# What is the sense of electronic configuration of an atom if an electron can be in any superposition of energy eigenfunctions?

I recently started learning abou the quantum mechanical explanation of hydrogen atom. In the derivation of the wave functions of hydrogen atom we find its energy eigenfunctions. But according to the Schrödinger equation, the electron can exist in any linear combination of the energy eigenfunctions. So what I don’t understand is that why do we say that the electrons in an atom belong to specific orbitals? Why do we always treat electrons in an atom as if their energy is fixed?

The answer that will satisfy you depend on how familiar you are with the mathematics and principles of quantum chemistry.

So, let's start small. Hydrogen atom. Exact expressions for eigenfunctions for the Hydrogen are known, they are the orbitals we know and love(?). Although the electron can be in any combination of eigenfunctions, it can only stay that way when you are not looking. When you do look, that is, take a measurement, it will collapse to a specific eigenfunction. That eigenfunction could be 1s or it could be 6h, but it turns out that 99.999% of the time, when you measure the energy of a Hydrogen atom, it will display the familiar 1s. That is the sense in which we can say that the electron in a Hydrogen atom is is in orbital 1s, what we really mean is that, typically, for the vast majority of time, the electron of the Hydrogen will be occupying the ground eigenstate.

Now that is for Hydrogen. For heavier elements, you no longer have orbitals, strictly speaking, you have more complicated eigenstates, obtained by solving the Schrodinger Equation to each atom separately. Currently, there is no analytic solution to the equation when you include a second electron. We approximate the ground state of each atom as though the electron were occupying different hydrogen-like orbitals (hydrogen-like because they differ from the pure hydrogen orbitals in a few parameters that are added).

Because electrons cannot occupy the same quantum state (considering spin), the way to do that is to assume that each electron is in a spin orbital. The way to distribute the electrons through the infinitely many hydrogen-like orbitals is to distribute them in the way you have learned, following the Pauli diagram. The resulting electron distribution is an approximation to the true ground state, using hydrogen-like orbitals, because the true ground state is not known.

O hope I made myself understood, but if you need clarification, please ask and I shall edit the answer as much as possible. I have fond memories of the time when I was asking the same questions :)

EDIT 1

You asked why is the probability of finding the electron in 1s orbital highest when we measure its energy?

The doubt comes because from your understanding these probability comes from writing the initial state as a linear combination of eigenbasis of our choice (in this case the energy eigenfunctions) and the initial state can be any function (normalisable and continuous)

Now, while in principle this is correct, in practice that is not so. When you have a vast sample of hydrogen atoms and you measure the energy of each one of them, the vast majority of them will return the ground state for most experimental conditions. That is because experimental setups that actually manage a high degree of superposition with excited states are very hard to prepare.

This means that the experimentally expected state for an electron in a Hydrogen atom is 1s for the vast majority of experimental setups. This is the relevant information. The preparation of an arbitrary superimposed state is not trivial, and they do not occur easily in nature.

This leads us to the approximation using hydrogen-like orbitals and the aufbau principle, according to which the ground state (the most experimentally relevant eigenstate) of an atom can be approximated as though each electron occupied one spin orbital around the nucleus, following the Pauli diagram of energies.

It is not so much a theoretical result as much as a way to make theory and practice agree. There is a way to reach the aufbau principle from first principles, but it involves doing statistical thermodynamics using Fermi Statistics. These calculations would tell you how electrons usually distribute through their available energy levels, and it would become clear that the aufbau principle is pretty sensible, from an experimental standpoint.

• Thank you for your answer! First of all can you explain why is the probability of finding the electron in 1s orbital highest when we measure its energy? Because from my understanding these probability comes from writing the initial state as a linear combination of eigenbasis of our choice (in this case the energy eigenfunctions) and the initial state can be any function (normalisable and continuous). – Kartikeya Badola Nov 30 '18 at 20:30
• My main problem is that I am not able to connect to the prior knowledge of orbitals and how electrons reside in them which brings out all the chemical properties of them to the new knowledge I acquired about wavefunctions and schrodinger equation. I have seen some your answers and I guess you are experienced in the field of quantum chemistry so can you suggest me some ways by which I can learn these things with understanding? – Kartikeya Badola Nov 30 '18 at 20:40
• Okay now I think I have understood. Just to confirm, I think you are saying that when we take a large sample of hydrogen atoms and measure its energy, almost all of the times we will find it in 1s state because thermodynamically it is not easy to prepare superposition of higher energy states. And aufbau principle comes as its consequence. Am I right? – Kartikeya Badola Dec 1 '18 at 6:07
• @Kartikeya yes, that is pretty much it :D – urquiza Dec 1 '18 at 9:20

Your question is in fact quite general in quantum mechanics and does not depend on H atoms or any atoms for that matter.

If a perfect experiment is repeated under the same conditions and in exactly the same way then what will be the outcome? Surely we would expect the same answer?

In general this is not true, and this is not due to experimental uncertainties because we perform an ideal, or perfect, experiment. Thus experiments in general do not have a right or wrong result, i.e in general we cannot predict the outcome, repeating the same experiment gives a different result.

What we can predict is what the range of results will be, what the the chance (probability) of observing a particular result is but we can always accurately calculate the average result via the expectation formula.

To be more exact we suppose that there is a 'system' that has a wavefunction $$\psi$$ and we operate on this with an 'operator' and so make a measurement. The 'system' is quite general and might be an H atom or a particle in a box and so forth. The measurement might be that of position or energy etc. If the operator is the Hamiltonian which is the total energy ( kinetic and potential) operator then the measurement is that of the energy. If the result of applying the operator $$\hat A$$ to the wavefunction is to produce a non-zero number multiplied by the wavefunction then we have an eigenvalue equation. We call the wavefunction that has this property the eigenfunction of the operator $$A$$ and label it $$f(x)$$, where $$x$$ is a coordinate, then the eigenvalue equation is $$\hat A f(x) = af(x)$$ where $$a$$ is the number we call the eigenvalue.

Returning now to our prediction of an experimental outcome, if the 'state' is in an eigenstate of the observable then the result of an experiment will be only the eigenvalue associated with that state, i.e. the number $$a$$. Thus if the state is the 1s wavefunction of the H atom then its eigenvalue will be measured every time an experiment is performed. In general if the eigenvalue equals, say, 2.1 then each repetition of the experiment will give the value 2.1 exactly.

If the 'system' is not in an eigenstate of an observable then the result will be one of the eigenstates associated with that observable, i.e. an eigenvalue will be measured but not always just the same one, i.e. eigenvalue 2.1 will still be measured some of the time but others as well. However, an important distinction now occurs because if we measure the system once and observe the eigenvalue, say, $$E_3$$, if we do the same experiment again on the same system then only eigenvalue $$E_3$$ will be observed no matter how many more times the experiment is done. The wavefunction has 'collapsed'. This is an unavoidable consequence of measurement (also called 'observation') of a quantum system, i.e. observation inevitably affects the system. As far as I am aware no one knows why or how wavefunction collapse occurs.

If the system is not in an eigenstate, i.e. is 'messed up' then to calculate the chance of observing each of the eigenvalues depends on the similarity or overlap of the 'messed up' state $$M$$ to the each eigenfunction $$f_n$$ belonging to each eigenvalue. The equation for the overlap is $$S_n=\int_\text{all space}M^*(x)f_n(x)\mathrm{d}x$$ where $$x$$ is the coordinate, and the probability of observing eigenvalue $$a_n$$ with eigenfunction $$f_n$$ is $$|S_n|^2$$. (By messed up I mean that the system is in a linear combination of eigenstates).

### Notes:

The wavefunctions $$\psi$$ are functions that have parameters and also variables that are the coordinates of the systems components. Thus for a particle in a box the wavefunctioon depends on the position (variable) in the box and its length (parameter) and its quantum number (parameter). In a given quantum state the parameter cannot vary. If it does then it has become another state.

The Born interpretaion is that the squared modulus of the wavefunction gives the probability density, i.e. $$\rho(x)=|\psi(x)|^2$$.

The probability of finding the particle at a particular place is found by integrating the probability density between two limits say $$a \to b$$. Thus $$P_{a\to B}=\int_a^b|\psi(x)|^2dx$$

In an experiment we operate of the wavefunction to produce an 'observable' which is the result of an experiment. Each and every observable has its associated operator. The operator always operates on the wavefunction. Examples of observables are position, momentum, energy. A position operator $$\hat x=x$$ and a momentum operator along $$x$$ is $$\displaystyle \hat p_x= -i\hbar\frac{\partial}{\partial x}$$. The Schroedinger equation has both kinetic $$\hat K$$ and potential energy $$\hat V$$ operators, these form the Hamiltonian, which is the total energy operator, $$\hat H= \hat K + \hat V$$

The average of many measurements gives the expectation value which is just an average. If the operator is $$\hat A$$ it has the form

$$\langle A\rangle =\int_\text{all space} \psi^*(q)\,\hat A\,\psi(q)\mathrm{d}q$$

where $$q$$ represents whatever the coordinates are, these are just $$x$$ for a particle in a 1D box. The * after the wavefunction means take the complex conjugate; if the wavefunction is complex replace $$i$$ with $$-i$$.

• Thank you for your answer! Your answer explained how observations affect a quantum state. But I was asking that why electrons in an atom are not in the “messed up” state? Why do we always assume that electrons have wavefunctions which are energy eigenstates, not linear combination of them? – Kartikeya Badola Dec 3 '18 at 18:29
• These are what we get from a direct solution of the Schroedinger equation, and it makes life a lot easier if we use them rather than some messed up state. In practice any messed up state is going to have more energy than the lowest eigenvalue so it is usually easier experimentally to make a single eigenstate, but not always so and so experimental methods have to work hard to achieve this, such as cooling isolated molecules to a few K in high vacuum. – porphyrin Dec 4 '18 at 12:59