Radial Distribution Function (RDF) versus Radial component

[sgstudent]

Homework Statement

For the RDF, we take the square of the radial component multiplied by 4pi r² (the surface area of a sphere) and this gives us the probability density of finding an electron r distance away. Whats the point in multiplying it by r²?

Homework Equations

RDF= r₂[R(r)]²

The Attempt at a Solution

I feel that when I just square the radial component we would get the probability density of finding an electron r distance away. Since the radial component already takes into account the distance from the nucleus (r), should it's squared value should tell us the probability density of finding an electron at a certain distance of r. Similar to how the full wavefunction squared would give us the probability density of finding the electron within a volume stipulated by r, theta, and phi, I feel like squaring the radial component alone would achieve the RDF.

So I can't wrap my head around the idea of multiplying it by the area of the sphere at r distance away as well.

I am a first year chemistry student learning about the Schrodinger's equation for the first time so I apologize if I am unclear about the topic.

 

[Orodruin]

First of all, what do you mean by the radial component? The wave function is a scalar and does not have components.

To answer your question, the wave function squared gives the probability per volume. In the radial distribution function you are not looking for this, you are looking for the probability to find the electron between r and r+dr so naturally you need to integrate the probability density over the entire volume. What is the volume of a thin shell of radius r and thickness dr?

 

[sgstudent]

First of all, what do you mean by the radial component? The wave function is a scalar and does not have components.

To answer your question, the wave function squared gives the probability per volume. In the radial distribution function you are not looking for this, you are looking for the probability to find the electron between r and r+dr so naturally you need to integrate the probability density over the entire volume. What is the volume of a thin shell of radius r and thickness dr?

The volume of the thin shell would be 4pi r² dr. What does it mean to integrate it over the entire volume?

I suppose integrating over the entire volume means we have to get the volume of the sphere then integrating it with respects to r to get the probability of finding r and dr right? But why do we need to multiply it by the volume in order to do so? Can't it just be integrated with respects to r without multiplying it with the volume of that thin layer?

 

[Orodruin]

The volume of the thin shell would be 4pi r2 dr

Exactly. So if the probability per volume is a number $p$. What is the probability to be within that shell?

 

[sgstudent]

Exactly. So if the probability per volume is a number $p$. What is the probability to be within that shell?

It would be p(4pi r² dr)

Would the R(r)^2 term be probability per unit volume?

 

[Orodruin]

Would the R(r)^2 term be probability per unit volume?

the wave function squared gives the probability per volume

 

[sgstudent]

I presume that's a yes?

So that's the reason why we multiply it by a hypothetical volume of 4pi r^2 dr to get the probability of an electron at radius r? So essentially the RDF is the probability and not probability density?

 

[Orodruin]

You need to divide by dr to get the probability per radial distance. The result of that is a distribution function, but in radius and not in volume.

Seen in a different way, the volume element in spherical coordinates is $r^2 \sin\theta\, dr\,d\theta\,d\varphi$. Integrating out the angles gives you the factor of 4pi.

 

[sgstudent]

You need to divide by dr to get the probability per radial distance. The result of that is a distribution function, but in radius and not in volume.

Seen in a different way, the volume element in spherical coordinates is $r^2 \sin\theta\, dr\,d\theta\,d\varphi$. Integrating out the angles gives you the factor of 4pi.

What do you mean by it? Do you mean 4pi r^2 [R(r)]^2 dr has to be divided by dr to get probability per radial distance?

So it would be accurate to say that if we kept the dr it would just be straight up probability of finding an electron in that sphere of 4 pi r^2? But if we actually plotted that against r it would be just a plot of very small numbers (by virtue of dr being really small) so we divide by a constant dr to get something more "useable"?

 

[Orodruin]

So it would be accurate to say that if we kept the dr it would just be straight up probability of finding an electron in that sphere of 4 pi r^2?

... and thickness dr.

But if we actually plotted that against r it would be just a plot of very small numbers (by virtue of dr being really small) so we divide by a constant dr to get something more "useable"?

The probability of finding it per radial distance is just that, the probability of finding it in a thin shell divided by the thickness of that shell.

 

[sgstudent]

The probability of finding it per radial distance is just that, the probability of finding it in a thin shell divided by the thickness of that shell.

Is there a reason why we don't just keep the dr in the RDF as in RDF= 4pi r^2 [R(r)]^2 dr and plotting it against r?

 

[Orodruin]

dr is infinitesimal and arbitrary. It conveys more information to normalise the distribution as it does not depend on your arbitrary choice of dr. The result also interpretation in terms of probability per radius and including dr it does not,

 

[sgstudent]

dr is infinitesimal and arbitrary. It conveys more information to normalise the distribution as it does not depend on your arbitrary choice of dr. The result also interpretation in terms of probability per radius and including dr it does not,

Am I correct to say that the actual radial distribution function 4pi r²[R(r)]^2 represents probability of finding an electron over a small thickness dr?

What does |Ψ|₂ represent actually? It tells us the probability density but when I substituted a value for r, θ and ∅ into that, what significance does it tell me? The probability per unit volume? What would the unit volume here represent and mean?

 

[Orodruin]

Am I correct to say that the actual radial distribution function 4pi r2[R(r)]^2 represents probability of finding an electron over a small thickness dr?

It represents the probability per unit length. To find the probability of finding the electron between $r_1$ and $r_2$, you need to integrate the radial distribution function from $r_1$ to $r_2$.

What does |Ψ|2 represent actually? It tells us the probability density but when I substituted a value for r, θ and ∅ into that, what significance does it tell me? The probability per unit volume? What would the unit volume here represent and mean?

I assume you mean $|\psi|^2$ and φ (not the empty set ∅). It is the probability per volume, yes. It means that if you integrate it over some volume, you get the probability of finding the electron in that volume.

 

[sgstudent]

probability per unit length

What does the probability per unit length mean? Sorry, but I can't really imagine what it represents very well yet.

It means that if you integrate it over some volume, you get the probability of finding the electron in that volume.

If I didn't integrate it over some volume and just plugged in some values for r, θ, and ∅, what would that number represent also?

Thanks for the patience with me.

 

[Orodruin]

What does the probability per unit length mean?

Exactly what I said. That it is what you have to integrate to get the probability to find the electron between two values of the radius.

If I didn't integrate it over some volume and just plugged in some values for r, ∅ and ∅, what would that number represent also?

As a density, its utter purpose is to be integrated. Compare to regular mass density, which tells you how much mass per volume there is. In order to find the actual mass inside a volume, you have to integrate it over that volume. For a small enough volume dV such that the probability density is effectively constant inside it, the probability of finding the electron inside that small volume is the probability density multiplied by dV.

 

[sgstudent]

That it is what you have to integrate to get the probability to find the electron between two values of the radius.

So similar to what you said about the density, it's meaningless without integrating it over r and r' right? In that case, why do people plot a graph of 4pi r2[R(r)]^2 against r and say that the maximum is at n^2a_o for instance? Is there any significance when we plotted it in a graph?

Thanks again for the help

 

[Orodruin]

The density does have a significance. It essentially tells you how likely it is to be found at different r. Where it has its maximal value, it is more likely to be found within a small r interval where the probability density is large than it is to find it in an r interval of the same width where the probability density is small.

 

[sgstudent]

The density does have a significance. It essentially tells you how likely it is to be found at different r. Where it has its maximal value, it is more likely to be found within a small r interval where the probability density is large than it is to find it in an r interval of the same width where the probability density is small.

I managed to find this link https://imgur.com/a/2CmuD9q which states that "we take the radial distribution function to be R(r)^2x4pi r^2, the probability of finding an electron on a spherical shell of 4pi r^2 at a distance of r from the nucleus." They mean that by taking the integral of that equation from r=0 (at nucleus) to any distance r we get the probability of finding an electron at distance r right? And not by subbing in values of r into that equation we get the probability?

And it's also stated that "the radial distribution function is not the probability of finding an electron at radius r, that is the square of the radial part of the wavefunction R(r)^2." Do they also refer to the integration of R(r)^2 with respects to volume, and from that volume we determine the r used to calculate that volume, so we get the probability of finding an electron within a specific r?

 

[Orodruin]

They mean that by taking the integral of that equation from r=0 (at nucleus) to any distance r we get the probability of finding an electron at distance r right?

No. They mean finding it on the actual sphere, not inside the sphere. Of course, the probability of having exactly that radius is zero, but the interpretation is in terms of a probability density as we have already discussed. The text you are referring to is quite badly formulated.

 

[sgstudent]

Where it has its maximal value, it is more likely to be found within a small r interval where the probability density is large than it is to find it in an r interval of the same width where the probability density is small.

So in the case of the RDF, when we have a maximum point it's not really probability density but actually more like probability/length. But the same idea is used, right? Where the greater the RDF value, the more likely we would find the particle when we integrated over that distance?

No. They mean finding it on the actual sphere, not inside the sphere. Of course, the probability of having exactly that radius is zero, but the interpretation is in terms of a probability density as we have already discussed.

What do you mean by this?

 

[Charles Link]

If I can join this discussion=In your previous "link" from the OP, they don't mention at all how the $Y_l ^m(\theta, \phi)$ are normalized. $\\$ And for the OP=one thing you seem to be confusing is the probability of finding a particle in $dV=dx \, dy \, dz$ is $\Psi^{\dagger}(x,y,z) \Psi(x,y,z) \, dx \, dy \, dz$. The wave function $\Psi(x,y,z)$ can be written at a location in spherical coordinates and its value is unchanged. $\\$ When they refer to the probability of finding the particle at radius $r$, they are really saying they want the probability that it is found between $r$ and $r+dr$, and that is given by $\Psi^{\dagger}(r, \theta, \phi) \Psi(r, \theta, \phi) \, 4 \pi \, r^2 \, dr$ , and this should be a somewhat obvious result. (I'm assuming here that $\Psi$ is spherically symmetric,(independent of $\theta$ and $\phi$), as it is for the $s$ orbitals. If not, an integral of $\Psi^{\dagger}(r, \theta, \phi) \Psi(r, \theta, \phi)$ is done over $d \Omega= \sin{\theta} \, d \theta \, d \phi$ for a given $r$, (between $r$ and $r +dr$, the $r$ is not integrated here, just the $\theta$ and $\phi$), that will give $4 \pi$ when any $\theta$ and $\phi$ dependence of the wave function are integrated).

 

[sgstudent]

When they refer to the probability of finding the particle at radius rr r , they are really saying they want the probability that it is found between rr r and r+drr+dr r+dr , and that is given by Ψ†(r,θ,ϕ)Ψ(r,θ,ϕ)4πr2drΨ†(r,θ,ϕ)Ψ(r,θ,ϕ)4πr2dr \Psi^{\dagger}(r, \theta, \phi) \Psi(r, \theta, \phi) \, 4 \pi \, r^2 \, dr

In this case since dr is so small, if we just sub in r into Ψ^†(r,θ,ϕ)Ψ(r,θ,ϕ)4πr² dr, dr won't we just get 0?

 

[Charles Link]

That is probably why they say it is the probability of finding it at $r$, as shorthand for "finding it between $r$ and $r+ dr$ , and then need to put on the $dr$ on the function. $\\$ I think you need to put more thought into your questions though=I think @Orodruin has given you some pretty good explanations, and I thought I could add a little that might help clarify things.

 

[sgstudent]

That is probably why they say it is the probability of finding it at $r$, as shorthand for "finding it between $r$ and $r+ dr$ , and then need to put on the $dr$ on the function. $\\$ I think you need to put more thought into your questions though=I think @Orodruin has given you some pretty good explanations, and I thought I could add a little that might help clarify things.

I see this makes sense.

So in the case of the RDF of 4πr²[R(r)]²dr, when we have a maximum point it's not really probability density because we are not integrating it with respects to volume but instead distance so it's not really density but actually more like probability/length?

But the same idea is used, where the greater the RDF value, the more likely we would find the particle when we integrated over that distance (r and r + dr where dr actually is some number that is not infinitesimally small)?

 

[Charles Link]

Yes, it's a probability per radial distance. $\int\limits_{0}^{+\infty} f(r) \, dr=1$, and $f(r)$ is the probability density function for the radial distance $r$. The probability that the particle is between $r$ and $r+\Delta r$ is $f(r) \, \Delta r$. The probability that the particle is between $r_1$ and $r_2$ is $\int\limits_{r_1}^{r_2} f(r) \, dr$. $\\$ If you were taking measurements, and you found the particle at some $r$ and some $\theta$ and some $\phi$, with this particular type of statistics, you simply record the $r$ value, and throw away the $\theta$ and $\phi$ information. You will have the same $r$ for all kinds of $\theta$ and $\phi$, but all that this particular process is interested in is $r$. $\\$ When they ask, what is the probability that the particle is at radius $r$ ?, they are doing this type of process, but they really don't spell it out in detail like this. They sort of expect you to understand what they mean, and it can be confusing the first time or two that you see it like this. $\\$ It may also interest you that the math teachers show this kind of thing in a Probability class with what is called a "probability distribution function" $F(r)$, where $F(r)$ is the probability $P(R \leq r)$, where $R$ is the variable describing the radial value that is measured. Then $F(r)=\int\limits_{0}^{r} f(r) \, dr$, and $\frac{dF(r)}{dr}=f(r)$. The function $f(r)$ is known (by the math teachers) as a "probability density function". With a little study of these functions, and some basic calculus, you should be able to see that $f(r) \, dr =F(r+dr)-F(r)$ is the probability that $r< R \leq r+dr$. $\\$ (The probability density function $f(r)$ is often referred to by physicists and others as a "probability distribution function", but I think this usage sometimes causes the math teachers some consternation. The math teachers clearly distinguish $F(r)$ and $f(r)$ by the names that they give them).

 

[sgstudent]

Yes, it's a probability per radial distance. $\int\limits_{0}^{+\infty} f(r) \, dr=1$, and $f(r)$ is the probability density function for the radial distance $r$. The probability that the particle is between $r$ and $r+\Delta r$ is $f(r) \, \Delta r$. The probability that the particle is between $r_1$ and $r_2$ is $\int\limits_{r_1}^{r_2} f(r) \, dr$. $\\$ If you were taking measurements, and you found the particle at some $r$ and some $\theta$ and some $\phi$, with this particular type of statistics, you simply record the $r$ value, and throw away the $\theta$ and $\phi$ information. You will have the same $r$ for all kinds of $\theta$ and $\phi$, but all that this particular process is interested in is $r$. $\\$ When they ask, what is the probability that the particle is at radius $r$ ?, they are doing this type of process, but they really don't spell it out in detail like this. They sort of expect you to understand what they mean, and it can be confusing the first time or two that you see it like this. $\\$ It may also interest you that the math teachers show this kind of thing in a Probability class with what is called a "probability distribution function" $F(r)$, where $F(r)$ is the probability $P(R \leq r)$, where $R$ is the variable describing the radial value that is measured. Then $F(r)=\int\limits_{0}^{r} f(r) \, dr$, and $\frac{dF(r)}{dr}=f(r)$. The function $f(r)$ is known (by the math teachers) as a "probability density function". With a little study of these functions, and some basic calculus, you should be able to see that $f(r) \, dr =F(r+dr)-F(r)$ is the probability that $r< R \leq r+dr$. $\\$ (The probability density function $f(r)$ is often referred to by physicists and others as a "probability distribution function", but I think this usage sometimes causes the math teachers some consternation. The math teachers clearly distinguish $F(r)$ and $f(r)$ by the names that they give them).

I think I've learned quite a bit from this thread. This another problem I'm facing with regards to the topic:

In RDF, 4πr^2[R(r)]^2 dr relates to the probability/radial distance so integrating it over a certain length gives us the probability of finding the electron within that range. So when we integrate it around the r=0, we get 0 probability to find the electron there. So for a 1s orbital, plotting RDF against radial distance, it starts from 0 and goes up to a peak and back down.

Wavefunction^2 relates to the probability/volume, so in order to make any sense of it, we have to integrate it against volume. So assuming the volume of the electron's orbit is a sphere, when we integrate it in the range of a small volume (basically r=0), we get a big number. So again for the 1s orbital, plotting Wavefunction^2 against radial distance, it starts from a large number and goes down.

Why is this so? I understand that we're integrating it against volume in Wavefunction^2, but if we plot Wavefunction^2 against radial distance instead of volume shouldn't we get a plot that resembles RDF?

Why is it not good enough to simply use a plot of Wavefunction^2 vs radial distance and when we want the probability over a certain radial distance, just integrate it with respects to volume, then getting the two radial distances from the 2 volume values specificed. So we know the probability over the radial distance.

 

[Orodruin]

The point is that there is a very small volume with $r$ close to zero. This is where the factor of $r^2$ comes from. For the 1s orbital, the probability density is highest at $r = 0$, but the volume of a ball with radius $dr$ at $r = 0$ is $4\pi (dr)^3/3$. Compare this with the volume of a spherical shell of thickness $dr$ anywhere else, which is $4\pi r^2 dr$, the ratio between those volumes is $dr^2/(3r^2)$! If you compare the volumes of spherical shells of thickness $dr$ at radii $r_1$ and $r_2$, the ratio is $r_1^2/r_2^2$. In other words, the volumes corresponding to shells of small radii is much smaller so the probability of finding the electron at a small radius, given by $R(r)^2$ multiplied by the volume, becomes small - not because the probability per volume is small, but because the volume itself is small.

 

[sgstudent]

For the 1s orbital, the probability density is highest at r=0

From the ψ² dV, if we integrated it with respects to volume we get probability and if we were to integrate it against a small volume would the probability = 0? Just like for 4πr^2[R(r)]^2 dr if we integrated it with respects to radial distance and if we were to integrate it against a very small radial distance would the probability = 0 as well?

It's just that we're used to seeing either probability density against radial length in the case of a plot of ψ² vs radial distance or probability/radial length against radial length in the case of the RDF. So the magnitudes of the probability density and probability/radial length are really different. But if we considered actual probabilities with the same conditions (e.g. of being close vicinity to the nucleus) in both cases they would yield the same results right?

Edit:
So at small radial distances, the probability density is huge because it's just ψ² but the probability/radial distance is small because it's 4πr^2[R(r)]^2 nothing complicated about it?