Let us consider a system composed of qubits, with subsystems $\mathcal{A}$ and $\mathcal{B}$. For any measurement, described by an Observable $O=O_{\mathcal{A}}\otimes I$ that can be viewed as putting a slit on the bloch-sphere representation, are there observables other than ones with eigenvalues $\pm1$ (apart from trivial 0)?

If you consider one qubit, indeed a measurement (in any basis) can be described by orthonormal states $|0_a\rangle$ and $|1_a\rangle$ and the corresponding observable $O_a=(+1)|0_a\rangle\langle 0_a|+ (-1)|1_a\rangle\langle 1_a|$. Also for 2 qubits, an observable $O_a\otimes I$ has the eigenvalues $\pm1$.

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    $\begingroup$ What do you mean? An operator of the form $c\, |0\rangle\langle 0|$ has $0$ and $c \in \mathbb R$ as eigenvalues. $\endgroup$ Commented Jun 27, 2022 at 21:06
  • $\begingroup$ Youre right. I should have said "other than the trivial 0". $\endgroup$
    – manuel459
    Commented Jun 28, 2022 at 9:22
  • $\begingroup$ The eigenvalues have to be different for an observable on a qubit to be non-trivial, but they don't have to be $\pm 1$. $\endgroup$
    – alanf
    Commented Jun 28, 2022 at 9:29
  • $\begingroup$ Why is that? Apparently (with $O_a$ above) one is able to do all measurements. $\endgroup$
    – manuel459
    Commented Jun 28, 2022 at 9:46
  • $\begingroup$ Well, take $O=c_1 |0\rangle \langle 0| + c_2 |1\rangle\langle 1|$ with $c_1,c_2 \in \mathbb R \setminus \{0\}$. Sorry, I still don't understand. $\endgroup$ Commented Jun 28, 2022 at 10:12

2 Answers 2


If you have a qubit then you have a 2-dimensional Hilbert space. This means that you cannot have more than 2 mutually orthogonal states. This means that if these are eigenstates of some observable then you can have at most 2 eigenvalues. So if the eigenvalues are $\pm 1$ then you do NOT have a further possibility of value 0 for that observable. (If you did then you would have a qutrit not a qubit).

The pair of eigenvalues can have any values. Just to make the point, here is a list of possibilities: $$ +1, -1 \\ 0, 1 \\ 0, 0 \\ -2.3435, 7.3438 $$ where in the third example I picked a degenerate case and in the fourth I picked a pair of random numbers. The eigenvalues you get depend on the observable you measure. This is no big deal. It is like one person choosing to measure a length $x$, and the next person introducing an observable $y$ which is defined by $$ y = 9.6873 x - 2.3435 $$ so when $x$ is 0 then $y$ will be $-2.3435$ and when $x$ is 1 then $y$ will be $7.3438$. The purpose of this example is simply to show that the values of eigenvalues can in principle be anything you like, depending on how the quantity you are measuring is defined. This has nothing to do with the qubit itself. All the qubit itself dictates is that you can have at most two eigenvalues for a given Hermitian operator.


Observables can have arbitrary (real) eigenvalues. Any Hermitian operator is an observable (which is a tautological statement: an observable is defined as a Hermitian operator). This holds for arbitrary quantum systems, and thus in particular for many-qubit quantum systems (which are those of dimension $2^n$ for some $n\in\mathbb N$).

If you consider local observables on many-qubit systems, that is, for two-qubit systems, observables of the form $O\otimes I$ or $I\otimes O$, then the eigenvalues will always be (at least) double-degenerate. That's trivially because you can always find eigenvectors of $O\otimes I$ to be simple tensors, i.e. vectors of the form $u\otimes v$, and moreover $v$ can be chosen arbitrarily.

If you want a specific example, consider for example the one-qubit observable defined as $O=\pi |0\rangle\!\langle 0|+\frac{1}{\pi^e}|1\rangle\!\langle 1|$. This has eigenvalues different than $\pm1$.

  • $\begingroup$ Note that Any Hermitian operator is an observable is a disputable point. Is e.g. the density operator an observable? I know this was discussed somewhere here, but I can't find the link to the post. $\endgroup$ Commented Jun 28, 2022 at 16:19
  • $\begingroup$ @JasonFunderberker sure it is. An observable is by definition an Hermitian operator. Granted, an observable can be hard or impossible to implement in a given physical architecture, or represent a trivial measurement, or things like that, but that's a different issue (one thing this is assuming is to deal with finite-dimensional systems thogh; more generally observables are symmetric operators) $\endgroup$
    – glS
    Commented Jun 28, 2022 at 16:33
  • $\begingroup$ I'd disagree with your first sentence, but okay, I did not want to start the discussion (I've found the post, see also the links therein) again. I just wanted to point out a subtle and or interesting point. $\endgroup$ Commented Jun 28, 2022 at 16:44

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