Calculating Probabilities of Particle in a L-box After Wall Movement

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SUMMARY

The discussion focuses on calculating the probabilities of measuring the energy of a particle in an infinite square well (L-box) after the wall is moved from L to 2L. The probability of measuring the energy in the ground state of the new box is approximately 0.36, while the probability of measuring the energy in the first excited state is exactly 1/2. The eigenstates and eigenvalues for both the original and new L-box configurations are crucial for these calculations, highlighting the importance of understanding wave functions in quantum mechanics.

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  • Understanding of quantum mechanics concepts, particularly infinite square wells.
  • Familiarity with wave functions and their role in determining probabilities.
  • Knowledge of eigenstates and eigenvalues in quantum systems.
  • Basic proficiency in mathematical expressions involving sine functions and constants like ℏ (reduced Planck's constant).
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  • Study the derivation of eigenstates for infinite square wells in quantum mechanics.
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mcheung4
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Homework Statement



A particle is in the ground state of a L-box. At t = 0 the wall at x = L is suddenly moved to 2L.

(a) If an energy measurement is made after the wall is moved, what is the probability of measuring the energy to be that of the ground state of the new box? (given solution ≈ 0.36)

(b) What is the probability of measuring the energy to be that of the first excited state of the new box? (given solution = 1/2)

*L-box is the infinite square well with zero potential for 0≤x≤L.


Homework Equations



eigenstates ψ_{n} = √(2/L) sin (n*pi*x/L)

eigenvalues (energy) E_{n} = ((n*pi*ℏ)^2)/(2mL^2)

The Attempt at a Solution



So initially the ground state (of L-box) ψ_{1} = √(2/L) sin (pi*x/L) and E_{1} = ((pi*ℏ)^2)/(2mL^2), so I think this energy is the maximum energy that the particle can have, before and after the change in size of the box.

For the 2L-box its ground state ψ'_{1} = √(1/L) sin (pi*x/2L) and E'_{1} = ((pi*ℏ)^2)/(8mL^2); its 1st excited state ψ'_{2} = √(1/L) sin (pi*x/L) and E'_{2} = ((pi*ℏ)^2)/(2mL^2).

So the particle can only be in either of these states (it doesn't have enough energy to move to other states). But since I wasn't given the wave function in this new box, how am I supposed to calculate the probablities?

My guess was that there should be no preference on the 2 states in the new box, so the probabilities are both 1/2, but which are not exactly correct according to the solution
 
Physics news on Phys.org
It's not the energy that remains constant; it's the wave function that's the same immediately before and after the box is lengthened.
 

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