228s13-l17

Thus the particle is in the box and we require x 0 at

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Unformatted text preview: rom 0 to L in x, y, and z directions. Again, as in the 1D case, outside the box U = ∞, so the differential equation is solved if the wave function ψ(x) = 0. Thus the particle is in the box and we require ψ(x) = 0 at x (or y or z) = 0 or x (or y or z) = L. In the box U(x) = 0, so S.E. is simply: −￿2 2 ∇ ψ (￿ ) = E ψ (￿ ) x x 2m We again need a wave function that has the same form as its 2nd derivative, and the right boundary conditions. It is similar to the 1D case, and we can guess the result: ψ(x) = (2/L)3/2 sin(qπx/L) sin(rπy/L) sin(sπz/L) Monday, April 1, 2013 Note the new indices: q in x, r in y, and s in z directions. Infinite 3D Square Well Energy Levels In the 1D case, we had k = qπ/L, ψ(x) = (2/L)1/2 sin(qπx/L), and Eq = q2ħ2π2/2mL2, with q = 1, 2, 3... What happens now with the 3D case where ψ(x) = (2/L)3/2 sin(qπx/L) sin(rπy/L) sin(sπz/L)? The three directions are independent, so we will have kx = qπ/L, ψx(x) = (2/L)1/2sin(qπx/L), and Eqx = q2ħ2π2/2mL2, with q = 1, 2,...
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This document was uploaded on 02/18/2014 for the course PHYS 228 at Rutgers.

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