The wavefunction of $2p$ orbitals with $m_l=\pm1$ have the form: $$\Psi_{p_{\pm1}}=\pm \frac{1}{\sqrt{2}}r\sin\theta\cdot e^{\pm i\phi}f(r)$$ We can make linear combinations and get the $p_x$ and $p_y$ orbitals: $$\Psi_{p_x}=- \frac{1}{\sqrt{2}}(p_{_{+1}}-p_{_{-1}})=r\sin\theta \cos\phi \cdot f(r)=xf(r)$$ $$\Psi_{p_y}=\frac{1}{\sqrt{2}}(p_{_{+1}}+p_{_{-1}})=r\sin\theta \cos\phi \cdot f(r)=yf(r)$$ I understand that because the Hamiltonian operator is linear then every linear combination of wavefunctions is also a solution. What I don't understand is why we use $p_x$ and $p_y$ orbitals (e.g. chemistry) in order to visualize probability densities. They are real-valued functions but they clearly have different probability densities than $p_{_{+1}}$ and $p_{_{-1}}$.
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asked Mar 1, 2020 at 15:50
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1$\begingroup$ In short: many of the chemists that don’t struggle so much with complex numbers end up doing a physics degree. $\endgroup$– ZeroTheHeroMar 1, 2020 at 18:11
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1 Answer
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The $p_x$ and $p_y$ orbitals are standing waves. They are not degenerate in energy when spherical symmetry is broken, as in molecules. Then the angular momentum is not constant, and $m_\ell$ is not a good quantum number.
It is the same with the $3d$ orbitals in crystals.
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1$\begingroup$ Are you assuming that the symmetry breaking perturbation is diagonal in the $p_x,p_y$ basis? Otherwise I see no reason to use these. So for perturbation applied along an axis (eg a uniform electric field) this would be a sensible choice I think? $\endgroup$ Mar 1, 2020 at 16:23
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$\begingroup$ @jacob1729 As a first approximation, the $p_x$ and $p_y$ orbitals will also be good starting points and intuitive when the perturbation is not diagonal in that basis, when $p_x$ and $p_y$ are not exact eigenstates. But for numerical calculations maybe there is no advantage. $\endgroup$– user137289Mar 1, 2020 at 16:36