3
$\begingroup$

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).

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.