Klein–Gordon equation

Question

It is said that Klein-Gordon equation is a relativistic version of the Schrodinger equation. In Schrodinger equation, it is straightforward to include potential energy. But for K-G eqn things seem to be more complicated.

To put it specifically, how can one find the energy eigenfunctions of a particle in a finite square well via K-G eqn?

This post imported from StackExchange Physics at 2014-04-01 16:21 (UCT), posted by SE-user skywaddler

Comments

In terms of the logical role it plays in QFT it is not true that the K-G equation is a relativistic version of the Schrodinger equation and the sooner you get this idea our of your head the better off you will be when you learn QFT.

This post imported from StackExchange Physics at 2014-04-01 16:21 (UCT), posted by SE-user pho

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BUt in terms of relativistic particle mechanics in which the number of particles is fixed, it is true to think of it as the relativistic version of the Schrodinger equation. But thinking of it that way introduced problems, such as negative probabilities, which Weisskopf and Pauli got rid of when they re-interpreted the K-G equation as not a quantum eq., but a classical one, which they then quantised to get a quantum field, which introduces all the usual problems of the infinities of Quantum Field Theory etc.

This post imported from StackExchange Physics at 2014-04-01 16:21 (UCT), posted by SE-user joseph f. johnson

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For a connection between Schr. eq. and Klein-Gordon eq, see e.g. A. Zee, QFT in a Nutshell, Chap. III.5, and this Phys.SE post plus links therein.

This post imported from StackExchange Physics at 2014-04-01 16:21 (UCT), posted by SE-user Qmechanic

Answer 1

K-G is a relativistic equation. So if you want to couple it to external fields you have to use a vector potential $A_\mu$ which will modify the partial-derivative, so that the Klein-Gordon equation:

$$ \left( \partial^\mu \partial_\mu + m^2 \right)\phi = 0 $$

becomes:

$$ \left ( D^\mu D_\mu + m^2 \right) \phi = 0 $$

where $ D_\mu = \partial_\mu + \imath A_\mu $.

Then it remains to figure out what $A_\mu$ will correspond to a given potential - finite square well or otherwise. For this you will likely have to fix a gauge before you can identify the potential as such.

A square well potential will correspond to $ A_\mu = (V(x), 0,0,0)$, where $ V(x) = V_0 \,\forall |x| > \pm a $ and $V(x) = 0, x \in [-a,a] $. The resulting behavior is a little counter-intuitive, yielding something known as the Klein paradox.

For further details see ch. 4 for this excellent thesis on this problem by D. Schaffer.

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Edit: This page also has excellent information along these lines.

This post imported from StackExchange Physics at 2014-04-01 16:21 (UCT), posted by SE-user user346

Comments

Just a remark (nothing wrong with your answer): you don't need a vector potential. This is only one possible theory (electromagnetic actually). You could also insert a term such as $\bar \Psi \phi \Psi$ (Yukawa interaction) coupling to fermions and lots of others (actually the only limitation is that the theory be Lorentz-invariant on this level).

This post imported from StackExchange Physics at 2014-04-01 16:21 (UCT), posted by SE-user Marek

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very nice answer!

This post imported from StackExchange Physics at 2014-04-01 16:21 (UCT), posted by SE-user Robert Filter

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Thanks Robert. Its one of those questions with an "obvious" answer that's not so obvious when you think about it !

This post imported from StackExchange Physics at 2014-04-01 16:21 (UCT), posted by SE-user user346

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@space_cadet: That is indeed true. For me, nothing is ever really clear; at a time, there is only a certain way one understands something. For thousands of years it was clear that the acceleration of a rock is much higher than for a feather (ok, this one was really ornate...) :)

This post imported from StackExchange Physics at 2014-04-01 16:21 (UCT), posted by SE-user Robert Filter

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Good correction, Marek, +1.

This post imported from StackExchange Physics at 2014-04-01 16:21 (UCT), posted by SE-user Luboš Motl

Answer 2

Schroedinger discovered the so-called Klein-Gordon equation first, before he heard about Heisenberg's work and before he thought of the Schroedinger equation. The reason he did this is because he knew relativity was true, and so thought that it would be stupid to look for a non-relativistic equation, all he had to do was modify de Broglie's relativistic equation to include a potential. This is why it is sometimes called the Schroedinger-Klein-Gordon-Fock equation, and sometimes the Schroedinger-Fock equation, and sometimes the Klein-Gordon equation. He didn't publish because, not knowing about spin, he thought it should describe an electron rotating around the proton, and so he plugged in the mass of the electron, and although the basic energy levels were right, the relativistic fine splitting was wrong, out of the bounds of experimental error, so he gave it up. Dirac later commented he should have had more confidence in his equation and published anyway. IN the 50's it was found to more or less correctly describe one kind of meson in the Coulomb field of a proton.

After he heard about Heisenberg's work, noticing it was non-relativistic, he worked out a non-relativistic approximation to his original equation and published that, thinking to himself, oh well at any rate I can salvage something from that idea. That published equation is now called the Schroedinger equation. But I suspect he had a guilty conscience about it since he thought he knew that its relativistic version was false...

Now, as to your specific eigenvalue-problem query. Pauli assigns that exact problem in his lectures on Wave Mechanics. The method of solution is the same as for the Scrhoedinger eq. Solve the free equ. in the half-infinite region on the left of the potential well. Solve it on the right. In the well, since the potential is constant, the same method (plug in an exponential function and solve for the parameters) works. Now match boundary conditions at the two interfaces, and see which values of the parameters allow exponential decay in the infinite regions so that the wave function is $L^2$. Once you choose a parameter in the left region, the parameters in the well and the right region are forced on you by matching the boundary conditions at the interfaces.

This post imported from StackExchange Physics at 2014-04-01 16:21 (UCT), posted by SE-user joseph f. johnson