
# Harmonic Oscillator

## Quantum Harmonic Oscillator

Recall that in a classical harmonic oscillator, we defined $\omega=\sqrt{\frac{k}{m}}$ and the total mechanical energy was $E=\frac{p^2}{2m}+\frac{1}{2}kx^2$. Similarly, the quantum mechanical Hamiltonian for the harmonic oscillator is: $$H=\frac{\hat{p}^2}{2m}+\frac{1}{2}m\omega^2\hat{x}^2$$ As usual, when provided with a potential energy system, we want to solve the energy eigenvalue equation $H\ket{E}=E\ket{E}$ for the allowed energies in the system. We can do this with a power series solution or the operator method.
We define the lowering operator: $a=\sqrt{\frac{m\omega}{2\hbar}}(x+\frac{ip}{m\omega})$ and the raising operator: $\ud{a}=\sqrt{\frac{m\omega}{2\hbar}}(x-\frac{ip}{m\omega})$. $$\ud{a}a+\frac{1}{2}=\frac{m\omega}{2\hbar}\bigg(x^2+\frac{p^2}{m^2\omega^2}\bigg)$$ \begin{align*} H&=\hbar\omega\bigg[\frac{m\omega}{2\hbar}\bigg(x^2+\frac{p^2}{m^2\omega^2}\bigg)\bigg]\\ &=\hbar\omega(\ud{a}a+\frac{1}{2}) \end{align*}
$$H=\hbar\omega(\ud{a}a+\frac{1}{2})\\ [a,\ud{a}]=a\ud{a}-\ud{a}a=1$$
Applying the raising and lowering operators on an energy eigenstate $\ket{E}$, $$H(a\ket{E})=(E-\hbar\omega)(a\ket{E})\\ H(\ud{a}\ket{E})=(E+\hbar\omega)(\ud{a}\ket{E})$$ Since we cannot have negative energies, we define the base case of $\ket{E_0}$. So, $a\ket{E_0}=0$ and $$E_n=\hbar\omega(n+\frac{1}{2})$$ \begin{align*} H\ket{n}&=E_n\ket{n}\\ \hbar\omega(\ud{a}a+\frac{1}{2})\ket{n}&=\hbar\omega(n+\frac{1}{2})\ket{n}\\ \ud{a}a\ket{n}&=n\ket{n} \end{align*} We define the number operator $N=\ud{a}a$. $$H=\hbar\omega(N+\frac{1}{2})$$

### Eigenvalue Equation

$$N\ket{n}=n\ket{n}$$

## Wavefunctions

Ground state wavefunction: $$\phi_0(x)=\bigg(\frac{m\omega}{\pi\hbar}\bigg)^\frac{1}{4}e^{-\frac{m\omega x^2}{2\hbar}}$$ To find the other energy eigenstates, we apply the raising operator to the ground state.
$$\ud{a}\ket{n}=\sqrt{n+1}\ket{n+1}$$ $$a\ket{n}=\sqrt{n}\ket{n-1}$$ $$\ket{n}=\frac{1}{\sqrt{n!}}(\ud{a})^n\ket{0}$$ $$\phi_n(x)=\frac{1}{\sqrt{n!}}\bigg(\sqrt{\frac{m\omega}{2\hbar}}(x-\frac{\hbar}{m\omega}\frac{d}{dx})\bigg)^n\phi_0(x)$$

## Position and Momentum Operators

$$x=\sqrt{\frac{\hbar}{2m\omega}}(a+\ud{a})$$ $$p=i\sqrt{\frac{\hbar m\omega}{2}}(\ud{a}-a)$$
Ex 1. Calculate $\angle{x^2}$ for $\phi_3(x)$.
Using the integral form of the wavefunction is too complex. Instead, we use the operator method. \begin{align*} \braketo{3}{x^2}{3}&=\braketo{3}{\frac{\hbar}{2m\omega}(a+\ud{a})^2}{3}\\ &=\frac{\hbar}{2m\omega}\braketo{3}{(a^2+a\ud{a}+\ud{a}a+\ud{a}^2)}{3}\\ \end{align*} $$*a^2\ket{3}\propto\ket{1}\implies\braket{3}{1}=0\\ *\ud{a}^2\ket{3}\propto\ket{5}\implies\braket{3}{5}=0\\ *\ud{a}a\ket{3}=\ud{a}\sqrt{3}\ket{2}=3\ket{3}\\ *a\ud{a}\ket{3}=a(2\ket{4})=4\ket{3}$$ \begin{align*} \braketo{3}{x^2}{3}&=\frac{\hbar}{2m\omega}(4\braket{3}{3}+3\braket{3}{3})\\ &=\frac{7\hbar}{2m\omega} \end{align*}