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?


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.


Not the answer you're looking for? Browse other questions tagged or ask your own question.