the ends x = ±L/2 are to be glued to form the circle.

Question

We take the interval of length L to be in the range

-L/2 < x < L/2

The ends x = ±L/2 are to be glued to form the circle.

-normalize Ψ

-what is the probability that the particle in the ring has momentum p=0

-If p = 0 is obtained in an experiment, which is Ψ , normalized, immediately after the

experiment?

-what is the probability that the particle in a ring is in some position with x>0

how much is p(x>0) ?

Answer 1

To find the wave function and probabilities for a particle in a ring of length L with ends glued together, we can use the following steps:

1. Normalize the wave function: We assume that the wave function has the form: Ψ(x) = A exp(i k x) where A is a normalization constant and k is the wave vector given by k = 2πn/L, where n is an integer. Since the ring is a closed loop, we need to ensure that the wave function is periodic and continuous at the ends of the loop. This implies that Ψ(-L/2) = Ψ(L/2) and Ψ'(-L/2) = Ψ'(L/2), where Ψ' is the derivative of Ψ with respect to x. Using these boundary conditions, we can determine the normalization constant to be A = 1/√L.

Therefore, the normalized wave function for a particle in a ring is: Ψ(x) = (1/√L) exp(i 2πn x/L)

1. Probability that the particle has momentum p=0: The momentum operator in quantum mechanics is given by: p = -iħ d/dx If the momentum of the particle is zero, then pΨ(x) = 0, which implies that Ψ(x) is a constant function. Therefore, the probability of finding the particle with zero momentum is the squared magnitude of the wave function averaged over the ring: P(p=0) = |∫_0^L Ψ(x) dx|^2 = |1/L ∫_0^L exp(i 2πn x/L) dx|^2 = 1/L^2 |[L sin(2πn/2)]_0^L|^2 = 0

So the probability of finding the particle with zero momentum is zero.

1. Wave function immediately after measuring momentum p=0: If the momentum of the particle is measured to be zero, the wave function collapses into an eigenstate of the momentum operator with zero momentum. This means that immediately after the measurement, the wave function is a constant function that represents an eigenstate of the momentum operator with zero momentum.
2. Probability that the particle is in a position with x>0: The probability of finding the particle in a region of the ring between x=a and x=b is given by: P(a < x < b) = ∫_a^b |Ψ(x)|^2 dx If we want to find the probability of finding the particle in a region with x>0, we can integrate the squared magnitude of the wave function over the region: P(x>0) = ∫_0^(L/2) |Ψ(x)|^2 dx Substituting the wave function, we get: P(x>0) = 1/L |∫_0^(L/2) exp(i 2πn x/L) dx|^2 Using the integral ∫_0^a sin(ax) dx = [1/a] (1-cos(a)), we get: P(x>0) = 1/L^2 |[L/2 - L/(2πn) sin(2πn/2)]_0^(L/2)|^2 Simplifying, we get: P(x>0) = (1/4)[1 - cos(πn)]^2

Therefore, the probability of finding the particle in a position with x>0 is (1/4)[1 - cos(πn)]^2, where n is an integer. The value of p(x>0) depends on the value of n. For n=1, the probability of finding the particle in a position with x>0 is (1/4)[1 - cos(π)]^2 = 0.18. For n=2, the probability is (1/4)[1 - cos(2π)]^2 = 0, and for n=3, the probability is (1/4)[1 - cos(3π)]^2 = 0.18, and so on.

So, the value of p(x>0) oscillates between 0 and 0.25, with a period of 2, and its maximum value is obtained for values of n such that cos(πn) = -1. These correspond to odd multiples of n, such as n=1, 3, 5, etc. In general, the probability of finding the particle in a region of the ring depends on the quantum number n, which determines the allowed energy levels of the system.