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## Harmonic oscillator Hamiltonian

### Function

$A=\kappa*\left( -\dfrac{1}{\kappa}\left(L\dfrac{\partial}{\partial x}\right)^2+\kappa\,\left(\dfrac{x-x_0}{L}\right)^2 \right)$

#### Discussion

##### 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.

##### Interpretation

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

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

$a = u \left(\dfrac{x-x_0}{l} + l\,\dfrac{\partial}{\partial x}\right)$

$a^\dagger = v \left(\dfrac{x-x_0}{l} - l\,\dfrac{\partial}{\partial x}\right)$

we find

$A = \kappa*\left(u\,v\,a^\dagger a + 1\right)$

(as $\left[\frac{\partial}{\partial x} , x \right] \psi(x) = (x\,\psi(x))' - x\,\psi'(x) = \psi(x)$)

We choose $u,v$ with $u\,v=\frac{1}{2}$, e.g. $u=v=\frac{1}{\sqrt{2}}$ so that

$A = \kappa*2*\left(a^\dagger a + \dfrac{1}{2}\right)$

$A = \kappa*\left(1+2\,a^\dagger a\right)$

The eigenstates are the sum of the two systems, although, again, the model for the potential is trivial.

##### Eigenstate

So the kernel of $a$ are eigenstates of $A$ and with eigenvalue $\kappa$. Consider a normalized one, $\left|0\right\rangle$. With

$\phi(x)=\exp\left({-\dfrac{1}{2}\left(\dfrac{x-x_0}{l}\right)^2}\right)$

we get

$l\,\dfrac{\partial}{\partial x} \phi(x) = -\dfrac{x-x_0}{l} \phi(x)$

and so $\left|0\right\rangle \propto \phi(x)$.

Note that thus $a^\dagger$ acts as a multiplication operator with factor $2\dfrac{1}{\sqrt{2}}\dfrac{x-x_0}{l}$ (we get Hermite polynomials),

and it remains to demonstrate how those are eigenstates.
todo: Compute $[A,a^\dagger]$, which is $a^\dagger$