**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. 🙂

**Original post**:

In my previous post, I mentioned that it was *not* so obvious (both from a *physical *as well as from a *mathematical *point of view) to write the wavefunction for electron orbitals – which we denoted as ψ(* x*,

*t*), i.e. a function of

*two*variables (or four: one time coordinate and three space coordinates) – as the product of two

*other*functions in

*one*variable only.

[…] OK. The above sentence is difficult to read. Let me write in math. 🙂 It is *not *so obvious to write ψ(* x*,

*t*) as:

ψ(* x*,

*t*) =

*e*

^{−i·(E/ħ)·t}·ψ(

*)*

**x**As I mentioned before, the physicists’ use of the same symbol (ψ, *psi*) for both the ψ(* x*,

*t*) and ψ(

*) function is quite confusing – because the two functions are*

**x***very*different:

- ψ(
,**x***t*) is a complex-valued function of*two*(real)and*x**t*. Or four, I should say, because= (**x***x*,*y*,*z*) – but it’s probably easier to think ofas one*x**vector*variable – a*vector-valued argument*, so to speak. And then*t*is, of course, just a*scalar*variable. So… Well… A function of*two*variables: the position in space (), and time (*x**t*). - In contrast, ψ(
) is a**x***real-valued*function of*one*(vector) variable only:, so that’s the position in space only.*x*

Now you should cry foul, of course: ψ(* x*) is

*not*necessarily real-valued. It

*may*be complex-valued. You’re right. You know the formula:Note the derivation of this formula involved a switch from Cartesian to polar coordinates here, so from

**= (**

*x**x*,

*y*,

*z*) to

*= (*

**r***r*, θ, φ), and that the function is also a function of the two quantum numbers

*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 Y

*(θ, φ) and F*

_{l,m}*(*

_{l,m}*r*) respectively. F

*(*

_{l,m}*r*) was a real-valued function alright, but the Y

*(θ, φ) had that*

_{l,m}*e*

^{i·m·φ}factor in it. So… Yes. You’re right: the Y

*(θ, φ) function is real-valued if – and*

_{l,m}*only*if –

*m*= 0, in which case

*e*

^{i·m·φ}= 1. Let me copy the table from Feynman’s treatment of the topic once again:The P

_{l}*(cosθ) 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 P*

^{m}

_{l}*(*

^{m}*cos*θ) is a

*real-valued*function. The point is the following:the ψ(

*,*

**x***t*) is a

*complex-valued*function because – and

*only*because – we multiply a

*real-valued*envelope function – which depends on

*position*only – with

*e*

^{−i·(E/ħ)·t}·

*e*

^{i·m·φ}=

*e*

^{−i·[(E/ħ)·t − }

^{m·φ]}.

[…]

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

- As mentioned in my previous post, the
*e*^{i·m·φ}factor just gives us phase shift: just a re-set of our zero point for measuring time, so to speak, and the whole*e*^{−i·[(E/ħ)·t − }^{m·φ]}factor just disappears when we’re calculating probabilities. - The envelope function gives us the basic amplitude – in the
*classical*sense of the word: the*maximum*displacement from the zero value. And so it’s that*e*^{−i·[(E/ħ)·t − }^{m·φ]}that ensures the whole expression somehow captures the*energy*of the oscillation.

Let’s first look at the envelope function again. Let me copy the illustration for *n* = 5 and *l *= 2 from a *Wikimedia Commons* article. Note the symmetry planes:

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

The first symmetry plane – or symmetry *line*, really (i.e. the *z*-axis) – should not surprise us, because the azimuthal angle φ is conspicuously absent in the formula for our envelope function if, as we are doing in this article here, we merge the *e*^{i·m·φ} factor with the *e*^{−i·(E/ħ)·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. OK. Clear enough – I hope. 🙂 But why is the the *xy*-plane a symmetry plane too? We need to look at that monstrous formula for the P_{l}* ^{m}*(

*cos*θ) function here: just note the

*cos*θ argument in it is being

*squared*before it’s used in all of the other manipulation. Now, we know that

*cos*θ =

*sin*(π/2 − θ). So we can define some

*new*angle – let’s just call it α – which is measured in the way we’re used to measuring angle, which is

*not*from the

*z*-axis but from the

*xy*-plane. So we write:

*cos*θ =

*sin*(π/2 − θ) =

*sin*α. 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

*cos*θ argument in the P

_{l}*(*

^{m}*cos*θ) function for

*sin*α =

*sin*(π/2 − θ). Now, if the

*xy*-plane is a symmetry plane, then we must find the same value for P

_{l}*(*

^{m}*sin*α) and P

_{l}*[*

^{m}*sin*(−α)]. Now, that’s not obvious, because

*sin*(−α) = −

*sin*α ≠

*sin*α. However, because the argument in that P

_{l}*(*

^{m}*x*) function is being squared before any other operation (like subtracting 1 and exponentiating the result), it is OK: [−

*sin*α]

^{2}= [

*sin*α]

^{2 }=

*sin*

^{2}α. […] 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

*d*

^{l+m}/

*dx*

^{l+m}operator, and so you should check what happens with the minus sign there. 🙂

[…] 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:

- A
*definite*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*envelope*function in three dimensions, really – which has the*z*-axis – i.e. the vertical axis – as a symmetry*line*and the xy-plane as a symmetry*plane*. - The
*e*^{−i·[(E/ħ)·t − }^{m·φ]}factor gives us the oscillation*within*the envelope function. As such, it’s this factor that, somehow, captures the*energy*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 *e*^{i·m·φ} factor just gives us phase shift: just a re-set of our zero point for measuring time, so to speak. Interesting, weird – but probably less relevant than the *e*^{−i·[(E/ħ)·t} 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 Blue *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 *force *on a unit charge, the real and imaginary part of our wavefunction must also represent some rapidly changing *force *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 *mass *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 *of mass, but then I am not sure how to define that unit right now (it’s probably *not *the kilogram!).

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

Let me just add one more note on the *e*^{i·m·φ} factor. It sort of defines the *geometry *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 *n *= 4 and *l *= 3. Look at the red hues in particular – or the blue – whatever: focus on one color only, and see how how – for *m* *= *±1, we’ve got one appearance of that color only. For *m* *= *±1, the same color appears at two ends of the ‘tubes’ – or *tori *(plural of *torus*), I should say – just to sound more professional. 🙂 For *m* *= *±2, the torus consists of *three* parts – or, in mathematical terms, we’d say the order of its *rotational symmetry* is equal to 3. Check that Wikimedia Commons article for higher values of *n *and *l*: the shapes become very convoluted, but the observation holds. 🙂

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

**Post scriptum**: You should do some thinking on whether or not these *m *= ±1, ±2,…, ±*l *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 not a *real *difference, is it? 🙂 As with other stuff, I’ll let you think about this for yourself.

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