I’ve talked about electron orbitals in a couple of posts already – including a fairly recent one, which is why I put the (II) after the title. However, I just wanted to tie up some loose ends here – and do some more thinking about the concept of a *definite* energy state. What is it really? We know the wavefunction for a definite energy state can always be written as:

ψ(* x*,

*t*) =

*e*

^{−i·(E/ħ)·t}·ψ(

*)*

**x**Well… In fact, we should probably formally *prove *that but… Well… Let us just *explore *this formula in a more intuitive way – for the time being, that is – using those electron orbitals we’ve derived.

First, let me note that ψ(* x*,

*t*) and ψ(

*) are*

**x***very*different functions and, therefore, the choice of the same

*symbol*for both (the Greek

*psi*) is – in my humble opinion – not very fortunate, but then… Well… It is

*the*choice of physicists – as copied in textbooks all over – and so we’ll just have to live with it. Of course, we can appreciate why they choose to use the same symbol – ψ(

*) is like a time-*

**x***in*dependent wavefunction now, so that’s nice – but… Well… You should note that it is

*not*so obvious to write some function as the product of two other functions. To be complete, I’ll be a bit more explicit here: if some function in two variables – say F(

*x*,

*y*) – can be written as the product of two functions in one variable – say f(

*x*) and g(

*y*), so we can write F as F(

*x*,

*y*) = f(

*x*)·g(y) – then we say F is a

*separable*function. For a full overview of what that means, click on this link. And note mathematicians

*do*choose a different symbol for the functions F and g. It would probably be interesting to explore what the conditions for separability actually imply in terms of

*properties*of… Well… The wavefunction and its argument, i.e. the space and time variables. But… Well… That’s stuff for another post. 🙂

Secondly, note that the *momentum *variable (** p**) – i.e. the

**p**

*in our*

*elementary*wavefunction

*a*·

*e*

^{i·(p·x−E·t)/ħ}has sort of vanished: ψ(

*) is a function of the position*

**x***only*. Now, you may think it should be

*somewhere*there – that, perhaps, we can write something like ψ(

*) = ψ[*

**x***),*

**x***(*

**p****)]. But… No. The momentum variable has effectively vanished. Look at Feynman’s solutions for the electron orbitals of a hydrogen atom:The Y**

*x**(θ, φ) and F*

_{l,m}*(ρ) functions here are functions of the (polar) coordinates ρ, θ, φ*

_{n,l}*only*. So that’s the

*position*only (these coordinates are

*polar*or

*spherical*coordinates, so ρ is the radial distance, θ is the polar angle, and φ is the azimuthal angle). There’s no idea whatsoever of any momentum in one or the other spatial

*direction*here. I find that rather remarkable. Let’s see how it all works with a simple example.

The functions below are the Y* _{l,m}*(θ, φ) for

*l*= 1. Note the symmetry: if we swap θ and φ for -θ and -φ respectively, we get the other function: 2

^{-1/2}·

*sin*(-θ)·

*e*

^{–i(-φ)}= -2

^{-1/2}·

*sin*θ·

*e*

^{iφ}.

To get the probabilities, we need to take the absolute square of the whole thing, including *e*^{−i·(E/ħ)}, but we know |*e*^{i·δ}|^{2} = 1 for any value of δ. Why? Because the absolute square of *any *complex number is the product of the number with its complex *conjugate*, so |*e*^{i·δ}|^{2} = *e*^{i·δ}·*e*^{–i·δ }= *e*^{i·0 }= 1. So we only have to look at the absolute square of the Y* _{l,m}*(θ, φ) and F

*(ρ) functions here. The F*

_{n,l}*(ρ) function is a*

_{n,l}*real*-valued function, so its absolute square is just what it is: some real number (I gave you the formula for the

*a*

_{k}coefficients in my post on it, and you shouldn’t worry about them: they’re real too). In contrast, the Y

*(θ, φ) functions are complex-valued – most of them are, at least. Unsurprisingly, we find the probabilities are also symmetric:*

_{l,m}P = |-2^{-1/2}·*sin*θ·*e*^{iφ}|^{2} = (-2^{-1/2}·*sin*θ·*e*^{iφ})·(-2^{-1/2}·*sin*θ·*e*^{–iφ})

= (2^{-1/2}·*sin*θ·*e*^{–iφ})·(2^{-1/2}·*sin*θ·*e*^{iφ}) = |2^{-1/2}·*sin*θ·*e*^{–iφ}|^{2} = (1/2)·*sin*^{2}θ

Of course, for *m *= 0, the probability is just *cos*^{2}θ. The graphs below are the polar graphs for the *cos*^{2}θ and (1/2)·*sin*^{2}θ functions respectively.

These polar graphs are not so easy to interpret, so let me say a few words about them. The points that are plotted combine (a) some *radial* *distance* from the center – which I wrote as P because this distance *is*, effectively,* *a probability – with (b) the polar angle θ (so that’s one of the three coordinates). To be precise, the plot gives us, for a given ρ, all of the (θ, P) combinations. It works as follows. To calculate the probability for some ρ and θ (note that φ can be any angle), we must take the absolute square of that ψ* _{n,l,m,}* = Y

*(θ, φ)·F*

_{l,m}*(ρ) product. Hence, we must calculate |Y*

_{n,l}*(θ, φ)·F*

_{l,m}*(ρ)|*

_{n,l}^{2}= |F

*(ρ)|*

_{n,l}^{2}·

*cos*

^{2}θ for

*m*= 0, and (1/2)·|F

*(ρ)|*

_{n,l}^{2}·

*sin*

^{2}θ for

*m*= ±1. Hence, the value of ρ determines the value of F

*(ρ), and that F*

_{n,l}*(ρ) value then determines the shape of the polar graph. The three graphs below – P =*

_{n,l}*cos*

^{2}θ, P = (1/2)·

*cos*

^{2}θ and P = (1/4)·

*cos*

^{2}θ – illustrate the idea. Note that we’re measuring θ

*from the z-axis*here, as we should. So that gives us the right orientation of this volume, as opposed to the other polar graphs above, which measured θ from the x-axis. So… Well… We’re getting there, aren’t we? 🙂

Now you’ll have two or three – or even more – *obvious* questions. The first one is: where is the third lobe? That’s a good question. Most illustrations will represent the p-orbitals as follows:Three lobes. Well… Frankly, I am not quite sure here, but the equations speak for themselves: the probabilities only depend on ρ and θ. Hence, the azimuthal angle φ can be anything. So you just need to rotate those P = (1/2)·*sin*^{2}θ and P = *cos*^{2}θ curves about the the *z*-axis. In case you wonder how to do that, the illustration below may inspire you.The second obvious question is about the size of those lobes. That 1/2 factor must surely matter, right? Well… We still have that F* _{n,l}*(ρ) factor, of course, but you’re right: that factor does

*not*depend on the value for

*m*: it’s the same for

*m*= 0 or ± 1. So… Well… Those representations above – with the three lobes, all of the same volume – may

*not*be accurate. I found an interesting site – Atom in a Box – with an

*app*that visualizes the atomic orbitals in a fun and exciting way. Unfortunately, it’s for Mac and iPhone only – but this YouTube video shows how it works. I encourage you to explore it. In fact, I need to explore it – but what I’ve seen on that YouTube video (I don’t have a Mac nor an iPhone) suggests the three-lobe illustrations may effectively be wrong: there’s some asymmetry here – which we’d expect, because those p-orbitals are actually supposed to be asymmetric! In fact, the most accurate pictures may well be the ones below. I took them from

*Wikimedia Commons*. The author explains the use of the color codes as follows: “The depicted rigid body is where the probability density exceeds a certain value. The color shows the complex phase of the wavefunction, where blue means real positive, red means imaginary positive, yellow means real negative and green means imaginary negative.” I must assume he refers to the sign of

*a*and

*b*when writing a complex number as

*a*+

*i·b*

The third obvious question is related to the one above: we should get some cloud, right? Not some rigid body or some surface. Well… I think you can answer that question yourself now, based on what the author of the illustration above wrote: if we *change *the cut-off value for the probability, then we’ll give a different shape. So you can play with that and, yes, it’s some cloud, and that’s what the mentioned *app* visualizes. 🙂

The fourth question is the most obvious of all. It’s the question I started this post with: **what are those definite energy states?** We have

*uncertainty*, right? So how does

*that*play out? Now

*that*is a question I’ll try to tackle in my next post.

**🙂**

*Stay tuned !***Post scriptum**: Let me add a few remarks here so as to – hopefully – contribute to an even better interpretation of what’s going on here. As mentioned, the key to understanding is, obviously, the following basic functional form:

ψ(* r*,

*t*) =

*e*

^{−i·(E/ħ)·t}·ψ(

*)*

**r**Wikipedia refers to the *e*^{−i·(E/ħ)·t }factor as a time-dependent *phase factor* which, as you can see, we can separate out because we are looking at a *definite *energy state here. Note the *minus* sign in the exponent – which reminds us of the minus sign in the exponent of the *elementary * wavefunction, which we wrote as:

*a·e*^{−i·θ} = *a·e*^{−i·[(E/ħ)·t − (p/ħ)∙x]} = *a·e*^{i·[(p/ħ)∙x − (E/ħ)·t]} = *a·e*^{−i·(E/ħ)·t}*·e*^{i·(p/ħ)∙x}

We know this *elementary *wavefunction is problematic in terms of interpretation because its absolute square gives us some *constant *probability P(* x*, t) = |

*a·e*

^{−i·[(E/ħ)·t − (p/ħ)∙x]}|

^{2 }=

*a*

^{2}. In other words, at any point in time, our electron is equally likely to be

*anywhere*in space. That is

*not*consistent with the idea of our electron being

*somewhere*at some point in time.

The other question is: what reference frame do we use to measure E and **p**? Indeed, the value of E and **p** = (p* _{x}*, p

*, p*

_{y}*) depends on our reference frame: from the electron’s own point of view, it has no momentum whatsoever:*

_{z}**p**=

**0**. Fortunately, we do have a point of reference here: the nucleus of our hydrogen atom. And our own position, of course, because you should note, indeed, that

*both*the subject

*and*the object of the observation are necessary to define the Cartesian

*=*

**x***x*,

*y*,

*z*– or, more relevant in this context – the polar

*= ρ, θ, φ coordinates.*

**r**This, then, defines some finite or infinite *box in space *in which the (linear) momentum (**p**) of our electron vanishes, and then we just need to solve Schrödinger’s *diffusion equation *to find the solutions for ψ(* r*). These solutions are more conveniently written in terms of the radial distance ρ, the polar angle θ, and the azimuthal angle φ:

The functions below are the Y* _{l,m}*(θ, φ) functions for

*l*= 1.

The interesting thing about these Y* _{l,m}*(θ, φ) functions is the

*e*

^{i·φ}and/or

*e*

^{−i·φ}factor. Indeed, note the following:

- Because the
*sin*θ and*cos*θ factors are*real**-valued*, they only define some*envelope*for the ψ() function.**r** - In contrast, the
*e*^{i·φ}and/or*e*^{−i·φ}factor define some*phase shift*.

Let’s have a look at the *physicality *of the situation, which is depicted below.

The nucleus of our hydrogen atom is at the center. The polar angle is measured from the *z*-axis, and we know we only have an amplitude there for *m *= 0, so let’s look at what that *cos*θ factor does. If θ = 0°, the amplitude is just what it is, but when θ > 0°, then |*cos*θ| < 1 and, therefore, the probability P = |F* _{n,l}*(ρ)|

^{2}·

*cos*

^{2}θ will diminish. Hence, for the same radial distance (ρ), we are

*less*likely to find the electron at some angle θ > 0° than on the

*z*-axis itself. Now

*that*makes sense, obviously. You can work out the argument for

*m*= ± 1 yourself, I hope. [The axis of symmetry will be different, obviously!] In contrast, the

*e*

^{i·φ}and/or

*e*

^{−i·φ}factor work very differently. These just give us a

*phase shift*, as illustrated below. A re-set of our zero point for measuring time, so to speak, and the

*e*

^{i·φ}and/or

*e*

^{−i·φ}factor effectively disappears when we’re calculating probabilities, which is consistent with the fact that this angle clearly doesn’t influence the

*magnitude*of the amplitude fluctuations.So… Well… That’s it, really. I hope you enjoyed this ! 🙂

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