User Tools

Site Tools


equations:schroedinger_equation

Differences

This shows you the differences between two versions of the page.

Link to this comparison view

Both sides previous revision Previous revision
Next revision
Previous revision
Last revision Both sides next revision
equations:schroedinger_equation [2018/05/12 12:59]
jakobadmin [Concrete]
equations:schroedinger_equation [2020/04/10 11:44]
109.81.208.52 [Concrete]
Line 26: Line 26:
  
 <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. 
Line 33: Line 33:
  
 \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} 
  
 ---- ----
Line 64: Line 67:
  
 $$ \phi(t) = A e^{-Et/​\hbar} $$ $$ \phi(t) = A e^{-Et/​\hbar} $$
-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}
Line 82: Line 85:
 \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}
Line 95: Line 97:
 **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
Line 103: Line 105:
 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.
Line 156: Line 160:
 \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 $n$.))
    
    
Line 190: Line 194:
 \[ \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