The wave function of the $2\mathrm{p}_z$ orbital is

$$Ψ = \frac{1}{4\sqrt{2π}}\left(\frac Z a\right)^{5/2} r \mathrm e^{-Zr/a}\cos θ.$$

I'm confused if this function will be a three-dimensional function or a four-dimensional function?

My professor told me that

$$z/r = \cos θ,$$

where $θ$ is the angle made by $r$ with the $z$ axis. So, I guess it will be four-dimensional since we would have to then plug in

$$r = \sqrt{x^2 + y^2 + z^2}$$

in the equation.

But when he plotted the function, he plotted a three-dimensional one! What is the correct answer and why?

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    $\begingroup$ The wave function is a function of $x, y$ and $z$. That is all. Why would you create a fourth dimension ? $\endgroup$ – Maurice Jan 26 at 13:21
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    $\begingroup$ The expression is in spherical coordinates, the coordinates as (r,theta,phi) - en.wikipedia.org/wiki/Spherical_coordinate_system . $\endgroup$ – Ian Bush Jan 26 at 13:25
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    $\begingroup$ You have to start with a simpler question: is $\sin x$ a one-dimensional or two-dimensional function? $\endgroup$ – Ivan Neretin Jan 26 at 14:03
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    $\begingroup$ I think we have found the misconception. Typically, when we refer to the dimension of a function, we are describing how many input variables it depends upon. In the case of $\sin(x)$, this is just 1. To plot the value of the function, we need one additional dimension, but that's not the same as saying $\sin(x)$ is two dimensional. $\sin(x)$ has a value for every point in one dimensional space. The pz orbital has a value for every point in 3 dimensional space. $\endgroup$ – Tyberius Jan 26 at 15:15
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    $\begingroup$ That's not the way we use these words. A function has 3 variables, hence it is 3-dimensional. A graph of it is 4D, but that's another story. $\endgroup$ – Ivan Neretin Jan 26 at 19:21

The wavefunction covers 3-dimensional space. Formally, it maps each point in space to a complex number. That is, it is a function that takes a point in 3-dimensional space as input and returns a complex number as output: $\Psi:\mathbb{R}^{3}\rightarrow \mathbb{C}$.

There are many ways to represent 3-dimensional space: cartesian coordinates $(x,y,z)$, spherical coordinates $(r,\theta,\phi)$, cylindrical coordinates $(z, r, \theta)$, to name the most common. We have these different ways because functions might look "nice" in one coordinate system and not nice in another.

As it turns out, the wave function for the $p_{z}$ orbital looks "nice" in spherical coordinates, that is, $\Psi(r, \theta, \phi)$. It looks less nice in cartesian coordinates, but you can do the replacement for $\theta$ and $\phi$ as noted in the question. However, people do not usually mix coordinates systems together because they are redundant. For example, no one writes $r\sqrt{x^{2}+y^{2}}$. They write either $r^{2}$ or $x^{2}+y^{2}$.

To conclude, the $p_{z}$ orbital you have written is a function of 3 variables: $r, \theta, \phi$. It just happens not to depend on $\phi$, so it looks like just a function of $r$ and $\theta$. Forget this concept of dimensionality because it's not doing you any favors.

  • $\begingroup$ you have written $r$ as the square root of $x^2+y^2$. Shouldn't it be the square root of $x^2+y^2+z^2$ $\endgroup$ – coconutmercury Jan 27 at 4:19
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    $\begingroup$ Only in spherical coordinates. In cylindrical coordinates, it's just a function of x and y. Honestly, though, the point is don't mix coordinates. If you're hung up on what the definition of r is, then you've missed the point. $\endgroup$ – Zhe Jan 27 at 13:04

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