Superconductivity and flux quantization

This post continues my mini-series on Feynman’s Seminar on Superconductivity. Superconductivity is a state which produces many wondrous phenomena, but… Well… The flux quantization phenomenon may not be part of your regular YouTube feed but, as far as I am concerned, it may well be the most amazing manifestation of a quantum-mechanical phenomenon at a macroscopic scale. I mean… Super currents that keep going, with zero resistance, are weird—they explain how we can trap a magnetic flux in the first place—but the fact that such fluxes are quantized is even weirder.

The key idea is the following. When we cool a ring-shaped piece of superconducting material in a magnetic field, all the way down to the critical temperature that causes the electrons to condense into a superconducting fluid, then a super current will emerge—think of an eddy current, here, but with zero resistance—that will force the magnetic field out of the material, as shown below. This current will permanently trap some of the magnetic field, even when the external field is being removed. As said, that’s weird enough by itself but… Well… If we think of the super current as an eddy current encountering zero resistance, then the idea of a permanently trapped magnetic field makes sense, right? In case you’d doubt the effect… Well… Just watch one of the many videos on the effect on YouTube. 🙂 The amazing thing here is not the permanently trapped magnetic field, but the fact that it’s quantized.

trapped flux

To be precise, the trapped flux will always be an integer times 2πħ/q. In other words, the magnetic field which Feynman denotes by Φ (the capitalized Greek letter phi), will always be equal to:

Φ = 2πħ/q, with = 0, 1, 2, 3,…

Hence, the flux can be 0, 2πħ/q, 4πħ/q, 6πħ/q , and so on. The fact that it’s a multiple of 2π shows us it’s got to do with the fact that our piece of material is, effectively, a ring. The nice thing about this phenomenon is that the mathematical analysis is, in fact, fairly easy to follow—or… Well… Much easier than what we discussed before. 🙂 Let’s quickly go through it.

We have a formula for the magnetic flux. It must be equal to the line integral of the vector potential (A) around a closed loop Τ, so we write:

flux

Now, we can choose the loop Τ to be well inside the body of the ring, so that it never gets near the surface, as illustrated below. So we know that the current J is zero there. [In case you doubt this, see my previous post.]

curve

One of the equations we introduced in our previous post, ħθ = m·v + q·A, will then reduce to:

ħθ = q·A

Why? The v in the m·v term (the velocity of the superconducting fluid, really), is zero. Remember the analysis is for this particular loop (well inside the ring) only. So… Well… If we integrate the expression above, we get:

integral

Combining the two expressions with the integrals, we get:

integral 2Now, the line integral of a gradient from one point to another (say from point 1 to point 2) is the difference of the values of the function at the two points, so we can write:

integral 3

Now what constraints are there on the values of θ1 and θ2? Well… You might think that, if they’re associated with the same point (we’re talking a closed loop, right?), then the two values should be the same, but… Well… No. All we can say is that the wavefunction must have the same value. We wrote that wavefunction as:

ψ = ρ(r)1/2eθ(r)

The value of this function at some point r is the same if θ changes by 2π. Hence, when doing one complete turn around the ring, the ∫∇θ·ds integral in the integral formulas we wrote down must be equal to 2π. Therefore, the second integral expression above can be re-written as:

result

That’s the result we wanted to explain so… Well… We’re done. Let me wrap up by quoting Feynman’s account of the 1961 experiment which confirmed London’s prediction of the effect, which goes back to 1950! It’s interesting, because… Well… It shows how up to date Feynman’s Lectures really are—or were, back in 1963, at least!feynman overview of experiment

Feynman’s Seminar on Superconductivity (2)

We didn’t get very far in our first post on Feynman’s Seminar on Superconductivity, and then I shifted my attention to other subjects over the past few months. So… Well… Let me re-visit the topic here.

One of the difficulties one encounters when trying to read this so-called seminar—which, according to Feynman, is ‘for entertainment only’ and, therefore, not really part of the Lectures themselves—is that Feynman throws in a lot of stuff that is not all that relevant to the topic itself but… Well… He apparently didn’t manage to throw all that he wanted to throw into his (other) Lectures on Quantum Mechanics and so he inserted a lot of stuff which he could, perhaps, have discussed elsewhere. :-/ So let us try to re-construct the main lines of reasoning here.

The first equation is Schrödinger’s equation for some particle with charge q that is moving in an electromagnetic field that is characterized not only by the (scalar) potential Φ but also by a vector potential A:

schrodinger

This closely resembles Schrödinger’s equation for an electron that is moving in an electric field only, which we used to find the energy states of electrons in a hydrogen atom: i·ħ·∂ψ/∂t = −(1/2)·(ħ2/m)∇2ψ + V·ψ. We just need to note the following:

  1. On the left-hand side, we can, obviously, replace −1/i by i.
  2. On the right-hand side, we can replace V by q·Φ, because the potential of a charge in an electric field is the product of the charge (q) and the (electric) potential (Φ).
  3. As for the other term on the right-hand side—so that’s the −(1/2)·(ħ2/m)∇2ψ term—we can re-write −ħ2·∇2ψ as [(ħ/i)·∇]·[(ħ/i)·∇]ψ because (1/i)·(1/i) = 1/i2 = 1/(−1) = −1. 🙂
  4. So all that’s left now, is that additional −q·A term in the (ħ/i)∇ − q·A expression. In our post, we showed that’s easily explained because we’re talking magnetodynamics: we’ve got to allow for the possibility of changing magnetic fields, and so that’s what the −q·A term captures.

Now, the latter point is not so easy to grasp but… Well… I’ll refer you that first post of mine, in which I show that some charge in a changing magnetic field will effectively gather some extra momentum, whose magnitude will be equal to p = m·v = −q·A. So that’s why we need to introduce another momentum operator here, which we write as:

new-momentum-operator

OK. Next. But… Then… Well… All of what follows are either digressions—like the section on the local conservation of probabilities—or, else, quite intuitive arguments. Indeed, Feynman does not give us the nitty-gritty of the Bardeen-Cooper-Schrieffer theory, nor is the rest of the argument nearly as rigorous as the derivation of the electron orbitals from Schrödinger’s equation in an electrostatic field. So let us closely stick to what he does write, and try our best to follow the arguments.

Cooper pairs

The key assumption is that there is some attraction between electrons which, at low enough temperatures, can overcome the Coulomb repulsion. Where does this attraction come from? Feynman does not give us any clues here. He just makes a reference to the BCS theory but notes this theory is “not the subject of this seminar”, and that we should just “accept the idea that the electrons do, in some manner or other, work in pairs”, and that “we can think of thos−e pairs as behaving more or less like particles”, and that “we can, therefore, talk about the wavefunction for a pair.”

So we have a new particle, so to speak, which consists of two electrons who move through the conductor as one. To be precise, the electron pair behaves as a boson. Now, bosons have integer spin. According to the spin addition rule, we have four possibilities here but only three possible values:− 1/2 + 1/2 = 1; −1/2 + 1/2 = 0; +1/2 − 1/2 = 0; −1/2 − 1/2 = − 1. Of course, it is tempting to think these Cooper pairs are just like the electron pairs in the atomic orbitals, whose spin is always opposite because of the Fermi exclusion principle. Feynman doesn’t say anything about this, but the Wikipedia article on the BCS theory notes that the two electrons in a Cooper pair are, effectively, correlated because of their opposite spin. Hence, we must assume the Cooper pairs effectively behave like spin-zero particles.

Now, unlike fermions, bosons can collectively share the same energy state. In fact, they are likely to share the same state into what is referred to as a Bose-Einstein condensate. As Feynman puts it: “Since electron pairs are bosons, when there are a lot of them in a given state there is an especially large amplitude for other pairs to go to the same state. So nearly all of the pairs will be locked down at the lowest energy in exactly the same state—it won’t be easy to get one of them into another state. There’s more amplitude to go into the same state than into an unoccupied state by the famous factor √n, where n−1 is the occupancy of the lowest state. So we would expect all the pairs to be moving in the same state.”

Of course, this only happens at very low temperatures, because even if the thermal energy is very low, it will give the electrons sufficient energy to ensure the attractive force is overcome and all pairs are broken up. It is only at very low temperature that they will pair up and go into a Bose-Einstein condensate. Now, Feynman derives this √n factor in a rather abstruse introductory Lecture in the third volume, and I’d advise you to google other material on Bose-Einstein statistics because… Well… The mentioned Lecture is not among Feynman’s finest. OK. Next step.

Cooper pairs and wavefunctions

We know the probability of finding a Cooper pair is equal to the absolute square of its wavefunction. Now, it is very reasonable to assume that this probability will be proportional to the charge density (ρ), so we can write:

|ψ|= ψψ* ∼ ρ(r)

The argument here (r) is just the position vector. The next step, then, is to write ψ as the square root of ρ(r) times some phase factor θ. Abstracting away from time, this phase factor will also depend on r, of course. So this is what Feynman writes:

ψ = ρ(r)1/2eθ(r)

As Feynman notes, we can write any complex function of r like this but… Well… The charge density is, obviously, something real. Something we can measure, so we’re not writing the obvious here. The next step is even less obvious.

In our first post, we spent quite some time on Feynman’s digression on the local conservation of probability and… Well… I wrote above I didn’t think this digression was very useful. It now turns out it’s a central piece in the puzzle that Feynman is trying to solve for us here. The key formula here is the one for the so-called probability current, which—as Feynman shows—we write as:

probability-current-2

This current J can also be written as:

probability-current-1

Now, Feynman skips all of the math here (he notes “it’s just a change of variables” but so he doesn’t want to go through all of the algebra), and so I’ll just believe him when he says that, when substituting ψ for our wavefunction ψ = ρ(r)1/2eθ(r), then we can express this ‘current’ (J) in terms of ρ and θ. To be precise, he writes J as: current formulaSo what? Well… It’s really fascinating to see what happens next. While J was some rather abstract concept so far—what’s a probability current, really?—Feynman now suggests we may want to think of it as a very classical electric current—the charge density times the velocity of the fluid of electrons. Hence, we equate J to J =  ρ·v. Now, if the equation above holds true, but J is also equal to J = ρ·v, then the equation above is equivalent to:

moment

Now, that gives us a formula for ħθ. We write:

ħθ = m·v + q·A

Now, in my previous post on this Seminar, I noted that Feynman attaches a lot of importance to this m·v + q·A quantity because… Well… It’s actually an invariant quantity. The argument can be, very briefly, summarized as follows. During the build-up of (or a change in) a magnetic flux, a charge will pick up some (classical) momentum that is equal to p = m·v = −q·A. Hence, the m·v + q·A sum is zero, and so… Well… That’s it, really: it’s some quantity that… Well… It has a significance in quantum mechanics. What significance? Well… Think of what we’ve been writing here. The v and the A have a physical significance, obviously. Therefore, that phase factor θ(r) must also have a physical significance.

But the question remains: what physical significance, exactly? Well… Let me quote Feynman here:

“The phase is just as observable as the charge density ρ. It is a piece of the current density J. The absolute phase (θ) is not observable, but if the gradient of the phase (θ) is known everywhere, then the phase is known except for a constant. You can define the phase at one point, and then the phase everywhere is determined.”

That makes sense, doesn’t it? But it still doesn’t quite answer the question: what is the physical significance of θ(r). What is it, really? We may be able to answer that question after exploring the equations above a bit more, so let’s do that now.

Superconductivity

The phenomenon of superconductivity itself is easily explained by the mentioned condensation of the Cooper pairs: they all go into the same energy state. They form, effectively, a superconducting fluid. Feynman’s description of this is as follows:

“There is no electrical resistance. There’s no resistance because all the electrons are collectively in the same state. In the ordinary flow of current you knock one electron or the other out of the regular flow, gradually deteriorating the general momentum. But here to get one electron away from what all the others are doing is very hard because of the tendency of all Bose particles to go in the same state. A current once started, just keeps on going forever.”

Frankly, I’ve re-read this a couple of times, but I don’t think it’s the best description of what we think is going on here. I’d rather compare the situation to… Well… Electrons moving around in an electron orbital. That’s doesn’t involve any radiation or energy transfer either. There’s just movement. Flow. The kind of flow we have in the wavefunction itself. Here I think the video on Bose-Einstein condensates on the French Tout est quantique site is quite instructive: all of the Cooper pairs join to become one giant wavefunction—one superconducting fluid, really. 🙂

OK… Next.

The Meissner effect

Feynman describes the Meissner effect as follows:

“If you have a piece of metal in the superconducting state and turn on a magnetic field which isn’t too strong (we won’t go into the details of how strong), the magnetic field can’t penetrate the metal. If, as you build up the magnetic field, any of it were to build up inside the metal, there would be a rate of change of flux which would produce an electric field, and an electric field would immediately generate a current which, by Lenz’s law, would oppose the flux. Since all the electrons will move together, an infinitesimal electric field will generate enough current to oppose completely any applied magnetic field. So if you turn the field on after you’ve cooled a metal to the superconducting state, it will be excluded.

Even more interesting is a related phenomenon discovered experimentally by Meissner. If you have a piece of the metal at a high temperature (so that it is a normal conductor) and establish a magnetic field through it, and then you lower the temperature below the critical temperature (where the metal becomes a superconductor), the field is expelled. In other words, it starts up its own current—and in just the right amount to push the field out.”

The math here is interesting. Feynman first notes that, in any lump of superconducting metal, the divergence of the current must be zero, so we write:  ∇·J = 0. At any point? Yes. The current that goes in must go out. No point is a sink or a source. Now the divergence operator (∇·J) is a linear operator. Hence, that means that, when applying the divergence operator to the J = (ħ/m)·[θ − (q/ħ)·A]·ρ equation, we’ll need to figure out what ∇·θ =   = ∇2θ and ∇·A are. Now, as explained in my post on gauges, we can choose to make ∇·A equal to zero so… Well… We’ll make that choice and, hence, the term with ∇·A in it vanishes. So… Well… If ∇·J equals zero, then the term with ∇2θ has to be zero as well, so ∇2θ has to be zero. That, in turn, implies θ has to be some constant (vector).

Now, there is a pretty big error in Feynman’s Lecture here, as it notes: “Now the only way that ∇2θ can be zero everywhere inside the lump of metal is for θ to be a constant.” It should read: ∇2θ can only be zero everywhere if θ is a constant (vector). So now we need to remind ourselves of the reality of θ, as described by Feynman (quoted above): “The absolute phase (θ) is not observable, but if the gradient of the phase (θ) is known everywhere, then the phase is known except for a constant. You can define the phase at one point, and then the phase everywhere is determined.” So we can define, or choose, our constant (vector) θ to be 0.

Hmm… We re-set not one but two gauges here: A and θ. Tricky business, but let’s go along with it. [If we want to understand Feynman’s argument, then we actually have no choice than to go long with his argument, right?] The point is: the (ħ/m)·θ term in the J = (ħ/m)·[θ − (q/ħ)·A]·ρ vanishes, so the equation we’re left with tells us the current—so that’s an actual as well as a probability current!—is proportional to the vector potential:

currentNow, we’ve neglected any possible variation in the charge density ρ so far because… Well… The charge density in a superconducting fluid must be uniform, right? Why? When the metal is superconducting, an accumulation of electrons in one region would be immediately neutralized by a current, right? [Note that Feynman’s language is more careful here. He writes: the charge density is almost perfectly uniform.]

So what’s next? Well… We have a more general equation from the equations of electromagnetism:

A and J

[In case you’d want to know how we get this equation out of Maxwell’s equations, you can look it up online in one of the many standard textbooks on electromagnetism.] You recognize this as a Poisson equation… Well… Three Poisson equations: one for each component of A and J. We can now combine the two equations above by substituting in that Poisson equation, so we get the following differential equation, which we need to solve for A:

A

The λ2 in this equation is, of course, a shorthand for the following constant:

lambda

Now, it’s very easy to see that both e−λr as well as e−λr are solutions for that Poisson equation. But what do they mean? In one dimension, r becomes the one-dimensional position variable x. You can check the shapes of these solutions with a graphing tool.

graph

Note that only one half of each graph counts: the vector potential must decrease when we go from the surface into the material, and there is a cut-off at the surface of the material itself, of course. So all depends on the size of λ, as compared to the size of our piece of superconducting metal (or whatever other substance our piece is made of). In fact, if we look at e−λx as as an exponential decay function, then τ = 1/λ is the so-called scaling constant (it’s the inverse of the decay constant, which is λ itself). [You can work this out yourself. Note that for = τ = 1/λ, the value of our function e−λx will be equal to e−λ(1/λ) = e−1 ≈ 0.368, so it means the value of our function is reduced to about 36.8% of its initial value. For all practical purposes, we may say—as Feynman notes—that the field will, effectively, only penetrate to a thin layer at the surface: a layer of about 1/1/λ in thickness. He illustrates this as follows:

illustration

Moreover, he calculates the 1/λ distance for lead. Let me copy him here:

calculation

Well… That says it all, right? We’re talking two millionths of a centimeter here… 🙂

So what’s left? A lot, like flux quantization, or the equations of motion for the superconducting electron fluid. But we’ll leave that for the next posts. 🙂

Feynman’s Seminar on Superconductivity (1)

The ultimate challenge for students of Feynman’s iconic Lectures series is, of course, to understand his final one: A Seminar on Superconductivity. As he notes in his introduction to this formidably dense piece, the text does not present the detail of each and every step in the development and, therefore, we’re not supposed to immediately understand everything. As Feynman puts it: we should just believe (more or less) that things would come out if we would be able to go through each and every step. Well… Let’s see. Feynman throws a lot of stuff in here—including, I suspect, some stuff that may not be directly relevant, but that he sort of couldn’t insert into all of his other Lectures. So where do we start?

It took me one long maddening day to figure out the first formula:f1It says that the amplitude for a particle to go from to in a vector potential (think of a classical magnetic field) is the amplitude for the same particle to go from to b when there is no field (A = 0) multiplied by the exponential of the line integral of the vector potential times the electric charge divided by Planck’s constant. I stared at this for quite a while, but then I recognized the formula for the magnetic effect on an amplitude, which I described in my previous post, which tells us that a magnetic field will shift the phase of the amplitude of a particle with an amount equal to:

integral

Hence, if we write 〈b|a〉 for A = 0 as 〈b|aA = 0 = C·eiθ, then 〈b|a〉 in A will, naturally, be equal to 〈b|a〉 in A = C·ei(θ+φ) = C·eiθ·eiφ = 〈b|aA = 0 ·eiφ, and so that explains it. 🙂 Alright… Next. Or… Well… Let us briefly re-examine the concept of the vector potential, because we’ll need it a lot. We introduced it in our post on magnetostatics. Let’s briefly re-cap the development there. In Maxwell’s set of equations, two out of the four equations give us the magnetic field: B = 0 and c2×B = j0. We noted the following in this regard:

  1. The ∇B = 0 equation is true, always, unlike the ×E = 0 expression, which is true for electrostatics only (no moving charges). So the B = 0 equation says the divergence of B is zero, always.
  2. The divergence of the curl of a vector field is always zero. Hence, if A is some vector field, then div(curl A) = •(×A) = 0, always.
  3. We can now apply another theorem: if the divergence of a vector field, say D, is zero—so if D = 0—then will be the the curl of some other vector field C, so we can write: D = ×C.  Applying this to B = 0, we can write: 

If B = 0, then there is an A such that B = ×A

So, in essence, we’re just re-defining the magnetic field (B) in terms of some other vector field. To be precise, we write it as the curl of some other vector field, which we refer to as the (magnetic) vector potential. The components of the magnetic field vector can then be re-written as:

formula for B

We need to note an important point here: the equations above suggest that the components of B depend on position only. In other words, we assume static magnetic fields, so they do not change with time. That, in turn, assumes steady currents. We will want to extend the analysis to also include magnetodynamics. It complicates the analysis but… Well… Quantum mechanics is complicated. Let us remind ourselves here of Feynman’s re-formulation of Maxwell’s equations as a set of two equations (expressed in terms of the magnetic (vector) and the electric potential) only:

Wave equation for A

Wave equation for potential

These equations are wave equations, as you can see by writing out the second equation:

wave equation

It is a wave equation in three dimensions. Note that, even in regions where we do no have any charges or currents, we have non-zero solutions for φ and A. These non-zero solutions are, effectively, representing the electric and magnetic fields as they travel through free space. As Feynman notes, the advantage of re-writing Maxwell’s equations as we do above, is that the two new equations make it immediately apparent that we’re talking electromagnetic waves, really. As he notes, for many practical purposes, it will still be convenient to use the original equations in terms of E and B, but… Well… Not in quantum mechanics, it turns out. As Feynman puts it: “E and B are on the other side of the mountain we have climbed. Now we are ready to cross over to the other side of the peak. Things will look different—we are ready for some new and beautiful views.”

Well… Maybe. Appreciating those views, as part of our study of quantum mechanics, does take time and effort, unfortunately. 😦

The Schrödinger equation in an electromagnetic field

Feynman then jots down Schrödinger’s equation for the same particle (with charge q) moving in an electromagnetic field that is characterized not only by the (scalar) potential Φ but also by a vector potential A:

schrodinger

Now where does that come from? We know the standard formula in an electric field, right? It’s the formula we used to find the energy states of electrons in a hydrogen atom:

i·ħ·∂ψ/∂t = −(1/2)·(ħ2/m)∇2ψ + V·ψ

Of course, it is easy to see that we replaced V by q·Φ, which makes sense: the potential of a charge in an electric field is the product of the charge (q) and the (electric) potential (Φ), because Φ is, obviously, the potential energy of the unit charge. It’s also easy to see we can re-write −ħ2·∇2ψ as [(ħ/i)·∇]·[(ħ/i)·∇]ψ because (1/i)·(1/i) = 1/i2 = 1/(−1) = −1. 🙂 Alright. So it’s just that −q·A term in the (ħ/i)∇ − q·A expression that we need to explain now.

Unfortunately, that explanation is not so easy. Feynman basically re-derives Schrödinger’s equation using his trade-mark historical argument – which did not include any magnetic field – with a vector potential. The re-derivation is rather annoying, and I didn’t have the courage to go through it myself, so you should – just like me – just believe Feynman when he says that, when there’s a vector potential – i.e. when there’s a magnetic field – then that (ħ/i)·∇ operator – which is the momentum operator– ought to be replaced by a new momentum operator:

new-momentum-operator

So… Well… There we are… 🙂 So far, so good? Well… Maybe.

While, as mentioned, you won’t be interested in the mathematical argument, it is probably worthwhile to reproduce Feynman’s more intuitive explanation of why the operator above is what it is. In other words, let us try to understand that −qA term. Look at the following situation: we’ve got a solenoid here, and some current I is going through it so there’s a magnetic field B. Think of the dynamics while we turn on this flux. Maxwell’s second equation (∇×E = −∂B/∂t) tells us the line integral of E around a loop will be equal to the time rate of change of the magnetic flux through that loop. The ∇×E = −∂B/∂t equation is a differential equation, of course, so it doesn’t have the integral, but you get the idea—I hope.solenoid

Now, using the B = ×A equation we can re-write the ∇×E = −∂B/∂t as ∇×E = −∂(×A)/∂t. This allows us to write the following:

 ∇×E = −∂(×A)/∂t = −×(∂A/∂t) ⇔ E = −∂A/∂t

This is a remarkable expression. Note its derivation is based on the commutativity of the curl and time derivative operators, which is a property that can easily be explained: if we have a function in two variables—say x and t—then the order of the derivation doesn’t matter: we can first take the derivative with respect to and then to t or, alternatively, we can first take the time derivative and then do the ∂/∂x operation. So… Well… The curl is, effectively, a derivative with regard to the spatial variables. OK. So what? What’s the point?

Well… If we’d have some charge q, as shown in the illustration above, that would happen to be there as the flux is being switched on, it will experience a force which is equal to F = qE. We can now integrate this over the time interval (t) during which the flux is being built up to get the following:

0t F = ∫0t m·a = ∫0t m·dv/dt = m·vt= ∫0t q·E = −∫0t q·∂A/∂t = −q·At

Assuming v0 and Aare zero, we may drop the time subscript and simply write:

v = −q·A

The point is: during the build-up of the magnetic flux, our charge will pick up some (classical) momentum that is equal to p = m·v = −q·A. So… Well… That sort of explains the additional term in our new momentum operator.

Note: For some reason I don’t quite understand, Feynman introduces the weird concept of ‘dynamical momentum’, which he defines as the quantity m·v + q·A, so that quantity must be zero in the analysis above. I quickly googled to see why but didn’t invest too much time in the research here. It’s just… Well… A bit puzzling. I don’t really see the relevance of his point here: I am quite happy to go along with the new operator, as it’s rather obvious that introducing changing magnetic fields must, obviously, also have some impact on our wave equations—in classical as well as in quantum mechanics.

Local conservation of probability

The title of this section in Feynman’s Lecture (yes, still the same Lecture – we’re not switching topics here) is the equation of continuity for probabilities. I find it brilliant, because it confirms my interpretation of the wave function as describing some kind of energy flow. Let me quote Feynman on his endeavor here:

“An important part of the Schrödinger equation for a single particle is the idea that the probability to find the particle at a position is given by the absolute square of the wave function. It is also characteristic of the quantum mechanics that probability is conserved in a local sense. When the probability of finding the electron somewhere decreases, while the probability of the electron being elsewhere increases (keeping the total probability unchanged), something must be going on in between. In other words, the electron has a continuity in the sense that if the probability decreases at one place and builds up at another place, there must be some kind of flow between. If you put a wall, for example, in the way, it will have an influence and the probabilities will not be the same. So the conservation of probability alone is not the complete statement of the conservation law, just as the conservation of energy alone is not as deep and important as the local conservation of energy. If energy is disappearing, there must be a flow of energy to correspond. In the same way, we would like to find a “current” of probability such that if there is any change in the probability density (the probability of being found in a unit volume), it can be considered as coming from an inflow or an outflow due to some current.”

This is it, really ! The wave function does represent some kind of energy flow – between a so-called ‘real’ and a so-called ‘imaginary’ space, which are to be defined in terms of directional versus rotational energy, as I try to point out – admittedly: more by appealing to intuition than to mathematical rigor – in that post of mine on the meaning of the wavefunction.

So what is the flow – or probability current as Feynman refers to it? Well… Here’s the formula:

probability-current-2

Huh? Yes. Don’t worry too much about it right now. The essential point is to understand what this current – denoted by J – actually stands for:

probability-current-1

So what’s next? Well… Nothing. I’ll actually refer you to Feynman now, because I can’t improve on how he explains how pairs of electrons start behaving when temperatures are low enough to render Boltzmann’s Law irrelevant: the kinetic energy that’s associated with temperature can no longer break up electron pairs if temperature comes close to the zero point.

Huh? What? Electron pairs? Electrons are not supposed to form pairs, are they? They carry the same charge and are, therefore, supposed to repel each other. Well… Yes and no. In my post on the electron orbitals in a hydrogen atom – which just presented Feynman’s presentation on the subject-matter in a, hopefully, somewhat more readable format – we calculated electron orbitals neglecting spin. In Feynman’s words:

“We make another approximation by forgetting that the electron has spin. […] The non-relativistic Schrödinger equation disregards magnetic effects. [However] Small magnetic effects [do] occur because, from the electron’s point-of-view, the proton is a circulating charge which produces a magnetic field. In this field the electron will have a different energy with its spin up than with it down. [Hence] The energy of the atom will be shifted a little bit from what we will calculate. We will ignore this small energy shift. Also we will imagine that the electron is just like a gyroscope moving around in space always keeping the same direction of spin. Since we will be considering a free atom in space the total angular momentum will be conserved. In our approximation we will assume that the angular momentum of the electron spin stays constant, so all the rest of the angular momentum of the atom—what is usually called “orbital” angular momentum—will also be conserved. To an excellent approximation the electron moves in the hydrogen atom like a particle without spin—the angular momentum of the motion is a constant.”

To an excellent approximation… But… Well… Electrons in a metal do form pairs, because they can give up energy in that way and, hence, they are more stable that way. Feynman does not go into the details here – I guess because that’s way beyond the undergrad level – but refers to the Bardeen-Coopers-Schrieffer (BCS) theory instead – the authors of which got a Nobel Prize in Physics in 1972 (that’s a decade or so after Feynman wrote this particular Lecture), so I must assume the theory is well accepted now. 🙂

Of course, you’ll shout now: Hey! Hydrogen is not a metal! Well… Think again: the latest breakthrough in physics is making hydrogen behave like a metal. 🙂 And I am really talking the latest breakthrough: Science just published the findings of this experiment last month! 🙂 🙂 In any case, we’re not talking hydrogen here but superconducting materials, to which – as far as we know – the BCS theory does apply.

So… Well… I am done. I just wanted to show you why it’s important to work your way through Feynman’s last Lecture because… Well… Quantum mechanics does explain everything – although the nitty-gritty of it (the Meissner effect, the London equation, flux quantization, etc.) are rather hard bullets to bite. 😦

Don’t give up ! I am struggling with the nitty-gritty too ! 🙂