# How can the commutator of two observables be realized in an experiment?

Let's say we have two observables $$A$$ and $$B$$, and that we have an experimental setup in which we are able to measure both observables (not necessarily simultaneously). My question consists of two related questions:

• What is the meaning of the observable $$BA$$ in such an experiment? Is it simply measuring $$A$$ first, $$B$$ immediately afterwards, and then multiplying the two measurements?

• What is the meaning of $$[A,B]$$ in the experiment? If $$AB$$ and $$BA$$ are indeed obtained by measuring the observables in different orders, how could we possibly measure their linear combination, given that we can't revert the state after measuring either $$AB$$ or $$BA$$?

Examples of experiments would be appreciated. Especially for the case where $$A$$ and $$B$$ are momentum and displacement, though other observables are also fine, as long as they don't commute.

• The commutator is in general not an observable if $A$ and $B$ are observables (hermitian), because it is not hermitian, except in the trivial case where both observabels commute and the commutator is zero. Commented Jan 22, 2023 at 9:19
• And to add: $AB$ as well as $BA$ are not observables unless $A$ and $B$ commute. Commented Jan 22, 2023 at 9:26
• @TobiasFunke I think the question should that have been asked is how to observe $\hat C=\mathfrak j [\hat A,\hat B]$, which is Hermitean, for an arbitrary pair of observables not just $\hat x ,\hat p$, see below answer of Hyperon. Commented Jan 22, 2023 at 15:54
• @TobiasFunke your comment is fair but despite its ambiguity the question is not quite about a random operator. For argument's sake, let us just assume that we knew how to measure $\hat A$ or $\hat B$ and also knew that the issue of "measuring" $\hat C$ would be quite central to the various controversies in QM, then maybe help Vercassivelauonos to rephrase to something unambiguous and not just dismiss it so easily. Commented Jan 22, 2023 at 16:05
• I've deleted my comment. Anyway, can you give an example? Commented Jan 22, 2023 at 16:07

The product $$AB$$ of two hermitian operators $$A$$ and $$B$$ does, in general, not correspond to an observable, unless $$[A,B]=0$$. However, the operators $$C=i[A,B]$$ and $$D=AB+BA$$ are hermitian, thus corresponding (in principle) to observable quantities. Examples: $$A=X$$ and $$B=P$$ with $$i[X,P]=-\hbar \mathbf{1}$$ and $$XP+PX$$ (related to dilatations) or $$A= \sigma_x , \, B= \sigma_y$$ with $$i[\sigma_x, \sigma_y]=-2 \sigma_z$$ and $$\{\sigma_x, \sigma_y\}=0$$.

Edit: In view of the comments above, a few more remarks on subsequent measurements of two observables $$A$$ and $$B$$. Suppose a measurement of $$A$$ is performed on a state described by the density operator $$\rho$$ with outcome $$a \in \sigma(A)$$, where $$\sigma(A)$$ denotes the spectrum of $$A$$. As a result of the measurement, we find the system in a new state described by the density operator $$\rho_a = P_a \rho P_a/{\rm Tr}(P_a \rho P_a)$$, where $$P_a$$ is the projection operator on the eigenspace of the eigenvalue $$a$$ of $$A$$. Performing now a measurement of the observable $$B$$ with outcome $$b \in \sigma(B)$$ changes the state $$\rho_a$$ into a new state $$\rho_{ab}= Q_b \rho_a Q_b/{\rm Tr}(Q_b \rho_a Q_b)= Q_b P_a \rho P_a Q_b/{\rm Tr}(Q_b P_a \rho P_a Q_b)$$, where $$Q_b$$ denotes the projection operator on the eigenspace of the eigenvalue $$b$$ of $$B$$.

Although the product $$BA$$ cannot be interpreted as an operator describing the subsequent measurement of the observables $$A$$ and $$B$$, the product $$Q_b P_a$$ of the two projection operators (in general not an observable either) plays this role in the sense described above: Starting with a state $$\rho$$, the measurement of $$A$$ with outcome $$a$$ followed by a measurement of $$B$$ with outcome $$b$$ results in a state $$\rho_{ab}=\mathcal{N}_{ab} \, Q_b P_a \rho P_a Q_b$$ with a normalization factor $$\mathcal{N}_{ab}= 1/{\rm Tr}(Q_b P_a \rho P_a Q_b)$$.

Starting from the same state $$\rho$$, but inverting the order of the measurements ($$B$$ followed by $$A$$) leads to the state $$\rho_{ba}= \mathcal{N}_{ba} P_a Q_b \rho Q_b P_a$$. In general $$\rho_{ab} \ne \rho_{ba}$$, unless $$P_a$$ and $$Q_b$$ commute.

Let's say we have two observables $$A$$ and $$B$$, and that we have an experimental setup in which we are able to measure both observables (not necessarily simultaneously). My question consists of two related questions:

• What is the meaning of the observable $$BA$$ in such an experiment?

$$BA$$ is not generally an "observable." The fact that $$B$$ and $$A$$ are observables, per the statement of your question, indicates that $$BA$$ is not an observable (unless $$A$$ and $$B$$ commute).

If $$C=BA$$ then $$C^\dagger = AB$$ because $$A$$ and $$B$$ are Hermitian. So $$C^\dagger \neq C$$ unless $$A$$ and $$B$$ commute, which you wrote you don't want to assume.

Is it simply measuring $$A$$ first, $$B$$ immediately afterwards, and then multiplying the two measurements?

No.

• What is the meaning of $$[A,B]$$ in the experiment? If $$AB$$ and $$BA$$ are indeed obtained by measuring the observables in different orders...

But $$AB$$ and $$BA$$ aren't "indeed obtained by measuring the observables in different orders..." So, the rest of the question is moot.

Here's one example.

As others have mentioned, the commutator is not directly observable. But, $$i$$ times the commutator certainly can be.

For example, Angular momenta obey the rule $$\vec L\times \vec L = i\vec L\;,$$ so we see that $$i$$ times the commutator is just another component of the angular momentum.