# 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

## 1 Answer

In the same way that the expectation value of the position operator is the average position you'd get if you measured a bunch of identically-prepared states, the expectation value of the Hamiltonian operator is the average value of the Hamiltonian that you'd get if you measured a bunch of identically-prepared states. In most of the elementary situations you'll be looking at,* the value of the Hamiltonian is equivalent to the total energy, so the expectation value of the Hamiltonian is the average value of the energy that you'd get if you measured a bunch of identically-prepared states.

*The question of when the Hamiltonian is equivalent to the total energy is a complicated one, and depends, in part, on what you define "the total energy" to be in the first place, but until you get into Hamiltonians involving the electromagnetic field, you can usually take the two to be equivalent.

• Yes, that makes good sense until you treat Hamiltonian as a measurable quantity. It doesn't if you consider it as an operator. My statement "Average value of an operator (A machine that changes something to something else)" won't make logical sense. – Sameer Dambal Jul 14 '18 at 16:52
• @SameerDambal But you said in your question that there was no problem when you considered the average value of the position operator. So why is this different? – probably_someone Jul 14 '18 at 16:55
• 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
• @SameerDambal In order to make a measurement of an observable on a given state, we act on the state with the observable's associated operator. This operator (a machine that changes something to something else) changes the original state into a new state, whose form depends on the eigenvalues and eigenstates of that observer. When we make a measurement, we make a projection onto a random eigenstate. The probability of our projection onto each eigenstates is dictated in part by the form of the original state and in part by the form of the operator. – probably_someone Jul 14 '18 at 17:43
• @SameerDambal The expectation value of an operator is simply the average of the operator's eigenvalues, weighted by the probabilities determined above. In this way, when we say "the expectation value of an operator," what we mean is "the expectation value of the eigenvalues of that operator, when we act on a given state." – probably_someone Jul 14 '18 at 17:45