What is the exact definition of the radial distribution function?












2














I have been very confused by the radial distribution function which is often used in chemistry to predict the probability of finding an electron at a distance from the nucleus.



From Atkins' Physical Chemistry:




Now consider the total probability of finding the electron anywhere between the two walls of a spherical shell of thickness $mathrm dr$ at radius $r$. The... volume of the shell... is now $4pi r^2 mathrm dr$. ... The probability that the electron will be found within the inner and outer surfaces of the shell is the probability density at the radius $r$ multiplied by volume... i.e $|psi|^2times 4 pi r^2 mathrm dr$. $$P(r) = 4pi r^2|psi(r)|^2$$ The function $P(r)$ is called the radial distribution function (in this case, for an s orbital).




Immediately after this, I find:




It is also possible to devise a more general expression which applies to orbitals that are not spherically symmetrical. ... $$P(r)=r^2R(r)^2$$




From Housecroft and Sharpe's Inorganic Chemistry:




A useful way of depicting the probability density is to plot the radial distribution function ... $$text{Radial distribution function} = 4pi r^2R(r)^2$$




Again, from this page, the probability density turns out to be (for the 1s orbital) $$P(r)=4pi r^2|R(r)|^2$$ However, the integral that the answerer calculated $bigg(int limits _{theta =0}^{pi }int limits _{varphi =0}^{2pi }|Y(theta ,varphi)|^2sin theta ,mathrm {d} theta ,mathrm {d} varphi bigg)$ to get $4pi$ (for 1s orbital) is calculated to be 1 (for all orbitals) in Atkins', as "the spherical harmonics are normalized" (I don't know what that signifies).



My confusion is compounded by the fact that there is another function called the probability density which is $|psi|^2$. This function is maximum at the nucleus and falls exponentially (for an 1s orbital). The radial distribution function, however, is maximum at a certain distance from the nucleus.



So, my questions are:




  • What is the physical significance of probability distribution and that of radial distribution function?


There is a similar question here but the answer does not satisfy me. If there is no physical difference between these two functions, then why do we use the second to find out where the electron is most likely present? Why not the first function?




  • What is the correct expression for calculating radial distribution function?


Which begs the question, what is the definition of radial distribution function?










share|improve this question




















  • 2




    Please don't use ce for maths... (see edits).
    – orthocresol
    8 hours ago








  • 1




    "If there is no physical difference between these two functions, then why do we use the second to find out where the electron is most likely present? Why not the first function?": The former is $p(mathbf{r})$ and the latter is $p(r) = p(|mathbf{r}|)$; usually the latter is more relevant.
    – a-cyclohexane-molecule
    8 hours ago


















2














I have been very confused by the radial distribution function which is often used in chemistry to predict the probability of finding an electron at a distance from the nucleus.



From Atkins' Physical Chemistry:




Now consider the total probability of finding the electron anywhere between the two walls of a spherical shell of thickness $mathrm dr$ at radius $r$. The... volume of the shell... is now $4pi r^2 mathrm dr$. ... The probability that the electron will be found within the inner and outer surfaces of the shell is the probability density at the radius $r$ multiplied by volume... i.e $|psi|^2times 4 pi r^2 mathrm dr$. $$P(r) = 4pi r^2|psi(r)|^2$$ The function $P(r)$ is called the radial distribution function (in this case, for an s orbital).




Immediately after this, I find:




It is also possible to devise a more general expression which applies to orbitals that are not spherically symmetrical. ... $$P(r)=r^2R(r)^2$$




From Housecroft and Sharpe's Inorganic Chemistry:




A useful way of depicting the probability density is to plot the radial distribution function ... $$text{Radial distribution function} = 4pi r^2R(r)^2$$




Again, from this page, the probability density turns out to be (for the 1s orbital) $$P(r)=4pi r^2|R(r)|^2$$ However, the integral that the answerer calculated $bigg(int limits _{theta =0}^{pi }int limits _{varphi =0}^{2pi }|Y(theta ,varphi)|^2sin theta ,mathrm {d} theta ,mathrm {d} varphi bigg)$ to get $4pi$ (for 1s orbital) is calculated to be 1 (for all orbitals) in Atkins', as "the spherical harmonics are normalized" (I don't know what that signifies).



My confusion is compounded by the fact that there is another function called the probability density which is $|psi|^2$. This function is maximum at the nucleus and falls exponentially (for an 1s orbital). The radial distribution function, however, is maximum at a certain distance from the nucleus.



So, my questions are:




  • What is the physical significance of probability distribution and that of radial distribution function?


There is a similar question here but the answer does not satisfy me. If there is no physical difference between these two functions, then why do we use the second to find out where the electron is most likely present? Why not the first function?




  • What is the correct expression for calculating radial distribution function?


Which begs the question, what is the definition of radial distribution function?










share|improve this question




















  • 2




    Please don't use ce for maths... (see edits).
    – orthocresol
    8 hours ago








  • 1




    "If there is no physical difference between these two functions, then why do we use the second to find out where the electron is most likely present? Why not the first function?": The former is $p(mathbf{r})$ and the latter is $p(r) = p(|mathbf{r}|)$; usually the latter is more relevant.
    – a-cyclohexane-molecule
    8 hours ago
















2












2








2







I have been very confused by the radial distribution function which is often used in chemistry to predict the probability of finding an electron at a distance from the nucleus.



From Atkins' Physical Chemistry:




Now consider the total probability of finding the electron anywhere between the two walls of a spherical shell of thickness $mathrm dr$ at radius $r$. The... volume of the shell... is now $4pi r^2 mathrm dr$. ... The probability that the electron will be found within the inner and outer surfaces of the shell is the probability density at the radius $r$ multiplied by volume... i.e $|psi|^2times 4 pi r^2 mathrm dr$. $$P(r) = 4pi r^2|psi(r)|^2$$ The function $P(r)$ is called the radial distribution function (in this case, for an s orbital).




Immediately after this, I find:




It is also possible to devise a more general expression which applies to orbitals that are not spherically symmetrical. ... $$P(r)=r^2R(r)^2$$




From Housecroft and Sharpe's Inorganic Chemistry:




A useful way of depicting the probability density is to plot the radial distribution function ... $$text{Radial distribution function} = 4pi r^2R(r)^2$$




Again, from this page, the probability density turns out to be (for the 1s orbital) $$P(r)=4pi r^2|R(r)|^2$$ However, the integral that the answerer calculated $bigg(int limits _{theta =0}^{pi }int limits _{varphi =0}^{2pi }|Y(theta ,varphi)|^2sin theta ,mathrm {d} theta ,mathrm {d} varphi bigg)$ to get $4pi$ (for 1s orbital) is calculated to be 1 (for all orbitals) in Atkins', as "the spherical harmonics are normalized" (I don't know what that signifies).



My confusion is compounded by the fact that there is another function called the probability density which is $|psi|^2$. This function is maximum at the nucleus and falls exponentially (for an 1s orbital). The radial distribution function, however, is maximum at a certain distance from the nucleus.



So, my questions are:




  • What is the physical significance of probability distribution and that of radial distribution function?


There is a similar question here but the answer does not satisfy me. If there is no physical difference between these two functions, then why do we use the second to find out where the electron is most likely present? Why not the first function?




  • What is the correct expression for calculating radial distribution function?


Which begs the question, what is the definition of radial distribution function?










share|improve this question















I have been very confused by the radial distribution function which is often used in chemistry to predict the probability of finding an electron at a distance from the nucleus.



From Atkins' Physical Chemistry:




Now consider the total probability of finding the electron anywhere between the two walls of a spherical shell of thickness $mathrm dr$ at radius $r$. The... volume of the shell... is now $4pi r^2 mathrm dr$. ... The probability that the electron will be found within the inner and outer surfaces of the shell is the probability density at the radius $r$ multiplied by volume... i.e $|psi|^2times 4 pi r^2 mathrm dr$. $$P(r) = 4pi r^2|psi(r)|^2$$ The function $P(r)$ is called the radial distribution function (in this case, for an s orbital).




Immediately after this, I find:




It is also possible to devise a more general expression which applies to orbitals that are not spherically symmetrical. ... $$P(r)=r^2R(r)^2$$




From Housecroft and Sharpe's Inorganic Chemistry:




A useful way of depicting the probability density is to plot the radial distribution function ... $$text{Radial distribution function} = 4pi r^2R(r)^2$$




Again, from this page, the probability density turns out to be (for the 1s orbital) $$P(r)=4pi r^2|R(r)|^2$$ However, the integral that the answerer calculated $bigg(int limits _{theta =0}^{pi }int limits _{varphi =0}^{2pi }|Y(theta ,varphi)|^2sin theta ,mathrm {d} theta ,mathrm {d} varphi bigg)$ to get $4pi$ (for 1s orbital) is calculated to be 1 (for all orbitals) in Atkins', as "the spherical harmonics are normalized" (I don't know what that signifies).



My confusion is compounded by the fact that there is another function called the probability density which is $|psi|^2$. This function is maximum at the nucleus and falls exponentially (for an 1s orbital). The radial distribution function, however, is maximum at a certain distance from the nucleus.



So, my questions are:




  • What is the physical significance of probability distribution and that of radial distribution function?


There is a similar question here but the answer does not satisfy me. If there is no physical difference between these two functions, then why do we use the second to find out where the electron is most likely present? Why not the first function?




  • What is the correct expression for calculating radial distribution function?


Which begs the question, what is the definition of radial distribution function?







physical-chemistry quantum-chemistry orbitals atomic-structure






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 7 hours ago









orthocresol

37.9k7111227




37.9k7111227










asked 9 hours ago









Shoubhik Raj Maiti

1,240529




1,240529








  • 2




    Please don't use ce for maths... (see edits).
    – orthocresol
    8 hours ago








  • 1




    "If there is no physical difference between these two functions, then why do we use the second to find out where the electron is most likely present? Why not the first function?": The former is $p(mathbf{r})$ and the latter is $p(r) = p(|mathbf{r}|)$; usually the latter is more relevant.
    – a-cyclohexane-molecule
    8 hours ago
















  • 2




    Please don't use ce for maths... (see edits).
    – orthocresol
    8 hours ago








  • 1




    "If there is no physical difference between these two functions, then why do we use the second to find out where the electron is most likely present? Why not the first function?": The former is $p(mathbf{r})$ and the latter is $p(r) = p(|mathbf{r}|)$; usually the latter is more relevant.
    – a-cyclohexane-molecule
    8 hours ago










2




2




Please don't use ce for maths... (see edits).
– orthocresol
8 hours ago






Please don't use ce for maths... (see edits).
– orthocresol
8 hours ago






1




1




"If there is no physical difference between these two functions, then why do we use the second to find out where the electron is most likely present? Why not the first function?": The former is $p(mathbf{r})$ and the latter is $p(r) = p(|mathbf{r}|)$; usually the latter is more relevant.
– a-cyclohexane-molecule
8 hours ago






"If there is no physical difference between these two functions, then why do we use the second to find out where the electron is most likely present? Why not the first function?": The former is $p(mathbf{r})$ and the latter is $p(r) = p(|mathbf{r}|)$; usually the latter is more relevant.
– a-cyclohexane-molecule
8 hours ago












1 Answer
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oldest

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The atomic orbitals (wavefunctions) $psi(r,theta,phi)$ are comprised of a radial component $R_{n,l}(r)$, as well as an angular component $Y_{l,m}(theta,phi)$. These are obtained by separately solving the radial and angular parts of the Schrödinger equation, the details of which can be found in any text. Often these are referred to as the radial and angular wavefunctions, and the angular wavefunctions are also called spherical harmonics.



As a concrete example, the $2mathrm p_z$ orbital (${n,l,m} = {2,1,0}$) has the radial component $R_{2,1}$ and the angular component $Y_{1,0}$:



$$R_{2,1} = sqrt{frac{1}{24}}left(frac{Z}{a}right)^{3/2}rhomathrm e^{-rho/2} qquad Y_{1,0} = sqrt{frac{3}{4pi}}costheta tag{1}$$



where $a$ is a collection of constants ($a equiv 4pivarepsilon_0hbar^2/mu e^2$), and $rho$ is related to $r$ by another collection of constants ($rho equiv 2Zr/na$). The product of these two expressions is the wavefunction corresponding to the $2mathrm p_z$ orbital.



The atomic orbitals themselves are normalised in the sense that



$$int_{text{all space}} |psi(r,theta,phi)|^2 mathrm dV = int_0^{2pi} !!!!int_0^pi!!!int_0^infty |R(r)|^2|Y(theta,phi)|^2 r^2sintheta,mathrm dr,mathrm dtheta,mathrm dphi = 1 tag{2}$$



Conventionally, both the radial and angular wavefunctions are separately normalised. This means that



$$int_0^infty |R(r)|^2 r^2mathrm dr = 1; qquad int_0^{2pi}!!!!int_0^pi |Y(theta,phi)|^2 sintheta,mathrm dtheta,mathrm dphi = 1 tag{3}$$



(by multiplying these two together, you recover the original normalisation of the entire atomic orbital.)



Normally, since $psi = psi(r,theta,phi)$, the probability density $|psi|^2$ depends on all three coordinates. The probability density (technically, $|psi|^2,mathrm dV$) gives you the probability of finding an electron in an infinitesimal cube located at the point $(r,theta,phi)$. By summing these $|psi|^2,mathrm dV$ over every single point in space, i.e. by integrating over all three variables, we get the total probability of finding an electron in all space, which is $1$.



However, often we are not interested in the angular dependency, and so what we can do is to sum up all the tiny cubes which have one value of $r$, but different values of $theta$ and $phi$. That would give us the probability of finding the electron in a tiny spherical shell of radius $r$. Mathematically, we need to integrate over $theta$ and $phi$. By virtue of the angular wavefunction being normalised, the integral over $theta$ and $phi$ evaluates to $1$, and hence we have



$$|psi|^2,mathrm dV = (|R|^2r^2,mathrm dr)cdot (|Y|^2sintheta,mathrm dtheta,mathrm dphi) overset{text{integrate over }theta,phi}{longrightarrow} |R|^2r^2,mathrm dr tag{4}$$



So, this quantity $|R|^2r^2,mathrm dr$ gives us the probability of finding the electron in a spherical shell of radius $r$. The corresponding probability density function is the function which controls how this probability varies with $r$, and we give it a special name, the radial distribution function (RDF); hence Atkins defines the RDF to be equal to $|R|^2r^2$.



This expression is valid regardless of the form of the angular wavefunction, as no matter what it is, when integrating over $theta$ and $phi$, it comes out to be $1$. The widely-cited alternative definition $4pi r^2 |psi|^2$ is only really relevant for spherically symmetric orbitals (in other words, s-orbitals). In this case, we have ${l, m} = {0,0}$ and



$$Y_{0,0} = sqrt{frac{1}{4pi}} tag{5}$$



(you may check that this is normalised according to equation $(3)$.) Hence



$$|psi|^2 = |R|^2|Y^2| = frac{1}{4pi}|R^2| tag{6}$$



and it therefore follows that $|R|^2r^2 = 4pi r^2|psi|^2$, so both definitions are equivalent in the case of spherically symmetric orbitals (s-orbitals). Notice that in the other linked question it is an s-orbital under consideration, and so the second definition $4pi r^2|psi|^2$ is used (or $4pi r^2 rho$, where the probability density $rho$ is simply $|psi|^2$). However, for anything that is not spherically symmetric, these are not the same and the more general expression $|R|^2r^2$ should be used.



Based on the above arguments this is my take on the matter:





  • $|R^2|r^2$ is the best definition of the RDF;


  • $4pi r^2|psi|^2$ is equivalent to that for spherically symmetric orbitals and agrees with a-cyclohexane-molecule's answer here, but does not work for other orbitals (p, d, ...);

  • I still don't fully understand why Housecroft has an additional factor of $4pi$ – something which I've noticed many years ago. There is probably some motivation behind it, but I don't know what it is. In Feodoran's answer $Y_{0,0}$ is taken to be equal to $1$, which leads to an extra factor of $4pi$. However, I think it is more conventional to have $Y_{0,0}$ normalised as $sqrt{1/(4pi)}$.






share|improve this answer























  • Since you're working in spherical coordinates, $|psi|^2,mathrm dV$ represents probability not in a cube: instead it's in a frustum-like shape, which is especially different from a cube near the origin ($rapprox0$).
    – Ruslan
    46 mins ago











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1 Answer
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active

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4














The atomic orbitals (wavefunctions) $psi(r,theta,phi)$ are comprised of a radial component $R_{n,l}(r)$, as well as an angular component $Y_{l,m}(theta,phi)$. These are obtained by separately solving the radial and angular parts of the Schrödinger equation, the details of which can be found in any text. Often these are referred to as the radial and angular wavefunctions, and the angular wavefunctions are also called spherical harmonics.



As a concrete example, the $2mathrm p_z$ orbital (${n,l,m} = {2,1,0}$) has the radial component $R_{2,1}$ and the angular component $Y_{1,0}$:



$$R_{2,1} = sqrt{frac{1}{24}}left(frac{Z}{a}right)^{3/2}rhomathrm e^{-rho/2} qquad Y_{1,0} = sqrt{frac{3}{4pi}}costheta tag{1}$$



where $a$ is a collection of constants ($a equiv 4pivarepsilon_0hbar^2/mu e^2$), and $rho$ is related to $r$ by another collection of constants ($rho equiv 2Zr/na$). The product of these two expressions is the wavefunction corresponding to the $2mathrm p_z$ orbital.



The atomic orbitals themselves are normalised in the sense that



$$int_{text{all space}} |psi(r,theta,phi)|^2 mathrm dV = int_0^{2pi} !!!!int_0^pi!!!int_0^infty |R(r)|^2|Y(theta,phi)|^2 r^2sintheta,mathrm dr,mathrm dtheta,mathrm dphi = 1 tag{2}$$



Conventionally, both the radial and angular wavefunctions are separately normalised. This means that



$$int_0^infty |R(r)|^2 r^2mathrm dr = 1; qquad int_0^{2pi}!!!!int_0^pi |Y(theta,phi)|^2 sintheta,mathrm dtheta,mathrm dphi = 1 tag{3}$$



(by multiplying these two together, you recover the original normalisation of the entire atomic orbital.)



Normally, since $psi = psi(r,theta,phi)$, the probability density $|psi|^2$ depends on all three coordinates. The probability density (technically, $|psi|^2,mathrm dV$) gives you the probability of finding an electron in an infinitesimal cube located at the point $(r,theta,phi)$. By summing these $|psi|^2,mathrm dV$ over every single point in space, i.e. by integrating over all three variables, we get the total probability of finding an electron in all space, which is $1$.



However, often we are not interested in the angular dependency, and so what we can do is to sum up all the tiny cubes which have one value of $r$, but different values of $theta$ and $phi$. That would give us the probability of finding the electron in a tiny spherical shell of radius $r$. Mathematically, we need to integrate over $theta$ and $phi$. By virtue of the angular wavefunction being normalised, the integral over $theta$ and $phi$ evaluates to $1$, and hence we have



$$|psi|^2,mathrm dV = (|R|^2r^2,mathrm dr)cdot (|Y|^2sintheta,mathrm dtheta,mathrm dphi) overset{text{integrate over }theta,phi}{longrightarrow} |R|^2r^2,mathrm dr tag{4}$$



So, this quantity $|R|^2r^2,mathrm dr$ gives us the probability of finding the electron in a spherical shell of radius $r$. The corresponding probability density function is the function which controls how this probability varies with $r$, and we give it a special name, the radial distribution function (RDF); hence Atkins defines the RDF to be equal to $|R|^2r^2$.



This expression is valid regardless of the form of the angular wavefunction, as no matter what it is, when integrating over $theta$ and $phi$, it comes out to be $1$. The widely-cited alternative definition $4pi r^2 |psi|^2$ is only really relevant for spherically symmetric orbitals (in other words, s-orbitals). In this case, we have ${l, m} = {0,0}$ and



$$Y_{0,0} = sqrt{frac{1}{4pi}} tag{5}$$



(you may check that this is normalised according to equation $(3)$.) Hence



$$|psi|^2 = |R|^2|Y^2| = frac{1}{4pi}|R^2| tag{6}$$



and it therefore follows that $|R|^2r^2 = 4pi r^2|psi|^2$, so both definitions are equivalent in the case of spherically symmetric orbitals (s-orbitals). Notice that in the other linked question it is an s-orbital under consideration, and so the second definition $4pi r^2|psi|^2$ is used (or $4pi r^2 rho$, where the probability density $rho$ is simply $|psi|^2$). However, for anything that is not spherically symmetric, these are not the same and the more general expression $|R|^2r^2$ should be used.



Based on the above arguments this is my take on the matter:





  • $|R^2|r^2$ is the best definition of the RDF;


  • $4pi r^2|psi|^2$ is equivalent to that for spherically symmetric orbitals and agrees with a-cyclohexane-molecule's answer here, but does not work for other orbitals (p, d, ...);

  • I still don't fully understand why Housecroft has an additional factor of $4pi$ – something which I've noticed many years ago. There is probably some motivation behind it, but I don't know what it is. In Feodoran's answer $Y_{0,0}$ is taken to be equal to $1$, which leads to an extra factor of $4pi$. However, I think it is more conventional to have $Y_{0,0}$ normalised as $sqrt{1/(4pi)}$.






share|improve this answer























  • Since you're working in spherical coordinates, $|psi|^2,mathrm dV$ represents probability not in a cube: instead it's in a frustum-like shape, which is especially different from a cube near the origin ($rapprox0$).
    – Ruslan
    46 mins ago
















4














The atomic orbitals (wavefunctions) $psi(r,theta,phi)$ are comprised of a radial component $R_{n,l}(r)$, as well as an angular component $Y_{l,m}(theta,phi)$. These are obtained by separately solving the radial and angular parts of the Schrödinger equation, the details of which can be found in any text. Often these are referred to as the radial and angular wavefunctions, and the angular wavefunctions are also called spherical harmonics.



As a concrete example, the $2mathrm p_z$ orbital (${n,l,m} = {2,1,0}$) has the radial component $R_{2,1}$ and the angular component $Y_{1,0}$:



$$R_{2,1} = sqrt{frac{1}{24}}left(frac{Z}{a}right)^{3/2}rhomathrm e^{-rho/2} qquad Y_{1,0} = sqrt{frac{3}{4pi}}costheta tag{1}$$



where $a$ is a collection of constants ($a equiv 4pivarepsilon_0hbar^2/mu e^2$), and $rho$ is related to $r$ by another collection of constants ($rho equiv 2Zr/na$). The product of these two expressions is the wavefunction corresponding to the $2mathrm p_z$ orbital.



The atomic orbitals themselves are normalised in the sense that



$$int_{text{all space}} |psi(r,theta,phi)|^2 mathrm dV = int_0^{2pi} !!!!int_0^pi!!!int_0^infty |R(r)|^2|Y(theta,phi)|^2 r^2sintheta,mathrm dr,mathrm dtheta,mathrm dphi = 1 tag{2}$$



Conventionally, both the radial and angular wavefunctions are separately normalised. This means that



$$int_0^infty |R(r)|^2 r^2mathrm dr = 1; qquad int_0^{2pi}!!!!int_0^pi |Y(theta,phi)|^2 sintheta,mathrm dtheta,mathrm dphi = 1 tag{3}$$



(by multiplying these two together, you recover the original normalisation of the entire atomic orbital.)



Normally, since $psi = psi(r,theta,phi)$, the probability density $|psi|^2$ depends on all three coordinates. The probability density (technically, $|psi|^2,mathrm dV$) gives you the probability of finding an electron in an infinitesimal cube located at the point $(r,theta,phi)$. By summing these $|psi|^2,mathrm dV$ over every single point in space, i.e. by integrating over all three variables, we get the total probability of finding an electron in all space, which is $1$.



However, often we are not interested in the angular dependency, and so what we can do is to sum up all the tiny cubes which have one value of $r$, but different values of $theta$ and $phi$. That would give us the probability of finding the electron in a tiny spherical shell of radius $r$. Mathematically, we need to integrate over $theta$ and $phi$. By virtue of the angular wavefunction being normalised, the integral over $theta$ and $phi$ evaluates to $1$, and hence we have



$$|psi|^2,mathrm dV = (|R|^2r^2,mathrm dr)cdot (|Y|^2sintheta,mathrm dtheta,mathrm dphi) overset{text{integrate over }theta,phi}{longrightarrow} |R|^2r^2,mathrm dr tag{4}$$



So, this quantity $|R|^2r^2,mathrm dr$ gives us the probability of finding the electron in a spherical shell of radius $r$. The corresponding probability density function is the function which controls how this probability varies with $r$, and we give it a special name, the radial distribution function (RDF); hence Atkins defines the RDF to be equal to $|R|^2r^2$.



This expression is valid regardless of the form of the angular wavefunction, as no matter what it is, when integrating over $theta$ and $phi$, it comes out to be $1$. The widely-cited alternative definition $4pi r^2 |psi|^2$ is only really relevant for spherically symmetric orbitals (in other words, s-orbitals). In this case, we have ${l, m} = {0,0}$ and



$$Y_{0,0} = sqrt{frac{1}{4pi}} tag{5}$$



(you may check that this is normalised according to equation $(3)$.) Hence



$$|psi|^2 = |R|^2|Y^2| = frac{1}{4pi}|R^2| tag{6}$$



and it therefore follows that $|R|^2r^2 = 4pi r^2|psi|^2$, so both definitions are equivalent in the case of spherically symmetric orbitals (s-orbitals). Notice that in the other linked question it is an s-orbital under consideration, and so the second definition $4pi r^2|psi|^2$ is used (or $4pi r^2 rho$, where the probability density $rho$ is simply $|psi|^2$). However, for anything that is not spherically symmetric, these are not the same and the more general expression $|R|^2r^2$ should be used.



Based on the above arguments this is my take on the matter:





  • $|R^2|r^2$ is the best definition of the RDF;


  • $4pi r^2|psi|^2$ is equivalent to that for spherically symmetric orbitals and agrees with a-cyclohexane-molecule's answer here, but does not work for other orbitals (p, d, ...);

  • I still don't fully understand why Housecroft has an additional factor of $4pi$ – something which I've noticed many years ago. There is probably some motivation behind it, but I don't know what it is. In Feodoran's answer $Y_{0,0}$ is taken to be equal to $1$, which leads to an extra factor of $4pi$. However, I think it is more conventional to have $Y_{0,0}$ normalised as $sqrt{1/(4pi)}$.






share|improve this answer























  • Since you're working in spherical coordinates, $|psi|^2,mathrm dV$ represents probability not in a cube: instead it's in a frustum-like shape, which is especially different from a cube near the origin ($rapprox0$).
    – Ruslan
    46 mins ago














4












4








4






The atomic orbitals (wavefunctions) $psi(r,theta,phi)$ are comprised of a radial component $R_{n,l}(r)$, as well as an angular component $Y_{l,m}(theta,phi)$. These are obtained by separately solving the radial and angular parts of the Schrödinger equation, the details of which can be found in any text. Often these are referred to as the radial and angular wavefunctions, and the angular wavefunctions are also called spherical harmonics.



As a concrete example, the $2mathrm p_z$ orbital (${n,l,m} = {2,1,0}$) has the radial component $R_{2,1}$ and the angular component $Y_{1,0}$:



$$R_{2,1} = sqrt{frac{1}{24}}left(frac{Z}{a}right)^{3/2}rhomathrm e^{-rho/2} qquad Y_{1,0} = sqrt{frac{3}{4pi}}costheta tag{1}$$



where $a$ is a collection of constants ($a equiv 4pivarepsilon_0hbar^2/mu e^2$), and $rho$ is related to $r$ by another collection of constants ($rho equiv 2Zr/na$). The product of these two expressions is the wavefunction corresponding to the $2mathrm p_z$ orbital.



The atomic orbitals themselves are normalised in the sense that



$$int_{text{all space}} |psi(r,theta,phi)|^2 mathrm dV = int_0^{2pi} !!!!int_0^pi!!!int_0^infty |R(r)|^2|Y(theta,phi)|^2 r^2sintheta,mathrm dr,mathrm dtheta,mathrm dphi = 1 tag{2}$$



Conventionally, both the radial and angular wavefunctions are separately normalised. This means that



$$int_0^infty |R(r)|^2 r^2mathrm dr = 1; qquad int_0^{2pi}!!!!int_0^pi |Y(theta,phi)|^2 sintheta,mathrm dtheta,mathrm dphi = 1 tag{3}$$



(by multiplying these two together, you recover the original normalisation of the entire atomic orbital.)



Normally, since $psi = psi(r,theta,phi)$, the probability density $|psi|^2$ depends on all three coordinates. The probability density (technically, $|psi|^2,mathrm dV$) gives you the probability of finding an electron in an infinitesimal cube located at the point $(r,theta,phi)$. By summing these $|psi|^2,mathrm dV$ over every single point in space, i.e. by integrating over all three variables, we get the total probability of finding an electron in all space, which is $1$.



However, often we are not interested in the angular dependency, and so what we can do is to sum up all the tiny cubes which have one value of $r$, but different values of $theta$ and $phi$. That would give us the probability of finding the electron in a tiny spherical shell of radius $r$. Mathematically, we need to integrate over $theta$ and $phi$. By virtue of the angular wavefunction being normalised, the integral over $theta$ and $phi$ evaluates to $1$, and hence we have



$$|psi|^2,mathrm dV = (|R|^2r^2,mathrm dr)cdot (|Y|^2sintheta,mathrm dtheta,mathrm dphi) overset{text{integrate over }theta,phi}{longrightarrow} |R|^2r^2,mathrm dr tag{4}$$



So, this quantity $|R|^2r^2,mathrm dr$ gives us the probability of finding the electron in a spherical shell of radius $r$. The corresponding probability density function is the function which controls how this probability varies with $r$, and we give it a special name, the radial distribution function (RDF); hence Atkins defines the RDF to be equal to $|R|^2r^2$.



This expression is valid regardless of the form of the angular wavefunction, as no matter what it is, when integrating over $theta$ and $phi$, it comes out to be $1$. The widely-cited alternative definition $4pi r^2 |psi|^2$ is only really relevant for spherically symmetric orbitals (in other words, s-orbitals). In this case, we have ${l, m} = {0,0}$ and



$$Y_{0,0} = sqrt{frac{1}{4pi}} tag{5}$$



(you may check that this is normalised according to equation $(3)$.) Hence



$$|psi|^2 = |R|^2|Y^2| = frac{1}{4pi}|R^2| tag{6}$$



and it therefore follows that $|R|^2r^2 = 4pi r^2|psi|^2$, so both definitions are equivalent in the case of spherically symmetric orbitals (s-orbitals). Notice that in the other linked question it is an s-orbital under consideration, and so the second definition $4pi r^2|psi|^2$ is used (or $4pi r^2 rho$, where the probability density $rho$ is simply $|psi|^2$). However, for anything that is not spherically symmetric, these are not the same and the more general expression $|R|^2r^2$ should be used.



Based on the above arguments this is my take on the matter:





  • $|R^2|r^2$ is the best definition of the RDF;


  • $4pi r^2|psi|^2$ is equivalent to that for spherically symmetric orbitals and agrees with a-cyclohexane-molecule's answer here, but does not work for other orbitals (p, d, ...);

  • I still don't fully understand why Housecroft has an additional factor of $4pi$ – something which I've noticed many years ago. There is probably some motivation behind it, but I don't know what it is. In Feodoran's answer $Y_{0,0}$ is taken to be equal to $1$, which leads to an extra factor of $4pi$. However, I think it is more conventional to have $Y_{0,0}$ normalised as $sqrt{1/(4pi)}$.






share|improve this answer














The atomic orbitals (wavefunctions) $psi(r,theta,phi)$ are comprised of a radial component $R_{n,l}(r)$, as well as an angular component $Y_{l,m}(theta,phi)$. These are obtained by separately solving the radial and angular parts of the Schrödinger equation, the details of which can be found in any text. Often these are referred to as the radial and angular wavefunctions, and the angular wavefunctions are also called spherical harmonics.



As a concrete example, the $2mathrm p_z$ orbital (${n,l,m} = {2,1,0}$) has the radial component $R_{2,1}$ and the angular component $Y_{1,0}$:



$$R_{2,1} = sqrt{frac{1}{24}}left(frac{Z}{a}right)^{3/2}rhomathrm e^{-rho/2} qquad Y_{1,0} = sqrt{frac{3}{4pi}}costheta tag{1}$$



where $a$ is a collection of constants ($a equiv 4pivarepsilon_0hbar^2/mu e^2$), and $rho$ is related to $r$ by another collection of constants ($rho equiv 2Zr/na$). The product of these two expressions is the wavefunction corresponding to the $2mathrm p_z$ orbital.



The atomic orbitals themselves are normalised in the sense that



$$int_{text{all space}} |psi(r,theta,phi)|^2 mathrm dV = int_0^{2pi} !!!!int_0^pi!!!int_0^infty |R(r)|^2|Y(theta,phi)|^2 r^2sintheta,mathrm dr,mathrm dtheta,mathrm dphi = 1 tag{2}$$



Conventionally, both the radial and angular wavefunctions are separately normalised. This means that



$$int_0^infty |R(r)|^2 r^2mathrm dr = 1; qquad int_0^{2pi}!!!!int_0^pi |Y(theta,phi)|^2 sintheta,mathrm dtheta,mathrm dphi = 1 tag{3}$$



(by multiplying these two together, you recover the original normalisation of the entire atomic orbital.)



Normally, since $psi = psi(r,theta,phi)$, the probability density $|psi|^2$ depends on all three coordinates. The probability density (technically, $|psi|^2,mathrm dV$) gives you the probability of finding an electron in an infinitesimal cube located at the point $(r,theta,phi)$. By summing these $|psi|^2,mathrm dV$ over every single point in space, i.e. by integrating over all three variables, we get the total probability of finding an electron in all space, which is $1$.



However, often we are not interested in the angular dependency, and so what we can do is to sum up all the tiny cubes which have one value of $r$, but different values of $theta$ and $phi$. That would give us the probability of finding the electron in a tiny spherical shell of radius $r$. Mathematically, we need to integrate over $theta$ and $phi$. By virtue of the angular wavefunction being normalised, the integral over $theta$ and $phi$ evaluates to $1$, and hence we have



$$|psi|^2,mathrm dV = (|R|^2r^2,mathrm dr)cdot (|Y|^2sintheta,mathrm dtheta,mathrm dphi) overset{text{integrate over }theta,phi}{longrightarrow} |R|^2r^2,mathrm dr tag{4}$$



So, this quantity $|R|^2r^2,mathrm dr$ gives us the probability of finding the electron in a spherical shell of radius $r$. The corresponding probability density function is the function which controls how this probability varies with $r$, and we give it a special name, the radial distribution function (RDF); hence Atkins defines the RDF to be equal to $|R|^2r^2$.



This expression is valid regardless of the form of the angular wavefunction, as no matter what it is, when integrating over $theta$ and $phi$, it comes out to be $1$. The widely-cited alternative definition $4pi r^2 |psi|^2$ is only really relevant for spherically symmetric orbitals (in other words, s-orbitals). In this case, we have ${l, m} = {0,0}$ and



$$Y_{0,0} = sqrt{frac{1}{4pi}} tag{5}$$



(you may check that this is normalised according to equation $(3)$.) Hence



$$|psi|^2 = |R|^2|Y^2| = frac{1}{4pi}|R^2| tag{6}$$



and it therefore follows that $|R|^2r^2 = 4pi r^2|psi|^2$, so both definitions are equivalent in the case of spherically symmetric orbitals (s-orbitals). Notice that in the other linked question it is an s-orbital under consideration, and so the second definition $4pi r^2|psi|^2$ is used (or $4pi r^2 rho$, where the probability density $rho$ is simply $|psi|^2$). However, for anything that is not spherically symmetric, these are not the same and the more general expression $|R|^2r^2$ should be used.



Based on the above arguments this is my take on the matter:





  • $|R^2|r^2$ is the best definition of the RDF;


  • $4pi r^2|psi|^2$ is equivalent to that for spherically symmetric orbitals and agrees with a-cyclohexane-molecule's answer here, but does not work for other orbitals (p, d, ...);

  • I still don't fully understand why Housecroft has an additional factor of $4pi$ – something which I've noticed many years ago. There is probably some motivation behind it, but I don't know what it is. In Feodoran's answer $Y_{0,0}$ is taken to be equal to $1$, which leads to an extra factor of $4pi$. However, I think it is more conventional to have $Y_{0,0}$ normalised as $sqrt{1/(4pi)}$.







share|improve this answer














share|improve this answer



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edited 7 hours ago

























answered 7 hours ago









orthocresol

37.9k7111227




37.9k7111227












  • Since you're working in spherical coordinates, $|psi|^2,mathrm dV$ represents probability not in a cube: instead it's in a frustum-like shape, which is especially different from a cube near the origin ($rapprox0$).
    – Ruslan
    46 mins ago


















  • Since you're working in spherical coordinates, $|psi|^2,mathrm dV$ represents probability not in a cube: instead it's in a frustum-like shape, which is especially different from a cube near the origin ($rapprox0$).
    – Ruslan
    46 mins ago
















Since you're working in spherical coordinates, $|psi|^2,mathrm dV$ represents probability not in a cube: instead it's in a frustum-like shape, which is especially different from a cube near the origin ($rapprox0$).
– Ruslan
46 mins ago




Since you're working in spherical coordinates, $|psi|^2,mathrm dV$ represents probability not in a cube: instead it's in a frustum-like shape, which is especially different from a cube near the origin ($rapprox0$).
– Ruslan
46 mins ago


















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