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similar in the y and z directions. Then
Eqrs = Eq + Er + Es = (q2+r2+s2)ħ2π2/2mL2.
The g.s. will have q=r=s=1, and E111 = 3ħ2π2/2mL2. Monday, April 1, 2013 iClicker Eqrs = Eq + Er + Es = (q2+r2+s2)ħ2π2/2mL2.
The g.s. will have q=r=s=1, and E111 = 3ħ2π2/2mL2.
What is the energy of the 1st excited state?
How many of them are there?
a) E2 = Egs, 1
b) E2 = 2Egs, 1
c) E2 = 4Egs, 1
d) E2 = 2Egs, 3
e) E2 = 4Egs, 3 Monday, April 1, 2013 Degenerate States
States of the same energy are called "degenerate states".
Is is possible to break the degeneracy by having all 3 sides
of the box different lengths, so that
Eqrs = Eqx + Ery + Esz = (q2+r2+s2)ħ2π2/2mL2 →
Eqrs = Eqx + Ery + Esz = (q2/Lx2 +r2/Ly2 +s2/Lz2) ħ2π2/2m. Monday, April 1, 2013 Probability Distributions Again in 3D QM, as in the 1D case, the probability of ﬁnding a
particle at some point x0, y0, z0 in space is given by ψ*ψ = |ψ|2 at
Note an odd feature of QM: the particle in, e.g., the ψ2,1,1 state
can never be found in the x=L/2 plane...
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