equations:schroedinger_equation
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equations:schroedinger_equation [2018/05/12 12:59] jakobadmin [Concrete] |
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} | ||
| - | \text{ the classical momentum } p_i \ &\rightarrow \ i \partial_{x_i} \, . | + | \text{ the classical momentum } p_i \ &\rightarrow \ {-i} \hbar \partial_{x_i} \, . |
| \end{align} | \end{align} | ||
| - | Formulated differently, the Hamiltonian operator is calculated from the classical energy $E= T +V$ by replacing the classical momentum $p_i$ with the momentum operator $ \hat{p}_i \equiv i \partial_{x_i}$: | + | Formulated differently, the Hamiltonian operator is calculated from the classical energy $E= T +V$ by replacing the classical momentum $p_i$ with the momentum operator $ \hat{p}_i \equiv {-i} \hbar \partial_{x_i}$: |
| \begin{equation} \hat H \equiv - \frac{\hbar^2}{2m} \Delta^2 + \hat V \hat{=} \frac{\hat{p}^2}{2m} + \hat V. \end{equation} | \begin{equation} \hat H \equiv - \frac{\hbar^2}{2m} \Delta^2 + \hat V \hat{=} \frac{\hat{p}^2}{2m} + \hat V. \end{equation} | ||
| Line 42: | Line 42: | ||
| It is conventional to denote operators by an additional hat above the classical symbol. | It is conventional to denote operators by an additional hat above the classical symbol. | ||
| - | The Hamiltonian is what is different for different systems. Formulated differently, the Hamiltonian characterizes the system in question. The rest of the Schrödinger equation stays the same for all systems. | + | The Hamiltonian is what is different for different systems. Formulated differently, the Hamiltonian characterizes the system in question. The rest of the Schrödinger equation stays the same for all systems. For example, the Hamiltonian for a [[models:basic_models:harmonic_oscillator|harmonic oscillator]] reads |
| + | |||
| + | \begin{equation} \hat H \equiv \frac{\hat{p}^2}{2m} - \frac{1}{2}k \hat{x}^2 . \end{equation} | ||
| ---- | ---- | ||
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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)). |
| <-- | <-- | ||
equations/schroedinger_equation.txt · Last modified: 2020/11/21 01:43 by 2a01:cb15:33b:c600:f4a9:8015:b3fa:6b19
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