# Expectation Value for system of identical particles

1. May 11, 2012

### kenphysics

Dear All:

I have a quite mysterious and cumbersome question concerning with the expectation values for a system of identical particles. For example, suppose I have a system of N identical bosons given by the wavefunction ψ(x1,x2,...xN), which is of course symmetrized. My concern is:

1. What is <x1> really means? I know mathematically it is given by:

∫dx1dx2...dxN x1 ψ*(x1,x2,...xN)ψ(x1,x2,...xN)

but I think this is unphysical? We cannot measure this value, right? since physical observable for identical particles should be symmetric, like x1+x2+x3, right? But can I say <x1>=<x2>=<x3> = 1/3<x1+x2+x3> due to symmetric argument?

2. Suppose I really want to compute the above quantity, how to write it in second quantized form, that is using field operators? Is it given by:

<x1> = ∫ x1 <ψdagger(x1)ψ(x1)> dx1 ? where ψ is now a field annihilation operator that annihilate a particle at the point x1. If yes, can someone give me some ideas on how to show it starting from the wavefunction description?

3. Continue the same idea, then what is <(x2-x1)^2> means? Also, how to write it in second quantized form?

Thank you for all

2. May 11, 2012

### Jano L.

Hello kenphysics,

welcome to PF.

Why do you think so? If we used this wave function to describe the ensemble of helium atoms, the expression you wrote for <x1> would most probably serve to calculate the expected average of the position of any helium atom. Due to symmetric property of the wave function, we would have <x1> = <x2> = ... and so on.

3. May 12, 2012

### kenphysics

OK! In fact, what trouble me a lot is that for second quantization, we always consider operators like:

$\sum_{i=1}^N f(x_i)$ (Sinlge-body operator)
$\sum_{i<j}^N V(x_i,x_j)$ (Two-body operator)

But we seldom consider operators, liked x1, x2 or (x2-x1)^2, (x3-x2)^2. So what I actually want to know is can I still write down <x1> and <(x2-x1)^2> in second quantized form?