Translation:On the derivation of Klein–Fock equation

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On the derivation of Klein–Fock equation (1926)
Lev Landau and Dmitri Ivanenko, translated from German by Wikisource

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Lev Landau and Dmitri Ivanenko4039286On the derivation of Klein–Fock equation1926Wikisource

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On the derivation of Klein–Fock equation

From D. Iwanenko and L. Landau in Leningrad

Received on 8 October 1926

It is shown that the generalized Schrödinger equation can be obtained by a transition from the relativistic analog of Hamilton's problem.

From the theory of relativity, the following expression is known for the differential of the action function:

(1) $dW=-mc\,ds+{\frac {e}{c}}\varphi _{k}dx^{k},$

where

$ds^{2}=-dx_{k}dx^{k}.$

The generalized momenta are given by

(2) $p_{k}={\frac {\partial L}{\partial {\dot {x}}^{k}}}={\frac {\partial (dW)}{\partial (dx^{k})}}=+mc\,{\frac {dx_{k}}{ds}}+{\frac {e}{c}}\varphi _{k},$

$\left(L={\frac {dW}{dt}}\,\,\,{\text{the Lagrangian function}}\right).$

It follows:

$p_{k}-{\frac {e}{c}}\varphi _{k}=mc\,{\frac {dx_{k}}{ds}},$

(3) $\left(p_{k}-{\frac {e}{c}}\varphi _{k}\right)\left(p^{k}-{\frac {e}{c}}\varphi ^{k}\right)+m^{2}c^{2}=0.$ ^[1]

On the other hand:

(4) $p_{k}={\frac {\partial W}{\partial x^{k}}}=W\cdot \chi ,$

also

(5) $W_{,k}W^{,k}-2{\frac {e}{c}}\varphi ^{k}W_{,k}+{\frac {e^{2}}{c^{2}}}\varphi _{k}\varphi ^{k}+m^{2}c^{2}=0.$

Let's make an assumption similar to Schrödinger's^[2] that equation (5) is the limit of a linear equation for

$\psi =e^{{\frac {i}{\hbar }}W}$ ^[3] as $\hbar \rightarrow 0$

with regards to, it follows:

$\psi \cdot \chi ={\frac {i}{\hbar }}\psi W\cdot \chi ,$

$\psi _{,k}^{,k}={\frac {-1}{\hbar ^{2}}}\psi W_{,k}W^{'k},$

(6) $\psi _{,k}^{,k}-2{\frac {e}{c}}{\frac {i}{\hbar }}\varphi ^{k}\psi _{,k}-{\frac {1}{\hbar ^{2}}}\left(m^{2}c^{2}+{\frac {e^{2}}{c^{2}}}\varphi _{k}\varphi ^{k}\right)\psi =0.$ ^[4]

This equation coincides with that of Klein-Fock,^[5] which also without the introduction of somewhat artificial fifth coordinate can be obtained.^[6]

Leningrad, Physikalisches Institut der Universität.

↑ See also P. A. M. Dirac, Proc. Roy. Soc. (A) 111, 405, 1926.
↑ E. Schrödinger, Ann. d. Phys. 79, 489, 1926.
↑ $\hbar$ denotes Planck's constant divided by $2\pi$ . Zeitschrift für Physik. Bd. XL.
↑ It should be noted that in the field-free case from (6) the speed of the de Broglie's phase waves can easily be obtained. This case yields $\psi =\psi 'e^{{\frac {i}{\hbar }}Et}$ , where $\psi '$ is independent of $t$ , because

$\nabla ^{2}\psi -{\frac {1}{c^{2}}}{\frac {\partial ^{2}\psi }{\partial t^{2}}}-{\frac {m^{2}c^{2}}{\hbar ^{2}}}\psi =0$

and

${\frac {\partial ^{2}\psi }{\partial t^{2}}}=-{\frac {E^{2}}{\hbar ^{2}}}\psi ,$

$\nabla ^{2}\psi -\left({\frac {1}{c^{2}}}-{\frac {m^{2}c^{2}}{E^{2}}}\right){\frac {\partial ^{2}\psi }{\partial t^{2}}}=0.$

Let's set

$E={\frac {mc^{2}}{\sqrt {1-{\frac {{\boldsymbol {b}}^{2}}{c^{2}}}}}},$

so that

$\nabla ^{2}\psi -{\frac {{\boldsymbol {b}}^{2}}{c^{4}}}{\frac {\partial ^{2}\psi }{\partial t^{2}}}=0,$

or, in other words, the required speed is $V={\frac {c^{2}}{\boldsymbol {b}}}$ , which could also be deduced directly from the relativistic Hamiltonian equation (5).
↑ O. Klein, ZS. f.Phys.37,895,1926; V. Fock, ZS. f. Phys.39, 226, 1926.
↑ Anmerknng bei der Korrektur. Meanwhile E. Schrödinger (Ann. d. Phys. 81, 132, 1926) has obtained the same equation successfully with the help of the operator method.

[1] See also P. A. M. Dirac, Proc. Roy. Soc. (A) 111, 405, 1926.

[2] E. Schrödinger, Ann. d. Phys. 79, 489, 1926.

[3] $\hbar$ denotes Planck's constant divided by $2\pi$ . Zeitschrift für Physik. Bd. XL.

[4] It should be noted that in the field-free case from (6) the speed of the de Broglie's phase waves can easily be obtained. This case yields $\psi =\psi 'e^{{\frac {i}{\hbar }}Et}$ , where $\psi '$ is independent of $t$ , because

$\nabla ^{2}\psi -{\frac {1}{c^{2}}}{\frac {\partial ^{2}\psi }{\partial t^{2}}}-{\frac {m^{2}c^{2}}{\hbar ^{2}}}\psi =0$

and

${\frac {\partial ^{2}\psi }{\partial t^{2}}}=-{\frac {E^{2}}{\hbar ^{2}}}\psi ,$

$\nabla ^{2}\psi -\left({\frac {1}{c^{2}}}-{\frac {m^{2}c^{2}}{E^{2}}}\right){\frac {\partial ^{2}\psi }{\partial t^{2}}}=0.$

Let's set

$E={\frac {mc^{2}}{\sqrt {1-{\frac {{\boldsymbol {b}}^{2}}{c^{2}}}}}},$

so that

$\nabla ^{2}\psi -{\frac {{\boldsymbol {b}}^{2}}{c^{4}}}{\frac {\partial ^{2}\psi }{\partial t^{2}}}=0,$

or, in other words, the required speed is $V={\frac {c^{2}}{\boldsymbol {b}}}$ , which could also be deduced directly from the relativistic Hamiltonian equation (5).

[5] O. Klein, ZS. f.Phys.37,895,1926; V. Fock, ZS. f. Phys.39, 226, 1926.

[6] Anmerknng bei der Korrektur. Meanwhile E. Schrödinger (Ann. d. Phys. 81, 132, 1926) has obtained the same equation successfully with the help of the operator method.

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