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Unformatted text preview: , this is the signed momentum, not the magnitude, and the
answer is 0. Since the particle in a box isn't going anywhere, its
average momentum must be 0, although the average magnitude of
the momentum is non0. Be careful about whether the average
momentum or the magnitude of the momentum is requested.
If you have this on a test, and are not sure, ASK!
Monday, April 1, 2013 3D Quantum Mechanics
When we go from 1D to 3D, we need to take
the 3D derivative. We need vector calculus
since the derivative can be different in
different directions. We use the gradient
o perator: = ∂ x+ ∂ y+ ∂ z
∇
ˆ
ˆ
ˆ
∂x
∂y
∂z
The timeindependent S.E. now becomes − 2
∇ ψ ( ) + U (x)ψ ( ) = E ψ ( )
x
x
x
2m
2 or, for a 3D square well potential we can write
−2 2
∇ ψ ( ) = E − U ( ) ψ ( )
x
x
x
2m where inside the box U0 = 0, and outside the box U0 = ∞.
Monday, April 1, 2013 Inﬁnite 3D Square Well
Let's consider a simple cubical box, each side
of length L, and the box extending f...
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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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