# Is is possible to have a pair commuting observables only in a single direction?

In quantum mechanics, for two observables to be compatible, successive measurements of the observables, say $$A$$ and $$B$$, should yield the same result as earlier, i.e if we do the measurements with the order $$A \to B \to A$$, the result from the first $$A$$ and the last $$A$$ should be the same, and similarly, for $$B \to A \to B$$.

However, is it possible for two observables to have the following relation;

If we measure $$A \to B \to A$$, the first and the last measurement of $$A$$ yield the same result, but if we measure $$B\to A \to B$$, the first and the last measurement of $$B$$ yields different measurements (in general).

• First caveat: Consecutive measurements of the same observable $$A$$ don't necessarily yield the same outcome, not even if no other observable $$B$$ is measured in between, because the system evolves according to the Schrödinger equation between the two $$A$$-measurements. (For example, the more precisely we try to measure a particle's position, the more rapidly it loses that precise precision after the measurement is done, because $$\Delta x\,\Delta p\geq \hbar/2$$.) With some idealization, though, we can say that two immediately-consecutive measurements of $$A$$ would yield the same outcome. The following answer uses this projective measurement idealization.

• One more caveat: Implicit in the projective-measurement idealization is the assumption that the observables in question are discrete, because only then do they have normalizable eigenstates. (This is implicit in the idea of a sharply-defined "outcome.") For that reason, the following answer considers only discrete observables. Since any continuous observable can be arbitrarily-well approximated by a discrete one, this should be sufficient for all practical purposes.

With these idealizations/caveats, here is the answer to the question:

• Suppose that observables $$A$$ and $$B$$ have the property that if we measure $$A$$-then-$$B$$-then-$$A$$ with no delay, the outcomes of the first and last $$A$$-measurement are always the same.

• This implies that if we measure $$B$$-then-$$A$$-then-$$B$$ with no delay, then the outcomes of the first and last $$B$$-measurement are always the same. In the OP's words, two observables cannot commute in only a single direction.

Proof: The observables' eigenvalues are not relevant to this question, so we can think of the discrete observable $$A$$ as a set $$A=\{A_1,A_2,...\} \tag{1}$$ of mutually commuting projection operators $$A_j$$ satisfying $$\sum_j A_j=1 \hskip2cm A_j A_k=0\text{ if }j\neq k \tag{2}$$ and similarly for the other observable $$B=\{B_1,B_2,...\}$$. Now, suppose that that whenever we measure $$A$$ then $$B$$ then $$A$$, the outcomes of the two $$A$$-measurements are always the same. If we start with any state-vector $$|\psi\rangle$$ and apply an $$A$$-measurement followed by a $$B$$-measurement, with outcomes $$A_j$$ and $$B_k$$, respectively, then the resulting state-vector is $$B_kA_j|\psi\rangle$$. We have assumed that the outcome of a subsequent $$A$$-measurement can only be $$A_j$$, which implies $$A_mB_k A_j|\psi\rangle=0 \hskip1cm \text{ for all k,\psi whenever } m\neq j. \tag{3}$$ This holds for all $$\psi$$, so $$A_mB_k A_j=0 \hskip1cm \text{ for all k whenever } m\neq j. \tag{4}$$ Equation (2) implies $$\sum_m A_mB_k A_j = B_k A_j \hskip1cm \text{for all }k,j \tag{5a}$$ and equation (4) implies $$\sum_m A_mB_k A_j = A_j B_k A_j \hskip1cm \text{for all }k,j. \tag{5b}$$ Combine equations (5a) and (5b) to get $$B_kA_j = A_jB_kA_j \hskip1cm \text{for all }k,j. \tag{6}$$ Taking the adjoint of both sides of (6) gives $$A_jB_k = A_jB_kA_j \hskip1cm \text{for all }k,j \tag{7}$$ because projection operators are self-adjoint and because the adjoint reverses the order of multiplication. Combine (6) and (7) to get $$A_jB_k=B_k A_j$$ for all $$j,k$$. This impiles that the observables $$A$$ and $$B$$ commute with each other, which completes the proof.

• However, when you measure, say $A$, when the system in the state $|x \lange$, then the state of the system is $A_j \phi_j$, where $\phi_j$ is a an eigenket of $A$; not necessarily $A_j \psi$, so are you assuming $\psi$ is an eigenket of $A$ ? – onurcanbektas Jan 19 at 6:13
• By the way, I'm confused with your notation, what is type of the object $A_i$, and what does $A_i A_j$ stand for ? – onurcanbektas Jan 19 at 6:15
• @onurcanbektas $A_j$ is a projection operator, which projects any state-vector into the $j$-th eigenspace of $A$. The notation $XY$ means the product of two operators: first apply $Y$ to the state, then apply $X$. If we measure $A$ in the state $|\psi\rangle$, then the result is $A_j|\psi\rangle$ for some $j$. I'm not assuming that $|\psi\rangle$ is an eigenket of $A$. For any state-vector $|\psi\rangle$ whatsoever, the state-vector $|\phi_j\rangle=A_j|\psi\rangle$ is an eigenket of $A$, because $A_j$ is the projection operator onto the $j$-th eigenspace of $A$. – Chiral Anomaly Jan 19 at 6:18
• I see. What about the statement that if the projection operators commute so the observables ? Why is this true ? – onurcanbektas Jan 19 at 6:22
• @onurcanbektas The observable $A$ is $A=\sum_j a_j A_j$, where the coefficients $a_j$ are real numbers and $A_j$ are the projection operators. Similarly, $B=\sum_k b_k B_k$ with real coefficients $b_k$. If all of the $A_j$s commute with all of the $B_k$s, then $A$ commutes with $B$ because $[A,B]=\sum_{j,k}a_j b_k\,[A_j,B_k]=0$. – Chiral Anomaly Jan 19 at 6:25