Is This Wavefunction Suitable for a Free Particle in Quantum Mechanics?

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The wavefunction psi = Ae^(i(kx+wt)) is not suitable for a free particle because it is not normalized and does not have finite energy, leading to oscillations between positive and negative infinity. To represent a free particle moving in the -x direction, the wavefunction should be modified to psi = Ae^(i(kx-wt)), where both k and w are positive and real. This adjusted wavefunction meets the necessary conditions for a physical state, including normalization and finite energy. The inclusion of the momentum term in the modified wavefunction correctly indicates the direction of motion. Thus, the original wavefunction fails to meet the criteria for admissibility in quantum mechanics.
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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?.
 
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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.
 
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