# A simple explanation of quantum-mechanical operators

I added an Annex to a paper that talks about all of the fancy stuff quantum physicists like to talk about, like scattering matrices and high-energy particle events. The Annex, however, is probably my simplest and shortest summary of the ordinariness of wavefunction math, including a quick overview of what quantum-mechanical operators actually are. It does not make use of state vector algebra or the usual high-brow talk about Gilbert spaces and what have you: you only need to know what a derivative is, and combine it with our realist interpretation of what the wavefunction actually represents.

I think I should do a paper on the language of physics. To show how (i) rotations (i, j, k), (ii) scalars (constants or just numerical values) and (iii) vectors (real vectors (e.g. position vectors) and pseudovectors (e.g. angular frequency or momentum)), and (iv) operators (derivatives of the wavefunction with respect to time and spatial directions) form ‘words’ (e.g. energy and momentum operators), and how these ‘words’ then combine into meaningful statements (e.g. Schroedinger’s equation).

All of physics can then be summed up in a half-page or so. All the rest is thermodynamics ðŸ™‚ JL

PS: You only get collapsing wavefunctions when adding uncertainty to the models (i.e. our own uncertainty about the energy and momentum). The ‘collapse’ of the wavefunction (let us be precise, the collapse of the (dissipating) wavepacket) thus corresponds to the ‘measurement’ operation. ðŸ™‚

PS2: Incidentally, the analysis also gives an even more intuitive explanation of Einstein’s mass-energy equivalence relation, which I summarize in a reply to one of the many ‘numerologist’ physicists on ResearchGate (copied below).

# Quantum-mechanical operators

I wrote a post on quantum-mechanical operatorsÂ some while ago but, when re-reading it now, I am not very happy about it, because it tries to cover too much ground in one go. In essence, I regret my attempt to constantly switch between theÂ matrixÂ representation of quantum physics – with the |Â stateÂ âŒª symbols –Â and theÂ wavefunctionÂ approach, so as to show how the operators work for both cases. But then that’s how Feynman approaches this.

However, let’s admit it: while Heisenberg’s matrixÂ approach is equivalent to SchrÃ¶dinger’sÂ wavefunction approach – and while it’s theÂ onlyÂ approach that works well for n-state systems – the wavefunction approach is more intuitive, because:

1. Most practical examples of quantum-mechanical systems (like the description of the electron orbitals of an atomic system) involve continuous coordinate spaces, so we have anÂ infiniteÂ number of states and, hence, we need to describe it using the wavefunction approach.
2. Most of us are much better-versed in using derivatives and integrals, as opposed to matrix operations.
3. A more intuitive statementÂ of the same argument above is the following: the idea of one state flowingÂ into another, rather than beingÂ transformedÂ through some matrix, is much more appealing. ðŸ™‚

So let’s stick to the wavefunction approach here. So, while you need to remember that there’s a ‘matrix equivalent’ for each of the equations we’re going to use in this post, we’re not going to talk about it.

## The operator idea

In classical physics – high schoolÂ physics, really – we would describe aÂ pointlike particle traveling in space by a functionÂ relating its positionÂ (x) to time (t): x = x(t). Its (instantaneous) velocity is, obviously, v(t) = dx/dt. Simple. Obvious. Let’s complicate matters now by saying that theÂ idea of a velocityÂ operatorÂ would sort of generalize the v(t) = dx/dtÂ velocity equation by making abstraction of the specificsÂ of the x = x(t) function.

Huh? Yes. We could defineÂ a velocity ‘operator’ as:

Now, you may think that’s a rather ridiculous way to describe what an operator does, but – in essence – it’s correct. We have some function – describing an elementary particle, or a system, or an aspect of the system – and then we have someÂ operator, which we apply to our function, to extractÂ the information from it that we want: its velocity, its momentum, its energy. Whatever.Â Hence, in quantum physics, we have an energyÂ operator, a positionÂ operator, aÂ momentumÂ operator, anÂ angularÂ momentum operator and… Well… I guess I listed the most important ones. ðŸ™‚

It’sÂ kindaÂ logical. Our velocity operator looks atÂ one particularÂ aspectÂ of whatever it is that’s going on: theÂ time rate of change of position. We do refer to that as theÂ velocity. Our quantum-mechanical operators do the same: they look at oneÂ aspectÂ of what’s being described by the wavefunction. [At this point, you may wonder what the other properties of our classical ‘system’ – i.e. other propertiesÂ than velocity – because we’re just looking at a pointlike particle here, but… Well… Think of electric charge and forces acting on it, so it accelerates and decelerates in all kinds of ways, and we have kinetic and potential energy and all that. Or momentum. So it’s just the same: the x = x(t) function may cover a lot of complexities, just like the wavefunction does!]

The Wikipedia article on the momentum operatorÂ is, for a change (I usually find Wikipedia quite abstruse on these matters), quite simple – and, therefore – quite enlightening here. It applies the following simple logic to the elementary wavefunctionÂ Ïˆ = eâˆ’iÂ·(Ï‰Â·t âˆ’ kâˆ™x), with the de BroglieÂ relations telling us thatÂ Ï‰ =Â E/Ä§ and k = p/Ä§:

Note we forget about the normalization coefficient a here. It doesn’t matter: we can always stuff it in later. The point to note is that we can sort of forget about Ïˆ (or abstract awayÂ from itâ€”as mathematicians and physicists would say) byÂ defining the momentum operator, which we’ll write as:

ItsÂ three-dimensional equivalent is calculated in very much the same way:

So this operator, when operating on a particular wavefunction, gives us the (expected)Â momentumÂ when we would actuallyÂ catchÂ our particle there,Â provided the momentum doesn’t vary in time. [Note that it may – and actually is likely toÂ –Â vary in space!]

So that’s the basic ideaÂ of an operator. However, the comparison goes further. Indeed, a superficial reading of what operators are all about gives you the impression we get all theseÂ observablesÂ (or propertiesÂ of the system) just by applying the operator to the (wave)function. That’s not the case. There is the randomness. The uncertainty. ActualÂ wavefunctions areÂ superpositionsÂ ofÂ several elementary waves with various coefficients representing their amplitudes.Â So we needÂ averages, or expected values: E[X] Even ourÂ velocityÂ operatorÂ âˆ‚/âˆ‚t – in the classical world – gives us an instantaneous velocity only. To get theÂ averageÂ velocity (in quantum mechanics, we’ll be interested in the theÂ averageÂ momentum, or theÂ averageÂ position, or the average energy – rather than the average velocity), we’re going to have the calculate theÂ totalÂ distance traveled. Now, that’s going to involve a line integral:

SÂ =Â âˆ«LÂ ds.

The principle is illustrated below.

You’ll say: this is kids stuff, and it is. Just note how we write the same integral in terms of the x and t coordinate, and using our new velocity operator:

Kids stuff. Yes. But it’s good to think about what itÂ representsÂ really. For example, the simplest quantum-mechanical operator is the positionÂ operator. It’s just xÂ for the x-coordinate,Â yÂ for theÂ y-coordinate, and z for the z-coordinate.Â To get theÂ averageÂ position of a stationary particle – represented by the wavefunctionÂ Ïˆ(r, t) – in three-dimensional space, we need to calculate the following volumeÂ integral:

Simple? Yes and no. The rÂ·|Ïˆ(r)|2Â integrand is obvious: we multiply each possibleÂ position (r) by its probability (or likelihood), which is equal to P(r) =Â |Ïˆ(r)|2. However, look at the assumptions: we already omitted the time variable. Hence, the particle we’re describing hereÂ mustÂ be stationary, indeed! So we’ll need to re-visit the whole subject allowing for averages to change with time. We’ll do that later. I just wanted to show you that those integralsÂ – even with very simple operators, like the position operator – can become very complicated. So you just need to make sure you know what you’re looking at.

## OneÂ wavefunctionâ€”or two? Or more?

There is another reason why, with the immeasurable benefit of hindsight, I now feel that my earlier post is confusing: I kept switching between theÂ positionÂ and theÂ momentumÂ wavefunction, which gives the impression we haveÂ differentÂ wavefunctions describingÂ different aspectsÂ of the same thing. That’s just not true. The position and momentum wavefunction describeÂ essentiallyÂ the same thing: we can go from one to the other, and back again, by a simple mathematical manipulation. So I should have stuck to descriptions in terms ofÂ Ïˆ(x, t), instead of switching back and forth between the Ïˆ(x, t) and Ï†(x, t) representations.

In any case, the damage is done, so let’s move forward. The key idea is that, when we know the wavefunction, we know everything. I tried to convey that by noting that the real and imaginary part of the wavefunction must, somehow, represent the total energy of the particle. The structural similarity between the mass-energy equivalence relation (i.e. Einstein’s formula:Â E = mÂ·c2) and the energy formulas for oscillators and spinning masses isÂ too obvious:

1. The energy of any oscillator is given by the E = mÂ·Ï‰02/2. We may want to liken the real and imaginary component of our wavefunction toÂ twoÂ oscillators and, hence, add them up. The E = mÂ·Ï‰02Â formula we get is then identical to the E = mÂ·c2Â formula.
2. The energy of a spinning mass is given by an equivalentÂ formula:Â E = IÂ·Ï‰2/2 (IÂ is the moment of inertia in this formula). The same 1/2 factor tells us our particle is, somehow, spinning in two dimensions at the same time (i.e. a ‘real’ as well as an ‘imaginary’ spaceâ€”but both are equally real, because amplitudes interfere), so we get the E = IÂ·Ï‰2Â formula.Â

Hence, the formulas tell us we should imagine an electron – or an electron orbital – as a very complicated two-dimensional standing wave. Now, when I writeÂ two-dimensional, I refer to theÂ realÂ andÂ imaginaryÂ component of our wavefunction, as illustrated below. What I am asking you, however, is to not only imagine these two components oscillating up and down, but also spinning about. Hence, if we think about energy as some oscillating mass – which is what the E = mÂ·c2Â formula tells us to do, we should remind ourselves we’re talking veryÂ complicated motions here: mass oscillates, swirls and spins, and it does so both in real as well as in imaginary space.Â Â

What I like about the illustration above is that it shows us – in a veryÂ obvious way – why the wavefunction depends on our reference frame. These oscillations do represent something inÂ absoluteÂ space, but how we measure it depends onÂ ourÂ orientation in that absolute space.Â But so I am writing this post to talk about operators, not about my grand theory about theÂ essenceÂ of mass and energy. So let’s talk about operators now. ðŸ™‚

In that post of mine, I showed how the position, momentum and energy operator would give us theÂ averageÂ position, momentum and energy of whatever it was that we were looking at, but I didn’t introduce theÂ angularÂ momentum operator. So let me do that now. However, I’ll first recapitulate what we’ve learnt so far in regard to operators.

## The energy, position and momentum operators

The equation below defines the energy operator, and also shows how we would apply it to the wavefunction:

To theÂ purists: sorry for not (always) using the hat symbol. [I explained why in that post of mine: it’s just too cumbersome.] The others ðŸ™‚ should note the following:

• EaverageÂ is also an expected value: EavÂ = E[E]
• The * symbol tells us to take theÂ complex conjugateÂ of the wavefunction.
• As for the integral, it’s an integral over some volume, so that’s what the d3r shows. Many authors use double or triple integral signs (âˆ«âˆ« or âˆ«âˆ«âˆ«) to show it’s a surface or a volume integral, but that makes things look veryÂ complicated, and so I don’t that.Â I could also have written the integral asÂ âˆ«Ïˆ(r)*Â·HÂ·Ïˆ(r) dV, but then I’d need to explain that theÂ dV stands for dVolume, not for any (differental) potential energy (V).
• WeÂ must normalize ourÂ wavefunction for these formulas to work, soÂ all probabilities over the volume add up to 1.

OK. That’s the energy operator. As you can see, it’s a pretty formidable beast, but then it just reflects SchrÃ¶dinger’s equation which, as I explained a couple of times already, we can interpret as an energy propagation mechanism, or an energy diffusion equation, so it is actuallyÂ notÂ that difficult to memorize the formula: if you’re able to remember SchrÃ¶dinger’s equation, then you’ll also have the operator. If not… Well… Then you won’t pass your undergrad physics exam. ðŸ™‚

I already mentioned that the position operator is a much simpler beast. That’s because it’s so intimately related to our interpretation of the wavefunction. It’s the oneÂ thing you know about quantum mechanics:Â the absolute square of the wavefunction gives us the probability density function. So, for one-dimensionalÂ space, the position operator is just:

The equivalent operator for three-dimensional space is equally simple:

Note how the operator, for the one- as well as for the three-dimensional case, gets rid of time as a variable. In fact, the idea itself of an average makes abstraction of the temporal aspect. Well… Here, at leastâ€”because we’re looking at some box in space, rather than some box in spacetime. We’ll re-visit that rather particular idea of an average, and allow for averages that change with time, in a short while.

Next, we introduced the momentumÂ operator in that post of mine. For one dimension, Feynman showsÂ this operator is given by the following formula:

Now that doesÂ notÂ look very simple. You might think that the âˆ‚/âˆ‚x operator reflects our velocity operator, but… Well… No: âˆ‚/âˆ‚t gives usÂ a time rate of change, while âˆ‚/âˆ‚x gives us theÂ spatialÂ variation. So it’sÂ not the same. Also, thatÂ Ä§/i factor is quite intriguing, isn’t it? We’ll come back to it in the next section of this post. Let me just give you the three-dimensional equivalent which, remembering that 1/i = âˆ’i, you’ll understand to be equal to the following vectorÂ operator:

Now it’s time to define the operator we wanted to talk about, i.e. theÂ angularÂ momentum operator.

## The angular momentum operator

The formula for the angular momentum operator is remarkably simple:

Why do I call this aÂ simpleÂ formula? Because it looks like the familiar formula of classical mechanics for the z-component of the classicalÂ angular momentumÂ L = rÂ Ã— p. I must assume you know how to calculate a vector cross product. If not, check one of my many posts on vector analysis. I must also assume you remember the L = rÂ Ã— pÂ formula. If not, the following animation might bring it all back. If that doesn’t help, check my post on gyroscopes. ðŸ™‚

Now, spin is a complicated phenomenon, and so, to simplify the analysis, we should think of orbitalÂ angular momentum only. This is a simplification, because electron spin is some complicated mix of intrinsic and orbital angular momentum. Hence, the angular momentum operator we’re introducing here is only the orbitalÂ angular momentum operator. However, let us not get bogged down in allÂ of the nitty-gritty and, hence, let’sÂ just go along with it for the time being.

I am somewhat hesitant to show you how we get that formula for our operator, but I’ll try to show you using anÂ intuitive approach, which uses onlyÂ bits and pieces of Feynman’s more detailed derivation. It will, hopefully, give you a bit of an idea of how theseÂ differential operatorsÂ work. Think about a rotation of our reference frame over an infinitesimally small angle – which we’ll denote asÂ Îµ – as illustrated below.

Now, the whole idea is that, because of that rotation of our reference frame, our wavefunction will look different. It’s nothing fundamental, but… Well… It’s just because we’re using a different coordinate system. Indeed, that’s where all these complicated transformation rulesÂ forÂ amplitudesÂ come in.Â  I’ve spoken about these at length when we were still discussingÂ n-state systems. In contrast, the transformation rules forÂ theÂ coordinatesÂ themselves are veryÂ simple:

Now, becauseÂ Îµ is an infinitesimally small angle, we may equate cos(Î¸) =Â cos(Îµ) to 1, and cos(Î¸) =Â sin(Îµ) to Îµ. Hence, x’ and y’ areÂ then written as x’ =Â xÂ + Îµy and y’ =Â yÂ âˆ’ Îµx, while z‘ remains z. Vice versa, we can also write the old coordinates in terms of the new ones:Â xÂ =Â x’Â âˆ’ Îµy, yÂ =Â y’Â +Â Îµx, and zÂ =Â z.Â That’s obvious. Now comes the difficult thing: you need to think about the two-dimensional equivalent of the simple illustration below.

If we have some function y = f(x), then we know that, for small Î”x, we have the following approximationÂ formula forÂ f(x +Â Î”x):Â f(x +Â Î”x)Â â‰ˆ f(x) + (dy/dx)Â·Î”x. It’s the formula you saw in high school: you would then take a limit (Î”xÂ â†’ 0), andÂ defineÂ dy/dxÂ as theÂ Î”y/Î”x ratioÂ forÂ Î”xÂ â†’ 0.Â You would this after re-writing theÂ f(x +Â Î”x)Â â‰ˆ f(x) + (dy/dx)Â·Î”xÂ formula as:

Î”y = Î”f = f(x +Â Î”x) âˆ’ f(x) â‰ˆ (dy/dx)Â·Î”x

Now you need to substitute f for Ïˆ, and Î”x for Îµ. There is only one complication here: Ïˆ is a function of twoÂ variables:Â x and y. In fact, it’s a function of three variables – x, y and z – but we keep zÂ constant. So think of moving from xÂ andÂ yÂ toÂ xÂ + ÎµyÂ = xÂ + Î”xÂ and to yÂ + Î”yÂ =Â yÂ âˆ’ Îµx. Hence, Î”xÂ = Îµy and Î”yÂ = âˆ’Îµx. It then makes sense to write Î”ÏˆÂ as:

If you agree with that, you’ll also agree we can write something like this:

Now that implies the following formula for Î”Ïˆ:

This looks great! You can see we get some sort of differential operatorÂ here, which is what we want. So the next step should be simple: we just letÂ Îµ go to zero and then we’re done, right? Well… No. In quantum mechanics, it’s always a bit more complicated. But it’s logicalÂ stuff. Think of the following:

1. We will want to re-write the infinitesimally small Îµ angle as a fraction of i, i.e. the imaginary unit.

Huh?Â Yes. This little iÂ represents many things. In this particular case, we want to look at it as a right angle. In fact, you know multiplication withÂ iÂ amounts to a rotation by 90 degrees. So we should replace Îµ by ÎµÂ·i. It’s like measuring Îµ in natural units. However, we’re not done.

2. We should also note that Nature measures angles clockwise, rather than counter-clockwise, as evidenced by the fact that the argument of our wavefunction rotates clockwise as time goes by. So our Îµ is, in fact, aÂ âˆ’Îµ. We will just bring the minus sign inside of the brackets to solve this issue.

Huh?Â Yes. Sorry. I told you this is a rather intuitive approach to getting what we want to get. ðŸ™‚

3. The third modification we’d want to make is to express ÎµÂ·iÂ as a multiple of Planck’s constant.

Huh?Â Yes. This is a veryÂ weird thing, but it should make senseâ€”intuitively: we’re talking angular momentum here, and its dimension is the same as that of physical action: NÂ·mÂ·s. Therefore, Planck’s quantum of action (Ä§ = h/2Ï€ â‰ˆ 1Ã—10âˆ’34Â JÂ·sÂ â‰ˆÂ 6.6Ã—10âˆ’16Â eVÂ·s) naturally appears as… Well… A natural unit, or a scalingÂ factor, I should say.

To make a long story short, we’ll want to re-write Îµ as âˆ’(i/Ä§)Â·Îµ. However, there is a thing called mathematical consistency, and so, if we want to do such substitutions and prepare for that limit situation (Îµ â†’ 0), we should re-write that Î”Ïˆ equation as follows:

So now – finally!Â – we do have the formula we wanted to find for our angular momentum operator:

The final substitution, which yields the formula we just gave you when commencing this section, just uses the formula for theÂ linearÂ momentum operator in the x– and y-direction respectively. We’re done! ðŸ™‚Â Finally!Â

Well… No. ðŸ™‚Â The question, of course, is the same as always: what does it all mean, really? That’s alwaysÂ a greatÂ question. ðŸ™‚ Unfortunately, the answer is rather boring: we can calculate theÂ average angular momentum in the z-direction,Â using a similar integral as the one we used to get the average energy, or the averageÂ linearÂ momentum in some direction. That’s basically it.

To compensate for thatÂ veryÂ boring answer, however, I will show youÂ something that is farÂ lessÂ boring. ðŸ™‚

## Quantum-mechanical weirdness

I’ll shameless copy from Feynman here. He notes that many classical equations get carried over into a quantum-mechanical form (I’ll copy some of his illustrations later). But then there are some that donâ€™t. As Feynman puts itâ€”rather humorously: “There had better be some that don’t come out right, because if everything did, then there would be nothing different about quantum mechanics. There would be no new physics.”Â He then looks at the following super-obvious equation in classical mechanics:

xÂ·pxÂ âˆ’Â pxÂ·x = 0

In fact, this equation is so super-obvious that it’s almost meaningless. Almost. It’s super-obviousÂ because multiplication isÂ commutativeÂ (for real as well for complex numbers). However, when we replace x and pxÂ by the position and momentumÂ operator, we get an entirely different result. You can verify the following yourself:

This is plain weird!Â What does it mean? I am not sure. Feynman’s take on it is nice but leaves us in the dark on it:

He adds: “If Planckâ€™s constant were zero, the classical and quantum results would be the same, and there would be no quantum mechanics to learn!” Hmm… What does it mean, really? Not sure. Let me make two remarks here:

1. We should not put any dot (Â·) between our operators, because they doÂ notÂ amount to multiplying one with another. We just apply operators successively. Hence, commutativity isÂ notÂ what we should expect.

2. Note that Feynman forgot to put the subscript in that quote. When doing the same calculations for the equivalent of the xÂ·pyÂ âˆ’Â pyÂ·x expression, we do get zero, as shown below:

These equations – zero or not – are referred to as ‘commutation rules’. [Again, I should not have used any dot between x and py, because there is no multiplication here. It’s just a separation mark.] Let me quote Feynman on it, so the matter is dealt with:

OK. So what do we conclude? What are we talking about?

## Conclusions

Some of the stuff above was really intriguing. For example, we found that the linear and angular momentum operators are differential operatorsÂ in the true sense of the word.Â The angular momentum operator shows us what happens to the wavefunction if weÂ rotateÂ our reference frame over an infinitesimally small angleÂ Îµ. That’s what’s captured by the formulas we’ve developed, as summarized below:

Likewise, the linear momentum operator captures what happens to the wavefunction for an infinitesimally smallÂ displacementÂ of the reference frame, as shown by the equivalent formulas below:

What’s the interpretation for theÂ positionÂ operator, and theÂ energyÂ operator? Here we are not so sure. The integrals aboveÂ make sense, but these integrals are used to calculate averages values, as opposed to instantaneous values. So… Well… There is not all that much I can say about the position and energy operator right now, except… Well… We now need to explore the question of howÂ averagesÂ could possibly change over time. Let’s do that now.

## Averages that change with time

I know: you are totally quantum-mechanicked out by now. So am I. But we’re almost there. In fact, this is Feynman’sÂ lastÂ LectureÂ on quantum mechanics and, hence, I think I should let the Master speak here. So just click on the link and read for yourself.Â It’s aÂ reallyÂ interesting chapter, as he shows us the equivalent of Newton’s Law in quantum mechanics, as well as the quantum-mechanical equivalent of other standard equations in classical mechanics. However, I need to warn you:Â Feynman keeps testing the limits of our intellectual absorption capacity by switching back and forth between matrix and wave mechanics. Interesting, but not easy. For example, you’ll need to remind yourself of the fact that the Hamiltonian matrix is equal to its own complex conjugate (or – because it’s a matrix – its own conjugate transpose.

Having said that, it’s all wonderful. The time rate of change of all those average values is denoted by using theÂ over-dotÂ notation. For example, theÂ time rate of change of the average position is denoted by:

Once you ‘get’ that new notation, you will quickly understand the derivations. They are not easy (what derivations are in quantum mechanics?), but we get very interesting results. Nice things to play with, or think aboutâ€”like this identity:

It takes a while, but you suddenly realize this is the equivalent of the classical dx/dt =Â v = p/m formula. ðŸ™‚

Another sweet result is the following one:

This isÂ the quantum-mechanical equivalent of Newton’s force law: F = mÂ·a. Huh? Yes.Â Think of it: the spatial derivative of the (potential) energy is the force. Now just think of the classical dp/dt = d(mÂ·v) = mÂ·dv/dt = mÂ·a formula. […] Can you see it now? Isn’t this justÂ Great Fun?

Note, however, that these formulas also showÂ the limits of our analysis so far, because they treat m as some constant. Hence, we’ll need to relativistically correct them. But that’s complicated, and so we’ll postpone that to another day.

[…]

Well… That’s it, folks!Â We’re really through! This was the last of the last of Feynman’s Lectures on Physics. So we’reÂ totallyÂ done now.Â Isn’t this great? What an adventure! I hope that, despite the enormous mental energy that’s required to digest all this stuff, you enjoyed it as much as I did. ðŸ™‚

Post scriptum 1: I just loveÂ Feynman but, frankly, I think he’s sometimes somewhat sloppy with terminology. In regard to what these operators really mean, we should make use of better terminology: anÂ averageÂ is something else than an expected value. Our momentum operator, for example, as such returns anÂ expectedÂ value –Â notÂ an average momentum. We need to deepen the analysis here somewhat, but I’ll also leave that for later.

Post scriptum 2:Â There is something really interesting about that iÂ·Ä§ or âˆ’(i/Ä§)Â scaling factor â€“ or whatever you want to call it â€“ appearing in our formulas. Remember the SchrÃ¶dinger equation can also be written as:

iÂ·Ä§Â·âˆ‚Ïˆ/âˆ‚t = âˆ’(1/2)Â·(Ä§2/m)âˆ‡2Ïˆ + VÂ·Ïˆ = HÏˆ

This is interesting in light of our interpretation of the SchrÃ¶dinger equation as an energy propagation mechanism. If we write SchrÃ¶dingerâ€™s equation like we write it here, then we have the energy on the right-hand side â€“ which is time-independent. How do we interpret the left-hand side now? Wellâ€¦ Itâ€™s kinda simple, but we just have the time rate of change of the real and imaginary part of the wavefunction here, and theÂ iÂ·Ä§ factor then becomes a sort of unitÂ in which we measure the time rate of change. Alternatively, you may think of â€˜splittingâ€™ Planckâ€™s constant in two: Planckâ€™s energy, and Planckâ€™s time unit, and then you bring the Planck energy unit to the other side, so weâ€™d express the energy in natural units. Likewise, the time rate of change of the components of our wavefunction would also be measured in natural time units if weâ€™d do that.

I know this is all veryÂ abstract but, frankly, itâ€™s crystal clear to me. This formula tells us that the energy of the particle thatâ€™s being described by the wavefunction is being carried by the oscillations of the wavefunction. In fact, the oscillationsÂ areÂ the energy. You can play with the mass factor, by moving it to the left-hand side too, or by using Einsteinâ€™s mass-energy equivalence relation. The interpretation remains consistent.

In fact, there is something really interesting here. You know that we usually separate out the spatial and temporal part of the wavefunction, so we write:Â Ïˆ(r, t) =Â Ïˆ(r)Â·eâˆ’iÂ·(E/Ä§)Â·t. In fact, it is quite common to refer to Ïˆ(r) â€“ rather than to Ïˆ(r, t) â€“ as the wavefunction, even if, personally, I find that quite confusing and misleading (see my page onSchrÃ¶dingerâ€™s equation). Now, we may want to think of what happens when weâ€™d apply the energy operator toÂ Ïˆ(r) rather than toÂ Ïˆ(r, t). We mayÂ think that weâ€™d get a time-independent value for the energy at that point in space, so energy is some function of position only, notÂ of time. Thatâ€™s an interesting thought, and we should explore it. For example, we then may think of energy as an average that changes with positionâ€”as opposed to the (average) position and momentum, which we like to think of as averages than change with time, as mentioned above. I will come back to this later â€“ but perhaps in another post or so. Not now. The only point I want to mention here is the following: you cannot use Ïˆ(r) in SchrÃ¶dingerâ€™s equation. Why? Wellâ€¦ SchrÃ¶dingerâ€™s equation is no longer valid when substituting Ïˆ forÂ Ïˆ(r), because the left-hand side is always zero, asÂ âˆ‚Ïˆ(r)/âˆ‚t is zero â€“ for anyÂ r.

There is another, related, point to this observation. If you think thatÂ SchrÃ¶dingerâ€™s equation implies that the operators on both sides of SchrÃ¶dingerâ€™s equationÂ must be equivalent (i.e. the same), youâ€™re wrong:

iÂ·Ä§Â·âˆ‚/âˆ‚t â‰  H = âˆ’(1/2)Â·(Ä§2/m)âˆ‡2Â + V

Itâ€™s a basic thing, really: SchrÃ¶dingerâ€™s equation is not valid for justÂ anyÂ function. Hence, it does notÂ work for Ïˆ(r). Only Ïˆ(r, t) makes it work, becauseâ€¦ Wellâ€¦ SchrÃ¶dingerâ€™s equation gaveÂ us Ïˆ(r, t)!

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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:

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:

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:

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:

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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