A wave packet for a particle has a small value of Ax the unc

A wave packet for a particle has a small value of Ax, the uncertainty in position. Briefly explain in terms of the formation of this wave packet why the uncertainty in momentum delta will be large (n.b., simply saying deltaxdeltap h/2 therefore deltap x 1/deltax is not the answer).

Solution

According to Classical mechanics, it is possible to determine all dynamical variables of a system to any desired degree of accuracy. This princple of determinism is fundmental in classical physics.

The position of plane wave is completely indeterminate as it is of infinite extent. so when waves are assigned to particles in motion an indeterminacy arises in the formulation beacuse of an electron wave of definite frequency is not localised.

As we know that micro scopic particle exhibits wave like behaviour under suitable conditions. According Schroedinger superposition of matter waves takesplace in space, on can locate the particle where constructive interference of matter waves takes place. This is called as wave packet. When the momentum of the particle is well defined, the wave can be of infinite extent, hence a free particle moving along x-axis with a well defined.   As the positon in uncertaninty is small, at larger region constructive interference of matter waves takes place.

An accumulation of waves of varying wavelengths can be combined to create an average wavelength through an interference pattern: this average wavelength is called the \"wave packet\". The more waves that are combined in the \"wave packet\", the more precise the position of the particle becomes and the more uncertain the momentum becomes because more wavelengths of varying momenta are added. Conversely, if we want a more precise momentum, we would add less wavelengths to the \"wave packet\" and then the position would become more uncertain. Therefore, there is no way to find both the position and momentum of a particle simultaneously.