# Re-visiting electron orbitals (III)

Pre-script (dated 26 June 2020): Our ideas have evolved into a full-blown realistic (or classical) interpretation of all things quantum-mechanical. In addition, I note the dark force has amused himself by removing some material. So no use to read this. Read my recent papers instead. ðŸ™‚

Original post:

In my previous post, I mentioned that it wasÂ not so obvious (both from a physicalÂ as well as from aÂ mathematicalÂ point of view) to write the wavefunction for electron orbitals – which we denoted as Ïˆ(x, t), i.e. a function of two variables (or four: one time coordinate and three space coordinates) –Â as the product of two other functions in one variable only.

[…] OK. The above sentence is difficult to read. Let me write in math. ðŸ™‚ It isÂ notÂ so obvious to write Ïˆ(x, t) as:

Ïˆ(x, t) = eâˆ’iÂ·(E/Ä§)Â·tÂ·Ïˆ(x)

As I mentioned before, the physicists’ use of the same symbol (Ïˆ, psi) for both the Ïˆ(x, t) and Ïˆ(x) function is quite confusing – because the two functions areÂ veryÂ different:

• Ïˆ(x, t) is a complex-valued function of twoÂ (real)Â variables: x and t. OrÂ four, I should say, because xÂ = (x, y, z) – but it’s probably easier to think of x as oneÂ vectorÂ variable – aÂ vector-valued argument, so to speak. And then t is, of course, just aÂ scalarÂ variable. So… Well… A function of twoÂ variables: the position in space (x), and time (t).
• In contrast, Ïˆ(x) is a real-valuedÂ function ofÂ oneÂ (vector) variable only: x, so that’s the position in space only.

Now you should cry foul, of course: Ïˆ(x) is notÂ necessarilyÂ real-valued. It mayÂ be complex-valued. You’re right.Â You know the formula:Note the derivation of this formula involved a switch from Cartesian to polar coordinates here, so from xÂ = (x, y, z) to rÂ = (r, Î¸, Ï†), and that the function is also a function of the twoÂ quantum numbersÂ l and m now, i.e. the orbital angular momentum (l) and its z-component (m) respectively. In my previous post(s), I gave you the formulas for Yl,m(Î¸, Ï†) and Fl,m(r) respectively. Fl,m(r) was a real-valued function alright, but the Yl,m(Î¸, Ï†) had that eiÂ·mÂ·Ï†Â factor in it. So… Yes. You’re right: the Yl,m(Î¸, Ï†) function is real-valued if – and onlyÂ if – m = 0, in which case eiÂ·mÂ·Ï†Â = 1.Â Let me copy the table from Feynman’s treatment of the topic once again:The Plm(cosÎ¸) functions are the so-called (associated) Legendre polynomials, and the formula for these functions is rather horrible:Don’t worry about it too much: just note the Plm(cosÎ¸)Â is aÂ real-valuedÂ function. The point is the following:theÂ Ïˆ(x, t) is a complex-valuedÂ function because – andÂ onlyÂ because – we multiply a real-valued envelope function – which depends on positionÂ only – with eâˆ’iÂ·(E/Ä§)Â·tÂ·eiÂ·mÂ·Ï†Â = eâˆ’iÂ·[(E/Ä§)Â·tÂ âˆ’Â mÂ·Ï†].

[…]

Please read the above once again and – more importantly – think about it for a while. ðŸ™‚ You’ll have to agree with the following:

• As mentioned in my previous post,Â the eiÂ·mÂ·Ï†Â factor just gives us phase shift: just aÂ re-set of our zero point for measuring time, so to speak, and the whole eâˆ’iÂ·[(E/Ä§)Â·tÂ âˆ’Â mÂ·Ï†]Â factor just disappears when weâ€™re calculating probabilities.
• The envelope function gives us the basic amplitude – in theÂ classicalÂ sense of the word:Â the maximum displacement fromÂ theÂ zeroÂ value. And so it’s that eâˆ’iÂ·[(E/Ä§)Â·tÂ âˆ’Â mÂ·Ï†]Â that ensures the whole expression somehow captures the energyÂ of the oscillation.

Let’s first look at the envelope function again. Let me copy the illustration forÂ n = 5 and lÂ = 2 from aÂ Wikimedia CommonsÂ article.Â Note the symmetry planes:

• Any plane containing theÂ z-axis is a symmetry plane – like a mirror in which we can reflect one half of theÂ shape to get the other half. [Note that I am talking theÂ shapeÂ only here. Forget about the colors for a while – as these reflect the complex phase of the wavefunction.]
• Likewise, the plane containingÂ bothÂ the x– and the y-axis is a symmetry plane as well.

The first symmetry plane – or symmetryÂ line, really (i.e. theÂ z-axis) – should not surprise us, because the azimuthal angle Ï† is conspicuously absent in the formula for our envelope function if, as we are doing in this article here, we merge theÂ eiÂ·mÂ·Ï†Â factor with the eâˆ’iÂ·(E/Ä§)Â·t, so it’s just part and parcel of what the author of the illustrations above refers to as the ‘complex phase’ of our wavefunction.Â OK. Clear enough – I hope. ðŸ™‚ But why is theÂ the xy-plane a symmetry plane too? We need to look at that monstrous formula for the Plm(cosÎ¸) function here: just note the cosÎ¸ argument in it is being squaredÂ before it’s used in all of the other manipulation. Now, we know that cosÎ¸ = sin(Ï€/2Â âˆ’Â Î¸). So we can define someÂ newÂ angle – let’s just call it Î± – which is measured in the way we’re used to measuring angle, which is notÂ from the z-axis but from the xy-plane. So we write: cosÎ¸ = sin(Ï€/2Â âˆ’Â Î¸) = sinÎ±. The illustration below may or may not help you to see what we’re doing here.So… To make a long story short, we can substitute the cosÎ¸ argument in the Plm(cosÎ¸) function for sinÎ± = sin(Ï€/2Â âˆ’Â Î¸). Now, if the xy-plane is a symmetry plane, then we must find the same value for Plm(sinÎ±) and Plm[sin(âˆ’Î±)]. Now, that’s not obvious, because sin(âˆ’Î±) = âˆ’sinÎ± â‰ Â sinÎ±. However, because the argument in that Plm(x) function is being squared before any other operation (like subtracting 1 and exponentiating the result), it is OK: [âˆ’sinÎ±]2Â = [sinÎ±]2Â =Â sin2Î±. […] OK, I am sure the geeks amongst my readers will be able to explain this more rigorously. In fact, I hope they’ll have a look at it, because there’s also that dl+m/dxl+mÂ operator, and so you should check what happens with the minus sign there. ðŸ™‚

[…] Well… By now, you’re probably totally lost, but the fact of the matter is that we’ve got a beautiful result here. Let me highlight the most significant results:

• AÂ definiteÂ energy state of a hydrogen atom (or of an electron orbiting around some nucleus, I should say) appears to us as some beautifully shaped orbital – an envelopeÂ function in three dimensions, really – whichÂ has the z-axis – i.e. the vertical axis – as a symmetry line and the xy-plane as a symmetry plane.
• The eâˆ’iÂ·[(E/Ä§)Â·tÂ âˆ’Â mÂ·Ï†]Â factor gives us the oscillation within the envelope function. As such, it’s this factor that, somehow,Â captures the energyÂ of the oscillation.

It’s worth thinking about this. Look at the geometry of the situation again – as depicted below. We’re looking at the situation along the x-axis, in the direction of the origin, which is the nucleus of our atom.

The eiÂ·mÂ·Ï†Â factor just gives us phase shift: just aÂ re-set of our zero point for measuring time, so to speak. Interesting, weird – but probably less relevant than the eâˆ’iÂ·[(E/Ä§)Â·tÂ factor, which gives us the two-dimensional oscillation that captures the energy of the state.

Now, the obvious question is: the oscillation of what, exactly? I am not quite sure but – as I explained in my Deep BlueÂ page – the real and imaginary part of our wavefunction are really like the electric and magnetic field vector of an oscillating electromagnetic field (think of electromagnetic radiation – if that makes it easier). Hence, just like the electric and magnetic field vector represent some rapidly changing forceÂ on a unit charge, the real and imaginary part of our wavefunction must also represent some rapidly changingÂ forceÂ on… Well… I am not quite sure on what though. The unit charge is usually defined as the charge of a proton – rather than an electron – but then forces act on some mass, right? And the massÂ of a proton is hugely different from the mass of an electron. The same electric (or magnetic) force will, therefore, give a hugely different acceleration to both.

So… Well… My guts instinct tells me the real and imaginary part of our wavefunction just represent, somehow, a rapidly changing force on some unit ofÂ mass, but then I am not sure how to define that unit right now (it’s probably notÂ the kilogram!).

Now, there is another thing we should note here: we’re actually sort of de-constructing a rotationÂ (look at the illustration above once again) in two linearly oscillating vectors – one along the z-axis and the other along the y-axis.Â Hence, in essence, we’re actually talking about something that’s spinning.Â In other words, we’re actually talking someÂ torqueÂ around the x-axis. In what direction? I think that shouldn’t matter – that we can write E or âˆ’E, in other words, but… Well… I need to explore this further – as should you! ðŸ™‚

Let me just add one more note on the eiÂ·mÂ·Ï†Â factor. It sort of defines the geometryÂ of the complex phase itself. Look at the illustration below. Click on it to enlarge it if necessary – or, better still, visit the magnificent Wikimedia Commons article from which I get these illustrations. These are the orbitals nÂ = 4 and lÂ = 3. Look at the red hues in particular – or the blue – whatever: focus on one color only, and see how how – for mÂ = Â±1, we’ve got one appearance of that color only. For mÂ = Â±1, the same color appears at two ends of the ‘tubes’ – or toriÂ (plural of torus), I should say – just to sound more professional. ðŸ™‚ For mÂ = Â±2, the torus consists of three parts – or, in mathematical terms, we’d say the order of its rotational symmetryÂ is equal to 3.Â Check that Wikimedia Commons article for higher values ofÂ nÂ andÂ l: the shapes become very convoluted, but the observation holds. ðŸ™‚

Have fun thinking all of this through for yourself – and please do look at those symmetries in particular. ðŸ™‚

Post scriptum: You should do some thinking on whether or not theseÂ mÂ =Â Â±1, Â±2,…, Â±lÂ orbitals are really different. As I mentioned above, a phase difference is just what it is: a re-set of the t = 0 point. Nothing more, nothing less. So… Well… As far as I am concerned, that’s notÂ aÂ realÂ difference, is it? ðŸ™‚ As with other stuff, I’ll let you think about this for yourself.

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# Quantum-mechanical operators

WeÂ climbed a mountainâ€”step by step, post by post. ðŸ™‚ We have reached the top now, and the view is gorgeous. WeÂ understand SchrÃ¶dinger’s equation, which describes how amplitudes propagate through space-time. It’s the quintessential quantum-mechanical expression. Let’s enjoy now, and deepen our understanding by introducing the concept of (quantum-mechanical) operators.

#### The operator concept

We’ll introduce the operator concept using SchrÃ¶dinger’s equation itself and, in the process, deepen our understanding of SchrÃ¶dinger’s equation a bit. You’ll remember we wrote it as:

However, you’ve probably seen it like it’s written on his bust, or on his grave, or wherever, which is as follows:

It’s the same thing, of course. The ‘over-dot’ is Newton’s notation for the time derivative. In fact, if you click on the picture above (and zoom in a bit), then you’ll see that the craftsman who made the stone grave marker, mistakenly, also carved a dot above the psi (Ïˆ)Â on the right-hand side of the equationâ€”but then someone pointed out his mistake and so the dot on the right-hand side isn’t painted. ðŸ™‚ The thing I want to talk about here, however, is the H inÂ thatÂ expression above, which is, obviously, the following operator:

That’s a pretty monstrous operator, isn’t it? It is what it is, however: an algebraicÂ operator (it operates on a numberâ€”albeit a complexÂ numberâ€”unlikeÂ a matrixÂ operator, which operates on a vectorÂ or another matrix). As you can see, it actuallyÂ consists ofÂ twoÂ other (algebraic) operators:

1. TheÂ âˆ‡2Â operator, which you know: it’s a differentialÂ operator. To be specific, it’s theÂ LaplaceÂ operator, which is the divergence (âˆ‡Â·) of the gradient (âˆ‡) of a function:Â âˆ‡2Â = âˆ‡Â·âˆ‡ = (âˆ‚/âˆ‚x, âˆ‚/âˆ‚yÂ , âˆ‚/âˆ‚z)Â·(âˆ‚/âˆ‚x, âˆ‚/âˆ‚yÂ , âˆ‚/âˆ‚z) = âˆ‚2/âˆ‚x2 Â +Â âˆ‚2/âˆ‚y2Â + âˆ‚2/âˆ‚z2. ThisÂ too operates on our complex-valued function wavefunctionÂ Ïˆ, and yields some other complex-valued function, which we then multiply byÂ âˆ’Ä§2/2m to get the first term.
2. The V(x, y, z) ‘operator’, whichâ€”in this particular contextâ€”just means: “multiply with V”. Needless to say, V isÂ the potentialÂ here, and so it captures the presence of external force fields. Also note that V is a real number, just likeÂ âˆ’Ä§2/2m.

Let me say something about the dimensions here. On the left-hand side of SchrÃ¶dinger’s equation, we have the product ofÂ Ä§ and a time derivative (iÂ is just the imaginary unit, so that’s just a (complex) number). Hence, the dimension there is [JÂ·s]/[s] (the dimension of a time derivative is something expressed per second). So the dimension of the left-hand side isÂ joule. On the right-hand side, we’ve got two terms. The dimension of that second-order derivative (âˆ‡2Ïˆ) is something expressedÂ per square meter, but then we multiply it with âˆ’Ä§2/2m, whose dimension is [J2Â·s2]/[J/(m2/s2)]. [Remember: m = E/c2.] So that reduces to [JÂ·m2]. Hence, the dimension of (âˆ’Ä§2/2m)âˆ‡2Ïˆ is joule. And the dimension of V isÂ jouleÂ too, of course. So it all works out.Â In fact, now that we’re here, it may or may not be useful to remind you of that heat diffusion equation we discussed when introducing the basic concepts involved in vector analysis:

That equation illustrated theÂ physicalÂ significance of the Laplacian. We were talking about the flow ofÂ heatÂ in, say, a block of metal, as illustrated below. TheÂ qÂ in the equation above is theÂ heat per unit volume, and the h in the illustration below was the heat flow vector (so it’s got nothingÂ to do with Planck’s constant), which depended on the material, and which we wrote asÂ hÂ = â€“Îºâˆ‡T, with T the temperature, and ÎºÂ (kappa) theÂ thermal conductivity. In any case, the point is the following: the equation below illustrates the physical significance of the Laplacian. We let itÂ operateÂ on the temperature (i.e. a scalarÂ function) and its product with some constant (just think of replacing ÎºÂ byÂ âˆ’Ä§2/2mÂ gives us theÂ time derivativeÂ of q, i.e. the heat per unit volume.

In fact, we know that qÂ is proportional to T, so if we’d choose an appropriate temperature scale â€“ i.e. choose the zero point such that qÂ =Â kÂ·TÂ (your physics teacher in high school would refer to kÂ as the (volume)Â specific heat capacity) â€“ then we could simple write:

âˆ‚T/âˆ‚t = (Îº/k)âˆ‡2T

From aÂ mathematicalÂ point of view, that equation is just the same asÂ âˆ‚Ïˆ/âˆ‚t = â€“(iÂ·Ä§/2m)Â·âˆ‡2Ïˆ, which is SchrÃ¶dinger’s equation for V = 0.Â In other words, you can â€“ and actually shouldÂ â€“ alsoÂ think of SchrÃ¶dinger’s equation as describing the flowÂ of… Well… What?

Well… Not sure. I am tempted to think of something like a probabilityÂ densityÂ in space, butÂ Ïˆ represents a (complex-valued) amplitude. Having said that, you get the ideaâ€”I hope! ðŸ™‚ If not, let me paraphrase Feynman on this:

“WeÂ can think of SchrÃ¶dinger’s equationÂ as describing the diffusion of a probability amplitude from one point to another. In fact, the equation looks something like the diffusion equation we introduced when discussing heat flow, or the spreading of a gas. But there is one main difference: the imaginary coefficient in front of the time derivative makes the behavior completely different from the ordinary diffusion such as you would have for a gas spreading out. Ordinary diffusion gives rise to real exponential solutions, whereas the solutions of SchrÃ¶dinger’s equationÂ are complex waves.”

That says it all, right? ðŸ™‚ In fact, SchrÃ¶dinger’s equation â€“ as discussed here â€“ was actually being derived when describing the motion of an electron along a line of atoms, i.e. for motion inÂ one directionÂ only, but you can visualize what it represents in three-dimensional space. The real exponential functions Feynman refer to exponentialÂ decayÂ function: as the energy is spread over an ever-increasing volume, the amplitude of the wave becomes smaller and smaller. That may be the case for complex-valued exponentials as well. The keyÂ difference between a real- and complex-valued exponential decay function is that aÂ complexÂ exponential is a cyclical function. Now, I quickly googled to see how we could visualize that, and I like the following illustration:

The dimensional analysis ofÂ SchrÃ¶dinger’s equation is also quite interesting because… Well… Think of it: that heat diffusion equation incorporates the same dimensions: temperature is a measure of the average energy of the molecules. That’s really something to think about. These differential equations are not onlyÂ structurallyÂ similar but, in addition, they all seem to describe someÂ flow of energy. That’s pretty deep stuff: it relates amplitudes to energies, so we should think in terms of Poynting vectors and all that. But… Well… IÂ need to move on, and so I willÂ move onâ€”so you can re-visit this later. ðŸ™‚

Now that we’ve introduced theÂ conceptÂ of an operator, let me say something about notations, because that’s quite confusing.

#### Some remarks on notation

Because it’s an operator, we should actually use the hat symbolâ€”in line with what we did when we were discussing matrixÂ operators: we’d distinguish the matrix (e.g. A) from its use as an operator (Ã‚). You may or may not remember we do the same in statistics: the hatÂ symbol is supposed to distinguish the estimator (Ã¢) â€“ i.e. some function we use to estimateÂ a parameter (which we usually denoted by some Greek symbol, likeÂ Î±)Â â€“Â from a specific estimateÂ of the parameter, i.e. the valueÂ (a) we get when applying Ã¢Â to a specific sample or observation. However, if you remember the difference, you’ll also rememberÂ thatÂ hat symbol was quickly forgotten, because theÂ contextÂ made it clear what was what, and so we’d just write a(x) instead of Ã¢(x). So… Well… I’ll be sloppy as well here, if only because the WordPressÂ editor only offers veryÂ fewÂ symbols with a hat! ðŸ™‚

In any case, this discussion on the use (or not) of thatÂ hatÂ is irrelevant. In contrast, whatÂ isÂ relevant is to realizeÂ this algebraicÂ operatorÂ H here is very different from that other quantum-mechanical Hamiltonian operator we discussed when dealing with a finite set of base states: thatÂ HÂ was the Hamiltonian matrix, but used in an ‘operation’ on some state. So we have theÂ matrixÂ operator H, and theÂ algebraicÂ operator H.

Confusing?

Yes and no. First, we’ve got the context again, and so you alwaysÂ knowÂ whether you’re looking at continuous or discrete stuff:

1. If your ‘space’ is continuous (i.e. if states are to defined with reference to an infiniteÂ set of base states), then it’s the algebraic operator.
2. If, on the other hand, your states are defined by some finite set ofÂ discreteÂ base states, then it’s the Hamiltonian matrix.

There’s another, more fundamental, reason why there should be no confusion. In fact, it’s the reason why physicists use the same symbol H in the first place: despite the fact that they look so different,Â these two operators (i.e. HÂ the algebraic operator and H the matrix operator) are actually equivalent. Their interpretation is similar, as evidenced from the fact that both are being referred to as the energy operatorÂ in quantum physics. The only difference is that one operates on a (state)Â vector, while the other operates on a continuous function.Â It’s just the difference betweenÂ matrix mechanicsÂ as opposed to wave mechanicsÂ really.

But…Â Well… I am sure I’ve confused you by nowâ€”and probably very much soâ€”and so let’s start from the start. ðŸ™‚

#### Matrix mechanics

Let’s start with the easy thing indeed: matrix mechanics. The matrix-mechanical approach is summarized in that set of Hamiltonian equations which, by now, you know so well:

If we have nÂ base states, then we haveÂ nÂ equations like this: one for eachÂ iÂ = 1, 2,… n. As for the introduction of the Hamiltonian, and the other subscript (j), just think of the description of a state:

So… Well… Because we had used iÂ already, we had to introduceÂ j. ðŸ™‚

Let’s think about |ÏˆâŒª. It is theÂ stateÂ of a system, like theÂ ground stateÂ of a hydrogen atom, or one of its manyÂ excitedÂ states. But… Well… It’s a bit of a weird term, really. It all depends on what you want to measure: when we’re thinking of the ground state, or an excited state, we’re thinking energy. That’s something else than thinking its position in space, for example. Always remember: a state is defined by a set ofÂ baseÂ states, and so those base states come with a certainÂ perspective: when talking states, we’re only looking at someÂ aspectÂ of reality, really. Let’s continue with our example ofÂ energyÂ states, however.

You know that the lifetime of a system in an excited state is usually short: some spontaneous or induced emission of a quantum of energy (i.e. a photon) will ensure that the system quickly returns to a less excited state, or to the ground state itself. However, you shouldn’t think of that here: we’re looking at stableÂ systems here. To be clear: we’re looking at systems that have someÂ definiteÂ energyâ€”or so we think: it’s just because of the quantum-mechanical uncertainty that we’ll always measure some other different value. Does that make sense?

If it doesn’t… Well… Stop reading, because it’s only going to get even more confusing. Not my fault, however!

#### Psi-chology

The ubiquity of that Ïˆ symbol (i.e. the Greek letterÂ psi) is really something psi-chological ðŸ™‚ and, hence, veryÂ confusing, really.Â In matrix mechanics, ourÂ Ïˆ would just denote a state of a system, like the energy of an electron (or, when there’s only one electron, our hydrogen atom). If it’s an electron, then we’d describe it by its orbital. In this regard, I found the following illustration from Wikipedia particularly helpful: the green orbitals show excitationsÂ of copperÂ (Cu) orbitals on a CuO2Â plane. [The two big arrows just illustrate the principle of X-ray spectroscopy, so it’s an X-ray probingÂ the structure of the material.]

So… Well… We’d write Ïˆ as |ÏˆâŒª just to remind ourselves we’re talking of some state of the system indeed. However,Â quantum physicists always want to confuse you, and so they will also use the psiÂ symbol toÂ denote something else: they’ll use it to denote a very particular CiÂ amplitude (or coefficient) in that |ÏˆâŒª = âˆ‘|iâŒªCiÂ formula above. To be specific, they’d replace the base states |iâŒª by the continuous position variable x, and they would write the following:

CiÂ = Ïˆ(i =Â x) = Ïˆ(x) = CÏˆ(x) = C(x) =Â âŒ©x|ÏˆâŒª

In fact, that’s just like writing:

Ï†(p) = âŒ© mom p | Ïˆ âŒª =Â âŒ©p|ÏˆâŒª =Â CÏ†(p) = C(p)

What they’re doing here, is (1) reduce the ‘system‘ to a ‘particle‘ once more (which is OK, as long as you know what you’re doing) and (2) they basically state the following:

If a particle is in some state |ÏˆâŒª, then we can associate someÂ wavefunction Ïˆ(x) orÂ Ï†(p)â€”with it, and that wavefunction will represent the amplitudeÂ for the system (i.e. our particle) to be at x, or to have a momentum that’s equal toÂ p.

So what’s wrong with that? Well… Nothing. It’s just that… Well… Why don’t they use Ï‡(x) instead of Ïˆ(x)? That would avoid a lot of confusion, I feel: one should notÂ use the same symbol (psi) for the |ÏˆâŒª state and the Ïˆ(x) wavefunction.

Huh?Â Yes. Think about it.Â The point is:Â theÂ positionÂ or the momentum, or even the energy, are properties of the system, so to speak and, therefore, it’s really confusing to use the same symbol psiÂ (Ïˆ) to describe (1) the stateÂ of the system, in general, versus (2)Â the positionÂ wavefunction, which describes… Well… Some very particularÂ aspect (or ‘state’, if you want) of the same system (in this case: its position). There’s no such problem withÂ Ï†(p), so… Well… Why don’t they useÂ Ï‡(x) instead of Ïˆ(x) indeed? I have only one answer:Â psi-chology. ðŸ™‚

In any case, there’s nothing we can do about it and… Well… In fact, that’s what this post is about: it’s about how to describe certainÂ propertiesÂ of the system. Of course,Â we’re talking quantum mechanics here and, hence,Â uncertainty, and, therefore, we’re going to talk about theÂ average position, energy, momentum, etcetera that’s associated with a particular stateÂ of a system, orâ€”as we’ll keep things very simpleâ€”the properties of a ‘particle’, really. Think of an electron in some orbital, indeed! ðŸ™‚

So let’s now look at that set of Hamiltonian equations once again:

Looking at it carefullyÂ â€“ so just look at it once again! ðŸ™‚Â â€“ and thinking aboutÂ what we did when going from the discrete to the continuous setting, we can now understand we should write the following for the continuous case:

Of course, combiningÂ SchrÃ¶dinger’s equation with the expression above implies the following:

Now how can we relate that integral to the expression on the right-hand side? I’ll have to disappoint you here, as it requires a lot of math to transform that integral. It requires writing H(x, x’) in terms of rather complicated functions, including â€“ you guessed it, didn’t you?Â â€“ Dirac’s delta function. Hence, I assume you’ll believe me if I say that the matrix- and wave-mechanical approaches are actually equivalent. In any case, if you’d want to check it, you can always read Feynman yourself. ðŸ™‚

Now, I wrote this post to talk about quantum-mechanicalÂ operators, so let me do that now.

#### Quantum-mechanical operators

You know the concept of an operator. As mentioned above, we should put a littleÂ hatÂ (^) on top of our Hamiltonian operator, so as to distinguish it from the matrix itself. However, as mentioned above, the difference is usually quite clear from the context. Our operators were all matrices so far, and we’d write the matrix elements of, say, some operator A, as:

AijÂ â‰¡Â âŒ© iÂ | A |Â jÂ âŒª

The whole matrix itself, however, would usually not act on a base state but… Well… Just on some more general state Ïˆ, to produce some new state Ï†, and so we’d write:

| Ï†Â âŒª = AÂ | ÏˆÂ âŒª

Of course, we’d have toÂ describeÂ | Ï†Â âŒª in terms of the (same) set of base states and, therefore, we’d expand this expression into something like this:

You get the idea. I should just add one more thing. You know this important property of amplitudes: the âŒ© ÏˆÂ | Ï† âŒª amplitude is the complex conjugateÂ of the âŒ© Ï† |Â Ïˆ âŒª amplitude. It’s got to do with time reversibility, because the complex conjugate of eâˆ’iÎ¸Â = eâˆ’i(Ï‰Â·tâˆ’kÂ·x)Â is equal to eiÎ¸Â = ei(Ï‰Â·tâˆ’kÂ·x),Â so we’re just reversing the x- andÂ tdirection.Â We write:

Â âŒ©Â Ïˆ |Â Ï† âŒª =Â âŒ© Ï† | Ïˆ âŒª*

Now what happens if we want to take the complex conjugate when we insert a matrix, so when writing âŒ© Ï† | A | Ïˆ âŒª instead of âŒ© Ï† | Ïˆ âŒª, this rules becomes:

âŒ© Ï† | A | Ïˆ âŒª* =Â âŒ© Ïˆ | Aâ€  | Ï†Â âŒª

TheÂ daggerÂ symbol denotes theÂ conjugate transpose, so Aâ€  is an operator whose matrix elements are equal to Aijâ€  = Aji*. Now, it may or may not happen that theÂ Aâ€  matrix is actually equal to the original A matrix. In that caseÂ â€“ andÂ onlyÂ in that caseÂ â€“ we can write:

âŒ© Ïˆ | A | Ï†Â âŒª = âŒ© Ï† | A | Ïˆ âŒª*

We thenÂ say that A is a ‘self-adjoint’ or ‘Hermitian’ operator. That’s just a definition of a property, which the operator may or may not haveâ€”but many quantum-mechanical operators are actually Hermitian. In any case, we’re well armed now to discuss some actualÂ operators, and we’ll start with that energyÂ operator.

#### The energy operator (H)

We know the state of aÂ systemÂ is described in terms of a set of base states. Now, our analysis of N-state systems showed we can always describe it in terms of aÂ specialÂ set of base states, which are referred to as the states of definite energyÂ because… Well… Because they’re associated with someÂ definiteÂ energy. In that post, we referred to these energy levels asÂ EnÂ (n = I, II,â€¦ N). We used boldface for the subscript n (so we wrote n instead of n) because of these RomanÂ numerals. With each energy level, we could associate a base state, of definite energyÂ indeed, that we wrote asÂ |nâŒª. To make a long story short, we summarized our results as follows:

1. The energiesÂ EI, EII,â€¦,Â En,â€¦, ENÂ are theÂ eigenvaluesÂ of the Hamiltonian matrix H.
2. The state vectors |nâŒª that are associated with each energy En, i.e. the set of vectorsÂ |nâŒª, are the corresponding eigenstates.

We’ll be working with some more subscripts in what follows, and these Roman numerals and the boldface notation are somewhat confusing (if only because I don’t want you to think of these subscripts as vectors), we’ll just denote EI, EII,â€¦,Â En,â€¦, ENÂ as E1, E2,â€¦,Â Ei,â€¦, EN, and we’llÂ numberÂ the states of definite energy accordingly, also using some Greek letter so as to clearly distinguish them from all ourÂ LatinÂ letter symbols: we’ll write these states as: |Î·1âŒª, |Î·1âŒª,… |Î·NâŒª. [If I say, ‘we’, I mean Feynman of course. You may wonder why he doesn’t write |EiâŒª, or |ÎµiâŒª. The answer is: writingÂ |EnâŒª would cause confusion, because this state will appear in expressions like: |EiâŒªEi, so that’s the ‘product’ of a state (|EiâŒª) and the associated scalarÂ (Ei). Too confusing. As for using Î· (eta) instead of Îµ (epsilon) to denote something that’s got to do withÂ energy… Well… I guess he wanted to keep the resemblance with theÂ n, and then the Ancient Greek apparently did use this Î· letter Â for a sound like ‘e‘ so… Well… Why not? Let’s get back to the lesson.]

Using these base states of definite energy, we can write the state of the system as:

|ÏˆâŒª =Â âˆ‘Â |Î·iâŒª CiÂ Â = âˆ‘Â |Î·iâŒªâŒ©Î·i|ÏˆâŒªÂ Â Â Â over allÂ iÂ (i = 1, 2,… , N)

Now, we didn’t talk all that much about what these base states actuallyÂ meanÂ in terms of measuring something but you’ll believe if I say that, when measuringÂ the energy of the system,Â we’ll always measure one orÂ the other E1, E2,â€¦,Â Ei,â€¦, ENÂ value. We’ll never measure something in-between: it’s eitheror. Now, as you know, measuring something in quantum physics is supposed to be destructive but… Well… Let usÂ imagineÂ we could make a thousand measurements to try to determine theÂ averageÂ energy of the system. We’d do so by counting the number of times we measureÂ E1Â (and of course we’d denote that number as N1),Â E2,Â E3, etcetera. You’ll agree that we’d measure the average energy as:

However, measurement is destructive, and we actuallyÂ knowÂ what theÂ expected valueÂ of this ‘average’ energy will be, because we know theÂ probabilitiesÂ of finding the system in a particular base state. That probability is equal to the absoluteÂ square of that CiÂ coefficient above, so we can use the PiÂ = |Ci|2Â formula to write:

âŒ©EavâŒª = âˆ‘ PiÂ EiÂ over allÂ iÂ (i = 1, 2,… , N)

Note that this is a rather general formula. It’s got nothing to do with quantum mechanics: if AiÂ represents the possibleÂ values of some quantity A, and PiÂ is the probability of getting that value, then (the expected value of) the average A will also be equal to âŒ©AavâŒª = âˆ‘ Pi Ai. No rocket science here! ðŸ™‚ But let’s now apply our quantum-mechanical formulas to that âŒ©EavâŒª = âˆ‘ PiÂ EiÂ formula. [Ohâ€”and I apologize for using the same angle brackets âŒ© andÂ âŒª to denote an expected value hereâ€”sorry for that! But it’s what Feynman doesâ€”and other physicists! You see: they don’t really want you to understand stuff, and so they often use very confusing symbols.] Remembering that the absolute square of a complex number equals the product of that number and its complex conjugate, we can re-write the âŒ©EavâŒª = âˆ‘ PiÂ EiÂ formula as:

âŒ©EavâŒª = âˆ‘ PiÂ EiÂ = âˆ‘ |Ci|2Â EiÂ =Â âˆ‘Â Ci*CiÂ EiÂ =Â âˆ‘Â CiÂ *CiÂ EiÂ = âˆ‘ âŒ©Ïˆ|Î·iâŒªâŒ©Î·i|ÏˆâŒªEiÂ =Â âˆ‘ âŒ©Ïˆ|Î·iâŒªEiâŒ©Î·i|ÏˆâŒª over allÂ i

Now, you know that Dirac’sÂ bra-ketÂ notation allows numerous manipulations. For example, what we could do is take out that ‘common factor’ âŒ©Ïˆ|, and so we may re-write that monster above as:

Huh?Â Yes. Note the difference betweenÂ |ÏˆâŒª =Â âˆ‘Â |Î·iâŒª CiÂ Â = âˆ‘Â |Î·iâŒªâŒ©Î·i|ÏˆâŒª and |Ï†âŒª = âˆ‘ |Î·iâŒªEiâŒ©Î·i|ÏˆâŒª. As Feynman puts it: Ï† is just some ‘cooked-up‘ state which you get by taking each of the base states |Î·iâŒª in the amount EiâŒ©Î·i|ÏˆâŒª (as opposed to the âŒ©Î·i|ÏˆâŒª amounts we took for Ïˆ).

I know: you’re getting tired and you wonder why we need all this stuff. Just hang in there. We’re almost done. I just need to do a few more unpleasant things, one of which is to remind you that this business of the energy states beingÂ eigenstatesÂ (and the energy levels beingÂ eigenvalues) of our Hamiltonian matrix (see my post on N-state systems) comes with a number of interesting properties, including this one:

HÂ |Î·iâŒª = Ei|Î·iâŒª =Â |Î·iâŒªEi

Just think about what’s written here: on the left-hand side, we’re multiplying a matrix with a (base) state vector, and on the left-hand side we’re multiplying it with aÂ scalar. So ourÂ |Ï†âŒª = âˆ‘ |Î·iâŒªEiâŒ©Î·i|ÏˆâŒª sum now becomes:

|Ï†âŒª =Â âˆ‘Â HÂ |Î·iâŒªâŒ©Î·i|ÏˆâŒª over allÂ iÂ (i = 1, 2,… , N)

Now we can manipulate that expression some more so as to get the following:

|Ï†âŒª =Â H âˆ‘|Î·iâŒªâŒ©Î·i|ÏˆâŒª = H|ÏˆâŒª

Finally, we can re-combine this now with theÂ âŒ©EavâŒª = âŒ©Ïˆ|Ï†âŒª equation above, and so we get the fantastic result we wanted:

âŒ©EavâŒª = âŒ© Ïˆ | Ï† âŒª =Â âŒ© Ïˆ | HÂ |Â Ïˆ âŒª

Huh?Â Yes!Â To get the average energy, you operate onÂ |ÏˆâŒª with H, and then you multiply the result with âŒ©Ïˆ|. It’s a beautiful formula. On top of that, theÂ new formula for the average energy is not only pretty but also useful, because now we donâ€™t need to say anything about any particular set of base states. We donâ€™t even have to know all of the possible energy levels. When we have to calculate the average energy of some system, we only need to be able to describe the stateÂ of that systemÂ in terms of some set of base states, and we also need to know the Hamiltonian matrix for that set, of course. But if we know that, we can calculate its average energy.

You’ll say that’s not a big deal because… Well… If you know the Hamiltonian, you know everything, so… Well… Yes. You’re right: it’s less of a big deal than it seems. Having said that, the whole development above is very interesting because of something else: we can easilyÂ generalizeÂ it for other physical measurements. I call it the ‘average value’ operator idea, but you won’t find that term in any textbook. ðŸ™‚ Let me explain the idea.

#### The average value operator (A)

The development above illustrates how we can relate a physical observable, like the (average) energy (E), to a quantum-mechanical operator (H). Now, the development above can easily be generalized to any observable that would be proportional to the energy.Â It’s perfectly reasonable, for example, to assume theÂ angular momentum â€“ as measured in some direction, of course, which we usually refer to as the z-direction â€“ would be proportional to the energy, and so then it would be easy to define a new operatorÂ Lz, which we’d define as the operator of the z-component of the angular momentum L. [I know… That’s a bit of a long name but… Well… You get the idea.] So we can write:

âŒ©LzâŒªav = âŒ© Ïˆ |Â LzÂ |Â Ïˆ âŒª

In fact, further generalization yields the following grand result:

If a physical observable A is related to a suitable quantum-mechanical operator Ã‚, then the average value of A for the stateÂ |Â Ïˆ âŒª is given by:

âŒ©AâŒªav = âŒ© Ïˆ | Ã‚Â |Â Ïˆ âŒª =Â âŒ© Ïˆ | Ï† âŒª with | Ï† âŒª = Ã‚Â |Â Ïˆ âŒª

At this point, you may have second thoughts, and wonder: what state |Â Ïˆ âŒª? The answer is: it doesn’t matter. It can be any state, as long as we’re able to describe in terms of a chosen set of base states. ðŸ™‚

OK. So far, so good. The next step is to look at how this works for the continuity case.

#### The energy operator for wavefunctionsÂ (H)

We can start thinking about the continuousÂ equivalent of theÂ âŒ©EavâŒª = âŒ©Ïˆ|H|ÏˆâŒª expression by first expanding it. We write:

You know the continuous equivalent of a sum like this is an integral, i.e. an infiniteÂ sum. Now, because we’ve gotÂ twoÂ subscripts here (i and j), we get the following doubleÂ integral:

Now, I did take my time to walk you through Feynman’s derivation of the energy operator for theÂ discreteÂ case, i.e. the operator when we’re dealing withÂ matrix mechanics, but I think I can simplify my life here by just copying Feynman’s succinct development:

Done! Given a wavefunction Ïˆ(x), we get the average energy by doing that integral above. Now, the quantity in the braces of that integral can be written as that operator we introduced when we started this post:

So now we can write that integral much more elegantly. It becomes:

âŒ©EâŒªav =Â âˆ«Â Ïˆ*(x)Â HÂ Ïˆ(x) dx

You’ll say that doesn’t look likeÂ âŒ©EavâŒª =Â âŒ© Ïˆ | HÂ |Â Ïˆ âŒª! It does. Remember that âŒ© Ïˆ | = |Â Ïˆ âŒª*. ðŸ™‚ Done!

I should add one qualifier though:Â the formula above assumes our wavefunction has been normalized, so all probabilities add up to one. But that’s a minor thing. The only thing left to do now is to generalize to three dimensions. That’s easy enough. Our expression becomes a volumeÂ integral:

âŒ©EâŒªav =Â âˆ«Â Ïˆ*(r)Â HÂ Ïˆ(r) dV

Of course, dV stands for dVolumeÂ here, not for any potential energy, and, of course, once again we assume all probabilities over the volume add up to 1, so all is normalized.Â Done! ðŸ™‚

We’re almost done with this post. What’s left is theÂ positionÂ andÂ momentumÂ operator. You may think this is going to another lengthy development but… Well… It turns out the analysis is remarkably simple. Just stay with me a few more minutes and you’ll have earned your degree. ðŸ™‚

#### The position operator (x)

The thing we need to solve here is really easy. Look at the illustration below as representing the probability density of some particle being at x. Think about it: what’s the average position?

Well? What? The (expected value of the) average position is just this simple integral: âŒ©xâŒªav =Â âˆ« xÂ P(x) dx, over all the whole range of possible values for x. ðŸ™‚ That’s all. Of course, because P(x) =Â |Ïˆ(x)|2Â =Ïˆ*(x)Â·Ïˆ(x), this integral now becomes:

âŒ©xâŒªav =Â âˆ«Â Ïˆ*(x) xÂ Ïˆ(x) dx

That looks exactlyÂ the same asÂ âŒ©EâŒªav =Â âˆ«Â Ïˆ*(x)Â HÂ Ïˆ(x) dx, and so we can look at xÂ as an operator too!

Huh?Â Yes. It’s an extremely simple operator: itÂ just means “multiply by x“. ðŸ™‚

I know you’re shaking your head now: is it thatÂ easy? It is. Moreover, the ‘matrix-mechanical equivalent’ is equally simple but, as it’s getting late here, I’ll refer you to Feynman for that. ðŸ™‚

#### The momentum operator (px)

Now we want to calculate the average momentum of, say, some electron. What integral would you use for that?Â […] Well… What?Â […] It’s easy: it’s the same thing as for x. We can just substitute replaceÂ xÂ forÂ pÂ in thatÂ âŒ©xâŒªav =Â âˆ« xÂ P(x) dxÂ formula, so we get:

âŒ©pâŒªav =Â âˆ« pÂ P(p) dp, over all the whole range of possible values for p

Now, you might think the rest is equally simple, and… Well… It actually isÂ simple but there’s one additional thing in regard to the need to normalize stuff here. You’ll remember we defined aÂ momentumÂ wavefunction (see my post on the Uncertainty Principle), which we wrote as:

Ï†(p) = âŒ© mom p | Ïˆ âŒª

Now, in the mentioned post, we related this momentum wavefunction to the particle’sÂ Ïˆ(x) = âŒ©x|ÏˆâŒª wavefunctionâ€”which we should actually refer to as theÂ positionÂ wavefunction, but everyone just calls itÂ the particle’sÂ wavefunction, which is a bit of a misnomer, as you can see now: a wavefunction describes someÂ propertyÂ of the system, and so we can associate severalÂ wavefunctions with the same system, really! In any case, we noted the following there:

• The two probability density functions,Â Ï†(p) and Ïˆ(x), look pretty much the same, but theÂ half-widthÂ (or standard deviation) of one was inversely proportionalÂ to the half-width of the other. To be precise, we found that the constant of proportionality was equal to Ä§/2, and wrote that relation as follows:Â ÏƒpÂ = (Ä§/2)/Ïƒx.
• We also found that, when using a regular normal distribution function for Ïˆ(x), we’d have to normalize the probability density functionÂ by inserting aÂ (2Ï€Ïƒx2)âˆ’1/2Â in front of the exponential.

Now, it’s a bit of a complicated argument, but the upshot is that we cannot just write what we usually write, i.e. PiÂ = |Ci|2 or P(x) =Â |Ïˆ(x)|2. No. We need to put a normalization factor in front, which combines the two factors I mentioned above. To be precise, we have to write:

So… Well… OurÂ âŒ©pâŒªav =Â âˆ« pÂ P(p) dpÂ integral can now be written as:

So that integral is totally like what we found for âŒ©xâŒªavÂ and so… We could just leave it at that, and say we’ve solved the problem. In that sense, itÂ isÂ easy. However, having said that, it’s obvious we’d want someÂ solution that’s written in terms ofÂ Ïˆ(x), rather than in terms of Ï†(p), and that requires some more manipulation. I’ll refer you, once more, to Feynman for that, and I’ll just give you the result:

So… Well… I turns out that the momentum operator â€“ which I tentatively denoted as pxÂ above â€“ is notÂ so simple as our positionÂ operator (x). Still… It’s notÂ hugelyÂ complicated either, as we can write it as:

pxÂ â‰¡ (Ä§/i)Â·(âˆ‚/âˆ‚x)

Of course, theÂ puristsÂ amongst you will, once again, say that I should be more careful and put aÂ hatÂ wherever I’d need to put one so… Well… You’re right. I’ll wrap this all up by copying Feynman’s overview of the operators we just explained, and so heÂ doesÂ use the fancy symbols. ðŸ™‚

Well, folksâ€”that’s it! Off we go! You know all about quantum physics now! We just need to work ourselves through the exercisesÂ that come with Feynman’s Lectures, and then you’re ready to go and bag a degree in physics somewhere. So… Yes… That’s what I want to do now, so I’ll be silent for quite a while now. Have fun! ðŸ™‚

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