# Consistent, complete, and generalized description of the quantum harmonic oscillator

Textbooks often introduce the quantum harmonic oscillator with the example of a mass on a spring, giving the Hamiltonian $$H = \frac{1}{2} k x^2 + \frac{1}{2m}p^2 \qquad [x,p] = i \hbar \, .$$ The raising/lowering operators are then introduced as $$a = \sqrt{\frac{m \omega_0}{2 \hbar}} \left( x + \frac{i}{m \omega} p \right) \, ,$$ where $\omega_0$ is the oscillation frequency $\omega_0 = \sqrt{k/m}$. This is rather messy and unenlightening.

Is there a useful physical interpretation of the prefactors in definition of $a$? In other words, is there a way to write the creation/annihilation operators and other useful relations in a way that reveals more physical insight?

• I would be interested to know why this has been downvoted twice (at the time of writing). – John Rennie Dec 31 '16 at 16:37

## General form

Let's write the Hamiltonian in the general form $$H = \frac{1}{2} \alpha u^2 + \frac{1}{2} \beta v^2 \qquad [u,v] = i \gamma \, .$$

## Dimensionless operators

First, we construct dimensionless operators $$X \equiv \frac{1}{\sqrt{2 \gamma}} \left( \frac{\alpha}{\beta} \right)^{1/4} u \qquad \text{and} \qquad Y \equiv \frac{1}{\sqrt{2\gamma}} \left( \frac{\beta}{\alpha} \right)^{1/4} v \, .$$ With these new operators, the Hamiltonian is $$H = \hbar \omega_0 \left( X^2 + Y^2 \right) \qquad [X,Y]=i/2 \, ,$$ where we've defined $\hbar \omega_0 \equiv \gamma \sqrt{\alpha \beta}$.

## Raising/lowering operators

Now we define the raising and lowering operators \begin{align} a = X + i Y &\qquad a^\dagger = X - i Y \qquad [a, a^\dagger] = 1 \\ X = \frac{1}{2} \left( a + a^\dagger \right) &\qquad Y = \frac{-i}{2} \left( a - a^\dagger \right) \, . \end{align} The Hamiltonian can then be written as $$H = \hbar \omega_0 \left(a^\dagger a + \frac{1}{2} \right) \, .$$

## Zero point motion

The zero point fluctuation in $X$ is $$X_\text{zpf}^2 \equiv \langle X^2 \rangle_0 \equiv \langle 0 | X^2 | 0 \rangle = \frac{1}{4}\langle 0 | a^2 + (a^\dagger)^2 + a a^\dagger + a^\dagger a | 0 \rangle = \frac{1}{4} \, .$$ From this, we find $$u_\text{zpf}^2 = \frac{1}{2} \gamma \sqrt{\beta / \alpha} \qquad v_\text{zpf}^2 = \frac{1}{2} \gamma \sqrt{\alpha / \beta} \, ,$$ and $$X = \frac{1}{2} \frac{u}{u_\text{zpf}} \qquad Y = \frac{1}{2} \frac{v}{v_\text{zpf}} \, .$$ We can also now relate $u$ and $v$ to the raising/lowering operators in a meaningful way: $$a = \frac{1}{2} \left( \frac{u}{u_\text{zpf}} + i \frac{v}{v_\text{zpf}} \right) \, ,$$ which also leads to $$u = u_\text{zpf} \left( a + a^\dagger \right) \qquad v = -i v_\text{zpf} \left( a - a^\dagger \right) \, .$$

Now we have some physical insight! The original operators $u$ and $v$ are equal to $a \pm a^\dagger$ with a dimensionful prefactor which is simply the zero point motion of each operator. This is very helpful; we like to use the raising/lowering operators in calculation because of their simple matrix elements, and we can now easily estimate the magnitude of various expectation values based on the zero point motion of the system.

Note, in particular, the trade-off between zero point fluctuation in $u$ and $v$. If we change the Hamiltonian in a way that reduces the fluctuation in $u$, then we get a corresponding increase in the fluctuation in $v$. Of course, this is essentially the Heisenberg uncertainty principle at work.

## Coherent state

We defined $X$ and $Y$ as we did so that they are the "coordinates" of a coherent state. Define $|\alpha \rangle$ as the coherent state satisfying $$a | \alpha \rangle = \alpha | \alpha \rangle \, .$$ Then we have $$\langle \alpha | X | \alpha \rangle = \frac{1}{2} \langle \alpha | a + a^\dagger | \alpha \rangle = \text{Re}(\alpha) \, .$$