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# The Schrodinger Equation
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# Die Schr"odinger Gleichung

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## Preface
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## Vorwort

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This article will cover the Schrodinger Equation, which represents a non-relativistic differential equation governing
quantum particles. Relativistic effects are not taken into account here.
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Dieser Artikel wird sich mit der sogenannten Schr"odinger Gleichung besch"aftigen, eine nicht-relativistische
Differenzialgleichung zur Beschreibung der Wellenfunktion von Quantenobjekten.

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## Derivation
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## Herleitung

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It should be mentioned that the Schrodinger Equation cannot be derived traditionally but rather represents a
semi-logical conclusion from a number of assumptions/axioms.
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Es soll erw"ahnt sein, dass die Schr"odinger Gleichung nicht traditionell hergeleitet werden kann und statdessen eine
halb-logische Folge einer Reihe an Annahmen und Axiomen darstellt.

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### Known properties of traditional waves
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### Bekannte Eigenschaften traditioneller Wellen

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In quantum mechanics, particles are assumed to be governed by a wave function.
[Louis-de-Broglie](https://de.wikipedia.org/wiki/Louis_de_Broglie) postulated that matter waves should have the same
wavelength as photons, whose wavelength depends on their respective impulse:
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In der Quantenmechanik werden Teilchen durch ihre jeweilige Wellenfunktion modelliert.
[Louis-de-Broglie](https://de.wikipedia.org/wiki/Louis_de_Broglie) postullierte, dass Materiewellen die selbe
wellenl"ange wie Photonen mit demselben Impuls aufweisen:

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$$\lambda = \frac{h}{p} = \frac{h}{m*v} $$

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A basic wave function for a travelling monochromatic wave can be written as:
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Eine einfache Wellenfunktion f"ur eine monochromatische bewegte Welle kann folgenderma"sen geschrieben werden:

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$$ \Psi = \cos(k*x - \omega * t) = e^{i*(k*x - \omega * t)} $$

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Schrodinger assumed that a matter wave should have the same wave form. From this, we can derive,
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Schrodinger nahm an, dass eine Materiewelle eine "ahnliche Wellenfunkoion aufweisen wu"rde. Aus dieser Wellenfunktion
l"asst sich mithilfe der oben erw"ahnten de-Broglie-Wellenl"ange herleiten, dass
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$k = \frac{2*\pi}{\lambda} = \frac{2 * \pi * p}{h} $
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using the de-Broglie-wavelength as mentioned above. With
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Mit dem Wissen, dass
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$ v = f * \lambda $
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for each wave, one can also conclude that
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f"ur Welle gilt, l"asst sich auch folgern, dass folgende Beziehung stimmt:
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$\omega = 2 * \pi * f = 2 * \pi * \frac{v}{\lambda} = \frac{2 * \pi * v * p}{h} $.
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We can rewrite the wave function as follows:
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Deshalb l"asst sich die Wellenfunktion folgenderma"sen schreiben:

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$$ \Psi = e^{i*(\frac{2 * \pi * p}{h}*x - \frac{2 * \pi * v * p}{h} * t)} = e^{\frac{i*p}{\hbar}*(x - v*t)} $$

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### Obtaining the Schrodinger equation through energy conversion

The total sum of energy should be a constant for physical particles. For this,
the sum of the kinetic Energy $E_{kin} = \frac{p^{2}}{2*m}$ and potential energy $E_{pot} = V$, where V describes some
arbitrary potential, which in practice will need to be multiplied by its respective "charge" of a particle (this could
be its mass for a gravitational potential or the charge for electric fields) needs to stay constant. Thusly, one can
write:

$$ E = E_{kin} + E_{pot} = \frac{p^{2}}{2*m} + V $$
$$ \hat{E} * \Psi = \frac{\hat{p}^{2}}{2*m} * \Psi + V * \Psi $$

$\hat{E}$ and $\hat{p}$ are some (yet unknown) operators on $\Psi$ which respectively should return the impulse and
total energy for the wave function. For wave function to be valid, this equation must be true. On the other hand, we can
obtain these operators using the basic wave function mentioned above.

Noting the factor $p$ in the exponent of $\Psi$, you might write $\frac{\partial{\Psi}}{\partial{x}} = \frac{i *
p}{\hbar}$ and obtain $p = -i*\hbar*\frac{\partial{\Psi}}{\partial{x}}$. Plugging this back into the energy conservation
equation, we obtain the time-independent schrodinger equation:

$$ \hat{E} * \Psi = -\frac{\hbar^2}{2*m} * \frac{\partial^2 \Psi}{\partial^2 x} + V*\Psi$$

You might also note that the total [energy of a photon](https://en.wikipedia.org/wiki/Photon_energy) can be expressed as
$E_{ph} = h * f = \frac{h*p*v}{h} = p * v$. Using the x-derivative is not convenient, but you might note that
$\frac{\partial \Psi}{\partial t} = \frac{-i*p*v}{\hbar}$. We can rewrite this as $E = p*v = i * \hbar * \frac{\partial
\Psi}{\partial t}$

Substituting this relation into the time-independent Schrodinger equation, the time-dependent can be obtained:

$$ i * \hbar * \frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2*m} * \frac{\partial^2 \Psi}{\partial^2 x} + V*\Psi$$