equations:schroedinger_equation
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equations:schroedinger_equation [2019/05/21 16:55] michael remove duplicated line in free particle example |
equations:schroedinger_equation [2020/04/10 11:44] 109.81.208.52 [Concrete] |
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| <tabbox Concrete> | <tabbox Concrete> | ||
| - | In the Schrödinger equation, the <color firebrick>wave-function</color> $\color{firebrick}{\Psi(\vec{x},t)}$ describes the state of the system. This means concretely that ${ \Psi(\vec{x},t)}$ encodes, for example, if a given system is excited or in its rest state. The wave-function is a [[basic_tools:complex_analysis|complex function]], which means that at a given point in space, say $ \vec{x}= (1,4,7)$, and time, say $t=9s$, the wave function is a complex number: ${\Psi(1,4,7,9)}= 5 + 3 i$. The wave-function at a given point in space and time yields a probability amplitude. By taking the absolute square of an probability amplitude we get the probability density and by integrating over some spatial volume we get the probability to the the particle within this volume. | + | In the Schrödinger equation, the <color firebrick>wave-function</color> $\color{firebrick}{\Psi(\vec{x},t)}$ describes the state of the system. This means concretely that ${ \Psi(\vec{x},t)}$ encodes, for example, if a given system is excited or in its rest state. The wave-function is a [[basic_tools:complex_analysis|complex function]], which means that at a given point in space, say $ \vec{x}= (1,4,7)$, and time, say $t=9s$, the wave function is a complex number: ${\Psi(1,4,7,9)}= 5 + 3 i$. The wave-function at a given point in space and time yields a probability amplitude. By taking the absolute square of an probability amplitude we get the probability density and by integrating over some spatial volume we get the probability of the particle being within this volume. |
| The left-hand side of the Schrödinger equation denotes the <color darkturquoise></color> of the wave function. We act with the <color darkturquoise>partial derivative</color> $\color{darkturquoise}{\partial_t}$ on our wave-function and the result: ${\partial_t}{ \Psi(\vec{x},t)}$ denotes how the wave function changes as time passes on. | The left-hand side of the Schrödinger equation denotes the <color darkturquoise></color> of the wave function. We act with the <color darkturquoise>partial derivative</color> $\color{darkturquoise}{\partial_t}$ on our wave-function and the result: ${\partial_t}{ \Psi(\vec{x},t)}$ denotes how the wave function changes as time passes on. | ||
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| $$ \phi(t) = A e^{-Et/\hbar} $$ | $$ \phi(t) = A e^{-Et/\hbar} $$ | ||
| - | and the second equation is known as the stationary Schrödnger equation. This means that for all systems where the Hamiltonian does not explicitly depend on the time, we known immediately how the time-dependence of the total wave function $\Psi(x,t)$ looks like, namely: $\Psi(x,t) = \phi(t) \psi(x) = A e^{-Et/\hbar} \psi(x)$. The only thing we then have to do is to solve the __stationary Schrödinger equation__ | + | and the second equation is known as the stationary Schrödnger equation. This means that for all systems where the Hamiltonian does not explicitly depend on the time, we know immediately what the time-dependence of the total wave function $\Psi(x,t)$ looks like, namely: $\Psi(x,t) = \phi(t) \psi(x) = A e^{-Et/\hbar} \psi(x)$. The only thing we then have to do is to solve the __stationary Schrödinger equation__ |
| \begin{equation} | \begin{equation} | ||
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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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