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
$$\langle \alpha | X | \alpha \rangle = \frac{1}{2} \langle \alpha | a + a^\dagger | \alpha \rangle = \text{Re}(\alpha) \, .$$
