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Suppose we have a He atom. I understand that its wavefunction can be written as product of 2 hydrogenic orbitals:

$$\Psi(r_1,r_2) = \psi(r_1)\psi(r_2),$$

which in this case, are two 1s wavefunctions with paired spin. I want to know whether it is possible for the wavefunctions of each individual 1s electron to have opposite phase? If yes, can these wavefunctions interfere with each other constructively or destructively? If not, is it correct to say that both of these wavefunctions are orthogonal to each other?

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    $\begingroup$ What does "opposite phase" actually mean in this context? Could you add a definition of what you think it does? Thanks, and Welcome to Stack Exchange! $\endgroup$ Commented Oct 2, 2022 at 1:00
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    $\begingroup$ @uhoh By "opposite phase" I was imagining, whether the two 1 electron orbitals could be "out of phase" with respect to each other, similar to how we can write $1σ^*$ as linear combination of two out of phase 1s orbitals. $\endgroup$ Commented Oct 2, 2022 at 3:02
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    $\begingroup$ Do their wave functions have any angular dependence? If $n=1$ then $l=0$, so $Y_l^m(\theta, \psi)$ which has $\exp(im \phi)$ and $P_l^m(\cos \theta)$ dependence should be constant. I'm not very good at this, but I don't see any place where a phase can be defined. Unless their spins are opposite I don't think there's any other way that they can differ unless you excite one of them to $n > 1$. $\endgroup$ Commented Oct 2, 2022 at 5:43

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I think the reason you're getting answers here that aren't directly addressing the question is because the question itself doesn't really make sense in a quantum world. The flaw underlying the question is reading too much into the use of the word "phase" (and maybe also "wave") and interpreting it/them as resembling their classical meanings.

In the quantum world, a wave function does not have a specific phase, nor do wave functions interfere constructively OR destructively. Rather, both phases coexist and wave functions interfere constructively AND destructively. For example, when two hydrogen atoms come together, molecular orbitals are formed from both the constructive (bonding) combination of atomic orbitals and the destructive (antibonding) combination.

So in your specific example of the He atom, since the 1s spatial orbital(s) do not have specific phase, they don't interfere with each other in that classical sense. As to whether they are orthogonal, the term "orthogonal" in reference to atomic and molecular orbitals is used only to refer to the spatial component, with the specific definition that the overlap integral over all space is zero. The spin component is not considered. In this treatment, spin is functionally considered as a fourth non-spatial dimension.

From a conceptual perspective, it is often most useful to consider that there is only one spatial orbital that contains two electrons of opposite spin rather than considering the two spin-spatial orbitals separately.

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You can only do what you suggest (make product) if the electrostatic repulsion between the two electrons is ignored. This means that there are two single electron equations (just as for hydrogen but with $Z=2$) with energy given by the Rydberg formula. Thus the energy of either is $E_1=E_2=-4hcR_\infty=-54.4$ eV so $-109$ in total.

The perturbation caused by electron-electron repulsion is the expectation energy

$$\frac{e^2}{4\pi\epsilon_o}\int_0^\infty\int_0^\infty \psi^*\frac{1}{r_{12}}\psi r_1^2dr_1r_2^2dr_2=34\;\mathrm{eV}$$

and $\psi=R_{1s}(r_1)R_{1s}(r_2)/4\pi$ and $R$ is the 1s hydrogen radial wavefunction but with $Z=2$. This amount of energy is far from a perturbation and so we have to use the Variation Method to get a much better approximation to the energy.

The spin and spatial (radial/angular) parts of the wavefunction are treated as a product as electrostatic energies do not depend on spin. The ground state will be a singlet with spin paired as both electrons have the same spatial wavefunction so only a symmetric solution to the spatial wavefunction exists. The total wavefunction is $\psi_{space}^A\psi_{spin}^S$ or $\psi_{space}^S\psi_{spin}^A$ where superscripts means symmetric or antisymmetric wrt electron-label exchange.

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This means to complete previous answers.

The first problem I appreciate is that you are thinking of the electrons as distinguishible independent entities but in a helium atom you can not distinguish them, nor are they independent.

Besides, there is a fundamental flaw in the wavefunction you suggest: it is not antisymmetric with respect to the exchange of the coordinates of two electrons, it does not comply with the Pauli exclusion principle.

This is

$$\Psi(r_1,r_2) = \Psi(r_2,r_1) $$

and it should read

$$\Psi(r_1,r_2) = -\Psi(r_2,r_1) $$

and your are forgetting the spin-coordinates.

To construct a wavefunction that conforms with the Pauli exclusion principle you can use an Slater determinant and for He you would obtain a wawefunction like:

$$\Psi = \frac{1}{\sqrt{2}} [\Psi_{1s}(r_1)\Psi_{1s}(r_2)] [\alpha(1)\beta(2) - \beta(2)\alpha(1)] $$

On this topic you may find helpful the book by D. A. McQuarrie and J. D. Simon, Physical Chemistry: A Molecular Approach, University Science Books (1997) or other book on Physcical Chemistry or Quantum Chemistry, or the following entrance of Chemistry LibreTexts on the topic: Antisymmetric Wavefunctions can be Represented by Slater Determinants

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