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equations:schroedinger_equation [2019/05/21 14:55]
michael remove duplicated line in free particle example
equations:schroedinger_equation [2019/05/21 15:12] (current)
michael Latex footnotes don't work, use docuwiki syntax of `((<footnote>))` instead
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 \begin{equation} \Psi(x,t) =   \big( C \sin(\sqrt{2mE}x) + D \cos(\sqrt{2mE}x) \big){\mathrm{e }}^{-i E t} \end{equation} \begin{equation} \Psi(x,t) =   \big( C \sin(\sqrt{2mE}x) + D \cos(\sqrt{2mE}x) \big){\mathrm{e }}^{-i E t} \end{equation}
  
-Next, we use that the wave-function must be a continuous function\footnote{If there are any jumps in the wave-function,​ the momentum of the particle $ \hat p_x \Psi = -i \partial_x \Psi$ is infinite because the derivative at the jumping point would be infinite.}. Therefore, we have the boundary conditions $\Psi(0)=\Psi(L) \stackrel{!}{=} 0$.+Next, we use that the wave-function must be a continuous function((If there are any jumps in the wave-function,​ the momentum of the particle $ \hat p_x \Psi = -i \partial_x \Psi$ is infinite because the derivative at the jumping point would be infinite.)). Therefore, we have the boundary conditions $\Psi(0)=\Psi(L) \stackrel{!}{=} 0$.
 We see that, because $\cos(0)=1$ we have $D\stackrel{!}{=}0$. Furthermore,​ we see that these conditions impose ​ We see that, because $\cos(0)=1$ we have $D\stackrel{!}{=}0$. Furthermore,​ we see that these conditions impose ​
 \begin{equation} \label{box:​quantbed} \sqrt{2mE}\stackrel{!}{=} \frac{n \pi}{L}, ​ \end{equation} \begin{equation} \label{box:​quantbed} \sqrt{2mE}\stackrel{!}{=} \frac{n \pi}{L}, ​ \end{equation}
- with arbitrary integer $n$, because for\footnote{Take note that we put an index $n$ to our wave-function,​ because we have a different solution for each $n$.}+ with arbitrary integer $n$, because for ((Take note that we put an index $n$ to our wave-function,​ because we have a different solution for each $n$.))
    
    
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 \[ \Phi(x,t) = A \Phi_1(x,t) +  B \Phi_2(x,t) + ...  \[ \Phi(x,t) = A \Phi_1(x,t) +  B \Phi_2(x,t) + ... 
 \] \]
-are solutions, too. These solutions have to be normalised again because of the probabilistic interpretation\footnote{A probability of more than $1=100\%$ doesn'​t make sense}.+are solutions, too. These solutions have to be normalised again because of the probabilistic interpretation((A probability of more than $1=100\%$ doesn'​t make sense)).
 <-- <--
  
equations/schroedinger_equation.txt · Last modified: 2019/05/21 15:12 by michael