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Hi Physics Stack Exchange,

I’m a student with first year university chemistry knowledge and high school mathematics and physics knowledge (e.g. am familiar with complex numbers). I’m having some difficulty defining and understanding the the concept of “radial probability density” i.e. the square of the wavefunction. Its definition seems less intuitive to me than the definition for radial probability distribution. So, what I am going to do here is:

  • Tell you what I know about I know about these quantities,
  • Highlight the areas which I have difficulty understanding,
  • Propose a definition for Psi^2 that makes sense to me, and
  • Ask you to either validate or correct my thinking as necessary.

Just a note before we start, the scope of my course only introduces the general form of the wave equation but does not require us to calculate solutions. I’ve done a bit of background reading about the wave equation and wave function while researching this question but am not equipped with a detailed mathematical understanding of either, so if I make any oversimplifications or mistakes please correct me 😊

I am given the definition of radial probability density as “the probability of finding an electron in a sphere of distance r from the nucleus”. The plot of Psi^2 for the hydrogen 1s orbital indicates probability density increases toward the nucleus and decreases away from the nucleus.

However, this appears to conflict with what I know about radial probability distribution. The definition I have for this is “the probability of finding an electron on the surface of a spherical shell at distance r from the nucleus”. I am familiar that this is expressed as Psi^2 *dV i.e. probability density * an infinitesimally small difference in volumes between two spheres, which equates to 4pir^2.
The plot of this quantity can then be drawn by function multiplication of Psi^2 and 4pir^2.

Herein lies my problem: the plot of radial probability distribution has a maximum (which I understand to be at r = the Bohr radius). However, radial probability density is greater closer to the nucleus, which seems to suggest you are more likely to find an electron at the nucleus! I think the source of my confusion may be due to the wording of the Psi^2 definition.

My proposed solution to this issue is a more detailed explanation of Psi^2 I obtained by working backwards from the definition of probability distribution and using language that makes more sense to me:

“Radial probability density as a function of r is equal to the sum of all radial probability distribution values for the infinitesimally number of surfaces between the nucleus and the surfaces at r, divided by the volume of the sphere with radius r.” i.e. it is the probability of finding an electron expressed as a density of the volume being examined.

(As radial probability distribution is a PDS i.e. has integral value 1, this would suggest that you could find Psi^2 by integrating and then dividing the result by the volume of the sphere with radius r.)

Please clarify whether my understanding of this concept is correct or not, and if not, provide a definition of Psi^2 that helps explain the perceived contradiction highlighted earlier. Thank you so much for taking the time to look over this question.

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2 Answers 2

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You have in fact given yourself the answer to your question. It all boils down to the $4 \pi r^2$ factor.

The volumic probability density $|\Psi|^2$ peaks at the center, but remains finite there.

The tridimentional integration of this quantity over all space gives one.

What is called the "radial probability density" is the volumic density already integrated in the infinitesimal spherical shell between two spheres, $4 \pi r^2 |\Psi|^2$. Its one dimensional integral over $r$ form 0 to $\infty$ covers the entire space and thus also equal to one.

The radial probability density is zero at the center because of the $4 \pi r^2$ factor and does peak around the Bohr radius.

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We speak here of three different quantities that are very precisely defined. I will describe them in the context of the hydrogen atom, to hopefully offer a clear example.

  • The radial probability amplitude, or radial function, $R_{nl}(r)$, is the solution of the radial Schrödinger equation for the hydrogen atom. It is obtained by seeking a particular type of solutions of the equation, namely solutions where the variables separate into a product between a radial and angular function of the form $\psi(\vec{r})=\psi_{nlm}(r,\theta,\varphi)=R_{nl}(r)Y_{lm}(\theta,\varphi)$. Here, $r$ is the radial distance from the origin of the coordinate system (assumed to be the center of the mass of the atom, so it is the relative separation between proton and electron), while $\theta$ and $\varphi$ denote the usual polar angles in spherical coordinates and $Y_{lm}$ are the spherical harmonics. $n$, $l$ and $m$ are the quantum numbers which identify the particular solution. This kind of solution is useful in the case of the Hydrogen atom due to the spherically symmetric nature of the Coulomb potential, which scales with the inverse of the distance between the proton and electron independent of their orientation in space.

  • The radial probability density is simply the squared modulus of the radial probability amplitude, namely $|R_{nl}(r)|^{2}=R_{nl}(r)R_{nl}^{*}(r)$, where by the star * I've denoted complex conjugation. In our particular case, this radial function is real, but the idea carries over from the general definition of an inner product over a Hilbert Space. Physically, it is interpreted as the probability density of finding the relative distance $r$ between the proton and electron for the state given by the quantum numbers $n,l$ upon performing a measurement. That is, if you were to multiply the total wave function $\psi_{nlm}$ with a small volume $dV$ around a point given by the coordinates $(r,\theta,\phi)$, it would yield the probability of finding that particular configuration upon measuring.

  • The radial distribution function is just $r^{2}|R_{nl}(r)|^{2}$, so it is the radial probability density multiplied by $r^{2}$. This factor arises out of the volume element of space written in spherical coordinates, which happens to be $dV=dr\times r d\theta \times r\sin(\theta) d\varphi$. This follows based on the definition of spherical polar coordinates as an orthogonal coordinate system. When multiplying the radial distribution function with a radial element, you obtain the probability of finding this particular configuration over an infinitesimal shell of radius $r$, so irrespective of direction. In other words, the radial distribution function is the probability per unit length of finding the configuration $r$. Again, this is useful due to the spherically symmetrical nature of the problem at hand, where all directions in space are equivalent. I can recommend the book "Quantum Mechanics" of Bransden and Joachain for a very clear discussion of this type of problems. It's an old book having gone through a somewhat recent revision, but precisely because it's an old book it tends to be very clear about details that more recent publications skip over.

Note that generally only the last 2 entries in the list have a direct physical interpretation. The first one is a necessary step within the theory in order to find quantities that are directly measurable. The reason why a probability amplitude on its own is usually physically ambiguous is because it can be defined only up to a complex phase. This phase vanishes upon considering the inner product of the amplitude. Let me know if further clarifications are needed and I will edit the answer.

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  • $\begingroup$ Thank you for your response! I have a few questions for clarification: 1. To confirm, the integral of Psi^2 plotted against r is 1? 2. Is the following an appropriate visualization of Psi^2: At a point with distance r from the nucleus, you draw a sphere around it and measure the probability density for finding an electron within that sphere. However, you give the sphere infinitesimally small volume, so you find probability density for just the point (r, θ, φ). I developed this visualization from the last sentence of your definition of radial probability density. $\endgroup$ Jan 10 at 1:33
  • $\begingroup$ 3. Assuming 2, what is the reason Psi^2 is greater at the nucleus? Density = quantity/volume, but if volume is infinitesimally small at each point you are measuring Psi^2, then for density to increase, the quantity (probability) must increase. But we know from the radial distribution function that the most probable distance to find an electron for a hydrogen 1s orbital is at the Bohr radius. These facts seem to conflict, so I’m assuming there is something wrong with my visualization. $\endgroup$ Jan 10 at 1:33
  • $\begingroup$ @TheLachNessMonster 1. The radial part of the wavefunction squared is indeed normalized to unity over the entire range of $r$, from 0 to $\infty$. The meaning being that if you integrate over the whole domain, the particle is definitely going to be somewhere, hence you get the certainty (the value 1). 2. What you are measuring is whether or not the particle was where you searched for it. In principle, if you probe space with such measurements you can create a map of how the probability is distributed, i.e. you map the probability density. $\endgroup$ Jan 10 at 21:35
  • $\begingroup$ @TheLachNessMonster 3. The way a wavefunction and related quantities look ultimately depends on what potential generates them and what kind of state you are studying (bound state, resonance, scattering state etc.). The lowest bound state for the Hydrogen atom happens to be a Gaussian centered at the Bohr radius, due to the constraints imposed by the structure of the Schrodinger equation, the Coulomb potential and the requirements of quantum mechanics (bound state vanishing at infinity etc.). $\endgroup$ Jan 10 at 21:41
  • $\begingroup$ Okay, thank you! One final question. In answer to my point 2 above, you say that Psi^2 (radial probability density) describes a map of where probability is distributed, and the plot of Psi^2 is greater as r becomes smaller, implying you are more likely to find an electron at the nucleus. However, we know, as you go on to mention, that the most likely location to find an electron is at the Bohr radius, not the nucleus (obtained from radial distribution function). These two results seem to contradict each-other. How can this be so? $\endgroup$ Jan 12 at 6:04

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