Why is the SE better than the KG equation for a hydrogen atom?

[snoopies622]

TL;DR Summary

Since the Schrodinger equation incorporates neither spin nor special relativity, why does it describe the hydrogen atom better than the Klein Gordon equation?

my premises:

— one can arrive at the Klein-Gordon equation by applying quantum mechanical operators to the special relativity dynamics equation E^2 = (mc^2)^2 + (pc)^2.

— Schrodinger arrived at this equation, but rejected it because it didn't correctly explain the behavior of an electron in a hydrogen atom.

— the reason it didn't is because it doesn't incorporate spin, so it only works for spin zero particles like the Higgs boson.

— Schrodinger then arrived at his famous "Schrodinger equation" which is based on plugging the deBroglie relation lambda=h/p into a generic wave equation, leaving out special relativity altogether, and he published it because, unlike the Klein-Gordon equation, it does work pretty well with the hydrogen atom.

— Dirac then "took the square root" of the Klein-Gordon equation to produce his Dirac equation, which works even better with the hydrogen atom because it both incorporates particle spin and is consistent with special relativity.

My question is: Since the Schrodinger equation incorporates neither spin nor special relativity, why does it describe the hydrogen atom better than the Klein-Gordon equation?

I realize that my premises probably aren't exactly correct, and would appreciate any feedback.

 

[Paul Colby]

Do you have a reference to the Klein Gordon hydrogen solutions? It's be nice to know how far off it is. Also, is it clear how one should add the electrostatic potential to the KG equation?

 

[Vanadium 50]

My question is: Since the Schrodinger equation incorporates neither spin nor special relativity, why does it describe the hydrogen atom better than the Klein-Gordon equation?

Who says it does better?

 

[topsquark]

Summary: Since the Schrodinger equation incorporates neither spin nor special relativity, why does it describe the hydrogen atom better than the Klein Gordon equation?

my premises:

— one can arrive at the Klein-Gordon equation by applying quantum mechanical operators to the special relativity dynamics equation E^2 = (mc^2)^2 + (pc)^2.

— Schrodinger arrived at this equation, but rejected it because it didn't correctly explain the behavior of an electron in a hydrogen atom.

— the reason it didn't is because it doesn't incorporate spin, so it only works for spin zero particles like the Higgs boson.

— Schrodinger then arrived at his famous "Schrodinger equation" which is based on plugging the deBroglie relation lambda=h/p into a generic wave equation, leaving out special relativity altogether, and he published it because, unlike the Klein-Gordon equation, it does work pretty well with the hydrogen atom.

— Dirac then "took the square root" of the Klein-Gordon equation to produce his Dirac equation, which works even better with the hydrogen atom because it both incorporates particle spin and is consistent with special relativity.

My question is: Since the Schrodinger equation incorporates neither spin nor special relativity, why does it describe the hydrogen atom better than the Klein-Gordon equation?

I realize that my premises probably aren't exactly correct, and would appreciate any feedback.

The S-equation can be used in any circumstance because it doesn't account for spin. The KG equation specifically works only for spin 0. Since the electron is spin 1/2 we need to use the Dirac equation if we are going to go Relativistic.

-Dan

 

[snoopies622]

The S-equation can be used in any circumstance because it doesn't account for spin. The KG equation specifically works only for spin 0.

This is puzzling to me since neither the SE nor the KGE use spin in their derivations. (I know "derivations" isn't really a proper word here, but neither assumes that spin exists in their starting premises.)

Do you have a reference to the Klein Gordon hydrogen solutions? It'd be nice to know how far off it is. Also, is it clear how one should add the electrostatic potential to the KG equation?

I don't know the answers Paul, but these are good questions. My recent thinking about this matter was inspired by this Dirac video.

 

[topsquark]

This is puzzling to me since neither the SE nor the KGE use spin in their derivations. (I know "derivations" isn't really a proper word here, but neither assumes that spin exists in their starting premises.)

The simplest way for me to explain this (and someone else may have a better answer) is that spin is a Relativistic property. So if the KG equation does not contain a spin component to its wavefunction then that spin component is a constant. The S-equation is non-relativistic and does not contain spin because, well, according to non-Relativistic theory, spin does not exist.

When trying to use the KG equation for electrons we get contradictions... the spin statistics are all wrong because the KG equation Mathematically describes bosons. I suppose you could consider it to be an approximation to an electron but it doesn't really do that good a job with electrons being fermions and all that. The S-equation describes "Boltzmann" particles (I think I just made up that term) and so we can simply "tack on" a spin 1/2 wavefunction to the S-equation wavefunctions and there is no problem.

-Dan

 

[PeterDonis]

spin is a Relativistic property

This is not correct. It's perfectly possible to model spin in non-relativistic QM.

What non-relativistic QM can't explain is the spin-statistics connection (which doesn't mean non-relativistic QM can't model it, just that non-relativistic QM has to adopt it as an additional postulate instead of deriving it) and the relativistic corrections to spin-related properties, like the gyromagnetic ratio.

we can simply "tack on" a spin 1/2 wavefunction to the S-equation wavefunctions and there is no problem.

Not quite; what you can do is add appropriate spin degrees of freedom to the Hilbert space and add appropriate spin-related terms to the Hamiltonian, as in the Schrodinger-Pauli equation:

https://en.wikipedia.org/wiki/Pauli_equation

You don't "tack on" anything to the wave function, you just get wave functions that are functions of additional degrees of freedom besides position/momentum when you solve the above equation.

 

[PeterDonis]

When trying to use the KG equation for electrons we get contradictions... the spin statistics are all wrong because the KG equation Mathematically describes bosons.

For the specific case asked about in the OP, the hydrogen atom, none of this matters, because there is only one electron, so statistics never comes into play. To see issues like this you need to model a system with at least two electrons; I believe the first two such systems were the H2 molecule and the helium atom.

 

[Paul Colby]

Klein-Gordon Wiki the section on scalar electrodynamics gives a reasonable interaction Lagrangian for a charged KG field which preserves gauge invariance. This would be where I would start if I wanted to set up the hydrogen problem with KG.

 

[renormalize]

My question is: Since the Schrodinger equation incorporates neither spin nor special relativity, why does it describe the hydrogen atom better than the Klein-Gordon equation?

A comprehensive comparison of the energy-levels of a single-electron atom using the Schrodinger-Pauli, Klein-Gordon (KG) and Dirac equations can be found in Paul Strange's Relativistic Quantum Mechanics (Cambridge U Press, 1998). He observes (pg. 240) the following regarding his derived expressions for the energies of the relativistic KG & Dirac wavefunctions (with my clarifications appearing in brackets):

"Clearly the first term [in eqs. (3.123) and (8.49)] is identical with the non-relativistic [Schrodinger] energy levels for the one-electron atom. The lifting of degeneracies comes at the order $E_{Ry}^{2}/mc^{2}$, where $E_{Ry}$ is the non-relativistic energy of the ground state of the one-electron atom."

and goes on to state:

"Equation (8.49) [for the Dirac energy levels] is very similar to equation (3.123) for the spinless [KG] atom. The only difference between the two is that $l+1/2$ in the denominator of the $n^{3}$ term is replaced by the combined spin-angular momentum quantum number $\kappa$. This is the leading order effect of spin on the energy levels of hydrogen. When the spectrum of atomic hydrogen is observed experimentally, splittings consistent with with equation (8.49), not with equation (3.123), are observed, hence the spectrum of hydrogen is strong evidence for the existence of electron spin."

So to the extent one can ignore electron spin, the complex relativistic KG equation is quite adequate for modelling a single-electron atom.

 

[vanhees71]

The hydrogen atom problem has so large symmetry that it is hard to get it wrong. That's why even Bohr's model gets the right energy levels. It's another funny coincidence that the relativistic treatment of Bohr's model, first done by Sommerfeld, gets the finestructure correct although, of course, it doesn't contain the notion of spin at all.

Schrödinger started his investigations using the equation we call nowadays Klein-Gordon equation, i.e., assuming spin 0 for the electron, and he got energy levels which were wrong concerning the fine structure, which was known to be a relativistic effect from the Bohr-Sommerfeld model. So he thought it's better to step back and first consider the non-relativistic case for the wave function first, and this leads to the correct energy levels neglecting the finestructure corrections.

Later it turned out that with the correct relativistic equation for spin-1/2 models, i.e., Dirac's equation the fine structure turned out correctly and coinciding with the result by Sommerfeld within the Bohr model.

 

[malawi_glenn]

I think this is treated in Gross "relativistic quantum mechanics and field theory" (Wiley)

 

[snoopies622]

Thanks, all! This has been very helpful indeed. The only thing I'm not clear on is what exactly is the connection (if any) between particle spin and special relativity. I mean, particles like electrons exhibit spin whether or not they're moving at relativistic velocities, no?

 

[PeterDonis]

The only thing I'm not clear on is what exactly is the connection (if any) between particle spin and special relativity. I mean, particles like electrons exhibit spin whether or not they're moving at relativistic velocities, no?

As I noted in post #7, it is indeed possible to model spin in non-relativistic QM. However, the spin-statistics connection (i.e., that particles of half-integer spin are fermions and obey the Pauli exclusion principle, while particles of integer spin are bosons) needs to be postulated in non-relativistic QM, whereas in relativistic QM it can be derived. Also, there are relativistic effects of spin that are captured in relativistic QM but which cannot be captured in non-relativistic QM.

 

[vanhees71]

Indeed, in non-relativistic QM spin is just something you put on top of the usual Heisenberg algebra.

To understand it better you can analyze the unitary ray representations of the Galilei group. It turns out that you get only a physically interpretable quantum mechanics when you use a central extension of the classical Galilei group with mass being a central charge of the corresponding Lie algebra. Also instead of the SO(3) rotation subgroup you have to use the covering group SU(2). Then you look for the unitary irreducible representations of this "quantum Galilei group". It turns out that as a complete set of compatible observables you can use the three momentum components and the spin-$z$ component. The spin $s \in \{0,1/2,1,\ldots \}$ defines the representation of the spatial rotations in the subspace with $\vec{p}=0$ (the socalled "little group"). The irrep. is then characterized by the value of the mass $m>0$ as a central charge and the spin quantum number $s$. Total angular momentum turns then out to be $\vec{J}=\vec{L}+\vec{S}$, where $\vec{L}$ is the usual orbital angular-momentum operator, and the position operator can always be constructed from the Galilei-boost generators in the obvious way, and spin is simply commuting with $\vec{x}$ and $\vec{p}$.

In the relativistic case there are more constraints to fulfill when constructing the observable algebra out of the symmetry group of spacetime, i.e., the proper orthochronous Poincare group. It turns out that in contradistinction to the Galilei group the Poincare group's Lie algebra does not have non-trivial central charges, i.e., without loss of generality you can simply look for the unitary irreps of the Poincare group, and any such irrep. is first of all characterized by the Casimir operators of the group representation, and that are the $p_{\mu} p^{\mu}=m^2$ (using natural units with $c=\hbar=1$) and the four-product square of the socalled Pauli-Ljubanski vector. The additional constraints come from the demand to find causal realizations of the dynamics of quantum systems, and as it turns out that can in the most simple way be realized in the quantum-field theoretical formulation. The "first-quantization formalism" which works in non-relativistic QM does not work here, because it turns out that there is no way to localize a single particle but in trying to do so due to the Heisenberg uncertainty relation for position and momentum, for localizing a particle better and better you have to confine it in a box, and the more the momentum must fluctuate. Squeezing the particle in a box with dimensions close to the particle's Compton wavelength usually particle anti-particle pairs are created, and you cannot ensure that you have only exactly one particle in the box (see Coleman's lectures on quantum field theory for a masterful explanation of this).

To construct the QFT now you can also analyze the unitary irreps of the Poincare group. In addition you can demand the microcausality constraint, i.e., that the local observables (like energy, momentum, and angular-momentum densities) commute at space-like separation of their arguments, which is achieved (at least for the free fields) when the fields and their derivatives commute (bosons) or anti-commute (fermions) at space-like separations of their arguments. This restricts (at least for the interacting case) the physically meaningful representations to either $m^2>0$ ("massive particles") or $m^2=0$ ("massless quanta").

For $m^2>0$ you can use $\vec{p}=0$ as the "standard momentum", and then in the corresponding subspace the Pauli-Ljubanski vector components obey the commutation relations of angular-momentum components (modulo a normalization). So in this case the modulus squared defines the spin of the particles as in the non-relativistic case, i.e., and the Pauli-Ljubanski components are the generators of the corresponding "little group", which keeps the momentum at $\vec{p}=0$, which of course is a representation of the rotation group with $s \in \{0,1/2,1,\ldots \}$. Observable is, however, only the total angular momentum, and the splitting in orbital and spin angular momentum components turns out to be rather arbitrary. In contradistinction to the non-relativistic case, the microcausality constraint and the demand that the energy should be bounded from below (by convention you choose the ground state to refer to $E=0$, and the corresponding vacuum state is invariant under Poincare transformations) lead to some remarkable conclusions

(a) the field operators must consist of specific superpositions of positive- as well as negative-freuquency energy-momentum eigenmodes. To have positive energies only one needs to use annihilation operators as coefficients of the postive-frequency modes and creation operators for the negative-frequency modes. Thus any quantum field with a given spin and mass $m^2>0$ describes particles and anti-particles, where you can always also have the special case, where the particles and antiparticles are the same ("strictly neutral particles").

(b) The fields for particles with integer spin must be quantized as bosons (i.e., with commutators for the equal-time canonical commutation relations) and those with half-integer spin as fermions (i.e., with anti-commutators for the equal-time canonical "commuation" relations). In contradistinction to the non-relativistic case this empirical finding thus can be derived from the space-time symmetry of Minkowski space and causality and stability constraints.

(c) Any such constructed microcausal ("local") relativistic QFT is invariant under the "grand reflection" CPT, i.e., given any scattering process you get another possible such process when substituting any particle by its anti-particle and doing a spatial reflection as well as using the time-reversed initial and final states. This is also very well confirmed empirically, while it is also confirmed that the weak interaction breaks all other kinds of these "reflection symmetries", i.e., P, T, C, CP, PT, CT.

The other possibility are the massless representations, i.e., those with $m^2=0$. There you cannot choose $\vec{p}=0$ as the standard momentum but only $p^{\mu}=(\pm p,0,0,p)$. The little group then turns out to be isomorphic to the symmetry group of the 2D Euclidean affine space, ISO(2), being generated by the SO(2) rotations around the origin and the two translations. The latter have the meaning of "null rotations", i.e., the "rotations" keeping the standard momentum $(\pm p,0,0,p)$ invariant but not the usual rotations around the $3$-axis. Since there are no particles with "continuous" spin, these null-rotations must be represented trivially and in order to get the usual integer and half-integer representations of the rotations on the entire space spanned by the momentum eigenvectors with $\vec{p} \neq 0$, the little group is characterized by the "helicity" $\lambda \in \{0,\pm 1/2,\pm 1, \ldots \}$, generated by the Pauli-Ljubanski-vector component $W^3$ but not by the Casimir operator $W_{\mu} W^{\mu}$, which is always 0 in these "massless representations". Also here the spin-statistics relation and CPT invariance follows from microcausality and stability constraints.

 

[Paul Colby]

To understand it better you can analyze the unitary ray representations of the Galilei group.

Do the Galilei group representations lead to the electrons gyro magnetic ratio of 2?

 

[vanhees71]

Only with a trick. If you take the free Pauli field (i.e., the non-relativistic field describing spin-1/2 particles) and write the Hamiltonian as
$$\hat{H}=\frac{1}{2m} (\hat{\vec{p}} \cdot \hat{\sigma})^2$$
and then use the "minimal substitution"
$$\vec{\nabla} \rightarrow \vec{\nabla}-\mathrm{i} q \vec{A}, \quad \partial_t \rightarrow \partial_t + \mathrm{i} q \Phi$$
you get the correct Pauli equation with the gyrofactor, $2$, but that's pretty arbitrary, because indeed
$$\hat{H}=\frac{1}{2m} \hat{\vec{p}}^2,$$
and if you make the minimal substitution there you don't get the gyrofractor of $2$.

For the Dirac equation it's more convincing, because there you have the $\gamma$-matrices explicitly in the Lagrangian/Hamiltonian, and together with Poincare invariance minimal substitution inevitably leads to the gyrofactor of 2.

 

[dextercioby]

Do the Galilei group representations lead to the electrons gyro magnetic ratio of 2?

Yes. Coupling a Galilei-invariant „wave-equation” to an electromagnetic field entails $g=2$.

 

[vanhees71]

Well, if you do the "minimal coupling" to the right form of the Hamiltonian, which is not unique (see #17).