Position and momentum

1. Nov 27, 2014

Frank Einstein

Good morning- afternoon.

First of all, apologize for my bad English.

After reading about how the expected value of an operator <q> is what we would measure in classical mechanics and that for the case in which we have various of them it is not trivial to deduce in which order these operators go for the lack of commutative propriety. <px> is not <xp> and that the true form to measure <xp>=(1/2)< xp+px>. I haven’t found how to calculate )< xp+px>

If anyone could point me a webpage or book where that is explained, it would be a great hel for me.

Thanks.

2. Nov 28, 2014

Orodruin

Staff Emeritus
You would evaluate it as you would the expectation value of any other operator. Apply the operator to the wave function, multiply by the complex conjugate of the wave function, and integrate.

3. Nov 29, 2014

naima

is it the same thing for E and B?

4. Nov 29, 2014

Frank Einstein

So would it be ∫dx Ψ* [(-iħ ∂/∂x)+(-iħ ∂/∂x)(x)] ψ = ∫dx Ψ* (-iħ ∂/∂x ψ) + ∫dx Ψ* (-iħ ∂/∂x)(x ψ) then?; with p = -iħ(∂/∂x) and position= x

5. Nov 29, 2014

Staff: Mentor

You omitted an x in the first term (from the xp).

6. Nov 29, 2014

Frank Einstein

Thanks for pinting that.