I have already seen similar questions asked in the site (like this or this), but I don't feel that my question has been fully addressed.

I understand that orbitals $np_x$ and $np_y$ are linear combinations of the wave functions with quantum numbers $(n,1,1)$ and $(n,1,-1)$ and thus they also satisfy the Schrödinger equation. However, why using one instead of the other? I get that the latter are complex-valued, but doesn't the Born interpretation already account for that? I mean, that is why we take $|\psi|^2$ to be the probability density, which is always real.

Furthermore, in Atkin's physical chemistry it is said "the $x, y, z$ designation is arbitrary, and it would be equally valid to use the complex forms of these orbitals" but why are both equally valid? I understand that the real-valued wavefunctions are more commonly used beacause they are easier to graph but the real-valued and the complex-valued correspond to different probability densities (since shielding depends on the shape of the orbital, do they even have the same energies?) and, in the complex ones, the angular momentum around the $z$-axis is non-zero. To me there is a big difference between both wavefunctions and I don't get why you can use them interchangeably.

As well, I believe that the more sensible choice would be using the wavefunction that agrees with the probability density that we observe by performing experiments so why is it valid to use real or complex wavefunctions and not just the one which is experimentally observed?

My question can be summarized as

Why is it fine to arbitrarily choose to use complex or real wavefunctions? Which of these is the one that is experimentally obtained?


2 Answers 2


I think the main stumbling block that is tripping you up is that the answer is different depending on whether the shell is full or only partially-filled.

  • If a shell is full, then it will contain electrons in all valid orbitals in the shell.

    This has a number of implications, but the simplest of these is that the charge density of the electrons in the shell is spherically symmetric. The way that this works is that you take the incoherent sum of all of the orbitals, i.e., you form the probability densities $|\psi(\mathbf r)|^2$ for all of the orbitals $\psi$ in the shell, and then you add up the probability densities. It is then a fairly simple exercise to show that the $p_x$, $p_y$ and $p_z$ orbitals, whose wavefunctions are of the form $\psi_{p_x}(\mathbf r) = xe^{-r/a}$, add up to a function of $r$ only.

    The full result is actually much stronger, though. In particular, when you have multiple electrons in multiple orbitals, the only quantum state which is truly physical is the global multi-electron state, which you can form from the individual orbitals via a Slater determinant. And once you do that, it is pretty easy to show that changing a set of orbitals to a linearly-independent combination does not change the global state. As a corollary of that, the multi-electron global state for a fully-filled $p$ shell is independent of whether you represent the orbitals in complex-valued spherical forms or in real-valued cartesian forms.

    (Indeed, it is a good exercise to show that the three-electron Slater determinant $\Psi(\mathbf r_1, \mathbf r_2, \mathbf r_3) = ⟨\mathbf r_1, \mathbf r_2, \mathbf r_3 | \psi_{p_x}\psi_{p_y}\psi_{p_z}⟩$ is proportional to the vector triple product $\mathbf r_1\cdot (\mathbf r_2\times \mathbf r_3)$. The result will extend, with modifications, once you include spin.)

    For a simpler version of this principle in action, see e.g. this answer.

  • If the shell is not full, on the other hand, things are quite different.

    For a partially-filled shell, you need to choose how you partially fill the shell, i.e., which states within the manifold of states in the shell are occupied and which ones are not. How do you choose? well, it depends! to a large extent, all combinations of states in the shell are valid physical states, so there is nothing to stop you from choosing the ones which are convenient to you.

    Of course, there is always the possibility that there will be additional factors of the experiment or physical situation that you want to describe which will impact what choices you take. For example, you might be studying the Zeeman splitting of a given level, in which case you want to make sure that your quantization axis is aligned with the external field and that you are calculating the global multi-electron states with well-defined total $L_z$, which is easiest to do using the complex orbitals. But, on the other hand, you might be studying the anisotropy of the dynamics that occurs after excitation from a linearly-polarized laser pulse, in which case you might be best served by the real orbitals, which might give a simpler description of the physics.

    That said, it's important to know that the physical state that you're describing (whether it be a one-electron system or a multi-electron one) can always be written down as a linear combination of either basis. The choice of which basis to use (real orbitals vs complex orbitals) is down to convenience alone, and there are no objective physical reasons to choose either one.

  • $\begingroup$ Exactly the answer I was looking for! Thank you so much! As a last small question, is there a particular reason why (at least in chemistry) real orbitals are used more frequently than the complex ones? $\endgroup$
    – efrenump
    Commented Aug 30, 2023 at 16:16
  • 1
    $\begingroup$ Convenience (plus potentially a fair degree of collective inertia). I suspect that it helps with numerical efficiency, since having everything be real-valued cuts down on the required memory storage by a factor of two. $\endgroup$ Commented Aug 30, 2023 at 16:23
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    $\begingroup$ (and, happy to help.) $\endgroup$ Commented Aug 30, 2023 at 16:24

"the $x$, $y$, $z$ designation is arbitrary, and it would be equally valid to use the complex forms of these orbitals" but why are both equally valid?

This all depends on what you want to do. Both the $\{|p_x\rangle,$ $|p_y\rangle,$ $|p_z\rangle\}$ and $\{|m=-1\rangle,$ $|m=0\rangle,$ $|m=1\rangle\}$ are valid choices of basis in the degenerate subspace of states with $n=2$. You can switch from one basis to another by a linear transformation. The former basis states are easier to visualize and see the oscillatory structure of the $\varphi$-dependent terms since the wavefunctions are real-valued, while the latter have definite values of $L_z,$ which may be more convenient in some computations.

Why is it fine to arbitrarily choose to use complex or real wavefunctions? Which of these is the one that is experimentally obtained?

This depends on how you prepared your atoms. E.g. if you excited your atoms to p-states and passed them through a Stern-Gerlach-like separator, then each of the 3 output beams will be in an eigenstate of $\hat L_z,$ so you'd have to use the complex-valued functions to describe them.


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