Quantum Chemistry?

Question

Calculate the average value of Φ for the particle-on-a-ring system. The wavefunctions are Ψ(Φ) = 1/√2 * e^(imΦ) m = 0,+/-1, +/-2...

I know I should set up an equation like this:
∫ (-&inf; --> &inf;) Ψ(Φ)* Φ(hat) Ψ(Φ) dΦ

Except I have no idea what Φ(hat) is..?

THANK YOU SO MUCH IN ADVANCE!!!!!!

Love,
Mary ♥

Answer 1

Regarding the other answer, I don't think Phi(hat) is a normalizing function. I believe your wavefunctions are already normalized, except the prefactor should be 1/sqrt(2 pi) instead of 1/sqrt(2).

I am not 100% sure what Φ(hat) is. From the wording of your question, it appears to be an operator that corresponds to an observable quantity, and you're being asked to determine the average value (expectation value) of measurements of that observable when the system is in state Ψ(Φ). Since your wavefunction is a function of the angular coordinate (plane polar coordinate) Φ, I suspect that Φ(hat) is the operator corresponding to that angular coordinate, and you're being asked to determine the average angular coordinate of the system (just like you might be asked to determine the average position x of a particle in a one-dimensional box when the particle is in state Ψ(x)).

If my assumption is true, then Φ(hat) and Ψ(Φ) satisfy an eigenvalue equation,

Φ(hat) Ψ(Φ) = Φ Ψ(Φ) ,

where Φ is the eigenvalue (angular coordinate of the particle in state Ψ(Φ). You want to find the expectation value

< Φ(hat) > = ∫ (0 --> 2 pi) Ψ*(Φ) Φ(hat) Ψ(Φ) dΦ
= ∫ (0 --> 2 pi) Ψ*(Φ) Φ Ψ(Φ) dΦ
= ∫ (0 --> 2 pi) |Ψ(Φ)|^2 Φ dΦ ,

where I have used the eigenvalue equation to replace Φ(hat) Ψ(Φ) with Φ Ψ(Φ). Remember that when determining expectation values you integrate over "all space," and for a one-dimensional ring problem the angular coordinate ranges from 0 to 2 pi. (To be strictly rigorous, one might still consider the ring to be in 3D space and one would still need to integrate over three coordinates, but the potential is infinite off the ring and the wavefuction is zero there, so the domain of the integral reduces to positions only on the ring.)

Note that since Ψ(Φ) = [1/sqrt(2 pi) exp(i m Φ)], Ψ*(Φ) = [1/sqrt(2 pi) exp(-i m Φ)], and |Ψ(Φ)|^2 = 1/(2 pi). Therefore,

< Φ(hat) > = ∫ (0 --> 2 pi) [1/(2 pi)] Φ dΦ
= 1/(2 pi) ∫ (0 --> 2 pi) Φ dΦ
= 1/(2 pi) {(2 pi)^2/2 - (0)^2/2}
= pi .

This position is halfway around the ring from the origin (Φ = 0). Note that the probability of finding the particle on the ring between position Φ and Φ + dΦ is |Ψ(Φ)|^2 dΦ = 1/sqrt(2 pi), which is independent of Φ. All positions on the ring are equally probable, no matter which ring energy eigenstate you're in (if you're in a superposition of energy eigenstates, some positions become more probable than others). If that's confusing (after all, we just calculated a specific average position of the particle and now I'm saying that all positions are equally likely), consider the fact that the choice of the origin of coordinates Φ = 0 is arbitrary. It's a ring -- you could choose the origin to be anywhere. Therefore it's a bit meaningless to ask what is the average position of a particle in a system that has no physically-defined origin.

Answer 2

Φ(hat) is the normalising function.

The average values will be 2m (lambda/2) where lambda is related to the radius.