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equations:schroedinger_equation [2018/05/12 12:59]
jakobadmin [Concrete]
equations:schroedinger_equation [2020/11/21 01:43] (current)
2a01:cb15:33b:c600:f4a9:8015:b3fa:6b19 Typo in the Hamiltonian
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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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 \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} \nabla^2 + \hat V \hat{=} \frac{\hat{p}^2}{2m} + \hat V. \end{equation}
  
 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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 $$ \phi(t) = A e^{-Et/​\hbar} $$ $$ \phi(t) = A e^{-Et/​\hbar} $$
-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{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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 **Wave Packets** **Wave Packets**
  
-{{ :​equations:​wavepacket3d.png?​nolink&​200|}}+{{ :​equations:​wavepacket3d.png?​nolink&​300|}}
  
 A generic solution of the stationary Schrödinger equation A generic solution of the stationary Schrödinger equation
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 is non-physical,​ since it describes a plane wave that oscillates till infinity and therefore can't be normalized. is non-physical,​ since it describes a plane wave that oscillates till infinity and therefore can't be normalized.
  
-{{ :​equations:​wavepacket2.png?​nolink&​200|}}+
  
 To get something that we can normalize we must use a superposition of such generic solutions. By adding plane waves with different oscillation frequencies we can build a wave packet that is localized in space and therefore can be normalized. ​ To get something that we can normalize we must use a superposition of such generic solutions. By adding plane waves with different oscillation frequencies we can build a wave packet that is localized in space and therefore can be normalized. ​
 +
 +{{ :​equations:​wavepacket2.png?​nolink&​200|}}
  
 One possibility is a Gaussian wave-packet,​ where $A(\vec p)$ is a Gauss distribution. One possibility is a Gaussian wave-packet,​ where $A(\vec p)$ is a Gauss distribution.
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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.1526122766.txt.gz · Last modified: 2018/05/12 10:59 (external edit)