Quantum Harmonic Oscillator

Quantum harmonic oscillator is the problem when the potential energy is a harmonic oscillating potential energy $$V(x) = \frac{1}{2}m\omega \hat{x}^2$$

Adjoint Operator

Let $\hat{\theta}$ and $\hat{\theta^{+}}$ be two operators. If $$\int f^*\hat{\theta^{+}}g dx = \int \left(\hat{\theta}f \right)^* g dx$$ then $\hat{\theta^{+}}$ is the adjoint operator of $\hat{\theta}$.

If $$\hat{\theta^{+}} = \hat{\theta}$$ then the operator is an Hermitian operator.

Ladder Operator

$$\hat{H} = \left(\frac{\hat{p}^2}{2m} + \frac{1}{2}m\omega \hat{x}^2\right)$$ Define $$x_0 = \sqrt{\frac{2\hbar}{m\omega}}$$ and $$p_0 = \sqrt{2\hbar\omega m}$$ So, the energy operator can be expressed as $$\hat{E} = \left[\left(\frac{\hat{x}}{x_0} \right)^2 + \left(\frac{\hat{p}}{p_0}\right)^2 \right]\hbar\omega$$ Factorize $$\begin{align} \left(\frac{\hat{x}}{x_0} \right)^2 + \left(\frac{\hat{p}}{p_0}\right)^2 &= \left(\frac{\hat{x}}{x_0} - i\frac{\hat{p}}{p_0} \right) \left(\frac{\hat{x}}{x_0} + i\frac{\hat{p}}{p_0} \right) - \frac{i}{x_0p_0}[x,p] \\ &= \left(\frac{\hat{x}}{x_0} - i\frac{\hat{p}}{p_0} \right) \left(\frac{\hat{x}}{x_0} + i\frac{\hat{p}}{p_0} \right) - \frac{i}{2\hbar}(i\hbar) \\ &= \left(\frac{\hat{x}}{x_0} - i\frac{\hat{p}}{p_0} \right) \left(\frac{\hat{x}}{x_0} + i\frac{\hat{p}}{p_0} \right) + \frac{1}{2} \\ \end{align}$$ Define $$\hat{a} = \frac{\hat{x}}{x_0} + i\frac{\hat{p}}{p_0}$$ Its adjoint operator is $$\hat{a^{+}} = \frac{\hat{x}}{x_0} - i\frac{\hat{p}}{p_0}$$ The energy operator can be written as $$\bbox[5px,border:2px solid #666]{ \hat{E} = \hbar \omega(\hat{a^{+}}\hat{a} + \frac{1}{2}) }$$

Commutator of Ladder Operator

The commutator of adjoint ladder operator and the ladder operator is $$\begin{align} [\hat{a},\hat{a^{+}}] &= \hat{a}\hat{a^{+}} - \hat{a^{+}}\hat{a} \\ &= \left(\frac{\hat{x}}{x_0} + i\frac{\hat{p}}{p_0}\right) \left(\frac{\hat{x}}{x_0} - i\frac{\hat{p}}{p_0}\right) - \left(\frac{\hat{x}}{x_0} - i\frac{\hat{p}}{p_0}\right) \left(\frac{\hat{x}}{x_0} + i\frac{\hat{p}}{p_0}\right) \\ &= \frac{1}{2} - \left(-\frac{1}{2}\right) \\ &= 1 \end{align}$$ The commutator of the energy operator and the ladder operator is $$\begin{align} [\hat{E},\hat{a}] &= \hat{E}\hat{a} - \hat{a}\hat{E} \\ &= \hbar \omega\left[(\hat{a^{+}}\hat{a} + \frac{1}{2})\hat{a} - \hat{a}(\hat{a^{+}}\hat{a} + \frac{1}{2})\right] \\ &= \hbar \omega \left(\hat{a^{+}}\hat{a} - \hat{a}\hat{a^{+}} \right)\hat{a} \\ &= \hbar \omega (-1) \hat{a} \\ &= -\hbar \omega \hat{a} \end{align}$$ The commutator of the energy operator and the adjoint ladder operator is $$\begin{align} [\hat{E},\hat{a^{+}}] &= \hat{E}\hat{a^{+}} - \hat{a^{+}}\hat{E} \\ &= \hbar \omega\left[(\hat{a^{+}}\hat{a} + \frac{1}{2})\hat{a^{+}} - \hat{a}(\hat{a^{+}}\hat{a^{+}} + \frac{1}{2})\right] \\ &= \hbar \omega \left(\hat{a^{+}}\hat{a} - \hat{a}\hat{a^{+}} \right)\hat{a^{+}} \\ &= \hbar \omega \hat{a^{+}} \end{align}$$

Finding Solutions

Let $\phi_E$ be an eigenstate of the energy operator. $$\hat{E} \phi_E = E\phi_E$$ Let $\psi$ be the resulting function after applying ladder operator on $\phi_E$ $$\hat{a}\phi_E = \psi$$ Apply the energy operator on $\psi$ $$\begin{align} \hat{E}\psi &= \hat{E}\hat{a}\phi_E \\ &= \left(\hat{E}\hat{a} - \hat{a}\hat{E} + \hat{a}\hat{E} \right)\phi_E \\ &= \left(-\hbar\omega\hat{a} + \hat{a}\hat{E} \right) \phi_E \\ &= (E - \hbar\omega)\hat{a}\phi_E \\ &= (E - \hbar\omega)\psi \end{align}$$ It turns out $\hat{a}\phi_E$ is also an eigenstate of the energy operator! As the energy value is lowered by $\hbar\omega$, $\hat{a}$ is called the lower ladder.

Similarly, if $$\hat{a}^{+}\phi_E = \psi'$$ Then, $$\hat{E}\psi' = (E + \hbar\omega)\psi'$$ Thus, $a^{+}$ is called the upper ladder.

Ground State

When $\hat{a}$ has subtracted the eigenvalue to zero $$\hat{a}\phi_0 = 0$$ we can solve for the eigenstate $$\phi_0 = Ce^{-\frac{\hat{x}}{x_0}}$$ Since $$\begin{align} < E>&= \int dx \phi^*\hat{E}\phi \\ &= \int dp|\tilde{\Psi}(p)|^2 \frac{p^2}{2m} + \int |\Psi(x)|^2\frac{m\omega^2}{2}x^2 \\ \end{align}$$ We know that energy expectation is non-negative. If the energy eigenvalue $E \neq (n)\hbar\omega$, for some non-negative integer $n$, such that $\hat{a}$ can never subtract $E$ to zero, the set of energy eigenvalues have no lower bound. The expectation value of $E$ cannot be positive, which violates our calculation. So, $E$ must be of $n \hbar \omega$ and $\hat{a}\phi_E$ must come to zero eventually. So, the set of the eigenstates found by the ladder operators are all the eigenstates of the harmonic oscillator.