How can I convince myself that wavefunctions of electrons on molecular orbitals are indeed standing waves?

Is it a consequence of the fact that electrons don't drift away from the molecule?

In other words, can one prove from the Schrödinger equation that, unless $\psi(x,t)$ can be represented as $\phi(x)\theta(t)$, then $\lim_{t \to \infty}\int_U |\psi(\bar x,t)|^2d\bar x=0$ for any bounded set $U\subset \mathbb R^3$ (or something along those lines)?

Or are there physical considerations that explain the standing waves?

Update. Apparently «standing wave» is an ambiguous/controversial term here, so let me reformulate my question in a more mathematical and unambiguous way without referring to standing waves.

Let a wavefunction $\psi$ correspond to a stationary state, i.e. $|\psi(x,t)|=\mathrm{const}(t)$. We can conclude, then, that $\psi(x,t)=\phi(x)\theta(x,t)$, where $|\theta(x,t)|=1$. In order to separate the variables and move on to the time-independent Schrödinger equation, we also need to establish that $\theta(x,t)$ doesn't depend on $x$. Where does this assumption follow from?

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    $\begingroup$ They don't have to be standing waves, and they aren't. However they do have to be time independant solutions otherwise your molecules would be changing with time. $\endgroup$ – John Rennie Sep 10 '15 at 16:53
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    $\begingroup$ @JohnRennie: what are time-independent solutions apart from standing waves? $\endgroup$ – user84106 Sep 10 '15 at 19:49

How can I convince myself that wavefunctions of electrons on molecular orbitals are indeed standing waves?

Actually, it's better not to. In modern Quantum Physics the idea of electrons as standing waves is increasingly seen as no more than an analogy and not a very good one either. In some cases like this system it's a rather compelling one but even there it's not necessary to think of bound particles as standing waves.

Instead look at the wave function $\psi$ as a mathematical function that contains all the information about the particle and with these properties.

Wave functions of bound particles are the eigenvalues of the Time Independent Schrödinger Equation, $\hat{H}\psi=E\psi$. $\psi$ contains information like the probability density distribution of the particles, so that orbital 'shapes' can be determined as iso-probability surfaces.

Looking at the electrons in orbitals as standing waves adds nothing to this approach.

As regards bound states and scattered states, I recommend this part of the Feynman Lectures on it.

  • $\begingroup$ I'm probably missing something, but isn't the Time Independent Schrödinger Equation the same as the standing wave assumption? Since the time-dependent solution resulting from it is $\psi(x,t)=e^{-iEt/\hslash}\phi(x)$, which looks like a standing wave to me. $\endgroup$ – user84106 Sep 10 '15 at 19:58
  • $\begingroup$ @RC: "looks like" are the operative words here. To borrow from another (more knowledgeable) member when answering almost the same question as your post's: (c'nued below). $\endgroup$ – Gert Sep 10 '15 at 21:26
  • $\begingroup$ @RC: "Quantum objects are not waves. Quantum objects are not classical point-like particles. They are quantum objects, which may show wave-like and particle-like properties. You may represent a quantum state by its "probability wave" or wave function, whose square gives the probability density to find the object "as a particle" at certain locations. It is not a wave in the classical sense that anything physical would be oscillating here, and the Schrödinger equation does not always look like a wave equation." $\endgroup$ – Gert Sep 10 '15 at 21:27
  • $\begingroup$ Fair enough, but I was asking about wavefunctions of quantum objects, not objects themselves. I think it's a legitimate question to ask whether a wavefunction (a solution of a wave equation) is a standing wave or not. $\endgroup$ – user84106 Sep 10 '15 at 21:50
  • $\begingroup$ @RC: Your question IS legitimate. But is it even meaningful to distinguish a ‘quantum object’ from the ‘wave function of that object’ when all we can possibly learn about the object comes from its wave function (1st Postulate of QM)? $\endgroup$ – Gert Sep 10 '15 at 22:11

How can I convince myself that wavefunctions of electrons on molecular orbitals are indeed standing waves?

It seems to me there is a confusion between a Bohr type model of atoms and molecules and the quantum mechanical framework with the orbitals.

One can design an orbit of an electron as a standing wave classical solution and then one has to postulate the stability, i.e. that only quantized states can exist.

The orbitals around atoms and molecules are not standing waves in space in the same sense. The wavefunction is sinusoidal, but the wave nature appears in the probability distribution, which is the complex square of the wavefunction, and can be verified over many measurements. In quantum mechanics one does not have a trajectory for the electron around the atom or molecule but an orbital, as you state.

hydr orbital

Here is a measurement of the hydrogen orbitals. Each dot is an individual measurement of a different electron, not a path for the same electron. It can only be seen as a probability distribution.

The figure at the top of this article shows the team's main result – the raw camera data for four measurements, where the hydrogen atoms were excited to states with zero, one, two and three nodes in the wavefunction for one of the parabolic coordinates. "If you look at the measured projections on the detector, you can easily recognize the nodes, and see their radial, ring-like structure," says Vrakking.

The wavefunction since it is sinusoidal, will have nodes and peaks , but it is a probability that is varying at the nodes and peaks , the electron itself is within the bounds of the uncertainty principle.

  • $\begingroup$ Thanks for the response, Anna. I was just trying to understand this bit from wikipedia: «The electron wavefunction oscillates according to the Schrödinger wave equation, and orbitals are its standing waves.» $\endgroup$ – user84106 Sep 10 '15 at 19:41
  • $\begingroup$ How do we know that they are standing waves? If I'm interpreting your answer correctly, you're saying that we know that from an experiment, and not as a theoretical result. But how can we know that these wavefunctions are standing waves in other atoms, or even complex organical molecules? Is this just a hypothesis/postulate that has been consistent with the observations so far, or can it be established directly from the other postulates of QM? $\endgroup$ – user84106 Sep 10 '15 at 19:41
  • $\begingroup$ That is what sinusoidal means in a steady state when one solves a boundary condition equation. Look at the Bohr atoms solutions a few pages down in the link above. Mathematics is similar but the interpretation of the functions is different. The wavefunction solutions are standing waves, not the electrons which have a probable (r,theta,phi)position. hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydwf.html#c3 In molecules it will be more complicated but the idea is the same. Otherwise there would be no stability, as with decays and interactions. $\endgroup$ – anna v Sep 11 '15 at 3:20

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