What Are the Correct Values for Probability and Momentum in a Quantum Well?

AI Thread Summary
In a quantum well where a particle is in the ground state, the probability of finding the particle moving in the positive x direction (P+) and its momentum (p) are under discussion. The potential well is defined as V(x) = 0 for |x|<L/2 and infinite otherwise. The possible answers for P+ and p include values of 0, 1/2, and 1 for probability, with corresponding momentum values of 0, π/2L, and π/L. The consensus indicates that P+ should be 1/2, but there is confusion regarding the momentum value, which does not align with standard units. The discussion highlights the need for clarity in the units used for momentum calculations.
PRASHANT KUMAR
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Homework Statement


A particle in one dimension is in the ground state of the potential well given by V(x)= 0 for |x|<L/2 and infinite otherwise. Let P+ be the probability that the particle is found to move along the positive x direction and p be the magnitude of the momentum for that state of motion. Then what will be the probability and momentum value?

answers:
(a) P+ = 0 and p = 0
(b) P+ =1/2 and p = pi/2L
(c) P+ = 1/2 and p= pi/L
(d) P+ = 1 and p = pi/L

Homework Equations

The Attempt at a Solution


from the theory of the particle in one dimensional potential well , the momentum eigen values are
given by = nh/2L and the probability of finding the particle in ground state in the positive direction should be half but my momentum value is not matched.
 
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I agree with your approach.

Without some non-standard units the answers offered don’t even have the right units for the momentum.
 
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