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1. [Particle in a box] If we shift the boundaries of the box to be at x = -a/2 and x = a/2, what will be the new form of the solution?

1. [Particle in a box] If we shift the boundaries of the box to be at x = -a/2 and x = a/2, what will be the new form of the solution?

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-a/2 1. [Particle in a box] If we shift the boundaries of the box to be at x = - and x = a/2, what will be the new form of the solution?
Transcribed Image Text:-a/2 1. [Particle in a box] If we shift the boundaries of the box to be at x = - and x = a/2, what will be the new form of the solution?
The Infinite Square Well Suppose 0≤x≤ a, V(x) = othewise Outside the well, x < 0 y(x) = a < x Inside the well, V For E ≥ 0 Let k = √2mE/ħ 10, = 0 and TISE, ħ² d² 2m dx² d² dx² (1) = ħ² d² 2m dx² + Vy = Ev = Εψ V(x) A Figure 2.1: The infinite square well potential (Equation 2.22). Particle in a box (2) (3) (4) 2mE ħ² -4
Transcribed Image Text:The Infinite Square Well Suppose 0≤x≤ a, V(x) = othewise Outside the well, x < 0 y(x) = a < x Inside the well, V For E ≥ 0 Let k = √2mE/ħ 10, = 0 and TISE, ħ² d² 2m dx² d² dx² (1) = ħ² d² 2m dx² + Vy = Ev = Εψ V(x) A Figure 2.1: The infinite square well potential (Equation 2.22). Particle in a box (2) (3) (4) 2mE ħ² -4
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