Suppose we have a quantum state $|\Psi\rangle = \alpha|0\rangle + \beta|1\rangle$.According to a measurement operator M, the projective measurement of $|\Psi\rangle$ is given by $\langle\Psi|M|\Psi\rangle$. Is this measurement always bounded between +1 and -1? If it is, why is that?


2 Answers 2


The measurement operators $\{M_i\}$ obey two conditions, firstly they are positive operators, $M_i\geq 0$, which is $\forall |\psi \rangle,\, \langle \psi | M_i |\psi \rangle \geq 0$.

Secondly, they must also obey $\sum_i M_i = I$. We can then deduce that $\langle \psi | M_j |\psi \rangle \leq \sum_i \langle \psi | M_i |\psi \rangle =\langle \psi | \sum_i M_i |\psi \rangle = \langle\psi | I|\psi \rangle =1$. Where the inequality comes from the fact that the sum contains the term on the left and additional non-negative terms.

Putting this all together, we have $0 \leq \langle \psi | M_i |\psi \rangle \leq 1$.

  • $\begingroup$ I don't get your answer... could you go further? $\endgroup$
    – aghin
    Commented Feb 8 at 18:21
  • $\begingroup$ @aghin00, I've added some more details about manipulating the inequality. Was that the part you had issues with? $\endgroup$
    – Ethan
    Commented Feb 9 at 12:07
  • $\begingroup$ I can agree with you that the upper bound is 1, but I don't think your last inequality is correct. If you choose your qubit to be, for example, $|-\rangle$ and your measurement operator X, then by doing a projective measurement you will get -1 as a result $\endgroup$
    – aghin
    Commented Feb 10 at 18:04
  • $\begingroup$ @Ethan the quantum expectation does not need to be limited to measurement operators (i.e. POVMs) for instance it is common to talk about the expectation value $\langle \psi|Z|\psi\rangle$. $\endgroup$
    – Condo
    Commented Feb 12 at 15:36
  • $\begingroup$ I think in the question, $M$ is corresponding to a projective measurement (or more generally a POVM). So that would rule out cases like $Z$ which is not positive. If $Z$ was allowed, then so would $\alpha Z$ and there would be no possible bounds. $\endgroup$
    – Ethan
    Commented Feb 12 at 21:39

The expectation value of an observable is not constrained to be between +1 and -1.

You can define $M$ to be any observable - i.e. a Hermitian operator. See https://physics.stackexchange.com/questions/27038/what-hermitian-operators-can-be-observables for example. If I have the eigendecomposition $M=\sum_\lambda \lambda |\lambda\rangle \langle \lambda|$, a single measurement value can be any of the $\lambda$ eigenvalues, that will be real due to $M$ being Hermitian.

$\langle \psi | M | \psi \rangle$ will give you the "expectation value" of these eigenvalues, which is a sum of each $\lambda$ weighted with the overlap of the eigenstate $|\lambda \rangle$ and $|\psi\rangle$.


  1. Think of a Hamiltonian, which has energy levels as the eigenvalues. Energy values are not restricted to be between +1 and -1.
  2. You could measure the observable 2Z - it is a perfectly fine observable, and has eigenvalues +2 and -2.
  • $\begingroup$ Maybe it is something about the operators M that I'm missing. In our course it is said that the measurement (projective) of a qubit $|\Psi\rangle$ is always bounded between +1 and -1 $\endgroup$
    – aghin
    Commented Feb 10 at 18:05
  • 2
    $\begingroup$ @aghin00 It is not true that the expectation is always between +1 and -1, unless you have additional assumptions on the matrix $M$. For instance, if you assume that $M=M^*$ and $M^2=I$, sometimes these are called binary observables, and indeed the expectation of such observables is always between $-1$ and $+1$. If you assume $M$ is a projection then you will get a value between 0 and 1. $\endgroup$
    – Condo
    Commented Feb 12 at 15:39
  • 1
    $\begingroup$ @Condo you are right! It was all about the assumptions on the M operator. Thank you! $\endgroup$
    – aghin
    Commented Feb 12 at 17:53

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