This post is about something I promised to write about *aeons *ago: how do we get those electron orbitals out of Schrödinger’s equation? So let me write it now – for the simplest of atoms: hydrogen. I’ll largely follow Richard Feynman’s *exposé* on it: this text just intends to walk you through it and provide some comments here and there.

Let me first remind you of what that famous Schrödinger’s equation actually represents. In its simplest form – i.e.* not *including any potential, so then it’s an equation that’s valid for free space only—no force fields!—it reduces to:

*i*·ħ∙∂ψ/∂t = –(1/2)∙(ħ^{2}/m_{eff})∙∇^{2}ψ

Note the enigmatic concept of the *efficient *mass in it (m_{eff}), as well as the rather awkward 1/2 factor, which we may get rid of by re-defining it. We then write: m_{eff}^{NEW} = 2∙m_{eff}^{OLD}, and Schrödinger’s equation then simplifies to:

- ∂ψ/∂t +
*i*∙(V/ħ)·ψ =*i*(ħ/m_{eff})·∇^{2}ψ - In free space (no potential): ∂ψ/∂t =
*i*∙(ħ/m_{eff})·∇^{2}ψ

In case you wonder where the minus sign went, I just brought the imaginary unit to the other side. Remember 1/*i *= −*i*. 🙂

Now, in my post on quantum-mechanical operators, I drew your attention to the fact that this equation is structurally similar to the heat diffusion equation – or to any diffusion equation, really. Indeed, assuming the heat per unit volume (q) is proportional to the temperature (T) – which is the case when expressing T in degrees Kelvin (K), so we can write q as q = k·T – we can write the heat diffusion equation as:

Moreover, I noted the similarity is not only structural. There is more to it: both equations model *energy* flows. *How exactly *is something I wrote about in my e-publication on this, so let me refer you to that. Let’s jot down the complete equation once more:

∂ψ/∂t + *i*∙(V/ħ)·ψ = *i*(ħ/m_{eff})·∇^{2}ψ

In fact, it is rather surprising that Feynman drops the *eff *subscript almost immediately, so he just writes:

Let me first remind you that ψ is a function of position in space and time, so we write: ψ = ψ(x, y, z, t) = ψ(**r**, t), with (x, y, z) = **r**. And m, on the other side of the equation, is what it always was: the *effective* electron mass. Now, we talked about the subtleties involved before, so let’s not bother about the definition of the *effective *electron mass, or wonder where that factor 1/2 comes from here.

What about V? V is the *potential* energy of the electron: it depends on the distance (*r*) from the proton. We write: V = −*e*^{2}/│**r**│ = −*e*^{2}/*r*. Why the minus sign? Because we say the potential energy is zero at large distances (see my post on potential energy). Back to Schrödinger’s equation.

On the left-hand side, we have ħ, and its dimension is J·s (or N·m·s, if you want). So we multiply that with a time derivative and we get J, the unit of energy. On the right-hand side, we have Planck’s constant *squared*, the mass factor in the denominator, and the Laplacian operator – i.e. ∇^{2 }= **∇**·**∇**, with **∇ = **(∂/∂x, ∂/∂y, ∂/∂z) – operating on the wavefunction.

Let’s start with the latter. The Laplacian works just the same as for our heat diffusion equation: it gives us a flux density, i.e. something expressed per square meter (1/m^{2}). The ħ^{2 }factor gives us J^{2}·s^{2}. The mass factor makes everything come out alright, if we use the mass-equivalence relation, which says it’s OK to express the mass in J/(m/s)^{2}. [The mass of an electron is usually expressed as being equal to 0.5109989461(31) MeV/*c*^{2}. That unit uses the E = m·*c*^{2 }mass-equivalence formula. As for the eV, you know we can convert that into *joule*, which is a rather large unit—which is why we use the *electronvolt *as a measure of energy.] To make a long story short, we’re OK: (J^{2}·s^{2})·[(m/s)^{2}/J]·(1/m^{2}) = J! Perfect. [As for the Vψ term, that’s *obviously *expressed in *joule *too.]

In short, Schrödinger’s equation expresses the energy conservation law too, and we may express it per square meter or per second or per cubic meter as well, if we’d wish: we can just multiply both sides by 1/m^{2} or 1/s or 1/m^{3 }or by whatever dimension you want. Again, if you want more detail on the Schrödinger equation as an *energy propagation *mechanism, read the mentioned e-publication. So let’s get back to our equation, which, taking into account our formula for V, now looks like this:

Feynman then injects one of these enigmatic phrases—enigmatic for *novices *like us, at least!

“We want to look for definite energy states, so we try to find solutions which have the form: ψ (**r**, t) = *e*^{−(i/ħ)·E·t}·ψ(**r**).”

At first, you may think he’s just trying to get rid of the relativistic correction in the argument of the wavefunction. Indeed, as I explain in that little booklet of mine, the –(p/ħ)·x term in the argument of the elementary wavefunction *a·**e*^{−i·θ} = *a·**e*^{−i·[(E/ħ)·t – (p/ħ)·x]} is there because the young *Comte Louis* *de Broglie*, back in 1924, when he wrote his groundbreaking *PhD* thesis, suggested the θ = ω∙t – **k**∙**x** = (E∙t – **p**∙**x**)/ħ formula for the argument of the wavefunction, as he knew that relativity theory had already established the invariance of the four-vector (dot) product p_{μ}x_{μ} = E∙t – **p**∙**x** = p_{μ}‘x_{μ}‘ = E’∙t’ – **p’**∙**x’**. [Note that Planck’s constant, as a physical constant, should obviously not depend on the reference frame either. Hence, if the E∙t – **p**∙**x** product is invariant, so is (E∙t – **p**∙**x**)/ħ.] So the θ = E∙t – **p**∙**x** and the θ = E_{0}∙t’ = E’·t’ are fully equivalent. Using *lingo*, we can say that the argument of the wavefunction is a *Lorentz scalar* and, therefore, invariant under a Lorentz boost. Sounds much better, doesn’t it? 🙂

But… Well. That’s not why Feynman says what he says. He just makes abstraction of *uncertainty *here, as he looks for states with a *definite *energy state, indeed. Nothing more, nothing less. Indeed, you should just note that we can re-write the *elementary a·**e*^{−i[(E/ħ)·t – (p/ħ)·x]} function as *e*^{−(i/ħ)·E·t}·*a·**e*^{i·(p/ħ)·x]}. So that’s what Feynman does here: he just eases the search for functional forms that satisfy Schrödinger’s equation. You should note the following:

- Writing the coefficient in front of the complex exponential as ψ(
**r**) =*a·**e*^{i·(p/ħ)·x]}does the trick we want it to do: we do*not*want that coefficient to depend on time: it should only depend on the size of our ‘box’ in space, as I explained in one of my posts. - Having said that, you should also note that the ψ in the ψ(
**r**, t) function and the ψ in the ψ(**r**) denote two different*beasts*: one is a function of*two*variables (**r**and t), while the other makes abstraction of the time factor and, hence, becomes a function of one variable only (**r**). I would have used another symbol for the ψ(**r**) function, but then the Master probably just wants to test your understanding. 🙂

In any case, the differential equation we need to solve now becomes:

** Huh? **How does

*that*work? Well… Just take the time derivative of

*e*

^{−(i/ħ)·E·t}·ψ(

**r**), multiply with the

*i*·ħ in front of that term in Schrödinger’s original equation and re-arrange the terms.

*[Just*

*do*it: ∂[

*e*

^{−(i/ħ)·E·t}·ψ(

**r**)]/∂t = −(

*i*/ħ)·E·

*e*

^{−(i/ħ)·E·t}·ψ(

**r**). Now multiply that with

*i*·ħ: the ħ factor cancels and the minus disappears because

*i*

^{2 }= −1.]

So now we need to solve that differential equation, i.e. we need to find functional forms for ψ – and please do note we’re talking ψ(**r**) here – *not *ψ(**r, t**)! – that satisfy the above equation. Interesting question: is our equation still Schrödinger’s equation? Well… It is and it isn’t. Any linear combination of the *definite energy *solutions we find will also solve Schrödinger’s equation, but so we limited the solution set here to those definite energy solutions *only*. Hence, it’s *not quite* the same equation. We removed the time dependency here – and in a rather interesting way, I’d say.

The next thing to do is to switch from Cartesian to *polar *coordinates. Why? Well… When you have a central-force problem – like this one (because of the potential) – it’s easier to solve them using polar coordinates. In fact, because we’ve got *three *dimensions here, we’re actually talking a *spherical *coordinate system. The illustration and formulas below show how spherical and Cartesian coordinates are related:

*x* = *r*·sinθ·cosφ; *y* = *r*·sinθ·sinφ; *z* = *r*·cosθ

As you know, θ (*theta*) is referred to as the *polar* angle, while φ (*phi*) is the *azimuthal* angle, and the coordinate transformation formulas can be easily derived. The rather simple differential equation above now becomes the following *monster*:

** Huh? **Yes, I am very sorry. That’s how it is. Feynman does this to help us. If you think you can get to the solutions by

*directly*solving the equation in Cartesian coordinates, please do let me know. 🙂 To tame the beast, we might imagine to first look for solutions that are

*spherically symmetric*, i.e. solutions that do

*not*depend on θ and φ. That means we could rotate the reference frame and none of the amplitudes would change. That means the ∂ψ/∂θ and ∂ψ/∂φ (partial) derivatives in our formula are equal to zero. These spherically symmetric states, or

*s*-states as they are referred to, are states with zero (orbital) angular momentum, but you may want to think about that statement before accepting it. 🙂 [It’s not that there’s

*no*angular momentum (on the contrary: there’s lots of it), but the

*total*angular momentum should obviously be zero, and so that’s what meant when these states are denoted as

*l*= 0 states.] So now we have to solve:

Now that looks somewhat less monstrous, but Feynman still fills two rather dense pages to show how this differential equation can be solved. It’s not only tedious but also complicated, so please check it yourself by clicking on the link. One of the steps is a switch in variables, or a *re-scaling*, I should say. Both *E* and *r* are now measured as follows:

The complicated-looking factors are just the Bohr radius (*r*_{B }= ħ^{2}/(m·e^{2}) ≈ 0.528 Å) and the Rydberg energy (*E*_{R }= m·e^{4}/2·ħ^{2} ≈ 13.6 eV). We calculated those long time ago using a rather heuristic model to describe an atom. In case you’d want to check the dimensions, note e^{2 }is a rather special animal. It’s got nothing to do with Euler’s number. Instead, e^{2 }is equal to k_{e}·q_{e}^{2}, and the k_{e} here is Coulomb’s constant: k_{e} = 1/(4πε_{0}). This allows to re-write the force between two electrons as a function of the distance: F = e^{2}/*r*^{2}. This, in turn, explains the rather weird dimension of e^{2}: [e^{2}] = N·e^{2 }= J·m. But I am digressing too much. The bottom line is: the various energy levels that fit the equation, i.e. the *allowable *energies, are fractions of the Rydberg energy, i.e. *E*_{R }=*m*·*e*^{4}/2·ħ^{2}. To be precise, the formula for the *n*^{th }energy level is:

*E*_{n }= − *E*_{R}/n^{2}.

The interesting thing is that the spherically symmetric solutions yield real-valued ψ(**r**) functions. The solutions for n = 1, 2, and 3 respectively, and their graph is given below.

As Feynman writes, all of the wave functions approach zero rapidly for large r (also, confusingly, denoted as ρ) after oscillating a few times, with the number of ‘bumps’ equal to n. Of course, you should note that you should put the time factor back in in order to correctly *interpret *these functions. Indeed, remember how we separated them when we wrote:

ψ(**r**, t) = *e*^{−i·(E/ħ)·t}·ψ(**r**)

We might say the ψ(**r**) function is sort of an *envelope function *for the whole wavefunction, but it’s not *quite *as straightforward as that. However, I am sure you’ll figure it out.

### States *with* an angular dependence

So far, so good. But what if those partial derivatives are *not *zero? Now the calculations become *really *complicated. Among other things, we need these *transformation matrices *for rotations, which we introduced a *very* long time ago. As mentioned above, I don’t have the intention to copy Feynman here, who needs another two or three dense pages to work out the logic. Let me just state the grand result:

- We’ve got a whole range of definite energy states, which correspond to orbitals that form an orthonormal basis for the
*actual*wavefunction of the electron. - The orbitals are characterized by three quantum numbers, denoted as
*l*,*n*and*m*respectively:- The
*l*is the quantum number of (total) angular momentum, and it’s equal to 0, 1, 2, 3, etcetera. [Of course, as usual, we’re measuring in units of ħ.] The*l*= 0 states are referred to as*s*-states, the*l*= 1 states are referred to as*p*-states, and the*l*= 2 states are*d*-states. They are followed by*f*,*g*,*h*, etcetera—for no particular good reason. [As Feynman notes: “The letters don’t mean anything now. They did once—they meant “sharp” lines, “principal” lines, “diffuse” lines and “fundamental” lines of the optical spectra of atoms. But those were in the days when people did not know where the lines came from. After f there were no special names, so we now just continue with g, h, and so on.] - The
*m*is referred to as the ‘magnetic’ quantum number, and it ranges from −*l*to +*l*. - The
*n*is the ‘principle’ quantum number, and it goes from*l*+ 1 to infinity (∞).

- The

How do these things actually *look *like? Let me insert two illustrations here: one from Feynman, and the other from Wikipedia.

The number in front just tracks the number of s-, p-, d-, etc. orbital. The shaded region shows where the amplitudes are large, and the plus and minus signs show the *relative *sign of the amplitude. [See my remark above on the fact that the ψ factor is real-valued, even if the wavefunction as a whole is complex-valued.] The Wikipedia image shows the same *density* plots but, as it was made some 50 years later, with some more color. 🙂

This is it, guys. Feynman takes it further by also developing the electron configurations for the next 35 elements in the periodic table but… Well… I am sure you’ll want to read the original here, rather than my summaries. 🙂

Congrats ! We now know all what we need to know. All that remains is lots of practical exercises, so you can be sure you master the material for your exam. 🙂

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