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In chemistry, I have come across the term " energy of orbital" several times but I am unable to understand it from a physics point of view.

1) is the energy of an orbital the electric potential energy of each electron or all the electrons present in the orbital? I am asking this because I want to know how the energy of an orbital changes with the addition of the 2nd electron. In the former case, the "orbital energy" would increase, while in the latter case, it would decrease.

2) Orbitals, as far as I understand, are surface plots of 3-d probability distribution of position and an electron inside an orbital can be at any position, so is this "energy" the average electric potential energy of the electron?

Note: I am aware that potential energy is a property of a system, not a single object, so when I refer to the electric p.e of an electron, I am talking about its electric binding energy or the sum of electric p.e's of all 2 body systems containing that electron and another charged body

Edit: I completely forgot about the K.E of the electron(s). I should probably use total mechanical energy instead, but I am more interested in the potential energy so I wont change it for now

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2 Answers 2

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The best way to understand this is to appreciate that except for single electron atoms the orbitals are approximations that do not really exist, and to understand what the energy of an orbital is requires an understanding of the approximation.

If we take a single electron atom like the hydrogen atom then the force on the single electron is central so the potential is spherically symmetric. In this case we can solve the Schrodinger equation analytically and we get the well known $1s$, $2s$, etc functions. Although these are often referred to as orbitals they are actually the wavefunctions that the single electron can have. For more on this read Emilio's answer to What do atomic orbitals represent in quantum mechanics? As wavefunctions these have well defined properties associated with them e.g. $|\psi_{1s}|^2$ gives us the probability distribution for the $1s$ state, and the energy of the $1s$ orbital is:

$$ E_{1s} \psi_{1s} = H \psi_{1s} $$

This energy is the total energy of the atom i.e. of the electron-nucleus system. You can split it into kinetic, $T$, and potential, $V$, energy parts and you'll find it obeys the virial theorem $-2T = V$. So far so good.

The problem is that with more than one electron we have forces acting between electrons as well as forces between the electrons and the nucleus. The forces between electrons mean the force is no longer central, and they entangle all the electrons so we cannot write the wavefunction as a product of separate functions for each electron. Instead we get a single wavefunction $\Psi(\mathbf r_1, \mathbf r_2, \cdots, \mathbf r_n)$, where $\mathbf r_i$ is the position of the $i$th electron, and the energy we get from this wavefunction is the total energy of the whole system of the nucleus and all $n$ electrons. Since the electrons are all entangled individual electrons do not have a well defined energy or a well defined probability distribution. We can find only the total energy and total electron probability distribution in the atom.

But it turns out that the interactions between pairs of electrons can be averaged out to produce an approximately central force, and using this approximation we can divide our total wavefunction into separate wavefunctions for each electron. So for lithium we could write:

$$ \Psi(\mathbf r_1, \mathbf r_2, \mathbf r_3) \approx \psi_{1s↑} ~ \psi_{1s↓} ~ \psi_{2s↑} $$

And these functions $\psi_{1s↑}$ etc are the atomic orbitals.

Each orbital has an energy and an electron density associated with them, but these are not the energy and probability distribution of that electron because as we discussed above the electrons are all entangled and cannot be described separately. However if we add up the energies of all the orbitals we will get the total energy of the atom, and likewise for the probability density.

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  • $\begingroup$ If I understand this correctly, when we try to represent the wavefunction of a polyelectronic atom through hydrogen-like wavefunctions (orbitals), we are essentially ignoring all interactions between electrons, so trying to find the effect of adding a 2nd electron to an orbital is pointless ? $\endgroup$ Jan 15, 2021 at 12:05
  • $\begingroup$ @OVERWOOTCH We are not ignoring e-e interactions but we are averaging them. Suppose you remove an electron to make a positive ion, then these averages all change so all the "orbitals" change as well. Even just exciting the atom to a new energy state means all the averages and therefore all the orbitals change. $\endgroup$ Jan 15, 2021 at 12:16
  • $\begingroup$ @JohnRennie I fear you are using the term "entanglement" too liberally. Do you have any references for the claim that the electrons in a multi-electron atom are all entangled? Clearly they should be correlated, but entanglement is a much more specific claim which seems hard to justify. $\endgroup$ Jan 15, 2021 at 23:46
  • $\begingroup$ @aquirdturtle I am using it in the sense that the n electron wavefunction cannot be decomposed into a product of n single electron wavefunctions. $\endgroup$ Jan 16, 2021 at 5:28
  • $\begingroup$ @JohnRennie this is always trivially true for identical electrons in first quantization because of the exchange symmetrization requirement. $\endgroup$ Jan 16, 2021 at 22:14
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The orbital picture of an electron in an atom is just another representation of the solutions of the Schrodinger equation describing the states of an electron inside an atom. The Schrodinger equation gives you different possible energy states of the electron in an atom along with their wavefunctions.

If you are familiar with the basics of quantum mechanics, we know that the wave function gives a complete description of the properties of electrons. One such property is the probability to find an electron at a given position in space. The popular drawings of the orbitals like $s, p, d$, and $f$ are the probability distributions and do not represent any potential in the space, and therefore do not correspond to its potential energy. The plots also do not give us any information about the kinetic energy of the electron either.

Coming to the commonly used phrase "energy of orbital", it only means the energy of the electron which is in the state represented by the s-orbital. For example, if one says that s-orbital energy of a Hydrogen atom is -13.6 eV, it means that an electron that has the wave function of an s-orbital has an energy of -13.6 eV.

Do remember that the "shapes of the orbitals" is a wrong term to use. Rather, the orbitals have a positional probability distribution in those shapes, and they represent, more accurately, the wavefunction of the electron in the atom.

Edit: Do note that I am only talking about a single electron in an atom. If there are more than 1 electrons, then the description in the other answer makes complete sense. I was only explaining the terms used, what energy means and clear any confusion.

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  • $\begingroup$ So in a fully filled 1s orbital of H- ion, the energy of the orbital will be the mechanical energy of each of 2 electrons and thus more positive( taking in account electron-electron repulsion) ? $\endgroup$ Jan 15, 2021 at 12:49
  • $\begingroup$ @OVERWOOTCH in exact terms, as Rennie mentioned, the orbital energy for two electrons is only approximate as the electron interaction energy is not considered properly. Also, I would refrain from using the term mechanical energy in this regard as it generally refers to classical systems. This energy is purely quantum mechanical in origin, and its due to the electromagnetic force here. $\endgroup$
    – lattitude
    Jan 15, 2021 at 12:58

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