# Does Heisenberg's uncertainty hold for any two quantum measurements?

Heisenberg's uncertainty principle is most commonly expressed in terms of the uncertainty in measurement of position and momentum of a particle, $$\Delta x\Delta p \geq \hbar$$and uncertainty in energy and time $\Delta E\Delta t$.

Does the principle hold for any quantum measurements? For example, will measurements of $\hat{L}^2$ then decrease the certainty in a measurement of $\hat{L}_z$?

What is the scope of the principle?

Heisenberg's uncertainty principle is in fact not a principle but a consequence of the operator formalism of QM.

If we associate to the operator $X$ the standard deviation

$$\Delta_X = \sqrt{ \langle{X^2}\rangle -\langle X \rangle^2}$$

it can be then shown that, given two operators $A,B$

$$\Delta_A \Delta_B \geq \frac{1}{2} \left| \langle [A,B]\rangle\right|$$

where $[A,B] = AB-BA$ is the commutator.

If we take the operators $x$ and $p$, we have

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

from which the relation everybody knows follows:

$$\Delta_x \Delta_{p_x} \geq \frac \hbar 2$$

Notice that this is true only for the $x$ component of position and momentum. For example $[x,p_y]=0$, so you can measure the $x$ component of the position vector and the $y$ component of the momentum with arbitrary precision.

Since $L^2$ and $L_z$ commute (their commutator is $0$), in this case you have

$$\Delta_{L^2} \Delta_{L_z} \geq 0$$

and you can measure both with arbitrary precision.

Notice that in non-relativistic quantum mechanics time is not an operator, so this relation is in principle not valid for $E$ and $t$. The origin of the time-energy "uncertainty relation" is different. See for example D.J.Griffiths, Introduction to Quantum Mechanics.