Quantum Harmonic Oscillator
- Quantum Mechanics 2.3
· 17 min 
from 「Modern Quantum Mechanics」: Sakurai, J. J.
Quantum Harmonic Oscillator
Quantum Harmonic Oscillator
Quantum Harmonic Oscillator is a system that has a harmonic potential. It is one of the most important systems in quantum mechanics, and it is also a system that can be solved exactly. Since it is applied to many situations and also used as a model for quantum field theory, it is important to deeply understand the QHO.
Let’s start with the Schrödinger equation for the 1-dimensional QHO.
\[ -\frac{\hbar^2}{2m} \odvn{2}{\psi}{x} + \frac{1}{2}m\omega^2x^2\psi = E\psi \]
We’re now going to solve this equation in two different ways.
Analytic Solution
First, let’s nondimensionalize the equation so that it is easier to solve.
\[ \xi = \sqrt{\frac{m\omega}{\hbar}}x ,\quad \epsilon = \frac{2E}{\hbar\omega} \nt \odvn{2}{\psi}{\xi} + (\epsilon-\xi^2)\psi = 0 \]
The equation is not easy to solve, so let’s try an approximation. We find that $\xi \gg 1 \to \psi \propto \exp(-\xi^2/2)$, so assume that $\psi$ has a following form;
\[ \psi = A H(\xi) \exp\left( -\frac{\xi^2}{2} \right) \]
where $H(\xi)$ is a function to be determined and $A$ is an normalization constant. Substituting this into the Schrödinger equation, we get
\[ \odvn{2}{H}{\xi} - 2\xi\odv{H}{\xi} + (\epsilon-1)H = 0 \]
We’ll use the power series method to solve this equation.
\[ H(\xi) = \sum_{k=0}^\infty a_k\xi^k \;\Rightarrow\; \sum_{k=0}^\infty \Big[ (k+2)(k+1)a_{k+2}-2ka_k+(\epsilon-1)a_k \Big]\xi^k = 0 \nt a_{k+2} = \frac{2k+1 -\epsilon}{(k+2)(k+1)}a_k \]
Since the solution has to not diverge, there must be a $k$ such that $a_k=0$ and after all. Therefore, $\epsilon=2n+1$ for non-negative integer $n$. Then we rewrite the differential equation for $H(\xi)$ as
\[ \odvn{2}{H}{\xi} - 2\xi\odv{H}{\xi} + 2nH = 0 \]
This is the Hermite differential equation. The solution is called Hermite polynomial $H_n(\xi)$ and are the orthogonal polynomials. The final solution after the normalization and rescaling is,
\[ \psi_n(x) = \frac{1}{\sqrt{2^n n!}} \left( \frac{m\omega}{\pi\hbar} \right)^{1/4} H_n\left( \sqrt{\frac{m\omega}{\hbar}}x\right ) \exp\left( -\frac{m\omega}{2\hbar}x^2 \right) \]
The energy eigenvalue is
\[ E_n = \hbar\omega\left( n+\frac{1}{2} \right) \]
We can see that the energy is quantized, and the energy difference between the adjacent levels is $\hbar\omega$. Also, the ground state energy is nonzero. This is a significant difference from the classical case.
Ladder Operator Method(Algebraic Solution)
However, an analytic approach is not the best way to understand the QHO. We can solve the QHO in a more elegant way using the ladder operator method.
First, we define the following operators where $\beta=\sqrt{m\omega/\hbar}$.
$ \gdef\ann{\hat{a}} \gdef\cre{\hat{a}^\dagger} $
\[ \begin{align*} \ann &= \frac{\beta}{\sqrt{2}}\left( \hat{x} + i\frac{\hat{p}}{m\omega} \right) \nl \cre &= \frac{\beta}{\sqrt{2}}\left( \hat{x} - i\frac{\hat{p}}{m\omega} \right) \end{align*} \]
These operators are called annihilation operator and creation operator respectively, and are called ladder operators together. We can easily check the commutation relation between them.
\[ \begin{align*} \ann\cre &= \frac{\beta^2}{2}\left( \hat{x}^2 + \frac{\hat{p}^2}{m^2\omega^2} - \frac{i}{m\omega}\comm{\hat{x}}{\hat{p}} \right) \nl &= \frac{1}{\hbar\omega} \left( \frac{\hat{p}^2}{2m} + \frac{1}{2}m\omega^2\hat{x}^2 \right) - \frac{i}{\hbar} \cdot i\hbar \nl &= \frac{1}{\hbar\omega} \hat{H} + \frac{1}{2} \nt \cre\ann &= \frac{1}{\hbar\omega} \hat{H} - \frac{1}{2} \end{align*} \]
\[ \therefore \comm{\ann}{\cre} = 1 \]
If classical, Hamiltonian would just be the product of $\ann$ and $\cre$. However, quantum mechanics works in a different way. $\ann$ and $\cre$ do not commute, and the Hamiltonian is not the product of them. To seek for the properties of ladder operators, let’s look for operators that commute with Hamiltonian.
\[ \comm{\hat{H}}{\hat{Q}} = \lambda\hat{Q} \nt \Rightarrow \hat{H}\hat{Q}\psi = (\lambda\hat{Q}+\hat{Q}\hat{H})\psi = (\lambda+E)\hat{Q}\psi \]
This means that $\hat{Q}\psi$ is an state with the energy $\lambda+E$ if $\psi$ is an state with the energy $E$. Now we see what ladder operators do.
\[ \comm{\hat{H}}{\ann} = -\hbar\omega\ann \nl \comm{\hat{H}}{\cre} = \hbar\omega\cre \]
This means that $\ann$ and $\cre$ lower and raise the energy of the state by $\hbar\omega$ respectively. Then, we can think of the ground state $\psi_0$ as the state that $\ann$ cannot lower anymore. Ground state wavefunction can be obtained by solving simple differential equation.
\[ \ann\psi_0 = 0 \nt \therefore \psi_0 = \left( \frac{m\omega}{\pi\hbar} \right)^{1/4} \exp\left( -\frac{m\omega}{2\hbar}x^2 \right) \]
Ground state energy is also obtained.
\[ \hat{H}\psi_0 = \hbar\omega\left(\cre\ann+\frac{1}{2}\right)\psi_0 = \frac{1}{2}\hbar\omega\psi_0 \nt \therefore E_0 = \frac{1}{2}\hbar\omega \]
Then, $n$-th excited state can be obtained by applying $\cre$ to the ground state $n$ times. By using Dirac notation,
\[ \ann\cre\ket{n} = \left( \frac{1}{\hbar\omega}\hat{H}-\frac{1}{2} \right)\ket{n} = n\ket{n} \nl \cre\ann\ket{n} = \left( \frac{1}{\hbar\omega}\hat{H}+\frac{1}{2} \right)\ket{n} = (n+1)\ket{n} \]
Let’s find the normalizing constants by using the property above.
\[ \brktop{n}{\cre\ann}{n} = n\brkt{n}{n} = \abs{c_n}^2\brkt{n-1}{n-1} \nl \brktop{n}{\ann\cre}{n} = (n+1)\brkt{n}{n} = \abs{d_n}^2\brkt{n+1}{n+1} \nt \ann\ket{n} = \sqrt{n}\ket{n-1} \nl \cre\ket{n} = \sqrt{n+1}\ket{n+1} \]
Finally, we can express the $n$-th excited state in terms of the ground state and creation operator.
\[ \ket{n} = \frac{1}{\sqrt{n!}}(\cre)^n\ket{0} \]
Expectation Values
\1. $ \Expct{x} = 0 $ \[ \begin{align*} \Expct{x} &= \brktop{n}{\hat{x}}{n} \nl &= \frac{1}{\sqrt{2}\beta}\brktop{n}{\ann+\cre}{n} = 0 \end{align*} \] Proof
\2. $ \Expct{p} = 0 $ \[ \begin{align*} \Expct{p} &= \brktop{n}{\hat{p}}{n} \nl &= \frac{1}{\sqrt{2}\beta} \frac{m\omega}{i} \brktop{n}{\ann-\cre}{n} = 0 \end{align*} \] Proof
\3. $\dps \Expct{x^2} = \frac{\hbar}{m\omega}\left(n+\frac{1}{2}\right) $ \[ \begin{align*} \Expct{x^2} &= \frac{1}{2\beta^2} \brkt{(\cre+\ann)n}{(\ann+\cre)n} \nl &= \frac{1}{2\beta^2} \left( \sqrt{n}^2 + \sqrt{n+1}^2 \right) \brkt{n}{n} \nl &= \frac{\hbar}{m\omega} \left( n+\frac{1}{2} \right) \end{align*} \] Proof
\4. $\dps \Expct{p^2} = m\omega\hbar\left(n+\frac{1}{2}\right) $ \[ \begin{align*} \Expct{p^2} &= -\frac{m^2\omega^2}{2\beta^2} \brkt{(\cre-\ann)n}{(\ann-\cre)n} \nl &= -\frac{m^2\omega^2}{2\beta^2} \left( -\sqrt{n}^2 - \sqrt{n+1}^2 \right) \brkt{n}{n} \nl &= m\omega\hbar \left( n+\frac{1}{2} \right) \end{align*} \] Proof
\5. $\dps \Expct{T} = \frac{1}{2} \hbar\omega\left(n+\frac{1}{2}\right) $ \[ \Expct{T} = \frac{\Expct{p^2}}{2m} \] Proof
\6. $\dps \Expct{V} = \frac{1}{2} \hbar\omega\left(n+\frac{1}{2}\right) $ \[ \Expct{V} = \frac{1}{2}m\omega^2\Expct{x^2} \] Proof
We can also check that classical virial theorem is also valid in quantum mechanics.
\[ \Expct{T} = \Expct{V} = \frac{1}{2}\Expct{E} \]
Uncertainty Principle
\[ \sigma_x = \sqrt{ \Expct{x^2} - \Expct{x}^2 } = \sqrt{ \frac{\hbar}{m\omega}\left(n+\frac{1}{2}\right) } \nl \sigma_p = \sqrt{ \Expct{p^2} - \Expct{p}^2 } = \sqrt{ m\omega\hbar\left(n+\frac{1}{2}\right) } \]
\[ \sigma_x\sigma_p = \hbar\left(n+\frac{1}{2}\right) \ge \frac{\hbar}{2} \]
We also checked that the uncertainty principle holds well, and the equality holds when the system is in the ground state.
3D Harmonic Oscillator
Since harmonic potential is separable, the solution of the Schrödinger equation is the product of three 1-dimensional harmonic oscillator solutions. Also, energy is additive, so the energy of the system is the sum of three 1-dimensional harmonic oscillator energies.
\[ E = \hbar\omega\left( n_x + n_y + n_z + \frac{3}{2} \right) \]
