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qLinusq
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Hello,
In the book physical chemistry (P. Atkins & Julio de Paula, 2009, 5 ED) the authors derive a justification of the Schrödinger equation.
1.) \frac{-\hbar^{2}}{2m} \frac{d^{2}\psi}{dx^{2}}+V(x)\psi=E \psi
The derivation goes as follows:
Derivation:
We can justify the form of the Schrödinger equation to a certain extent by showing that it implies the de Broglie relation for a freely moving particle.
By free motion we mean motion in a region where the potential energy is zero (V=0 everywhere).
If V=0, equation 1 simplifies to:
2.) \frac{-\hbar^{2}}{2m} \frac{d^{2}\psi}{dx^{2}}=E \psi
So far all good, however they then present a solution to equation 2. without showing how they obtained it.
The solution is:
\psi=sin(kx)
k=\frac{(2mE)^{2}}{\hbar}
I have no problem understanding that this is a valid solution however i would like to derive it myself.
Could you provide me with the derivation to the solution of equation 2?
/Thanks in advance,
Linus.
In the book physical chemistry (P. Atkins & Julio de Paula, 2009, 5 ED) the authors derive a justification of the Schrödinger equation.
1.) \frac{-\hbar^{2}}{2m} \frac{d^{2}\psi}{dx^{2}}+V(x)\psi=E \psi
The derivation goes as follows:
Derivation:
We can justify the form of the Schrödinger equation to a certain extent by showing that it implies the de Broglie relation for a freely moving particle.
By free motion we mean motion in a region where the potential energy is zero (V=0 everywhere).
If V=0, equation 1 simplifies to:
2.) \frac{-\hbar^{2}}{2m} \frac{d^{2}\psi}{dx^{2}}=E \psi
So far all good, however they then present a solution to equation 2. without showing how they obtained it.
The solution is:
\psi=sin(kx)
k=\frac{(2mE)^{2}}{\hbar}
I have no problem understanding that this is a valid solution however i would like to derive it myself.
Could you provide me with the derivation to the solution of equation 2?
/Thanks in advance,
Linus.
