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I've always thought that orbitals lead to a loss of symmetry, and have never been able to give myself a satisfactory answer to this.

I'll explain via an example:

Let's take an $\ce{N^3+}$ atom. It's perfectly spherical, and has no distinguishing 'up' and 'down'. There is no set of 'preferred coordinate axes' for it since it has spherical symmetry (except the nucleus, but I doubt that matters).

Now, let's give it three electrons. They arrange themselves in the $2p$ orbitals, one in each (by Hund's rule). Now, suddenly, the atom has lost its spherical symmetry — we have a distinct triplet of orthogonal directions separate from the others.

This leads to these questions: How can symmetry 'break' this way? Are the directions of the axes 'hidden' in the atom beforehand? Are they themselves wavefunctions (though a wavefunction of wavefunctions sound odd to me, this explanation makes sense-random events can break symmetries)

So, I'd like a clear explanation of how/why the symmetry breaks.

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  • $\begingroup$ As long as the wavefunction share the eigenstates of L^2 and L^z, we call it spherical symmetric. $\endgroup$ – Rodriguez Feb 25 '17 at 6:59
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Symmetry is not actually broken that badly. If you take any one symmetry axis (or plane or...), it exists in both cases.

Lets do a thought experiment:

  • In the spherical case you have only one electron. Your only symmetry axis goes through the electron and the nucleus.
  • In the case of three electrons, the symmetry axis goes through one of the electrons and the nucleus.
  • Doing the above collapses the system in to one possible configuration from the many possible ones (you're measuring quantum system, Heisenberg, etc.).

In other words, before you measure the location of one of the electrons, you don't know the symmetry axis and thus the two cases are equally symmetric. The probability of finding the electron number one at the "north pole" of the atom is equally probable in both cases.

Note that defining the symmetry axis is purely theoretical (mathematical) construction. The symmetry axis exists even if you don't do the measurement. See AcidFlask's comment below.

Also, I'd like to highlight, that measuring electron states is becoming possible: http://phys.org/news177582885.html.

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  • $\begingroup$ So the axes themselves are wavefunctions? $\endgroup$ – ManishEarth May 5 '12 at 0:19
  • $\begingroup$ Wait, if you don't know the symmetry axis before measurement, then the p orbital will be spherical as well, right? $\endgroup$ – ManishEarth May 5 '12 at 2:13
  • $\begingroup$ I haven't heard anyone calling the axes as wave functions, so I would not recommend it, but basically they act like vectors with spherical wave functions. Before measurement the p-orbital is still shaped like a p-orbital, but it exists in all directions (compare this to quantum states and superposition). After you measure the location of the electron (which is impossible with current knowledge), you can say in which direction the p-orbital points. $\endgroup$ – Juha May 7 '12 at 13:15
  • $\begingroup$ As a side note, remember also that the electrons can exist outside the orbitals. Finding electrons in the orbitals is just much more probable than finding them outside orbitals. Orbitals are just places where the electron is most likely to be found. $\endgroup$ – Juha May 7 '12 at 13:17
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    $\begingroup$ Placing a symmetry axis is not equivalent to making a measurement. Your answer conflates a purely mathematical issue of where to place axes in space with experiments that break the symmetry along one particular axis, thus singling out one spatial direction. $\endgroup$ – Jiahao Chen May 10 '12 at 22:11
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What you describe is unfortunately a very common misconception. Defining orbitals does not break spatial symmetry. It is still completely arbitrary which directions you define as $x$, $y$ and $z$, and thus it is entirely arbitrary how you orient the $2p$ orbitals in space. Therefore rotational invariance is still preserved.

The other thing to remember is that all the $p$ orbitals are (essentially) degenerate and so you can take any linear superposition of them you want. For example you can write down an orbital that looks like

$$\frac{1}{\sqrt{3}}\left(\phi_{2p_x}+\phi_{2p_y}+\phi_{2p_z}\right) $$

and placing an electron in this orbital does not break rotational symmetry.

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  • $\begingroup$ In my understanding the superposition principle says that the system is in one of the states, but you don't know which state until you measure the system. In other words, if you take one electron and excite it to one of the p-states, it physically will be on one of the orbitals and this orbital is geometrically asymmetric. And superposition claims that because you cannot say which state it is, you say that the electron is in all the states with probability X. Now I have to get my quantum book to verify this... $\endgroup$ – Juha May 11 '12 at 13:34
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    $\begingroup$ The p orbitals are degenerate in energy. Unless you make a measurement that breaks the degeneracy, an electron cannot be observed to be in a px, py or pz orbital. You may applied a magnetic field or something similar to break the degeneracy, which then allows such a distinction to be made, but then this privileges a particular direction in space which is the direction of the field around the atom. $\endgroup$ – Jiahao Chen May 11 '12 at 20:53
  • $\begingroup$ Are you saying that the electron is on all orbitals or just in one but is observed to be on all (because measrurement breaks...)? $\endgroup$ – Juha May 12 '12 at 16:10
  • $\begingroup$ @Juha The atom is in the state $\Psi_{\text{atom}}$, which one can choose to express as a LCAO. But the state is still $\Psi_{\text{atom}}$, that is, a single vector in Hilbert space. $\endgroup$ – CHM Aug 27 '12 at 3:25
  • $\begingroup$ I disagree, the particle is in orbital $\Psi_{p}$ with the probability $p = c (\Psi_{p})^2$. And if there is multiple orbitals, you cannot know in which orbital the atom is, until you measure it (and then the system collapses to one orbital and is destroyed etc.). In quantum system, you cannot know the location or orbitals of electrons (until recently, see the link in my answer). $\endgroup$ – Juha Aug 27 '12 at 7:53
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The following is to clarify the degeneracy issue debated in the comments.

Hypothesis: Electron orbitals are quantum wave functions (the truth value is left to reader).

The text below is an example adabted from Liboff: introductory quantum mechanics, pages 119 - 121.

The system is in degenerate state of

$\Psi = \frac{3\phi_2+4\phi_9}{\sqrt{25}} (5.17)$

Probability of finding energy $E_n$ is:

$P(E_n) = \langle\phi_n|\phi_n\rangle$

$P(E_2) = \langle\phi_2|\phi_2\rangle = \frac{9}{25}$

$P(E_9) = \langle\phi_9|\phi_9\rangle = \frac{16}{25}$

$P(E_(2+9)) = \langle\phi_2|\phi_9\rangle = 0$

$P(E_(a+b)) = \langle\phi_a|\phi_b\rangle = 0$, if $a \neq b$

Following is a direct quotation: "In an ensemble of 2500 identical one-dimensional boxes, each containing an identical particle in the same state $\phi(x,0)$ given by (5.17), measurement of $E$ at $t=0$ finds about 900 particles to have energy $E_2 = 4 E_1$ and 1600 particles to have energy $E_9 = 81E_1$."

So, the above states that the particle can be only in one state, not in many states and not in any linear combination of states. The ensemble average can have a linear combination of states.

Now, adapt this to orbitals. Electron can be only in one orbital. On average, electrons of bulk material are on linear combination of orbitals.

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  • $\begingroup$ iii) On average, electrons of bulk material can never be in linear superposition of orbitals, because a linear superposition is a coherent quantum state, and what you have is the analog of a classical probability distribution, called the density matrix. $\endgroup$ – perplexity Aug 28 '12 at 2:15
  • $\begingroup$ ii) According to standard Quantum Mechanics, the particle can be in any state of the Hilbert space including, of course, superposition states. If you adopt the view that they can only be in eigenstates of the Hamiltonian, then any individual molecule can not have any dynamics other than the trivial dynamics (evolution = $exp(-iEt/\hbar)$); also then you would need to think of what happens when you measure any other observable incompatible with energy, etc. $\endgroup$ – perplexity Aug 28 '12 at 2:15
  • $\begingroup$ I found several missconceptions in your last answer: i) Orbitals are not wave functions in any conceivable way (except, obviously, for the Hidrogen atom or any one-electron ion). $\endgroup$ – perplexity Aug 28 '12 at 2:15
  • $\begingroup$ i) true, but this is the hypothesis you have to make to get the orbitals (and this step is made in the comments). ii) Only if you consider ensemble averages. I want to see a reference that says "one particle can exist on a linear superposition of states". iii) Again, this is the step back from wavefunctions to orbitals. $\endgroup$ – Juha Aug 28 '12 at 12:02
  • $\begingroup$ In general, not until recently it is not experimentally clear where the electrons are on an atom. You have to make a step from theory to real world and back. This question is about real electrons, but people tend to consider it about wave functions (and I disagree how they interpret the superposition principle). $\endgroup$ – Juha Aug 28 '12 at 12:12

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