# Expectation Value of x

1. Nov 5, 2008

### pardesi

How does this follow from the defintion of the expectation value of x

2. Nov 5, 2008

### malawi_glenn

Just as in ordinary theory of probability. Just change wavefunction to probability density function.

3. Nov 5, 2008

### pardesi

i am not able to get the importance of identically prepared systems

4. Nov 5, 2008

### malawi_glenn

compare it with the dice example in statistics, the dice must be indentical in all the "experiments". i.e the probability density must be the same in all experiments.

What if you first have a dice with perfect 6 sides in the first experiment, then in the 4435th you switch to a dice which does not have 6 perfect sides, but prefers to show 4?

It's just basic theory and definition of probability theory.

5. Nov 5, 2008

### Fra

In case it isn't clear after malawi_glenn's response, another good question that might bother you is that, OK the statistics refers to the same dice, but the real question is how do you KNOW that dice is the same?

You might argue that in many realistic cases you don't.

Then the supposed solution to those cases is the density matrix formalism(http://en.wikipedia.org/wiki/Density_matrix), which can be thought of as a classical type probability distribution over the possible quantum states (Here the dice is the quantum state). So I think of it intuitively as a double dice setup. A classical dice, which you throw to see what quantum-dice you get. So each "face" of the classical dice, is actually a quantum-dice, which can then be thrown.

The case where you know (left alone HOW you acquired this knowledge with certainty) that the dice(really meaning quantum state) is determined, is called a pure state. If the dice is uncertain it's a mixed state.

/Fredrik

6. Nov 5, 2008

### clem

You have it backwards.
"The expectation value is the average of repeated measurements on an ensemble of identically prepared systems" is the definition.
From the definition, you can derive $$<\psi|x|\psi>$$ as one method of calculating the expectation value.

7. Nov 5, 2008

### Fredrik

Staff Emeritus
Just expand in the basis of eigenstates and note that the right-hand side is the weighted average of all the eigenvalues:

$$\langle\psi|A|\psi\rangle=\sum_a\langle\psi|A|a\rangle\langle a|\psi\rangle =\sum_a a\big|\langle a|\psi\rangle\big|^2$$