# Performing a projective measurement, is the resulting expectation value $\langle \Psi|M|\Psi\rangle$ bounded between $+1$ and $-1$?

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?

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 $$0 \leq \langle \psi | M_i |\psi \rangle \leq 1$$.

• I don't get your answer... could you go further? Commented Feb 8 at 18:21
• @aghin00, I've added some more details about manipulating the inequality. Was that the part you had issues with? Commented Feb 9 at 12:07
• 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 Commented Feb 10 at 18:04
• @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 $\langle \psi|Z|\psi\rangle$. Commented Feb 12 at 15:36
• 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. 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 $$|\psi\rangle$$.

Examples:

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.
• 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 Commented Feb 10 at 18:05
• @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. Commented Feb 12 at 15:39
• @Condo you are right! It was all about the assumptions on the M operator. Thank you! Commented Feb 12 at 17:53