Visual depiction of atomic orbitals

[hokhani]

TL;DR

Why do the orbitals' shape show the directional dependence?

The wavefunction of an atomic orbital like $p_x$-orbital is generally in the form $f(\theta)e^{i\phi}$ so the probability of the presence of particle is identical at all the directional angles $\phi$. However, it is dumbbell-shape along the x direction which shows $\phi$-dependence!

 

[PeroK]

Can you provide an example?

 

[hokhani]

Can you provide an example?

Sorry, I don't know which example do you mean? As far as I understand, the spherical harmonics $\gamma_l^{m}$ are atomic orbitals. For $l=1$ they present p orbitals which are introduced as dumbbell shaped in the literatures. In fact, I don't know how the atomic orbitals are represented in the real space? It seems that the probability is represented by radius length.

 

[PeroK]

Sorry, I don't know which example do you mean? As far as I understand, the spherical harmonics $\gamma_l^{m}$ are atomic orbitals. For $l=1$ they present p orbitals which are introduced as dumbbell shaped in the literatures. In fact, I don't know how the atomic orbitals are represented in the real space? It seems that the probability is represented by radius length.

You claim the literature shows shapes without a symmetry implied by the harmonic function they represent. Let's see an example of what you are talking about.

I don't believe there are any such examples.

 

[PeterDonis]

I don't know which example do you mean?

He means providing a reference to an actual textbook or peer reviewed paper that says this:

The wavefunction of an atomic orbital like $p_x$-orbital is generally in the form $f(\theta)e^{i\phi}$

 

[WernerQH]

The wavefunction of an atomic orbital like $p_x$-orbital is generally in the form $f(\theta)e^{i\phi}$ so the probability of the presence of particle is identical at all the directional angles $\phi$. However, it is dumbbell-shape along the x direction which shows $\phi$-dependence!

Wavefunctions are usually written down in a coordinate system where the z-axis is singled out, i.e. the azimuthal quantum number $m$ refers to the component of angular momentum in the z-direction. For p-orbitals having $l=1$ we have one wavefunction ($m=0$) proportional to $\cos \theta$ and two wavefunctions ($m =\pm 1$) proportional to $\sin \theta \ e^{\pm i \phi}$. The first wavefunction (the "$p_z$-orbital") has the familiar dumbbell shape, oriented in the z-direction. But since these states have the same energy, you can form linear combinations of the other two wavefunctions to produce "$p_x$" and "$p_y$" orbitals. Of course, $\theta$ and $\phi$ then refer to different angles.

 

[hokhani]

You claim the literature shows shapes without a symmetry implied by the harmonic function they represent. Let's see an example of what you are talking about.

I think you agree that the p-orbitals are spherical harmonics $\gamma_l^{m}(\theta, \phi) \propto P_l^{m}(cos \theta) e^{im\phi}$ with $l=1$ (https://en.wikipedia.org/wiki/Spherical_harmonics). For $m=1$, the orbital shape is a dumbbell along the x-axis (please see: https://upload.wikimedia.org/wikipedia/commons/8/8d/Schrodinger_model_of_the_atom.svg) while it is proportional to $e^{i\phi}$. In fact, I don't know what the dumbbell -shaped $p_x$ orbital shows? If it shows the presence probability at $(\theta, \phi)$, this probability must be independent of $\phi$ and the dumbbell doesn't make sense.

 

[hokhani]

He means providing a reference to an actual textbook or peer reviewed paper that says this:

The Spherical harmonics are in the form $\gamma_l^{m}(\theta, \phi) \propto P_l^{m}(cos \theta) e^{im\phi}$ (https://en.wikipedia.org/wiki/Spherical_harmonics).

 

[PeroK]

I think you agree that the p-orbitals are spherical harmonics $\gamma_l^{m}(\theta, \phi) \propto P_l^{m}(cos \theta) e^{im\phi}$ with $l=1$ (https://en.wikipedia.org/wiki/Spherical_harmonics). For $m=1$, the orbital shape is a dumbbell along the x-axis (please see: https://upload.wikimedia.org/wikipedia/commons/8/8d/Schrodinger_model_of_the_atom.svg) while it is proportional to $e^{i\phi}$. In fact, I don't know what the dumbbell -shaped $p_x$ orbital shows? If it shows the presence probability at $(\theta, \phi)$, this probability must be independent of $\phi$ and the dumbbell doesn't make sense.

Those look like the real and imaginary parts of the spherical harmonic.

 

[JimWhoKnew]

I think you agree that the p-orbitals are spherical harmonics $\gamma_l^{m}(\theta, \phi) \propto P_l^{m}(cos \theta) e^{im\phi}$ with $l=1$ (https://en.wikipedia.org/wiki/Spherical_harmonics). For $m=1$, the orbital shape is a dumbbell along the x-axis (please see: https://upload.wikimedia.org/wikipedia/commons/8/8d/Schrodinger_model_of_the_atom.svg) while it is proportional to $e^{i\phi}$. In fact, I don't know what the dumbbell -shaped $p_x$ orbital shows? If it shows the presence probability at $(\theta, \phi)$, this probability must be independent of $\phi$ and the dumbbell doesn't make sense.

$Y^m_l$ is a function of two variables that has complex values on a sphere. Following @WernerQH in post #6, $p_z$ is described by $Y^0_1$ which is real. It is visualized by the middle "dumbbell" in the second line here. It is not the spatial distribution! It uses the 3 dimensions to express $\theta, \phi, Y^0_1(\theta,\phi)$ simultaneously. Somewhat like the way in which we express the strength of the electrostatic field by the density of the force lines. The larger the distance of the dumbbell's surface from the origin, the larger $\left|Y^0_1\right|$ in that direction. So the dumbbell shape means that $\left|Y^0_1\right|$ is maximal at the poles of the sphere and vanishes on the equator (and independent of $\phi$ , of course). If you'll use colors on the sphere instead, like in a topographic map, it will look more like a physical dumbbell. If you want to express probability density in this way, draw $\left|Y^0_1\right|^2$ . If you'll rotate the dumbbell of $p_z$ by $\pm\pi/2$ around the y-axis, you'll get the dumbbell that visualizes $p_x$ (looks like the dumbbell on the right of the second line in the picture, although it visualizes something else). It is not independent of $\phi$ like $p_z$ , because $\phi$ is defined relative to the z-axis. However, you can easily see that the $p_x$ dumbbell is invariant under rotations around the x-axis.

 

[hokhani]

It uses the 3 dimensions to express $\theta, \phi, Y^0_1(\theta,\phi)$ simultaneously.

As you clearly outlined, $\gamma_l^m$ has complex values on a sphere. In the case of $p_z$ there is no problem because $\gamma$ is real. But generally, $\gamma=Re+iIm$ and so the complex number $\gamma_1^{+1(-1)}$ cannot be represented as dumbbell-shaped $p_{x(y)}$. Following @WernerQH in post #6, it seems that $p_{x(y)}=\frac{\gamma_1^1 +(-) \gamma_1^{-1}}{\sqrt(2)}$, right?

 

[JimWhoKnew]

Following @WernerQH in post #6, it seems that $p_{x(y)}=\frac{\gamma_1^1 +(-) \gamma_1^{-1}}{\sqrt(2)}$, right?

Divide your result corresponding to $p_y$ by $i$ (a global phase factor) and observe that now both expressions are real (and $\phi$-dependent, as expected).

 

[Cthugha]

For $m=1$, the orbital shape is a dumbbell along the x-axis (please see: https://upload.wikimedia.org/wikipedia/commons/8/8d/Schrodinger_model_of_the_atom.svg) while it is proportional to $e^{i\phi}$.

Just to add to all the correct comments that were made already: The wording above is dangerous because it mixes two different pictures.
In chemistry, often real-valued orbitals are preferred, while in physics often complex-valued ones are used. When you explicitly talk about m=1, this implies that you are considering the complex-valued orbitals commonly used in physics as the standard orbitals in Chemistry are either those for m=0 or superpositions of m=+/-1.

While it might be difficult to see this from written words, my students typically benefit strongly from playing with this applet:
Simulator for orbitals

You can switch back and forth between the chemistry and the physics convention at the top and you will see that there are px/py/pz in the chemistry version and explicit values for m in the physics convention. for n=2, l=1 and m=+/- one, you will actually also see the rotation in the phase that corresponds to the angular momentum if you rotate the view to look at the x-y-plane.

It also makes sense that there are two conventions. In physics, e.g. an isolated hydrogen atom in free space is quite interesting, while in chemistry you will usually rarely have isolated atoms, but rather some chemical bond along some preferred direction, which is what you get by px/py/pz orbitals.