism by which an electron shared by two protons provides, in effect, an attractive force between the two protons. So we’ve got a single electron actually holding two protons together, which chemists refer to as a “one-electron bond.”

So… Well… That explains why the energy EII = E− A is what it is, so that’s smaller than Eindeed, with the difference equal to the value A for an interproton distance of 1 Å. But how should we interpret E= E+ A? What is that higher energy level? What does it mean?

That’s a rather tricky question. There’s no easy interpretation here, like we had for our ammonia molecule: the higher energy level had an obvious physical meaning in an electromagnetic field, as it was related to the electric dipole moment of the molecule. That’s not the case here: we have no magnetic or electric dipole moment here. So, once again, what’s the physical meaning of E= E+ A? Let me quote Feynman’s enigmatic answer here:

“Notice that this state is the difference of the states |1⟩ and |2⟩. Because of the symmetry of |1⟩ and |2⟩, the difference must have zero amplitude to find the electron half-way between the two protons. This means that the electron is somewhat more confined, which leads to a larger energy.”

What does he mean with that? It seems he’s actually trying to do what I said we shouldn’t try to do, and that is to interpret what adding versus subtracting states actually means. But let’s give it a fair look. We said that the |I〉 = (1/√2)[|1〉 − |2〉] expression didn’t mean much: we should add a final state and write: 〈ψ|I〉 = (1/√2)[〈ψ|1〉 − 〈ψ|2〉], which is equivalent to 〈I|ψ〉 = (1/√2)[〈1|ψ〉 − 〈2|ψ〉]. That still doesn’t tell us anything: we’re still adding amplitudes, and so we should allow for interference, and saying that |1⟩ and |2⟩ are symmetric simply means that 〈1|ψ〉 − 〈2|ψ〉 = 〈2|ψ〉 − 〈1|ψ〉 ⇔ 2·〈1|ψ〉 = 2·〈2|ψ〉 ⇔ 〈1|ψ〉 = 〈2|ψ〉. Wait a moment! That’s an interesting reflection. Following the same reasoning for |II〉 = (1/√2)[|1〉 + |2〉], we get 〈1|ψ〉 + 〈2|ψ〉 = 〈2|ψ〉 + 〈1|ψ〉 ⇔ … Huh? No, that’s trivial: 0 = 0.

Hmm… What to say? I must admit I don’t quite ‘get’ Feynman here: state I, with energy E= E+ A, seems to be both meaningless as well as impossible. The only energy levels that would seem to make sense here are the energy of a hydrogen atom and a proton and the (lower) energy of an ionized hydrogen molecule, which you get when you bring a hydrogen atom and a proton together. 🙂

But let’s move to the next thing: we’ve added only one electron to the two protons, and that was it, and so we had an ionized hydrogen molecule, i.e. an H2+ molecule. Why don’t we do a full-blown H2 molecule now? Two protons. Two electrons. It’s easy to do. The set of base states is quite predictable, and illustrated below: electron a can be either one of the two protons, and the same goes for electron b.

base

We can then go through the same as for the ion: the molecule’s stability is shown in the graph below, which is very similar to the graph of the energy levels of the ionized hydrogen molecule, i.e. the H2+  molecule. The shape is the same, but the values are different: the equilibrium state is at an interproton distance of 0.74 Å, and the energy of the equilibrium state is like 5 eV (ΔE/E≈ −0.375) lower than the energy of two separate hydrogen atoms.

raph3The explanation for the lower energy is the same: state II is associated with some kind of molecular orbital for both electrons, resulting in “more space where the electron can have a low potential energy”, as Feynman puts it, so “the electron can spread out—lowering its kinetic energy—without increasing its potential energy.”

However, there’s one extra thing here: the two electrons must have opposite spins. That’s the only way to actually distinguish the two electrons. But there is more to it: if the two electrons would not have opposite spin, we’d violate Fermi’s rule: when identical fermions are involved, and we’re adding amplitudes, then we should do so with a negative sign for the exchanged case. So our transformation would be problematic:

〈II|ψ〉 = (1/√2)[〈1|ψ〉 + 〈2|ψ〉] = (1/√2)[〈2|ψ〉 + 〈1|ψ〉]

When we switch the electrons, we should get a minus sign. The weird thing is: we do get that minus sign for state I:

〈I|ψ〉 = (1/√2)[〈1|ψ〉 − 〈2|ψ〉] = −(1/√2)[〈2|ψ〉 − 〈1|ψ〉]

So… Well… We’ve got a bit of an answer there as to what that the ‘other’ (upper) energy level of E= E+ A actually means, in physical terms, that is. It models two hydrogens coming together with parallel electron spins. Applying Fermi’s rules  – i.e. the exclusion principle, basically – we find that state II is, quite simply, not allowed for parallel electron spins: state I is, and it’s the only one. There’s something deep here, so let me quote the Master himself on it:

“We find that the lowest energy state—the only bound state—of the H2 molecule has the two electrons with spins opposite. The total spin angular momentum of the electrons is zero. On the other hand, two nearby hydrogen atoms with spins parallel—and so with a total angular momentum —must be in a higher (unbound) energy state; the atoms repel each other. There is an interesting correlation between the spins and the energies. It gives another illustration of something we mentioned before, which is that there appears to be an “interaction” energy between two spins because the case of parallel spins has a higher energy than the opposite case. In a certain sense you could say that the spins try to reach an antiparallel condition and, in doing so, have the potential to liberate energy—not because there is a large magnetic force, but because of the exclusion principle.”

You should read this a couple of times. It’s an important principle. We’ll discuss it again in the next posts, when we’ll be talking spin in much more detail once again. 🙂 The bottom line is: if the electrons are parallel, then they won’t ‘share’ any space at all and, hence, they are really much more confined in space, and the associated energy level is, therefore, much higher.

Post scriptum: I said we’d ‘calculate’ the equilibrium interproton distance. We didn’t do that. We just gave them through the graphs, which are based on the results of a ‘detailed quantum-mechanical calculation’—or that’s what Feynman claims, at least. I am not sure if they correspond to experimentally determined values, or what calculations are behind, exactly. Feynman notes that “this approximate treatment of the H2molecule as a two-state system breaks down pretty badly once the protons get as close together as they are at the minimum in the curve and, therefore, it will not give a good value for the actual binding energy. For small separations, the energies of the two “states” we imagined are not really equal to E0, and a more refined quantum mechanical treatment is needed.”

So… Well… That says it all, I guess.