How Does a Rigid Wall Affect a Particle's Wavefunction in Quantum Mechanics?

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SUMMARY

The discussion focuses on the behavior of a free particle in one-dimensional space with a rigid wall at x = 0, specifically addressing the time-independent wavefunction and its relation to the Hamiltonian. The wavefunction is expressed as psi(x) = Ae^(ikx) + Be^(-ikx), where k is the wave vector, and the energy E is related to k by the equation E = (h^2 * k^2) / (2m). The wavefunction is confirmed to be an eigenfunction of the Hamiltonian, satisfying the time-independent Schrödinger equation. Additionally, the time-dependent state is given by psi(x,t) = Ae^(i(kx - Et)/h) + Be^(i(-kx - Et)/h).

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danai_pa
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I have a problems, help me please

a) A free particle of mass m moves in one-dimensional space in the interval 0 <= x, with energy E. There is a rigid wall at x = 0. Write down a time-independent wavefunction, which satisfies these conditions, in term of x and k wher k is the wave vector of the motion. State the relation between k and E for this wavefunction.


b) Show explicity that the wavefunction you have found in part a) is an eigenfunction of the Hamitonian for this system.

c) What is the time-dependent state, psi(x,t). corresponding to the wavefunction psi(x)?
 
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It shouldn't be too complicated,what did you do so far?

Daniel.
 


a) The time-independent wavefunction for a free particle in one-dimensional space with a rigid wall at x = 0 can be written as:

psi(x) = Ae^(ikx) + Be^(-ikx)

where A and B are constants and k is the wave vector of the motion. The relation between k and E for this wavefunction is given by:

E = (h^2 * k^2) / (2m)

where h is Planck's constant and m is the mass of the particle.

b) To show that this wavefunction is an eigenfunction of the Hamiltonian, we need to show that it satisfies the time-independent Schrödinger equation:

H * psi(x) = E * psi(x)

where H is the Hamiltonian operator. Substituting the wavefunction into this equation, we get:

H * (Ae^(ikx) + Be^(-ikx)) = (h^2 * k^2) / (2m) * (Ae^(ikx) + Be^(-ikx))

Expanding the Hamiltonian operator, we get:

(-h^2 / (2m)) * (d^2/dx^2) * (Ae^(ikx) + Be^(-ikx)) = (h^2 * k^2) / (2m) * (Ae^(ikx) + Be^(-ikx))

Simplifying, we get:

(-h^2 * k^2 / 2m) * (Ae^(ikx) + Be^(-ikx)) = (h^2 * k^2 / 2m) * (Ae^(ikx) + Be^(-ikx))

This shows that the wavefunction psi(x) is indeed an eigenfunction of the Hamiltonian, with eigenvalue E = (h^2 * k^2) / (2m).

c) The time-dependent state corresponding to the time-independent wavefunction psi(x) is given by:

psi(x,t) = Ae^(i(kx - Et)/h) + Be^(i(-kx - Et)/h)

where E is the energy and h is Planck's constant. This time-dependent state describes the probability amplitude of the particle at position x and time t.
 

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