4.C. Harmonic Oscillator


## Appendix 4.C. Harmonic Oscillator

Definition 1. Quantum harmonic oscillator is an operator on $\mathbb{R}$ $$L:=-\frac{1}{2}\partial_x^2 +\frac{1}{2} x^2 \label{equ-4.C.1}$$ It is defined in the space $L^2(\mathbb{R})$ of square integrable functions on $\mathbb{R}$.

Remark 1. Operator $$L_{\alpha\beta}:=-\frac{\alpha^2}{2}\partial_x^2 +\frac{\beta^2}{2} x^2 \label{equ-4.C.2}$$ can be reduced to (\ref{equ-4.C.1}) by change of variables $x:= x\gamma$ with $\gamma=\sqrt{\beta/\alpha}$ and division by $\sqrt{\alpha\beta}$.

Observe that $$L=\frac{1}{2}Z^*Z+\frac{1}{2}=\frac{1}{2}ZZ^*-\frac{1}{2} \label{equ-4.C.3}$$ with $$Z:= \partial_x +x ,\qquad Z^* =-\partial _x +x \label{equ-4.C.4}$$ and $Z^*$ is adjoint to $Z$: $(Zu,v)=(u,Z^* v)$.

Note that (\ref{equ-4.C.3}) implies that the lowest eigenvalue of $L$ is $\frac{1}{2}$ with eigenfunction which is "annihilated" by $Z$, i.e. $u_0(x):= e^{-\frac{x^2}{2}}$.

To find other eigenvalues and eigenfunctions observe that $[Z^*,Z]=-2$ and therefore $$LZ^* = Z^* (L+1),\qquad LZ = Z (L-1). \label{equ-4.C.5}$$ The first equality implies that if $u$ is an eigenfunction with an eigenvalue $\lambda$, then $Z^*u$ is an eigenfunction with an eigenvalue $\lambda+1$; therefore we have a sequence of eigenvalues $\lambda_n =(n+\frac{1}{2})$, $n=0,1,2,\ldots$ and eigenfunctions $u_n$ defined $$u_n = Z^* u_{n-1},\qquad n=1,2,\ldots. \label{equ-4.C.6}$$

Theorem 1.

1. There are no other than above eigenvalues and eigenfunctions;
2. $u_n(x)= H_n (x) e^{-\frac{x^2}{2}}$ where $H_n(x)$ is a polynomial of degree $n$ (and it is even/odd for even/odd $n$);
3. All $u_n(x)$ are orthogonal; $\|u_n\|=\sqrt{\pi n!}$.
4. System $\{u_n\}$ is complete.

Proof.

1. The second of equalities (\ref{equ-4.C.5}) implies that if $u$ is an eigenfunction with an eigenvalue $\lambda$, then $Zu$ is an eigenfunction with an eigenvalue $\lambda-1$; however, since eigenvalues start from $\frac{1}{2}$ there are no eigenvalues in $(\frac{1}{2}, \frac{3}{2})$; so the next eigenvalue is $\frac{3}{2}$ and if $u$ is a corresponding eigenfunction then $Z u=c u_0$. But then $Z^*Z u = cZ^* u_0$; but $Z^*Z u=(L-\frac{1}{2})u= u$ and $u=c Z^* u_0= c u_1$. Continuing these arguments we conclude that there are no eigenvalues in $(\frac{3}{2}, \frac{5}{2})$; so the next eigenvalue is $\frac{5}{2}$ and $u=c_2$ and so on.
2. By induction.
3. Due to $L^*=L$ functions are orthogonal; on the other hand \begin{multline*} \|u_n\|^2=\|Z^* u\|^2 = (Z^* u_{n-1}, Z^* u_{n-1})= \\(ZZ^* u,u)= ((L+\frac{1}{2}) u_{n-1},u_{n-1})= (\lambda+\frac{1}{2})\|u_{n-1}\|^2=n\|u\|^2 \end{multline*} and by induction it is equal to $n! \|u_0\|^2= n! \pi$.
Here we used the fact that $|u_0|^2=\int_{-\infty}^\infty e^{-x^2}\,dx=\pi$.
4. Beyond our reach.

Definition 2. Functions $u_n$ are Hermite functions, $H_n(x)$ are Hermite polynomials.

One can prove $$H_n(x) = n! \sum_{m=0}^{\lfloor \tfrac{n}{2} \rfloor} \frac{(-1)^m}{m!(n - 2m)!} (2x)^{n - 2m}. \label{equ-4.C.7}$$ Then \begin{align*} &H_0(x)=1,\\ &H_1(x)=2x,\\ &H_2(x)=4x^2-2,\\ &H_3(x)=8x^3-12x,\\ &H_4(x)=16x^4-48x^2+12,\\ &H_5(x)=32x^5-160x^3+120x,\\ &H_6(x)=64x^6-480x^4+720x^2-120,\\ &H_7(x)=128x^7-1344x^5+3360x^3-1680x,\\ &H_8(x)=256x^8-3584x^6+13440x^4-13440x^2+1680,\\ &H_9(x)=512x^9-9216x^7+48384x^5-80640x^3+30240x,\\ &H_{10}(x)=1024x^{10}-23040x^8+161280x^6-403200x^4+302400x^2-30240 \end{align*}

Remark 2. In the toy-model of QFT (Quantum Field Theory) $u_n$ is considered as $n$-particle state, in particular $u_0$ is a vacuum state; operators $a=\frac{1}{\sqrt{2}}Z$ and $a^+=\frac{1}{\sqrt{2}}Z^*$ are operators of annihilation and creation respectively, $N=a^+a= L-\frac{1}{2}$ is an operator of number of the particles (actually, it is true only for bosons).

#### References

See plots for Hermite polynomial and Hermite functions. Observe that $H_n(x)$ changes sign exactly $n$-times.