Question

If two operator do not commute what does it mean

Answer 1

From the definition

$$[A, B] = AB - BA \neq 0$$

i.e. there exists such state

$$A(B|\psi\rangle) \neq B(A|\psi\rangle)$$

Moreover, if A is hermitian, there must exists an eigenstate of A, that is not an eigenstate of B. Proof by contradiction: let any eigenstate of A with the eigenvalue a is an eigenstate of B with the eigenvalue b. Then

$$AB|\psi\rangle = Ab|\psi\rangle = ab|\psi\rangle = BA|\psi\rangle$$

and due to the fullness of the set of eigenvectors of a hermitian operator this equality is true for any state - wrong.

Hence there exists a state, in which both quantities A and B cannot be determined simultaneously.

Heisenberg's uncertainty principle: consider displacement

$$\Delta A = A - \langle A \rangle$$ $$\langle\Delta A^2 \rangle= \langle (A - \langle A \rangle)^2\rangle = \langle A^2 \rangle - \langle A \rangle^2$$

Then in any state

$$\langle \Delta A^2 \rangle \langle \Delta B^2 \rangle \geq\frac{\langle C \rangle^2}{4}$$

where we define

$$i C \equiv [A,B]$$

Hence C is hermitian:

$$-iC^\dagger = [A,B]^\dagger = (AB-BA)^\dagger = B^\dagger A^\dagger - A^\dagger B^\dagger = [B,A] = -[A,B] = -i C$$

Proof: consider

$$\langle (A -i\alpha B)\psi | (A - i\alpha B) \psi \rangle = \langle \psi | (A +i\alpha B) (A -i\alpha B) |\psi \rangle \geq 0$$ $$\langle \Delta A^2 - i \alpha [\Delta A, \Delta B] +\alpha^2\Delta B^2 \rangle \geq 0$$

using

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

and the condition that the discriminant of the inequality with respect to alpha is non-negative:

$$\langle C \rangle ^2 - 4 \langle \Delta A^2 \rangle \langle \Delta B^2 \rangle \geq 0$$ $$\langle \Delta A^2 \rangle \langle \Delta B^2 \rangle \geq\frac{\langle C \rangle^2}{4}$$

For example, it quantum mechanics

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

Here A = x, B = p and C = h, hence

$$\langle \Delta x^2 \rangle \langle \Delta p^2 \rangle \geq\frac{\hbar^2}{4}$$

in any state.