# What exactly does the expectation value of $x$ mean in quantum mechanics?

When I learn quantum mechanics （by reading Griffith's book Introduction to quantum mechanics 2ed edition (Page 15)）, I was confused by the concept of the expectation value of $x$, i.e. $\langle x\rangle=\int^{+\infty}_{-\infty}x|\Psi(x,t)|^2dx$. He said that,

In short, the expectation value is the average of repeated measurements on an ensemble of identically prepared systems, not the average of repeated measurements on one and the same system.

I can't understand that why he said we should have a whole ensemble of identically prepared systems. I hope you could explain that to me in detail.

• @yuggib you could make that as an answer – Oswald Jan 27 '16 at 15:37
• @TheGhostOfPerdition here you go ;-) – yuggib Jan 27 '16 at 16:05

## 1 Answer

In quantum systems it is not possible to perform a measurement without affecting the measured system. This is because, roughly speaking, the interaction with the instrument creates a correlation between the system and the instrument whose effective result is a modification of the system's state.

Therefore if you perform repeated measurements of the same observable in the same system and take the average, you would not get the same value as if you average the outcome of many measurements of the same observable in identical copies of the given initial state of such system. As a matter of fact, measuring the observable an $n$-th time in a system where a measurement has already been done, you would get exactly the same outcome as in the aforementioned previous measurement.

The expectation value is instead what you would get, in the limit of an infinite number of measurements, averaging the measured value of an observable in identical copies of the same initial configuration. Because of the above, apart from the case in which the initial state is particularly special (eigenstate of the observable), the two procedures do not coincide.