Jonginn Yun

Quantization And Noncommutativity

Canonical quantization replaces Poisson brackets by commutators:

$$\{\,\cdot\,,\,\cdot\,\} \quad\longrightarrow\quad \frac{1}{i\hbar}[\,\cdot\,,\,\cdot\,].$$

This means that canonical observables are not merely uncertain because of experimental imperfection. Their noncommutativity is part of the structure of the theory.

For position and momentum,

$$[x,p]=i\hbar.$$

This algebraic statement is the root of the uncertainty principle.

Fluctuation Operators

For an observable $A$, define the fluctuation operator

$$\Delta A = A-\langle A\rangle.$$

The variance is

$$(\sigma_A)^2 = \langle(\Delta A)^2\rangle.$$

Similarly,

$$\Delta B = B-\langle B\rangle, \qquad (\sigma_B)^2=\langle(\Delta B)^2\rangle.$$

Cauchy-Schwarz Argument

Apply Cauchy-Schwarz to the two state vectors

$$\Delta A|\psi\rangle, \qquad \Delta B|\psi\rangle.$$

Then

$$\langle(\Delta A)^2\rangle \langle(\Delta B)^2\rangle \ge \left| \langle \Delta A\,\Delta B\rangle \right|^2.$$

The product $\Delta A\Delta B$ can be decomposed into anticommutator and commutator pieces:

$$\Delta A\Delta B = \frac{1}{2}\{\Delta A,\Delta B\} + \frac{1}{2}[\Delta A,\Delta B].$$

Since constants commute,

$$[\Delta A,\Delta B]=[A,B].$$

Keeping only the commutator contribution gives the standard lower bound

$$\sigma_A\sigma_B \ge \frac{1}{2} \left| \langle[A,B]\rangle \right|.$$

For $A=x$ and $B=p$,

$$\sigma_x\sigma_p \ge \frac{\hbar}{2}.$$

Interpretation

The uncertainty principle is not just a statement about measurement disturbance. It follows from the impossibility of assigning a state sharply localized in two noncommuting observables.

The phase-space picture of a classical point therefore cannot survive unchanged in quantum mechanics. Coherent states are important precisely because they saturate this bound while retaining the most classical phase-space behavior available.