I was reading about the atomic orbital in my chemistry textbook. It says that the atomic orbital (psi) is a mathematical wave function that depends on the coordinates of the electron. It is also mentioned that psi does not carry any physical meaning. That part is understood, however I could not understand what a psi vs r graph represents.(r is the distance from the nucleus) Can someone please clarify ??
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$\begingroup$ You need to roll back your understanding, because it is wrong. $\psi$ does carry a great deal of physical meaning. $\endgroup$– Ivan NeretinCommented Dec 7, 2019 at 13:03
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$\begingroup$ Thanks for the response , but then my NCERT textbook says exactly that " The wave function is a mathematical function whose value depends upon the coordinates of the electron in the atom and does not carry any physical meaning." (psi ultimately refers to a wave function) How should I interpret that ? $\endgroup$– Aurav S TomarCommented Dec 7, 2019 at 13:08
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4$\begingroup$ Maybe the book will later talk about Born's interpretation, and will mention that physical meaning was later attributed to the square power of psi, which represents the orbital's electron (probability) density $\endgroup$– user32223Commented Dec 7, 2019 at 13:11
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$\begingroup$ Yes that is correct, it does say that psi squared refers to the probability density of the electron. However, I wanted to know how I can understand a psi vs r graph ?? $\endgroup$– Aurav S TomarCommented Dec 7, 2019 at 13:15
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$\begingroup$ I do understand that a psi squared vs r graph would represent the probability density of the electron in an orbital at a distance r from the nucleus, obviously. $\endgroup$– Aurav S TomarCommented Dec 7, 2019 at 13:24
1 Answer
$\psi$ does carry a great amount of physical meaning. The annoying part is, that it doesn't represent anything directly observable. The way (non-relativistic) quantum mechanics is constructed, makes $\psi(x_i, s_i, t)$ ($x_i$ - physical coordinates of $i$-th electron, $s_i$ - spin projection on $z$ of said electron) - a direct analogue to trajectory, or, in case $\psi$ doesn't depend on $t$ (time), orbit. *
Orbitals themselves, however, do not have direct physical meaning in that sense that they are nothing more than building blocks we use to construct collective $\psi$ of the electrons in the chemical system after we apply Born-Oppenheimer approximation (that we can ignore nucleus movement when we talk about electron movement). It's a good approximation, so it is almost always in effect in quantum chemistry.
In some special cases (hydrogen-like atoms) the electron wave-function represents exactly one electron of one atom and, ignoring spin contribution, orbital equals wave-function of that one-electron system.
Orbitals of hydrogen-like atoms can be represented ** as $\psi(x)=\psi(r,\theta,\phi)=\psi_r(r)\psi_{angle}(\theta,\phi)$. It is a useful representation, because even if $\psi_r$ changes from level to level, $\psi_{angle}$ doesn't, so you can see parallels between, say, $2p$ and $3p$ orbitals, which differ*** only by $\psi_r$. Again, neither part of the orbital is observable. What IS observable is "electron density" i.e. a probability to observe electron in this particular finite (small) volume. Those can be extracted from various electron or X-ray diffraction experiments. What, however, it is good at is to point out, how good outer electrons shield inner electrons and that chemical properties are mostly guarded by outermost reaches of outer shells.
$\psi_r$ on its own cannot be observed and has no simple interpretation. It has no inherent meaning on its own, and I'm unaware of obvious interpretation. It IS useful when one has to compare different orbitals between each other or to compare various approximations of the same orbital though.
$\psi^2_r$ is fairly useless, imho. It is observable for $s$ orbitals as electron density at given radius, but it isn't very useful. What is useful IMO is density integrated over spherical surface, which for $s$ orbitals is $ |rR|^2 = 4\pi r^2 |\psi_r|^2 $, which shows how much time electron spends at given distance from nucleus. It shows, how diffuse the orbital is, and when plotted against observed total electron density, it shows how good electron is shielded against outer influence.
I hope this helps.
* Note: Here is one of the non-obvious differences between quantum and classical systems lie: you CAN have a quantum system with large kinetic energy and $\psi$ independent on time.)
** Note: it is not the case for all mathematical functions, we are lucky that it is the case for atomic orbitals.
*** Note: orbitals on the same level CAN differ by $\psi_r$. The general rule is that orbitals on the same level have the same number of surfaces where $\psi$ is zero that is equal number of the level minus one. That means that we have only $s$ orbitals on the first level with zero such surfaces, but $2s$ and $2p$ orbitals. For $2p$ orbitals the "zero" surface is one of coordinate planes and for $2s$ orbital$ it is a sphere around the nucleus.
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$\begingroup$ Thanks for the answer. It does give some more clarity. Please do note that my main question was "how I can interpret the psi vs r graph ? " and it would be great if you can help with that. Also, what would be the difference between a psi squared vs r and a psi vs r graph ?? (basically that's where I got confused when I was reading my textbook) $\endgroup$ Commented Dec 7, 2019 at 14:34
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$\begingroup$ @AuravSinghTomar I edited in my opinion on the matter. $\endgroup$ Commented Dec 7, 2019 at 15:49
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$\begingroup$ Thanks for the edit. And the help too ! $\endgroup$ Commented Dec 7, 2019 at 19:56