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I was going through Prof. Leonard Susskind's lectures on Quantum Field Theory (Lec 2). Professor said that the commutator of two observables $AB-BA$, has nothing to do with the 'measurement'- B measured first and then A minus A measured first and then B. What does a commutator then mean?

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  • $\begingroup$ Well, it means the order in which you operate with A and B does not matter, if the commutator is zero, but it does matter if you get a non zero answer. But I am sure you know that already, and it's the measurement bit that's the problem. Someone with more knowledge than I should clear it up for both of us. $\endgroup$ – user108787 Sep 24 '16 at 17:29
  • $\begingroup$ It just means that $A$ and $B$ can be simultaneously diagonalized in some common basis, in which consequently the vectors are eigenstates of both operators. This means, as @CountTo10 said, that the order in which you apply the operators does not matter. The operation becomes associative: $A(B\mid \psi \rangle) = B(A \mid \psi \rangle)$. Is it still unclear? Which part? $\endgroup$ – Kyle Arean-Raines Sep 24 '16 at 18:32
  • $\begingroup$ Related: physics.stackexchange.com/q/130800/2451 and physics.stackexchange.com/q/9194/2451 $\endgroup$ – Qmechanic Sep 24 '16 at 18:34
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    $\begingroup$ As I understood this operators are observables which act upon the states. So, in my opinion it has something to do with measurement. I am now also waiting for the answer that clears up proffesors claim about measurement having nothing to do with the operators. $\endgroup$ – Žarko Tomičić Sep 24 '16 at 19:32
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I'm sorry but commutator has direct relation to the possibility of simultaneous measurements.

The observable being self-adjoint operator can be represented as a sum of self-adjoint orthogonal projectors on eigenspaces. \begin{equation} \hat{A}=\sum_k \lambda_k \mathcal{P}_{\lambda_k},\quad \hat{A}\mathcal{P}_{\lambda_k}|\psi\rangle=\lambda_k\mathcal{P}_{\lambda_k}|\psi\rangle,\quad \mathcal{P}_{\lambda_k}\mathcal{P}_{\lambda_m}=\delta_{km}\mathcal{P}_{\lambda_k},\quad \mathcal{P}_{\lambda_k}=\mathcal{P}_{\lambda_k}^\dagger \end{equation} In particular, if all eigenspaces are 1-dimensional we can write $\mathcal{P}_{\lambda_k}=|\lambda_k\rangle\langle\lambda_k|$

The ideal measurement in quantum mechanics is defined the following way. If you measure $A$ and get that it equals $\lambda_k$ then the state changes by projection on the corresponding eigenspace, \begin{equation} |\psi\rangle\mapsto \frac{1}{\sqrt{P_\psi(A=\lambda_k)}}\mathcal{P}_{\lambda_k}|\psi\rangle, \end{equation} with probability (in case of continous spectrum - probability density) given by, \begin{equation} P_\psi(A=\lambda_k)=\langle\psi|\mathcal{P}_{\lambda_k}^\dagger\mathcal{P}_{\lambda_k}|\psi\rangle=\langle\psi|\mathcal{P}_{\lambda_k}|\psi\rangle \end{equation} When you apply this to two consequetive measurements of two observables $A$ and $B$ it happens that you can't define simultaneous measurements without specifying the order in which you measure. That's because in general, \begin{equation} \langle\psi|\mathcal{P}_{A=\lambda_k}^\dagger\mathcal{P}_{B=\mu_m}^\dagger\mathcal{P}_{B=\mu_m}\mathcal{P}_{A=\lambda_k}|\psi\rangle\neq \langle\psi|\mathcal{P}_{B=\mu_m}^\dagger\mathcal{P}_{A=\lambda_k}^\dagger\mathcal{P}_{A=\lambda_k}\mathcal{P}_{B=\mu_m}|\psi\rangle \end{equation}

The only exception is when observables commute. That's because of the simultaneous diagonalizability - the Hilbert space happens to be a direct sum of eigenspaces $(\lambda_k,\mu_m)$ where all states are simultaneously eigenstates of $A$ and $B$. From that it follows that $[\mathcal{P}_{\lambda_k},\mathcal{P}_{\mu_m}]=0$ and you can define simultaneous measurements of $A$ and $B$ not caring about their order.

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I have just realized why the application of an operator is not a measurement. Let me update my answer accordingly.

One axiom of quantum mechanics is that a measurement will take a state to an eigenstate of the corresponding operator. If two operators commute, $[A, B] = 0$, then a basis can be found such that both operators are diagonal in it. Therefore states can be eigenstates of both operators at the same time.

Just applying an operator to a state does not collapse the wave function, however. So take the harmonic oscillator with eigenstates $|n\rangle$ as an example. My operator is $\hat n$, the occupation number operator. When I take a state which is not an eigenstate, like $|\psi\rangle = |1\rangle + |2\rangle$, applying the operator will give me the following: $$ \hat n |\psi\rangle = \hat n|1\rangle + \hat n|2\rangle = |1\rangle + 2 |2\rangle \,.$$ This is neither proportional to the original state, nor a pure eigenstate of $\hat n$. If an actual measurement was to be performed, then by the axioms of quantum mechanics it would have to be a pure eigenstate of $\hat n$. We would have a 50% chance that we had $n = 1$ or $n = 2$ out of the measurement. The system would then be in either $|1\rangle$ or $|2\rangle$ but not a linear combination any more.

Therefore the simple application of a hermitian operator, which is an observable, is not a measurement. The collapse of the wavefunction to a pure eigenstate of the operator is needed as well. And this is what is missing when one applies the commutator to a state.

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