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In many textbooks and lecture notes, it says that the Klein-Gordon equation was discarded first because when interpreting it as an equation for a single-particle wave function and trying to calculate the spectrum of the hydrogen atom, it gives wrong results.

But what exactly does go wrong when trying to calculate the spectrum of hydrogen from the Klein-Gordon equation instead of using the Schroedinger equation? How doe the wrong results look like?

Nowhere have I seen a detailled explanation of this issue.

2 Answers2

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For hydrogen-like atoms (an electron and a nucleus with atomic number $Z$) the difference between the solutions of the Dirac equation and of the Klein-Gordon equation is quite subtle. So we need to look at them in more detail.

Dirac equation

Solving the Dirac equation gives the following exact energy levels (see Wikipedia - Hydrogen-like atom - Solution to Dirac equation): $$E_{nj}=mc^2\left[1+\left(\frac{Z\alpha}{n-j-\frac 12+\sqrt{\left(j+\frac 12\right)^2-Z^2\alpha^2}}\right)^2\right]^{-1/2}$$ with the quantum numbers

  • $n=1,2,...$
  • $l=0,1,2,...,n-1$
  • $j=l+\frac 12, |l-\frac 12|$

and the fine-structrure constant $\alpha=\frac{1}{137.06}$.

Taylor-expansion for small $\alpha$ gives: $$E_{nj}\approx mc^2\left[1-\frac{Z^2\alpha^2}{2n^2}-\frac{Z^4\alpha^4}{2n^3}\left(\frac{1}{j+\frac 12}-\frac{3}{4n}\right)+\cdots\right]$$

The first few energy levels calculated from this are: $$\begin{array}{c|ccc|c} & n & l & j & E_{nj} \\ \hline 1S_{1/2} & 1 & 0 & \frac 12 & mc^2\left(1 - \frac 12 Z^2\alpha^2 - \frac 18 Z^4\alpha^4 + \cdots \right) \\ 2S_{1/2} & 2 & 0 & \frac 12 & mc^2\left(1 - \frac 18 Z^2\alpha^2 - \frac{5}{128}Z^4\alpha^4 + \cdots \right) \\ 2P_{1/2} & 2 & 1 & \frac 12 & mc^2\left(1 - \frac 18 Z^2\alpha^2 - \frac{5}{128}Z^4\alpha^4 + \cdots \right) \\ 2P_{3/2} & 2 & 1 & \frac 32 & mc^2\left(1 - \frac 18 Z^2\alpha^2 - \frac{1}{128}Z^4\alpha^4 + \cdots \right) \end{array} \tag{1}$$

Klein-Gordon equation

Solving the Klein-Gordon equation gives the following exact energy levels (see for example Bound state solutions of the Klein-Gordon equation by Fleischer and Soff, equation (3.2)): $$E_{nl}=mc^2\left[1+\left(\frac{Z\alpha}{n-l-\frac 12+\sqrt{\left(l+\frac 12\right)^2-Z^2\alpha^2}}\right)^2\right]^{-1/2}$$ with the quantum numbers

  • $n=1,2,...$
  • $l=0,1,2,...,n-1$

(This looks quite similar to the Dirac solution above, except here we have $l$ instead of $j$.)

Taylor-expansion for small $\alpha$ gives: $$E_{nl}\approx mc^2\left[1-\frac{Z^2\alpha^2}{2n^2}-\frac{Z^4\alpha^4}{2n^3}\left(\frac{1}{l+\frac 12}-\frac{3}{4n}\right)+\cdots\right]$$

The first few energy levels calculated from this are: $$\begin{array}{c|cc|c} & n & l & E_{nl} \\ \hline 1s & 1 & 0 & mc^2\left(1 - \frac 12 Z^2\alpha^2 - \frac 58 Z^4\alpha^4 + \cdots \right) \\ 2s & 2 & 0 & mc^2\left(1 - \frac 18 Z^2\alpha^2 - \frac{13}{128}Z^4\alpha^4 + \cdots \right) \\ 2p & 2 & 1 & mc^2\left(1 - \frac 18 Z^2\alpha^2 - \frac{7}{384}Z^4\alpha^4 + \cdots \right) \end{array} \tag{2}$$

Comparison with experiment

From tables (1) and (2) you see, the predicted energy levels are slightly different. Look at the fine-structure terms $\propto Z^4\alpha^4$.

Of course the correct energy levels (or more precisely: the differences between energy levels) were known with high precision long before the theories, from experimental spectroscopy of the hydrogen spectrum. So it turned out, that the Klein-Gordon results are wrong, and the Dirac results are correct.

Thomas Fritsch
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I'm just getting to relativistic QM in my AQM course. The textbook I use as a reference [1] derives the continuity equation for a free particle. The standard procedure to do so is:

$\phi^* \hat{KG}\phi - (\phi^* \hat{KG}\phi)^{\dagger}= \dot{\rho}+ div\vec{j}=0$

where $\hat{KG}\phi=0$ is the Klein Gorden equation. As the KG equation has the second derivative with time and this component isn't complex doing this procedure leads to:

$\rho= \frac{i\hbar}{2mc^{2}} (\phi^*\frac{\partial\phi}{\partial t}-\phi\frac{\partial\phi^*}{\partial t})$. (the usual result for TDSE is $\rho=|\phi^{*}\phi|$)

Now given that under certain I.Cs this does is not always definitely positive it cannot be interpreted as a probability density. However my lecturer, and Dirac himself [2] state that is was perfectly healthy back in the day to treat it as a current density.

In the introduction to the theory of the electron [2] he states: "The Gordon-Klein interpretation can answer such questions [probability of dynamical quantities taking values in certain limits] if they refer to the position of the electron (by the use of [charge/probability density]), but not if they refer to its momentum, or angular momentum or any other dynamical variable."

I guess this isn't an exact answer but gives an idea as to why it fails fundamentally. An extension is:

Does anyone have a good example of the KG failing for these other variables?

[1]- F.Schwabl Advanced Quantum Mechanic 4th edition [2]-Dirac, P.A "The quantum theory of the electron" [URL:https://www.rpi.edu/dept/phys/Courses/PHYS6520/DiracElectron.pdf]

J.R.Green
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