228s13-l17

Since the particle in a box isnt going anywhere its

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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 non-0. 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 time-independent 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.

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