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Given two (or more) observables $A, B$ which commute one can construct a third observable $C= A \circ B$. If $\psi$ is a common eigenvector of $A, B$ with eigenvalues $\lambda_1, \lambda_2$ then it is clear that the measurement of $C$ of the state $\psi$ gives the measurement result $\lambda =\lambda_1 \lambda_2$, i.e. the result of the measurement of the observable $C$ is the result from the measurement of $A$ times the result from the measurement $B$. But what if $\psi$ is an eigenvector of $C$, but not of $A$ and $B$? Is there any connection between the measurement results of $A$, $B$ and $C$?

Example: Let there be three observers which measure a spin state with the corresponding observables $A = \sigma_x \otimes \mathbb{I} \otimes \mathbb{I}$, $B=\mathbb{I} \otimes \sigma_y \otimes \mathbb{I}$ and $C=\mathbb{I} \otimes \mathbb{I} \otimes \sigma_y$. They commute and we can construct $D= A \circ B \circ C = \sigma_x \otimes \sigma_y \otimes \sigma_y$. Now the GHZ-state $\psi = \frac{1}{\sqrt{2}} ( | +z, +z, +z \rangle - | -z, -z, -z\rangle)$ is an eigenvector of $D$ with eigenvalue $\lambda =-1$ but it is not an eigenstate of $A, B$ or $C$.

Each of the observers will get a result $\pm 1$. Is there any connection between this individual results and the eigenvalue of $\psi$ (respectively the expectation value $\langle \psi | D | \psi \rangle = -1$)? Intuitively I would say that the product of the results should give the eigenvalue of $\psi$ but I can't see how this should follow from any quantum mechanical postulate or mathematical reasoning like in the case of the common eigenvector.

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Just to make sure we're on the same page, an observable is defined as a Hermitian linear operator whose eigenvectors suffice to form a basis. For such operators, or observables, $A,B$, it is not hard to show that they commute iff their product is an observable, which amounts to their product having a basis of eigenvectors. Thus in the formalism of q.mechanics at least, if they don't commute then their product ceases to be observable as its possible observations don't suffice to characterize the physical system at hand. I would agree, however, that this may not be a very satisfactory answer... –  ntropy Oct 31 '13 at 4:01
    
... to the kind of idea I think you're trying to get it. It certainly feels like there should still be something going on... –  ntropy Oct 31 '13 at 4:02
    
Yes, we are on the same page. But my question goes further. I suppose I stated my question not clear enough. Please look at my answer below for clarification. –  David Oct 31 '13 at 23:23
    
I realized a few days ago that some thoughts of my question, and therefore the question itself, are wrong. Please see my comment below my answer. –  David Nov 10 '13 at 19:25
    
@David That's actually okay - it's still useful if your answer explains what your misconception was and/or how to resolve it. –  David Z Mar 23 at 18:09
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2 Answers

Taking $C=A_1A_2....A_n$, the problem arises because the eigenvector subspace corresponding to a eigenvalue of $C$ does not correspond to the eigenvector subspaces corresponding to a eigenvalue of the $A_i$

To see that, we will take an example with $C=A_1A_2$, with $A_1 = \sigma_x \otimes Id, A_2 = Id \otimes\sigma_x $. So $C= \sigma_x \otimes\sigma_x$ We have the following array :

$$\begin{pmatrix} &\sigma_x \otimes Id&Id \otimes\sigma_x&\sigma_x \otimes\sigma_x \\(|00\rangle+|10\rangle)+(|01\rangle+11\rangle)&+&+&+ \\(|00\rangle+|10\rangle)-(|01\rangle+11\rangle)&+&-&- \\(|00\rangle+|01\rangle)-(|10\rangle+11\rangle)&-&+&- \\(|00\rangle+|11\rangle)-(|01\rangle+10\rangle)&-&-&+\end{pmatrix}$$

The first column is made of the common eingenvectors, and the other columns correspond to the eigenvalues ($\pm$ means $\pm1$).

The subspace corresponding to the eigenvalue $+1$ of $\sigma_x \otimes\sigma_x$ is $2-$dimensional and corresponds to the first and last eigenvectors.

Now, if we add the first and the last eigenvector, we get the state $|00\rangle+|11\rangle$, and because the first and the last eigenvectors have the same eigenvalue $+1$ for $\sigma_x \otimes\sigma_x$, then $|00\rangle+|11\rangle$ is also a eigenvector with eigenvalue $+1$ for $\sigma_x \otimes\sigma_x$.

But the problem is that the first and last eigenvector have not the same eigenvalue for $\sigma_x \otimes Id$ and $Id\otimes\sigma_x$, so any combination of these $2$ eigenvectors cannot be an eigenvector for $\sigma_x \otimes Id$ and $Id\otimes\sigma_x$. And this is indeed the case for $|00\rangle+|11\rangle$

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Firstly I think I did not state my question clearly enough: I know the mathematics behind commuting observables and their eigenvalues but I wanted to know what happens in a real physical experiment. I wanted to know if there is a connection between the eigenvalue $\lambda$ of a prepared eigenstate $\psi$ of $C=A_1 \ldots A_n$ and the individual results $\lambda_1, \ldots, \lambda_n$ of the measurements of $A_1, \ldots, A_n$. The situation is clear if $\psi$ is a common eigenvector of $A_1, \ldots, A_n$ but it was not clear to me what happens if $\psi$ is NOT a common eigenvector.

Trimok's answer did not answer my question directly but his/her example gave me an important insight which helped me to figure this out.

One of my misconceptions was the idea that an eigenstate $|\psi\rangle$ of $C=A_1 \ldots A_n$ does not change during a measurement of $C$ because of the mathematical relation $C|\psi\rangle = \lambda |\psi\rangle$. But this is not necessarily true. For example, the state $|00\rangle + |11\rangle$ changes during the measurement of $\sigma_x \otimes \sigma_x$: each observer will either get $|0\rangle$ or $|1\rangle$ so the state after the measurement will be $|00\rangle$ or $|11\rangle$ although $|00\rangle + |11\rangle$ is an eigenstate of $\sigma_x \otimes \sigma_x$ to the eigenvalue $+1$.

I think the connection between the different eigenvalues is as follows. Let $|\psi\rangle$ be an eigenstate of $C = A_1 \ldots A_n$ to the eigenvalue $\lambda$, but it does not necessarily have to be a common eigenvector of the observables $A_1, \ldots, A_n$. In general, $|\psi\rangle$ can be expressed as a linear combination of common eigenvectors: $$ |\psi\rangle =\sum_{\substack{\alpha_1, \ldots, \alpha_n \\ \alpha_1 \cdots \alpha_n=\lambda}} c_{\alpha_1, \ldots, \alpha_n} |\alpha_1, \ldots, \alpha_n\rangle, $$ where $|\alpha_1, \ldots, \alpha_n\rangle$ is an eigenvector of $A_1$ to the eigenvalue $\alpha_1$, etc. The product of the $\alpha_i$ has to be $\lambda$, because it is an abstract mathematical result that the set of eigenvalues of $C$ has this form.

Now, what happens if the observables $A_1, \ldots, A_n$ are measured separately (w.r.t. $|\psi\rangle)$? Each measurement will make the subspace where the measured state lives smaller: The measurement of $A_1$ will force the state $|\psi\rangle$ to change into a state $|\beta_1\rangle$ where $\beta_1$ is an eigenvalue of $A_1$. But it can still be expressed as a linear combination of common eigenvectors of $A_2, \ldots, A_n$. Then the measurement of $A_2$ will change the state into $|\beta_1, \beta_2\rangle$ and so on. After the measurement of $A_n$ we are left with a common eigenstate $|\beta_1, \ldots, \beta_n\rangle$ and $\beta_1 \cdots \beta_n = \lambda$ because $|\psi\rangle$ was originally a linear combination of common eigenvectors which fulfill this equation.

So the statement "The product of the individual results is equal to the eigenvalue $\lambda$ of the eigenstate $|\psi\rangle$ of $C$" still holds in the case, when $|\psi\rangle$ is not a common eigenvector. This intuitive result is now backed up by a correct argument.

Just as an interesting side note: When an observable $C$ is a composition of other observables, e.g. $C=A\circ B$ then this does not mean that one has to measure $B$ and $A$ consecutively in order to measure $C$. $C$ can represent a single measurement. For example, let's take $A=\sigma_x \otimes \sigma_y$ and $B=\sigma_y \otimes \sigma_x$, then $C = (\sigma_x \otimes \sigma_y) \circ (\sigma_y \otimes \sigma_x) = \sigma_z \otimes \sigma_z$. So $C$ can represent different experimental setups: either a consecutive measurement of $B$ and $A$ or a single measurement of two qubits along the z-axis. (I am new here and I don't know if such remarks are welcomed here or regarded as annoying. I remove them if somebody wishes.)

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My statement that a measurement of $C$ of a common eigenvector $\psi$ of $A_1, \ldots, A_n$ with eigenvalue $\lambda = \lambda_1 \ldots \lambda_n$ w.r.t. $C=A_1 \ldots A_n$ gives $\lambda$. That's wrong! For example, take $C= -\sigma_x \otimes \sigma_x$, $A_1=\sigma_y \otimes \sigma_y$, $A_2=\sigma_z \otimes \sigma_z$. A Bell state like in Trimok's example is a common eigenvector, but the measurment of $A_2$ reduces it to a vector, that is not an eigenvector of $A_1$ anymore! So after the measurement of $A_1$ the product of the results does not have to be equal to -1 (the eigenvalue w.r.t. C). –  David Nov 10 '13 at 19:41
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