Is there any operator in quantum mechanics that measure an observable with non-zero uncertainty?

What does a measurement do? The answer is: If the detector is designed to measure some observable O, it will leave the measured object, at least for an instant, in a zero-uncertainty state. I want to know, in the context of quantum mechanics and base on Hilbert space, can define an operator to measure an observable with non-zero uncertainty?

• The rare case is the reversed one. A measurement has zero uncertainty (variance) if and only if the state is an eigenstate of the observable.
– lcv
Commented Apr 18, 2020 at 16:04

The uncertainty of an observable depends on the state $$|\psi\rangle$$ of the system that is being measured. The expectation value of the observable $$A$$ is given by $$\langle A\rangle=\langle\psi|A|\psi\rangle$$ and the uncertainty is given by $$(\Delta A)^2=\langle (A-\langle A\rangle)^2\rangle=\langle A^2\rangle-\langle A\rangle^2$$

Therefore, you can have a state with non-zero uncertainty in $$A$$ when $$\langle \psi|(A-\langle A\rangle)^2|\psi\rangle\neq0$$. There are many examples of systems like these. For example, a Gaussian wave packet has non-zero uncertainties in both position and momentum.

In terms of measurement, of course if you measure an observable the system is in an eigenstate of the observable and hence has $$0$$ uncertainly for that observable. But that does not make the new state a state of "non-zero uncertainty" in general, because the state could have uncertainty with respect to other observables. More explicitly, if we measure $$A$$ for our state so that now $$|\psi\rangle=|a\rangle$$, we now have $$\Delta A=0$$, but there could be (definitely is?) another observable $$B$$ such that $$\Delta B\neq0$$ for this new state $$|a\rangle$$.

If you want to make an observable where state after measuring $$A$$ has a non-zero uncertainty in $$A$$, then that is impossible. This is because the state after measuring $$A$$ has to be an eigenstate of $$A$$, so then $$\Delta A=0$$ after measurement always.

• Thanks for your answer. If I understand your answer you say: if, for example, two operators don't commute the uncertainty appears, also, we always have zero uncertainty cause the measured state is an eigenstate of an observable operator. This is true cause this point that "measured state is an eigenstate of an observable operator" comes from zero uncertainty. But I want: measure and observable with uncertainty and knowing the state after this measure.
– ALIN
Commented Apr 19, 2020 at 7:01
• @ALIN If you measure a system then you know its state after the measurement. This is true regardless of what uncertainties are present. Commented Apr 19, 2020 at 18:01
• @BioPhysicist How did u go from $\langle (A-\langle A\rangle)^2\rangle$ to $\langle A^2\rangle-\langle A\rangle^2$? What were the intermediate steps? Commented May 20, 2021 at 7:58
• @Theoretical Just FOIL it out, keeping in mind that $\langle\dots\rangle$ is a linear operation. Commented May 20, 2021 at 11:09

Defining an observable $$M = \sum_m m P_m$$, with $$m$$ the eigenvalues and $$P_m$$ the eigenspace projectors, is equivalent to defining a set of projective measurements $$\{P_m\}$$ such that $$\sum_m P_m = I$$ and $$P_m P_{m'} = \delta_{mm'} P_m$$

Since the operators $$P_m$$ are idempotent ($$P_m^2 = P_m$$), two consecutive measurements, one immediately after the other, necessarily yield the same result. Any other set of measurements $$\{M_m\}$$ would be measuring something else.

On the other hand, it is possible to define a set of measurements $$\{M_m\}$$ such that two immediately consecutive measurements can yield different results. This can be expressed using the POVM (positive-operator-valued measures) formalism. In that case, the measurement operators no longer are orthogonal projectors.

To give an example: such measurements can be used to unambiguously distinguish non-orthogonal quantum states. By virtue of the no-cloning theorem, this is impossible to achieve with complete reliability. However we can have a measurement that is sometimes inconclusive but never makes an error of mis-identification.

Let's take the simple case of a system prepared in one of two states $$|\psi_1 \rangle = |0 \rangle$$ or $$| \psi_2 \rangle = \frac 1 {\sqrt 2}(|0 \rangle + |1 \rangle)$$. We will then apply the following POVM:

\begin{align} E_1 &= \frac {\sqrt 2} {1 + \sqrt 2} |1 \rangle \langle 1| \\ E_2 &= \frac {\sqrt 2} {1 + \sqrt 2} \frac {(|0 \rangle - |1 \rangle) (\langle 0| - \langle 1|)} 2 \\ E_3 &= \mathbf I - E_1 - E_2 \end{align}

We can see that if we have state $$|\psi_1 \rangle$$, there is zero probability of getting the result $$E_1$$: $$\langle \psi_1 | E_1 | \psi_1 \rangle = 0$$. Similarly, if we have state $$| \psi_2 \rangle$$, there is zero probability of getting the result $$E_2$$. Therefore, if we observe $$E_1$$, we know that the state was $$| \psi_2 \rangle$$ and vice versa.

However, in both cases, there is a non-zero probability of observing $$E_3$$, our "inconclusive" result: $$\langle \psi_1 | E_3 | \psi_1 \rangle = \langle \psi_2 | E_3 | \psi_2 \rangle = \frac 1 {\sqrt 2} \approx 0.71$$