On the physics site, I asked the difference between real & complex orbital. Real orbitals are the superposition of complex orbitals having definite magnetic quantum number states.

\begin{align} p_z &= p_0 \\ p_x &= \frac{1}{\sqrt{2}} \left(p_1 + p_{-1} \right) \\ p_y &= \frac{1}{i\sqrt{2}} \left( p_1 - p_{-1} \right)\\ \end{align} where $p_0 = R_{n1} Y_{10}$, $p_1 = R_{n1} Y_{11}$, and $p_{−1} = R_{n1} Y_{1−1}$, are the complex orbitals corresponding to $\ell = 1$.

My question is:

Why aren't the orbitals of hydrogenic atoms taken as complex orbitals instead of the real ones? They are valid states for an electron, aren't they? So, what is the problem in having complex atomic orbitals?

  • 2
    $\begingroup$ Quick and dirty answer: Because the complex orbitals don't have a direction, which likely helps calculation. But wait for one of the theoretical chemistry gurus to chime in here ;) $\endgroup$
    – Jan
    Jun 26, 2015 at 2:26
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    $\begingroup$ You can choose whichever you want or best suit for your purpose. For actual calculations often the complex ones used. I guess Chemistry textbooks prefer the one which is easier to draw and seems less complicated mathematically. $\endgroup$
    – Greg
    Jun 26, 2015 at 2:29
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    $\begingroup$ Indeed, one of the major pitfalls of using real orbitals is when students start asking what magnetic QNs they should assign to the p_y and p_x orbitals. I was never quite satisfied with the answer of "just pick one, it doesn't really matter." It wasn't until I took quantum that I really appreciated how ill-defined the question was. It still beats trying to explain complex orbitals at that level though. $\endgroup$
    – chipbuster
    Jun 26, 2015 at 3:36
  • $\begingroup$ AFAIK, when investigating a stationary state under Born-Oppenheimer, complex component may be separated into its own multiplier depending only on time and, since it has time-independent norm, it gives time-independent contribution for most stationary properties like energy. Since other multiplier depends only on space coordinates, for simplicity of math and faster computations, almost entire computational chemistry uses real math only. $\endgroup$
    – permeakra
    Jun 26, 2015 at 3:57

2 Answers 2


To add my 5 cents to the discussion, note that we also usually use real basis functions ("atomic orbitals") in electronic structure calculations, although, molecular orbital expansion coefficients may be complex. So, each and every MO $\phi_{k}$ is usually expressed as follows, $$ \phi_{k}(\vec{r}) = \sum\limits_{q=1}^{m} c_{qk} \chi_{q}(\vec{r}) \, , $$ where $\chi_{q}(\vec{r})$ are real functions and $c_{qk}$ may in general be complex numbers. The point is that there is no additional variational freedom in using both complex basis functions and complex expansion coefficients. Besides, the use of real basis functions reduces (due to permutational symmetry) the number of two-electron integrals to be calculated for a basis of $m$ real functions from $m^4$ to $M(M+1)^2$, where $M=m(m+1)/2$ is the number of distinct pairs of products of two basis functions.


The main disadvantage of complex atomic orbitals (AOs) is that many of them are hard to visualize. For example, try making a simple sketch explaining how $p_{-1}$ differs from $p_{1}$. Conversely the complex AOs have very few advantages over real MOs - they are all eigenfunctions of $\hat{L}_z$ but that's not really important for molecular and closed shell systems.

So with many key advantages and very few disadvantages real orbitals are understandably the most popular choice.

  • 1
    $\begingroup$ Totally agree, but I'd also add that because orbitals are not unique you can unitarily transform any real orbitals into complex orbitals. They only differ by a phase. Of course we must remember that orbitals are not physical but density is---which is always real---since the the inner product of the wave function eliminates any imaginary components. $\endgroup$
    – jjgoings
    Jun 26, 2015 at 23:18

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