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harmonic_oscillator_hamiltonian [2016/08/30 20:35] nikolaj |
harmonic_oscillator_hamiltonian [2016/08/31 15:32] (current) nikolaj |
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===== Harmonic oscillator Hamiltonian ===== | ===== Harmonic oscillator Hamiltonian ===== | ||
==== Function ==== | ==== Function ==== | ||
- | $A=w*\left( -\dfrac{1}{w}\left(L\dfrac{\partial}{\partial x}\right)^2+w\,\left(\dfrac{x-x_0}{L}\right)^2 \right)$ | + | $A=\kappa*\left( -\dfrac{1}{\kappa}\left(L\dfrac{\partial}{\partial x}\right)^2+\kappa\,\left(\dfrac{x-x_0}{L}\right)^2 \right)$ |
----- | ----- | ||
=== 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" $w$ in the "interaction term" with $w\cdot x$ 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 $L=\sqrt{w}\,l$ lets us pull out $w$ 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 | ||
- | $A = w*\left(u\,v\,a^\dagger a + 1\right)$ | + | $A = \kappa*\left(u\,v\,a^\dagger a + 1\right)$ |
(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 | ||
- | $A = w*2*\left(a^\dagger a + \dfrac{1}{2}\right)$ | + | $A = \kappa*2*\left(a^\dagger a + \dfrac{1}{2}\right)$ |
- | $A = w*\left(1+2\,a^\dagger a\right)$ | + | $A = \kappa*\left(1+2\,a^\dagger a\right)$ |
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 $w$. | + | 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 | ||