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harmonic_oscillator_hamiltonian [2016/08/30 20:36]
nikolaj old revision restored (2016/08/30 20:34)
harmonic_oscillator_hamiltonian [2016/08/31 15:32]
nikolaj
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 ===== Harmonic oscillator Hamiltonian ===== ===== Harmonic oscillator Hamiltonian =====
 ==== Function ==== ==== Function ====
-$A=\omega_0 ​*\left( -\dfrac{1}{\omega_0 ​}\left(L\dfrac{\partial}{\partial x}\right)^2+\omega_0 ​\,​\left(\dfrac{x-x_0}{L}\right)^2 \right)$+$A=\kappa*\left( -\dfrac{1}{\kappa}\left(L\dfrac{\partial}{\partial x}\right)^2+\kappa\,​\left(\dfrac{x-x_0}{L}\right)^2 \right)$
  
 ----- -----
 === Discussion === === Discussion ===
 +== Remark ==
 +Another "​quantum harmonical oscillator"​ is a model which looks similar, except $x$ is an operator $x(t)$ (and one a priori more general than right multiplication by $x$ as here) and where instead of $\dfrac{\partial}{\partial x}$ we consider $\dfrac{1}{L^2}x'​(t)$. In this case, we can may have $\kappa$ depend on $t$ too.
 +
 +It's basically a matter on how the $a$'s end up looking and in what way they relate to the ground state of the system.
 +
 == Interpretation == == Interpretation ==
-We describe a system with 1-dim degree of freedom, $x$, and a potential with no degrees of freedom. The "​spring constant"​ $\omega_0 ​$ in the "​interaction term" with $\omega_0 ​\cdot x$ quantifies the penalty for $x$ being away from $x_0$. ​+We describe a system with 1-dim degree of freedom, $x$, and a potential with no degrees of freedom. The "​spring constant"​ $\kappa$ in the "​interaction term" with $\kappa\cdot x$ quantifies the penalty for $x$ being away from $x_0$. ​
  
 == Completing the square == == Completing the square ==
-Introducing a new variable $l$ via $L=\sqrt{\omega_0 ​}\,l$ lets us pull out $\omega_0 ​$ as an overall multiplicative constant of the operator.+Introducing a new variable $l$ via $L=\sqrt{\kappa}\,l$ lets us pull out $\kappa$ as an overall multiplicative constant of the operator.
  
 Further, with Further, with
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 we find we find
  
-$A = \omega_0 ​*\left(u\,​v\,​a^\dagger a + 1\right)$+$A = \kappa*\left(u\,​v\,​a^\dagger a + 1\right)$
  
 (as $\left[\frac{\partial}{\partial x} , x \right] \psi(x) = (x\,​\psi(x))'​ - x\,​\psi'​(x) = \psi(x)$) (as $\left[\frac{\partial}{\partial x} , x \right] \psi(x) = (x\,​\psi(x))'​ - x\,​\psi'​(x) = \psi(x)$)
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 We choose $u,v$ with $u\,​v=\frac{1}{2}$,​ e.g.  $u=v=\frac{1}{\sqrt{2}}$ so that We choose $u,v$ with $u\,​v=\frac{1}{2}$,​ e.g.  $u=v=\frac{1}{\sqrt{2}}$ so that
  
-$A = \omega_0 ​*2*\left(a^\dagger a + \dfrac{1}{2}\right)$+$A = \kappa*2*\left(a^\dagger a + \dfrac{1}{2}\right)$
  
-$A = \omega_0 ​*\left(1+2\,​a^\dagger a\right)$+$A = \kappa*\left(1+2\,​a^\dagger a\right)$
  
 The eigenstates are the sum of the two systems, although, again, the model for the potential is trivial. The eigenstates are the sum of the two systems, although, again, the model for the potential is trivial.
  
 == Eigenstate== == Eigenstate==
-So the kernel of $a$ are eigenstates of $A$ and with eigenvalue $\omega_0 ​$. +So the kernel of $a$ are eigenstates of $A$ and with eigenvalue $\kappa$. 
 Consider a normalized one, $\left|0\right\rangle$. With Consider a normalized one, $\left|0\right\rangle$. With
  
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