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?

  • $\begingroup$ 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
    Commented Sep 24, 2016 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? $\endgroup$ Commented Sep 24, 2016 at 18:32
  • $\begingroup$ Related: physics.stackexchange.com/q/130800/2451 and physics.stackexchange.com/q/9194/2451 $\endgroup$
    – Qmechanic
    Commented Sep 24, 2016 at 18:34
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Sep 24, 2016 at 19:32

2 Answers 2


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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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