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).


I'll start with a couple of caveats and then try to answer the question.

  • 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.

  • $\begingroup$ 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 $A$ ? $\endgroup$ – onurcanbektas Jan 19 at 6:13
  • $\begingroup$ 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 ? $\endgroup$ – onurcanbektas Jan 19 at 6:15
  • $\begingroup$ @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 $A$. $\endgroup$ – Chiral Anomaly Jan 19 at 6:18
  • $\begingroup$ I see. What about the statement that if the projection operators commute so the observables ? Why is this true ? $\endgroup$ – onurcanbektas Jan 19 at 6:22
  • $\begingroup$ @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 $[A,B]=\sum_{j,k}a_j b_k\,[A_j,B_k]=0$. $\endgroup$ – Chiral Anomaly Jan 19 at 6:25

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