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a) Assume two operators $A$ and $B$.

1) Assume $$[A,B]=0 $$ and $$ ψ= \sum c_n u_n ~~~~\text a~ wavefunction~ describing~ the~ state~ of~ the~ system $$ with $$Aψ=a_n u_n $$ $$Bψ=b_n u_n$$ If we take two measurements, one after the other we have $$ABψ=a_n b_n u_n $$

2)Now let's assume that $$[A,B] \neq 0 $$If first we take $$Αψ=a_n u_n$$ and then we measure with $B$, shall we have $$B(a_n u_n)=b_m v_m $$ a different eigenvector of the system , that is: $$ψ-->a_n u_n --> b_m v_m $$

Does this result comes from the fact that for $B$, the vector $$u_n$$ is a superposition with some probabilities $P_n$ to have a result, after the act of B on $$u_n,$$ the $$b_m v_m$$ eigenvector of B?

b) Also, assume that we make a measurement with $A$. After a time $t$ from the measurement the system under consideration will return to a superposition state.

Does that happen because of the uncertainty principle or because of the internal interactions of the system? If the first is there a mathematical formulation to explain the evolution from the eigenstate to the superposition state? If the second, can we in theory represent the interactions through operators and the evolution that concludes to the superposition state (and if so, do these operators produce eigenstates of the system or different states)?

Note:I find the two parts of the question relevant because the subject under consideration is about operators and how they act on a system, or if you like is about understanding if every interaction can be described by operators and what a sum of measurements in a row can has as a result on the system. But if you think that the two parts should be two questions, please comment and answer only to the first part. I shall post the second part independently.

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  • $\begingroup$ I have no idea what your question in a) is, and to b) the "answer" is: All states obey the Schrödinger equation, so the mathematical formulation for the time evolution operator is $\mathrm{e}^{\mathrm{i}Ht}$. Uncertainty has nothing to do with it. $\endgroup$ – ACuriousMind Apr 18 '15 at 12:04
  • $\begingroup$ @ACuriousMind Thanks for replying on b). About a), i want to understand in what kind of state (or eigenstate) the system will be after measuring with B on the eigenstate produced by A and Does B act on a superposition or on the eigenstate of A? $\endgroup$ – Constantine Black Apr 18 '15 at 12:07
  • $\begingroup$ @ACuriousMind Is still unclear what I'm asking? If so what is wrong? $\endgroup$ – Constantine Black Apr 18 '15 at 12:19
  • $\begingroup$ To me, it's still a bit unclear - what is an "eigenstate of the system"? Eigenstates are of operators, not of systems. $\endgroup$ – ACuriousMind Apr 18 '15 at 12:20
  • $\begingroup$ @ACuriousMind Yes that's what I mean in my comment. If there is a mistake, where is it? What is it I don't understand and that's problematic in the question? $\endgroup$ – Constantine Black Apr 18 '15 at 12:29
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$\newcommand{\ket}[1]{\lvert #1 \rangle}$You seem to be confused about what measuring an operator means. Let $A,B$ be two commuting self-adjoint operators as in your question, and let $\{u_n\}$ be a basis of simultaneous eigenvectors, that is $$ A\ket{u_i} = a_i \ket{u_i} \ \vee \ B\ket{u_i} = b_i \ket{u_i}$$ Now, a generic state $\psi$ can be written as $$ \psi = \sum_i \psi_i \ket{u_i}$$ where the coefficients $\psi_i$ are just the inner product of $\psi$ with the basis vectors, i.e. $\psi_i = \langle \psi \vert u_i \rangle$.

The result of measuring $A$ or $B$ is now one of the $\ket{u_i}$ (the state has "collapsed" into an eigenstate of the measured operator), and the numerical result of the measurement is the eigenvalue $a_i$ or $b_i$ associated to $\ket{u_i}$. How exactly this measurement result comes about is the subject of much debate, I present one way of looking at it, the von Neumann measurement scheme in this answer.

However, it has to be said that the result of the measurement is not $A\psi$ or $B\psi$ unless $\psi$ was already an eigenstate, since that would be $$ A\psi = \sum_i \psi_i a_i \ket{u_i}$$ which is not an eigenvector of $A$, and hence not a measurement result, in general.

Regardless of how exactly we arrived at the measurement result as an eigenvector of $A$ or $B$, all such result will, afterwards, unitarily evolve in time by $\mathrm{e}^{\mathrm{i}Ht}$ just like any other state (in the Schrödinger picture).

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  • $\begingroup$ Really, much appreciated. I'll study the links you mentioned. When you have some time, explain if you can what you mean by However, it has to be said that the result of the measurement is not Aψ or Bψ unless ψ was already an eigenstate, since that would be $$Aψ=∑iψiai∣ui⟩$$ which is not an eigenvector of A, and hence not a measurement result, in general. $\endgroup$ – Constantine Black Apr 18 '15 at 13:39
  • $\begingroup$ @ConstantineBlack: I will happily explain if you tell me what you don't understand about that - I am saying that the result of a measurement is not $A\psi$, as your question seems to assume. $\endgroup$ – ACuriousMind Apr 18 '15 at 15:41
  • $\begingroup$ Is this the assumption you are referring to?:"If first we take $$Aψ=a_n u_n$$ and then we measure with B, shall we have $$B(a_n u_n)=b_m v_m$$ a different eigenvector of the system , that is: $$ψ−−>a_n u_n−−>b_m v_m$$" If so, what's the mistake here? Also with Aψ, do you mean the eigenvector of A? $\endgroup$ – Constantine Black Apr 18 '15 at 17:56
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    $\begingroup$ @ConstantineBlack: I see now where you think you say that in the question, but it is really not clear. Writing $A\psi = a_n u_n$ and saying that $\psi$ is an eigenvector begs the question why you didn't write $A\psi = a_n \psi$. If $A$ and $B$ have no eigenvalue with more than one eigenvector, then it follows that $\psi$ will also be an eigenvector of $B$. If they do have eigenvalues with multiple independent eigenvectors, then we cannot say that $BA\psi$ might be, in general. I still don't really see what you are geting at with this question. $\endgroup$ – ACuriousMind Apr 18 '15 at 18:26
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    $\begingroup$ @ConstantineBlack: Yes, I meant what not that. 1. I don't have Tannoudji. 2. If $A$ and $B$ do not commute, then, indeed the result of a measurement of $A$ (which is an eigenvector of $A$) will be a superposition of eigenvectors of $B$ in general. $\endgroup$ – ACuriousMind Apr 18 '15 at 19:41

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