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 expected value of momentum \( P \) in terms of the wave number \( k \) is given by \( \langle P \rangle = \hbar k \), where \( \hbar \) is the reduced Planck's constant.
The wave number \( k \) is related to the momentum \( P \) by the equation \( P = \hbar k \), where \( \hbar \) is the reduced Planck's constant.
The reduced Planck's constant \( \hbar \) acts as a proportionality constant between the wave number \( k \) and the momentum \( P \). It ensures that the units are consistent and provides the correct scaling factor.
Yes, the expected value of momentum can be negative. The sign of \( \langle P \rangle \) depends on the direction of the wave number \( k \). If \( k \) is negative, \( \langle P \rangle \) will also be negative.
To calculate the expected value of momentum for a given wave function \( \psi(x) \), you typically use the operator approach: \( \langle P \rangle = \int \psi^*(x) \left( -i \hbar \frac{\partial}{\partial x} \right) \psi(x) \, dx \). For plane waves, this simplifies to \( \langle P \rangle = \hbar k \).