Is Ae^i(kx+ωt) an admissible wavefunction for a free particle?

[danai_pa]

Consider the wavefunction

psi = Ae^i(kx+wt) ;w = omega

where k is real and w(omega) > 0 and is real. Is this wavefunction an admissible quantum state for a free particle?. Justify your answer is no, in what manner would you change the given function to describe a free particle moving in the -x direction?.

 

[StatusX]

How about showing some work?

 

[thepatientmental]

No, this wavefunction is not an admissible quantum state for a free particle. In order for a wavefunction to represent a physical state, it must satisfy certain conditions, such as being normalized and having a finite energy.

In this case, the given wavefunction is not normalized, as the constant A is not specified. Additionally, the wavefunction does not have a finite energy, as the term e^(iwt) will cause the energy to oscillate between positive and negative infinity.

To describe a free particle moving in the -x direction, we would need to change the given wavefunction to include a momentum term, which would determine the direction of motion. For example, we could modify the wavefunction to be:

psi = Ae^i(kx-wt) ; where k and w are both positive and real

This wavefunction satisfies the necessary conditions for a physical state and describes a free particle moving in the -x direction with a constant momentum.