# What is the physical meaning of expectation value of the Hamiltonian operator?

I've been studying David Griffiths' Introduction to Quantum Mechanics and int that, it was explained that the expectation value of position $x$ is the average of the positions of $N$ identically prepared particles. This makes sense but later on, they tried finding the expectation value of the Hamiltonian operator. What is the meaning of this? Average of an operator doesn't make sense.

• Is $\hat x$ not the position operator? – J. Murray Jul 14 '18 at 16:34
• @J.Murray Yes. But you see it is perfectly logical if x is not treated as an operator (from the identically prepared system). But if we make $\hat x$ an operator, its expectation value doesn't make sense physically (Just like $\hat H$ in my question). – Sameer Dambal Jul 14 '18 at 16:49
• The expectation value of the position operator is the average of the position measurements performed on a large number of identical systems. The expectation value of the Hamiltonian (i.e. energy) operator is the average of the energy measurements performed on a large number of identical systems. – J. Murray Jul 14 '18 at 16:59
• A discussion is going on this in the Answer section. Kindly go through my arguments there in order to refrain from double work. – Sameer Dambal Jul 14 '18 at 17:12

• I took care in writing 'x' and not $\hat x$. It's the operator that I've a problem with. – Sameer Dambal Jul 14 '18 at 17:10