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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.
 Spring '12
 RonGilman
 Energy, Work

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