While reading Atomic orbitals, I came before these two terms.

The 'real orbital' is given here:

Real orbitals

An atom that is embedded in a crystalline solid feels multiple preferred axes, but no preferred direction. Instead of building atomic orbitals out of the product of radial functions and a single spherical harmonic, linear combinations of spherical harmonics are typically used, designed so that the imaginary part of the spherical harmonics cancel out. These real orbitals are the building blocks most commonly shown in orbital visualizations.

In the real hydrogen-like orbitals, for example, n and ℓ have the same interpretation and significance as their complex counterparts$^1$, but m is no longer a good quantum number (though its absolute value is). The orbitals are given new names based on their shape with respect to a standardized Cartesian basis. The real hydrogen-like p orbitals are given by the following $$p_z = p_0 \\\\\ p_x = \frac{1}{\sqrt{2}} \left(p_1 + p_{-1} \right) \\\\ p_y = \frac{1}{i\sqrt{2}} \left( p_1 - p_{-1} \right)$$ where $p_0 = R_{n1} Y_{10},\quad p_1 = R_{n1} \quad Y_{11}, \quad \& \quad p_{−1} = R_{n1} Y_{1−1}$, are the complex orbitals corresponding to $ℓ = 1$.

My questions are:

  1. $^1$What is actually the difference between complex atomic orbitals & real atomic orbitals?

  2. Also, why is p-orbital written as these formulae? What is the reason?


1 Answer 1


To understand what these orbitals are, you first have to understand the notion of superposition in quantum mechanics. In regular classical physics, a particle or a system must be in a definite state. A car is at a particular mile marker on a highway, moving at a particular speed. The Moon orbits around the Earth with a particular velocity at a particular radius. Cats are either alive or dead.

In quantum mechanics, on the other hand, we find that particles and systems no longer necessarily have these definite properties; rather, they can exist in several different states at once. The famous example of this is, of course, Schrödinger's cat, which (after one half-life of its radioactive roommate) is neither completely alive, nor completely dead, but rather some weird combination of the two. While we have trouble envisioning this directly (or, at least, I do), it's pretty easy to mathematically describe this weird state of the cat. We use an abstract vector space, define one "direction" in this vector space to correspond to "alive", and the direction at right angles to "alive" to correspond to "dead". Call these vectors $\vec{a}$ and $\vec{d}$, respectively. The state of the cat after one half-life is then mathematically expressible as $$ \frac{1}{\sqrt{2}} (\vec{a} + \vec{d}). $$ The factor of $1/\sqrt{2}$ is because the states corresponding to vectors have to be unit vectors (or, more accurately, they can be taken to be unit vectors.) It's not a vector in either "direction", which means the cat is neither fully in the "alive" state nor in the "dead" state; rather, it's in a weird combination of the two.

So what does this have to do with orbitals? Well, when we solve the Schrödinger equation for the hydrogen atom, we find that the allowed wavefunctions of the electron are parameterized by three quantum numbers: $n$, $l$ (which is between 0 and $n$), and $m$ (which is between $-l$ and $+l$.) We can write these wavefunctions as something like $$ \psi_{n,l,m} (\vec{r}). $$ What's more, it happens that for a given $n$ and $l$, the wavefunctions with opposite $m$ values are complex conjugates of each other: $$ \psi_{n,l,-m} (\vec{r}) = \psi^*_{n,l,m} (\vec{r}) $$

That's all well and good, but what if we want a real-valued wave function? For example, let's take the set of wavefunctions with $n = 2$ and $l= 1$. By the above logic, $\psi_{2,1,0}$ is its own complex conjugate; so it's already real-valued. Let's call this wavefunction $p_z(\vec{r})$. The other two wavefunctions $\psi_{2,1,1}$ and $\psi_{2,1,-1}$ are complex-valued, unfortunately. However, we can write the following two combinations of these wave functions: $$ p_x(\vec{r}) = \frac{1}{\sqrt{2}}(\psi_{2,1,1} + \psi_{2,1,-1}) \qquad p_y(\vec{r}) = \frac{1}{\sqrt{2}i}(\psi_{2,1,1} - \psi_{2,1,-1}) $$ Both of these quantities are real (you should check this to satisfy yourself that this is true). So if the electron is in either of these superpositions, we can take its wavefunction to be entirely real-valued. In both cases, though, the electron no longer has a definite $m$ value; rather, it is partially in the $m = +1$ state and partially in the $m = -1$ state because it's in a superposition of these states of definite $m$ (just as Schrödinger's cat is not fully in the "alive" state or the "dead" state.)

I am of course glossing over a huge amount of subtlety and ambiguity here, but hopefully this explains what's going on with these real orbitals and why they can be written as sums of the complex orbitals.

  • $\begingroup$ Isn't there a typo in your answer, sir? $l$ can have values like $$l \in 0,1,2,3,\cdots (n-1)$$ & $l\neq n$. For $n=1$, $l$ can have only one value: $l= 0$. $\endgroup$
    – user36790
    Jun 22, 2015 at 19:02
  • $\begingroup$ Apart from this, the answer is satisfactory. Sir, cannot the orbital remain in either state where $m$ is definite? I mean to say why not $\psi_{(2,1,1)}$ or $\psi_{(2,1,-1)}$ only? $\endgroup$
    – user36790
    Jun 22, 2015 at 19:18
  • $\begingroup$ Those are also valid states for the electron. It's just that its wave-function won't be real-valued if it does. Moreover, if you're in a situation where there are preferred x, y, or z-axes, these combinations can be more useful in constructing the true orbitals of the electrons (i.e., the molecular orbitals under the influence of all the other atoms as well as the central atom.) $\endgroup$ Jun 22, 2015 at 19:26
  • $\begingroup$ Sir, thanks a lot. +1. Sir, I am new to the quantum arena. So, if you help me just briefly explaining how wavefunction is related to vector space & bases, I'll be very grateful. Also, a general wavefunction is a linear combination of all eigenfunctions & when we measure, the wavefunction collapses to a definite eigenfunction, right? How does it decide which eigenfunction to choose? I know they are bit irrelevent to the present context, but if you help, I'll be grateful to you:) $\endgroup$
    – user36790
    Jun 22, 2015 at 19:42
  • 1
    $\begingroup$ "How a wavefunction is related to vector space & bases" is a big topic, and one I can't fully address here. I would encourage you to start out by reading up on bra-ket notation and then posting a new question if there are things that still confuse you. As far as how the electron chooses which eigenfunction to collapse to: that's the measurement problem, and it doesn't have a universally accepted answer (yet). $\endgroup$ Jun 23, 2015 at 14:23

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