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equations:schroedinger_equation [2019/05/21 16:53] michael minus sign and hbar missing from momentum operator |
equations:schroedinger_equation [2019/05/21 17:12] michael Latex footnotes don't work, use docuwiki syntax of `((<footnote>))` instead |
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\begin{align} | \begin{align} | ||
H \psi(x)&= E\psi(x) \notag \\ | H \psi(x)&= E\psi(x) \notag \\ | ||
- | \frac{-\hbar \partial_x^2}{2m} \psi(x) &=E\psi(x) \notag \\ | ||
\frac{-\hbar \partial_x^2}{2m} \psi(x) &=E\psi(x) \notag | \frac{-\hbar \partial_x^2}{2m} \psi(x) &=E\psi(x) \notag | ||
\end{align} | \end{align} | ||
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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)). |
<-- | <-- | ||