I'm currently reading Quantum Computation and Quantum Information by Nielsen. I'm struggling to solve exercise 2.58. The problem is

Suppose we prepare a quantum system in an eigenstate $|\psi\rangle$ of some observable $M$, with corresponding eigenvalue m. What is the average observed value of $M$, and the standard deviation?

The average is easy to find, $$\langle M \rangle = \langle \psi | M | \psi \rangle = m \langle \psi | \psi \rangle = m.$$

The standard deviation is by definition, $$\Delta(M) = \sqrt{\langle M^2 \rangle - \langle M \rangle^2} = \sqrt{\langle M^2 \rangle - m^2}$$.

So you just need find the value of $\langle M^2 \rangle$, $$\langle M^2 \rangle = \langle \psi | M^2 | \psi \rangle = m \langle \psi | M | \psi \rangle = m^2,$$ which give the standard deviation of zero.

It's violating Heisenberg uncertainty principle so I believe that I've got something wrong. What is the correct answer to this exercise?


closed as off-topic by Kyle Kanos, stafusa, Jon Custer, JamalS, ZeroTheHero Oct 10 '17 at 0:02

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The correct answer is the one you obtained, i.e. 0. The Heisenberg uncertainty principle has to do with non-commuting observables. Read from the same book the treatment of non-commuting observables.


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