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Consider an isolated atom consisting of $N$ electrons. Does each electron have a wavefunction of its own (as is generally spoken about in fields like spectroscopy) or does our isolated atom have a single collective wavefunction?

Qmechanic
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prananna
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2 Answers2

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The $N$ electrons need to be described by a single collective wave function $$\Psi(\vec{x}_1,m_{s1},\dots,\vec{x}_N,m_{sN},t)$$ where $\vec{x}_i$ are the space coordinates of the electrons, $m_{si}$ are the spin quantum numbers ($+\frac{1}{2}$ or $-\frac{1}{2}$), and $t$ is time.

Notice also, because electrons are indistinguishable fermions, this wave function must be antisymmetric with respect to interchange of any two electrons $i$ and $j$. See also Wave function - Many particle states in 3d position space.

Nevertheless, in spectroscopy you often see factorized wave functions like $$\psi_1(\vec{x}_1,m_{s1})\ \psi_2(\vec{x}_2,m_{s2})$$

or (more correctly) Slater-determinant-like wave functions like $$\frac{1}{\sqrt{2}}\left(\psi_1(\vec{x}_1,m_{s1})\ \psi_2(\vec{x}_2,m_{s2})-\psi_2(\vec{x}_1,m_{s1})\ \psi_1(\vec{x}_2,m_{s2})\right)$$

But these wave functions are only approximative solutions to the multi-electron Schrödinger's equation which are valid only if the Coulomb correlation between the electrons can be neglected.

Thomas Fritsch
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  • Thanks for the reply. But in the fields such as spectroscopy, it is quite common to use statements such as "an electron has made a transition from level A to level B". In this case, isn't there an inherent assumption that each electron is being described by a separate wavefunction? – prananna Jan 10 '22 at 21:09
  • @prananna See my amended answer – Thomas Fritsch Jan 10 '22 at 21:21
  • This statement is incorrect: ‘which are valid only if the interaction between the electrons can be neglected.’ Probably what is meant is 'if the correlation between the electrons can be neglected.’ Even in this case, it should be noted that the Pauli correlation is accounted for.

    Amore exact solution that includes Coulomb correlation will involve multiple Slater determinants. The famous multi-configurational Hartree Fock atomic code by Charlotte Froese Fischer takes this into account. In some cases a single determinant is already a reasonable approximation.

     – my2cts Dec 25 '23 at 10:17
  • @my2cts Yes, of course. It should be Coulomb correlation (instead of interaction). – Thomas Fritsch Dec 25 '23 at 16:43
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Before answering this question some remarks need to be made. The commonly used term is 'orbital'. Secondly, electrons are indistinguishable fermions and in this sense all electrons occupy all orbitals. The total wave function is antisymmetric under permutation of electrons.

The simplest case is then that of a closed shell atom. As an example, for He a single Slater determinant, |1s$^2$| with doubly occupied (spin) 1s orbitals is already a reasonable approximation. Such a wave function includes electron repulsion and Pauli correlation, but leaves electrons of opposite spin uncorrelated. It lacks Coulomb correlation, which accounts for the fact that also opposite spin electrons avoid each other. This can be amended by admixing $|2p_x^2|$, $|2p_y^2|$ and $|2p_z^2|$. The resulting wave function has the n=2 electrons in oppositely polarised orbitals either $\alpha ~ 1s \pm \beta ~ 2p_x,y,z$ ($\beta << \alpha$). This means that the concept of orbital is only approximately applicable for an atom and, more generally, a many-electron system.

my2cts
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