Solving the Radial Equation for the Dirac Hydrogen Atom Solution

[topsquark]

TL;DR Summary

I'm trying to solve the radial equation for the Dirac Hydrogen atom.

I'm going to be a bit sketchy here, at least to start with. If you want me to show you exactly where I am I might post a pdf, if that's okay. (Only because it will simplify coding several pages of LaTeX.)

Briefly, what I'm trying to do is take this system of equations:
##F^{ \prime } + \dfrac{k}{ \rho } F = \left ( a - \dfrac{b}{ \rho } \right ) G##

##G^{ \prime } - \dfrac{k}{ \rho } G = \left ( a + \dfrac{b}{ \rho } \right ) F##

This is about half way through the Dirac Radial equation solution, just before we would take the large and small ##\rho## limits to show that we need to have ##e^{- \rho }## and ## \rho ^s## factors on F and G to keep the solutions finite. (And that just before we do a series solution.)

It's a mess, but what I'm trying to show is that the solution for F is
##F( \rho ) = e^{- \rho } \rho ^s \left ( A(2 \rho ) L_{n - k - 1}^{2s + 1} (2 \rho ) + B L_{n - k}^{2s - 1} (2 \rho ) \right ) ##
where the L are Laguerre polynomials.

My approach is to solve the top equation for G, take the derivative, and plug the G and G' into the second equation, leaving an equation for F. Then the goal is to substitute ##F = e^{ - \rho } \rho ^s H( \rho )## into it and show that we may reduce the equation for H to one that matches the solution.

Long story short I can't find a way to use the Laguerre differential equation to cancel things out.

Much more detail upon request, but at this point my question is merely, "Am I barking up the wrong tree?" Is there a better way to approach this?

Thanks!

-Dan

Addendum: Ignore those things below. I don't know how to get rid of them.
F′+kρ=(a−bρ)G

G′−kρ=(1a+bρ)F

 

[Vanadium 50]

Um, why?

If this is a mathematical exercise, OK, I get that. If you are actually trying to get an answer, the speed of a H electron is c/137, so relativistic corrections are small (and better handled via perturbation theory).

 

[topsquark]

Um, why?

If this is a mathematical exercise, OK, I get that. If you are actually trying to get an answer, the speed of a H electron is c/137, so relativistic corrections are small (and better handled via perturbation theory).

Pretty much just as a challenge. Conceptually it really isn't much different from the Schrodinger version and even though the derivation of the energy eigenvalue was a bit more "strenuous" than I had expected, it wasn't all that bad. The angular part is pretty straightforward as well. I just can't seem to finish out the radial part.

It would seem that most of the internet agrees with you: I can't find details anywhere. The energy eigenvalue is derived and "then a miracle occurs" and we suddenly have the radial solution.

-Dan

 

[Vanadium 50]

Well, the standard trick for solving differential equations is "assume a solution of the form".

It has been a very long time since I have done this. But since the angular part has worked out for you, I'd swittch to parabolic coordinates and take advantage of the radial/angular degeneracy. You have all the pieces, even though the remaining work is likely to be...strenuous.

 

[renormalize]

##F^{ \prime } + \dfrac{k}{ \rho } F = \left ( a - \dfrac{b}{ \rho } \right ) G##

##G^{ \prime } - \dfrac{k}{ \rho } G = \left ( a + \dfrac{b}{ \rho } \right ) F##

You should double check these equations. I looked in Sakurai, Advanced Quantum Mechanics (eq. 3.299) and at this UCSD site. Both agree that ##F##, ##G## satisfy (in your notation):$$F'-\frac{k}{\rho}F=\left(a-\frac{b}{\rho}\right)G$$$$G'+\frac{k}{\rho}G=\left(\frac{1}{a}+\frac{b}{\rho}\right)F$$I doubt the sign differences between these and your equations are significant since they can likely be made to agree by redefining the functions and constants. But the distinction between ##a## and ##1/a## seems crucial.

 

[topsquark]

You should double check these equations. I looked in Sakurai, Advanced Quantum Mechanics (eq. 3.299) and at this UCSD site. Both agree that ##F##, ##G## satisfy (in your notation):$$F'-\frac{k}{\rho}F=\left(a-\frac{b}{\rho}\right)G$$$$G'+\frac{k}{\rho}G=\left(\frac{1}{a}+\frac{b}{\rho}\right)F$$I doubt the sign differences between these and your equations are significant since they can likely be made to agree by redefining the functions and constants. But the distinction between ##a## and ##1/a## seems crucial.

Gah! Yes, it was simply a typo.

I'll double check the signs as well, but I think those are okay (as defined with the rest of my derivation.)

Thanks for the catch!

-Dan

 

[renormalize]

It's a mess, but what I'm trying to show is that the solution for F is
##F( \rho ) = e^{- \rho } \rho ^s \left ( A(2 \rho ) L_{n - k - 1}^{2s + 1} (2 \rho ) + B L_{n - k}^{2s - 1} (2 \rho ) \right ) ##
where the L are Laguerre polynomials.

...Long story short I can't find a way to use the Laguerre differential equation to cancel things out.

Much more detail upon request, but at this point my question is merely, "Am I barking up the wrong tree?" Is there a better way to approach this?

It's fairly straightforward (but somewhat lengthy!) to show that the Dirac radial equations for a single-electron atom are indeed satisfied by combinations of associated Laguerre polynomials.

To maintain contact with an existing QM textbook treatment, I begin by rewriting eq.(3.299) of Sakurai, Advanced Quantum Mechanics, using the OP's notation:$$F'-\frac{k}{\rho}F=\left(a-\frac{b}{\rho}\right)G,\;\; G'+\frac{k}{\rho}G=\left(\frac{1}{a}+\frac{b}{\rho}\right)F\tag{1a,b}$$The dimensionless parameters appearing in (1) are defined in terms of physical quantities by:$$\rho:=\frac{\sqrt{m^{2}c^{4}-E^{2}}}{\hbar c}r,\;\;a:=\sqrt{\frac{mc^{2}-E}{mc^{2}+E}},\;\;b:=Z\alpha,\;\;k:=\pm\left(j+\frac{1}{2}\right)\tag{2}$$(Here ##r## is the usual radial coordinate, ##E## and ##m## are the electron energy and mass, ##Z## is the number of positive charges on the atomic nucleus, ##\alpha## is the fine-structure constant, and ##j## is the quantum number of total angular momentum (one-half of an odd-integer).)

Next, in lieu of substituting (1a) into (1b) and vice versa to get second-order differential equations for ##F## and ##G##, I apply a clever ansatz originated by F.D. Pidduck in 1929. He turned (1a) and (1b) into two second-order equations for one unknown function ##P## by expressing ##F## and ##G## as linear combinations of ##P## with its derivative ##P'##:$$F=e^{-\rho}\rho^{s}\left(p_{1}P+p_{2}\thinspace\rho\thinspace P'\right),\;\;G=e^{-\rho}\rho^{s}\left(p_{3}P+p_{4}\thinspace\rho\thinspace P'\right)\tag{3a,b}$$Note that, by virtue of the overall factor ##e^{-\rho}\rho^{s}## appearing in (3), the natural boundary conditions for the function ##P## are: ##P(0)## must be nonzero finite and ##P(\rho)## must grow no faster than a power of ##\rho## as ##\rho\rightarrow\infty##.

The constants ##p_{2},p_{3},p_{4}## are now determined as follows. First, insert (3) into (1a,b) to get two distinct second-order differential equations, which I denote as ##D_{1a}^{(2)}P=0## and ##D_{1b}^{(2)}P=0##, for the one function ##P##. This is consistent only if the left sides of these two equations can be made proportional to one another:$$D_{1a}^{(2)}P=\lambda D_{1b}^{(2)}P\tag{4}$$I accomplish this by equating the coefficients of ##P,P',P''## appearing on each side of (4) and solving the resulting three simple algebraic equations to determine the unknown constants:$$\lambda=a,\;\;p_{2}=\frac{p_{1}a}{a\left(k+s\right)-b},\;\;p_{3}=\frac{p_{1}\left(ab-k+s\right)}{a\left(k+s\right)-b},\;\;p_{4}=\frac{p_{1}}{a\left(k+s\right)-b}\tag{5}$$Using (5) and discarding a nonessential multiplicative factor, I ultimately arrive at the single differential equation that ##P## must satisfy:$$\rho P''+\left(1+2s-2\rho\right)P'+\left(2\nu+\frac{s^{2}-\sigma^{2}}{\rho}\right)P=0\tag{6a}$$where:$$\sigma:=\sqrt{k^{2}-b^{2}},\;\;\nu:=\frac{b}{2a}-\frac{ab}{2}-s\tag{6b,c}$$Mathematica easily integrates (6a) to find its solution:$$P\left(\rho\right)=\rho^{\sigma-s}\left(c_{1}L_{\nu+s-\sigma}^{2\sigma}\left(2\rho\right)+c_{2}U\left(-\nu-s+\sigma,1+2\sigma,2\rho\right)\right)\tag{7}$$in terms of the associated Laguerre function ##L^{\beta}_{\mu}(z)## and the confluent hypergeometric function of the second kind ##U(x,y,z)##. Requiring nonzero finiteness at ##\rho=0##, I must set ##s=\sigma## and ##c_{2}=0## since ##U(x,y,0)## is singular. The remaining solution piece ##L^{2\sigma}_{\nu}(2\rho)## suffers from an unphysical essential singularity at ##\rho=\infty## unless ##\nu## is restricted to be a non-negative integer ##n##. With these choices (and putting ##c_{1}=1##), eq.(7) reduces to the physical solution:$$P\left(\rho\right)=L_{n}^{2s}\left(2\rho\right)\tag{8}$$where ##L_{n}^{2s}## is the ##n^{\text{th}}##-order associated Laguerrre polynomial. Putting (8) into (3) provides the sought-after solutions for ##F,G##:$$F\left(\rho\right)=p_{1}e^{-\rho}\rho^{s}\left(L_{n}^{2s}\left(2\rho\right)+\frac{2a\rho L_{n-1}^{2s+1}\left(2\rho\right)}{a\left(k+s\right)-b}\right)\tag{9a}$$$$G\left(\rho\right)=p_{1}e^{-\rho}\rho^{s}\left(\frac{\left(ab-k+s\right)L_{n}^{2s}\left(2\rho\right)-2\rho L_{n-1}^{2s+1}\left(2\rho\right)}{a\left(k+s\right)-b}\right)\tag{9b}$$$$s=\sqrt{k^{2}-b^{2}},\;\;n=\frac{b}{2a}-\frac{ab}{2}-\sqrt{k^{2}-b^{2}},\;\;n=0,1,2,\ldots\tag{9c,d,e}$$(Observe that the form for ##F## in eq.(9a) is apparently not the same as that guessed by the OP.) Also note that eq.(9d,e) is precisely the quantum condition for the energy-eigenvalues of a Dirac single-electron atom. This can be verified by substituting ##a,b,k## from (2) into (9d) and then solving for the energy ##E##, yielding:$$E=\frac{mc^{2}}{\sqrt{1+\frac{Z^{2}\alpha^{2}}{\left(n+\sqrt{\left(j+\frac{1}{2}\right)^{2}-Z^{2}\alpha^{2}}\right)^{2}}}}$$in agreement with Sakurai eq.(3.331).

I leave it to the reader to judge whether the above method of solving for ##F,G## directly in terms of the well-understood associated Leguerre polynomials is superior to (or merely different from) the classic QM textbook approach that relies on developing terminating power-series solutions. For the interesting history of this topic, I recommend the 2010 review article "Schrödinger and Dirac equations for the hydrogen atom, and Laguerre polynomials" by Mawhin and Ronveaux.

 

[topsquark]

It's fairly straightforward (but somewhat lengthy!) to show that the Dirac radial equations for a single-electron atom are indeed satisfied by combinations of associated Laguerre polynomials.

To maintain contact with an existing QM textbook treatment, I begin by rewriting eq.(3.299) of Sakurai, Advanced Quantum Mechanics, using the OP's notation:$$F'-\frac{k}{\rho}F=\left(a-\frac{b}{\rho}\right)G,\;\; G'+\frac{k}{\rho}G=\left(\frac{1}{a}+\frac{b}{\rho}\right)F\tag{1a,b}$$The dimensionless parameters appearing in (1) are defined in terms of physical quantities by:$$\rho:=\frac{\sqrt{m^{2}c^{4}-E^{2}}}{\hbar c}r,\;\;a:=\sqrt{\frac{mc^{2}-E}{mc^{2}+E}},\;\;b:=Z\alpha,\;\;k:=\pm\left(j+\frac{1}{2}\right)\tag{2}$$(Here ##r## is the usual radial coordinate, ##E## and ##m## are the electron energy and mass, ##Z## is the number of positive charges on the atomic nucleus, ##\alpha## is the fine-structure constant, and ##j## is the quantum number of total angular momentum (one-half of an odd-integer).)

Next, in lieu of substituting (1a) into (1b) and vice versa to get second-order differential equations for ##F## and
##G##, I apply a clever ansatz originated by F.D. Pidduck in 1929. He turned (1a) and (1b) into two second-order equations for one unknown function ##P## by expressing ##F## and ##G## as linear combinations of ##P## with its derivative ##P'##:$$F=e^{-\rho}\rho^{s}\left(p_{1}P+p_{2}\thinspace\rho\thinspace P'\right),\;\;G=e^{-\rho}\rho^{s}\left(p_{3}P+p_{4}\thinspace\rho\thinspace P'\right)\tag{3a,b}$$Note that, by virtue of the overall factor ##e^{-\rho}\rho^{s}## appearing in (3), the natural boundary conditions for the function ##P## are: ##P(0)## must be nonzero finite and ##P(\rho)## must grow no faster than a power of ##\rho## as ##\rho\rightarrow\infty##.

The constants ##p_{2},p_{3},p_{4}## are now determined as follows. First, insert (3) into (1a,b) to get two distinct second-order differential equations, which I denote as ##D_{1a}^{(2)}P=0## and ##D_{1b}^{(2)}P=0##, for the one function ##P##. This is consistent only if the left sides of these two equations can be made proportional to one another:$$D_{1a}^{(2)}P=\lambda D_{1b}^{(2)}P\tag{4}$$I accomplish this by equating the coefficients of ##P,P',P''## appearing on each side of (4) and solving the resulting three simple algebraic equations to determine the unknown constants:$$\lambda=a,\;\;p_{2}=\frac{p_{1}a}{a\left(k+s\right)-b},\;\;p_{3}=\frac{p_{1}\left(ab-k+s\right)}{a\left(k+s\right)-b},\;\;p_{4}=\frac{p_{1}}{a\left(k+s\right)-b}\tag{5}$$Using (5) and discarding a nonessential multiplicative factor, I ultimately arrive at the single differential equation that ##P## must satisfy:$$\rho P''+\left(1+2s-2\rho\right)P'+\left(2\nu+\frac{s^{2}-\sigma^{2}}{\rho}\right)P=0\tag{6a}$$where:$$\sigma:=\sqrt{k^{2}-b^{2}},\;\;\nu:=\frac{b}{2a}-\frac{ab}{2}-s\tag{6b,c}$$Mathematica easily integrates (6a) to find its solution:$$P\left(\rho\right)=\rho^{\sigma-s}\left(c_{1}L_{\nu+s-\sigma}^{2\sigma}\left(2\rho\right)+c_{2}U\left(-\nu-s+\sigma,1+2\sigma,2\rho\right)\right)\tag{7}$$in terms of the associated Laguerre function ##L^{\beta}_{\mu}(z)## and the confluent hypergeometric function of the second kind ##U(x,y,z)##. Requiring nonzero finiteness at ##\rho=0##, I must set ##s=\sigma## and ##c_{2}=0## since ##U(x,y,0)## is singular. The remaining solution piece ##L^{2\sigma}_{\nu}(2\rho)## suffers from an unphysical essential singularity at ##\rho=\infty## unless ##\nu## is restricted to be a non-negative integer ##n##. With these choices (and putting ##c_{1}=1##), eq.(7) reduces to the physical solution:$$P\left(\rho\right)=L_{n}^{2s}\left(2\rho\right)\tag{8}$$where ##L_{n}^{2s}## is the ##n^{\text{th}}##-order associated Laguerrre polynomial. Putting (8) into (3) provides the sought-after solutions for ##F,G##:$$F\left(\rho\right)=p_{1}e^{-\rho}\rho^{s}\left(L_{n}^{2s}\left(2\rho\right)+\frac{2a\rho L_{n-1}^{2s+1}\left(2\rho\right)}{a\left(k+s\right)-b}\right)\tag{9a}$$$$G\left(\rho\right)=p_{1}e^{-\rho}\rho^{s}\left(\frac{\left(ab-k+s\right)L_{n}^{2s}\left(2\rho\right)-2\rho L_{n-1}^{2s+1}\left(2\rho\right)}{a\left(k+s\right)-b}\right)\tag{9b}$$$$s=\sqrt{k^{2}-b^{2}},\;\;n=\frac{b}{2a}-\frac{ab}{2}-\sqrt{k^{2}-b^{2}},\;\;n=0,1,2,\ldots\tag{9c,d,e}$$(Observe that the form for ##F## in eq.(9a) is apparently not the same as that guessed by the OP.) Also note that eq.(9d,e) is precisely the quantum condition for the energy-eigenvalues of a Dirac single-electron atom. This can be verified by substituting ##a,b,k## from (2) into (9d) and then solving for the energy ##E##, yielding:$$E=\frac{mc^{2}}{\sqrt{1+\frac{Z^{2}\alpha^{2}}{\left(n+\sqrt{\left(j+\frac{1}{2}\right)^{2}-Z^{2}\alpha^{2}}\right)^{2}}}}$$in agreement with Sakurai eq.(3.331).

I leave it to the reader to judge whether the above method of solving for ##F,G## directly in terms of the well-understood associated Leguerre polynomials is superior to (or merely different from) the classic QM textbook approach that relies on developing terminating power-series solutions. For the interesting history of this topic, I recommend the 2010 review article "Schrödinger and Dirac equations for the hydrogen atom, and Laguerre polynomials" by Mawhin and Ronveaux.

Very interesting. I'll have to play with that for a while. It's a lot more constructive than the approach I was working with.

-Dan

 

[malawi_glenn]

I think "Relativistic Quantum Mechanics and Field Theory" by Gross covers this
https://www.wiley.com/en-us/Relativistic+Quantum+Mechanics+and+Field+Theory-p-9783527617340

 

[dextercioby]

I think "Relativistic Quantum Mechanics and Field Theory" by Gross covers this
https://www.wiley.com/en-us/Relativistic+Quantum+Mechanics+and+Field+Theory-p-9783527617340

Yes, that is the series solution in there. But I love the full blown solution from post #7.

 

[malawi_glenn]

Yes, that is the series solution in there. But I love the full blown solution from post #7.

Great, it was about 15 years since last time I opened that book so I might have forgotten the details (too lazy to find it in my bookshelf hehe)