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(Following this answer on Chemistry.SE)

Let's define "dwell"/"reside" as the area in which the probability to find an electron is more than 90%, for example.

Calculated orbitals comprise a basis set, but they do not represent the actual "dwelling" of the electrons.

Are there actual "dwelling orbitals" which are the result of linear combination of these calculated orbitals? In other words, for a given set of calculated orbitals and coefficients, does the resulting well-define shape represent an actual "charge density blob" in which 2 electrons (out of all the electrons in the system) reside?

See for example in this paper ("Pentacene imaged with STM and NC-AFM"), in which it looks like the calculated orbitals are in fact where electrons reside (most of the time).

Pentacene imaged with STM and NC-AFM

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  • $\begingroup$ Why 2? And what, exactly, do you mean by reside. The actual state of any single electron is a linear superposition of orbitals, that's exactly what the answer over there says. $\endgroup$ – ACuriousMind Nov 14 '15 at 15:45
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    $\begingroup$ @ACuriousMind 2 because of Pauli's exclusion. By reside I mean probability of more than an arbitrary value. $\endgroup$ – Sparkler Nov 14 '15 at 15:48
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    $\begingroup$ @ACuriousMind you say 'linear superposition', which means you know the coefficients of the linear combination that makes up this superposition. If you do know these coefficients, it means you can draw the "actual" superpositioned-orbital, in which you will most probably find the electron. So the question is if it actually works like that. $\endgroup$ – Sparkler Nov 14 '15 at 15:56
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    $\begingroup$ @Sparkler It seems to me you are asking "What linear combination of atomic orbitals actually describe the typical states of electrons in atoms?" If so, you should make your question more clear, because the answers you are receiving are just clarifying things you may already know. $\endgroup$ – Todd R Nov 14 '15 at 16:34
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    $\begingroup$ @Sparkler Well don't you see then? I asked if it was part of your question, you says it is, but the person answering you 50 different ways just above said "No, I don't think so". He doesn't understand your question and so will keep answering whatever question he thinks you are asking. You need to clarify your question. $\endgroup$ – Todd R Nov 14 '15 at 16:55
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I add my comment as a follow up to the question in chemistry SE.

The problem is due the fact that typical explanation of electrons and orbitals depict them as "boxes" or "orbits" in which electrons reside. This is due to historical reasons, but it compromises the understanding of this topic.

Electrons don't "reside" in orbitals. Orbitals describe electrons in space. Saying that they reside in orbitals would be like saying that you reside in a box labeled "human". The reality is that you are a human.

An orbital is nothing but one of many convenient 3d (let's not involve spin for simplicity) functions that has nice properties. They all belong to a Hilbert space, which is simply a bunch of functions. These functions share interesting properties, such as that they are orthogonal when centered on the same point in space, that they are not null, and that they solve one specific problem exactly: the problem of a single positive charge interacting with a single negative charge.

That's it. They are nothing more than the x and the y axis on a plane. You choose them orthogonal because it's practical, and you use them to decompose whatever vector you might handle in a x component and a y component. But the vector you handle is not "residing" on the x axis or the y axis. If you had a vector lying down on the x axis (or close to), you would say "it is pointing in the positive x direction" or "it's well described by the x axis" because it has a strong x component and barely no y component, exactly as you would say a given electronic distribution is well described by an s orbital because it has mostly s orbital component. This is what is meant when chemists say "it's in an s orbital", but it's deceiving and for all purposes incorrect.

Are there actual "dwelling orbitals" which are the result of linear combination of these calculated orbitals?

There are charge (and probability) distributions of a single electron, yes. You can compute it if you want, and it might make sense or not make sense, depending on the context, and this charge distribution may or may fit a physical shape that you expect or not, depending on the molecule shape, external factors (fields, light),

In other words, for a given set of calculated orbitals and coefficients, does the resulting well-define shape represent an actual "charge density blob" in which 2 electrons (out of all the electrons in the system) reside?

Only the total sum (the total density). If you want the one-electron density, you might have some degrees of freedom in how you extract it, meaning that there are multiple ways (within limits) of decomposing that total density into one-electron densities, and they are equivalent, exactly as 5 can be decomposed in 4+1 or 2+3.


Edit: in the article you added, you are seeing an image obtained from tunneling effect. What happens there is that you basically "tune" the tip of the AFM to "resonate" so that electrons can jump from/to the "orbital" with the same energy.

Note the massive use of quotes. What happens there is that you are probing electron density at a given energy level and in every specific point in space. Electron density is what you measure, and yes, electron density is, on strictly speaking inaccurate but correct practical terms, determined by the orbital. However, once again, it's not so simple. The reality is that you can use the orbital (what you obtain from solving the eigenvalue problem) to express that electron density, and that expression is pretty much spot on. However, depending on the methodology you use, you can transform the orbitals into an equivalent set that has a specific property. For example, you might combine them to provide an absolutely equivalent set that is instead localized, meaning that you won't get orbitals similar to that shape. You will instead get orbitals that describe a C-C bond, for example, or a C-H bond.

So, your point is somehow valid. There are physical electronic spatial distributions for a given energy that resemble the orbitals that you obtain from a computation, but what you are probing is a physical entity at a specific energy. Orbitals are a mathematical construct that, within some constraints, match the physical data you obtain from the AFM imaging, but it is due to the nature of the experiment, and the nature of the method.

The difference seems to be academic and pretty much pointless. It is indeed a subtle difference, but it's far from pointless.

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  • $\begingroup$ Please see the photo I added. Your input would be appreciated. $\endgroup$ – Sparkler Nov 15 '15 at 2:47
  • $\begingroup$ @sparkler Ok, the photo is very interesting. I will add an additional comment to this answer. $\endgroup$ – Stefano Borini Nov 15 '15 at 19:20
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I think the answer of Stefano Borini, in its well-meaning zeal to correctly dispel some of the misconceptions surrounding the concept of electron orbitals especially in Chemistry, is itself quite reductionist and somewhat misleading.

The importance of (valence) atomic and molecular orbitals in Quantum Chemistry (bond theory, essentially) is that they show the electron probability density of the atoms/molecules in space. Take this wonderful (but also deceptive) rendition of the three $\text{2p}$ hydrogen orbitals:

2p hydrogen orbitals

(Source.)

The shiny surfaces are iso-(electron) probability surfaces: that is, these values are observables. Simply put: $P(R)$ values calculated from the Schrödinger equation's eigenfunctions. Parity is represented by colour.

Or take this ORCA generated molecular orbital rendition of benzene ($\mathrm{C_6H_6}$), again the surfaces are iso-(electron) probability surfaces:

Benzene electron density.

(Source: Chemistry SE).

These very Real values allow chemists to determine which chemical bonds are broken and which are formed during chemical reactions.

It's true that these calculated iso-probabilities should not be equated with the orbitals but for practical purposes they can't really be separated from them either.

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  • $\begingroup$ so in your terminology, iso-probability surface is the result of superposition of orbitals? $\endgroup$ – Sparkler Nov 14 '15 at 16:47
  • $\begingroup$ @Sparkler those probabilities are obtained by the all-electron wavefunction, which in turn is derived from the one electron basis set. However, you can obtain that probability from different combinations. That's why you can see the same problem with valence orbitals (covalent + ionic) or with molecular orbitals (sigma + sigma^*), and it would not change a bit. They are just different functions that combine correctly into the same probability (and thus the same overall charge distribution) $\endgroup$ – Stefano Borini Nov 14 '15 at 16:51
  • $\begingroup$ @Sparkler: firstly, it's not my 'terminology', please look up 'Born interpretation of $\Psi^2$'. And no: no superposition going on here: a $\mathrm{2p}$ orbital is a pure state. In the case of benzene, there is resonance going on between the three $\pi$ orbitals but that's not exactly superposition either. $\endgroup$ – Gert Nov 14 '15 at 16:53
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    $\begingroup$ @Sparkler: no they don't really 'reside' anywhere, one can only calculate a probability of finding one somewhere. The surfaces are geometrical places of the same probability density. Electrons have also probability inside and outside of the surface drawn. $\endgroup$ – Gert Nov 14 '15 at 17:03
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    $\begingroup$ @Sparkler: I'd rather not input, without studying the paper in its entirety. I will caution strongly against interpreting 'images' of orbitals without a fuller understanding of the QM formalism on e-orbitals. Having said that, have a look at these hydrogen orbital images: geek.com/science/… $\endgroup$ – Gert Nov 15 '15 at 3:03

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