Yes.Kyuubi said:Homework Statement: Just need to clarify this point.
Relevant Equations: $$ \langle p \rangle = \int_{-\infty}^{\infty}\bar{\phi}(p,t) p \phi(p,t)dp $$
Now if I'm given a ##\phi(k)##, and I'm asked to find ##\langle p \rangle##, ##\langle p^2 \rangle##, etc. Am I justified to say that ##\langle p \rangle = \hbar \langle k \rangle## and that ##\langle p^2 \rangle = \hbar^2 \langle k^2 \rangle## ?
The formula for calculating the expected value of momentum P in terms of k is E(P) = ∫ P(k) * f(k) dk, where P(k) is the momentum as a function of k and f(k) is the probability density function of k.
The expected value of momentum P is related to the uncertainty principle in that it represents the average momentum of a particle in a given system. The uncertainty principle states that the more precisely the momentum of a particle is known, the less precisely its position can be known, and vice versa.
Yes, the expected value of momentum P can be negative. This indicates that there is a higher probability for the particle to have a negative momentum in the given system.
The expected value of momentum P can change with different values of k depending on the specific system and the probability density function of k. In some cases, the expected value of momentum may increase with increasing values of k, while in others it may decrease.
Calculating the expected value of momentum P in terms of k allows us to gain a better understanding of the behavior and properties of particles in a given system. It can also help us make predictions and calculations related to the motion and interactions of particles.