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

**Original post**:

In my previous post, I mentioned that it wasÂ *not* so obvious (both from a *physicalÂ *as well as from aÂ *mathematicalÂ *point of view) to write the wavefunction for electron orbitals – which we denoted as Ïˆ(* x*,

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

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

*other*functions in

*one*variable only.

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

*t*) as:

Ïˆ(* x*,

*t*) =

*e*

^{âˆ’iÂ·(E/Ä§)Â·t}Â·Ïˆ(

*)*

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

*t*) and Ïˆ(

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

**x***veryÂ*different:

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

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

*notÂ*necessarilyÂ real-valued. It

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

**= (**

*xÂ**x*,

*y*,

*z*) to

*= (*

**r**Â*r*, Î¸, Ï†), and that the function is also a function of the twoÂ quantum numbers

*Â 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

*(Î¸, Ï†) and F*

_{l,m}*(*

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

*(*

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

*(Î¸, Ï†) had that*

_{l,m}*e*

^{iÂ·mÂ·Ï†}Â factor in it. So… Yes. You’re right: the Y

*(Î¸, Ï†) function is real-valued if – and*

_{l,m}*onlyÂ*if –

*m*= 0, in which case

*e*

^{iÂ·mÂ·Ï†}Â = 1.Â Let me copy the table from Feynman’s treatment of the topic once again:The P

_{l}*(cosÎ¸) functions are the so-called (associated) Legendre polynomials, and the formula for these functions is rather horrible:Don’t worry about it too much: just note the P*

^{m}

_{l}*(*

^{m}*cos*Î¸)Â is aÂ

*real-valuedÂ*function. The point is the following:theÂ Ïˆ(

*,*

**x***t*) is a

*complex-valuedÂ*function because – andÂ

*onlyÂ*because – we multiply a

*real-valued*envelope function – which depends on

*positionÂ*only – with

*e*

^{âˆ’iÂ·(E/Ä§)Â·t}Â·

*e*

^{iÂ·mÂ·Ï†}Â =

*e*

^{âˆ’iÂ·[(E/Ä§)Â·tÂ âˆ’Â }

^{mÂ·Ï†]}.

[…]

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

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

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

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

The first symmetry plane – or symmetryÂ *line*, really (i.e. theÂ *z*-axis) – should not surprise us, because the azimuthal angle Ï† is conspicuously absent in the formula for our envelope function if, as we are doing in this article here, we merge theÂ *e*^{iÂ·mÂ·Ï†}Â factor with the *e*^{âˆ’iÂ·(E/Ä§)Â·t}, so it’s just part and parcel of what the author of the illustrations above refers to as the ‘complex phase’ of our wavefunction.Â OK. Clear enough – I hope. ðŸ™‚ But why is theÂ the *xy*-plane a symmetry plane too? We need to look at that monstrous formula for the P_{l}* ^{m}*(

*cos*Î¸) function here: just note the

*cos*Î¸ argument in it is being

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

*cos*Î¸ =

*sin*(Ï€/2Â âˆ’Â Î¸). So we can define someÂ

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

*notÂ*from the

*z*-axis but from the

*xy*-plane. So we write:

*cos*Î¸ =

*sin*(Ï€/2Â âˆ’Â Î¸) =

*sin*Î±. 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*Î¸ argument in the P

_{l}*(*

^{m}*cos*Î¸) function for

*sin*Î± =

*sin*(Ï€/2Â âˆ’Â Î¸). Now, if the

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

_{l}*(*

^{m}*sin*Î±) and P

_{l}*[*

^{m}*sin*(âˆ’Î±)]. Now, that’s not obvious, because

*sin*(âˆ’Î±) = âˆ’

*sin*Î± â‰ Â

*sin*Î±. 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: [âˆ’

*sin*Î±]

^{2}Â = [

*sin*Î±]

^{2Â }=Â

*sin*

^{2}Î±. […] 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}Â operator, and so you should check what happens with the minus sign there. ðŸ™‚

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

- AÂ
*definiteÂ*energy state of a hydrogen atom (or of an electron orbiting around some nucleus, I should say) appears to us as some beautifully shaped orbital – an*envelopeÂ*function in three dimensions, really – whichÂ has the*z*-axis – i.e. the vertical axis – as a symmetry*line*and the xy-plane as a symmetry*plane*. - The
*e*^{âˆ’iÂ·[(E/Ä§)Â·tÂ âˆ’Â }^{mÂ·Ï†]}Â factor gives us the oscillation*within*the envelope function. As such, it’s this factor that, somehow,Â captures the*energy*Â 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*^{iÂ·mÂ·Ï†}Â factor just gives us phase shift: just aÂ re-set of our zero point for measuring time, so to speak. Interesting, weird – but probably less relevant than the *e*^{âˆ’iÂ·[(E/Ä§)Â·t}Â factor, which gives us the two-dimensional oscillation that captures the energy of the state.

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

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

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

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

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

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

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