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Alex Ceballos
Dr.Cole
4/21/11
PHY 361 TEST #2
1.
The book states in pg. 200 that the slope of the wavefunction must be continuous
wherever U(x) has a finite value. Since at x=0 and x=L the potential U(x) equals infinity,
also known as a nonfinite value, the discontinuity of the slope of the wavefunction does
not violate the requirement imposed on the WF.
2.
(This is addressing Q3) [p] is fuzzy not because it’s changing in magnitude in time, but
because the sign flips from positive to negative as the particle bounces off the potential
walls over and over again. By squaring the momentum operator, we get rid of the sign
changing and hence obtain an eigenvalue for our eigenfunction. Thus [p
2
] is sharp.
3.
(This is addressing Q4) No, the normalization coefficient remains the same. We must
remember that physically speaking the system is still the same as before, the overall
length, or width of the box remains L. So when we impose the normalization condition,
the second term of the integrand, the cos(2x) contributes nothing to the integral, so all
you get is just 1, evaluated from [L/2, L/2], which turns out to be the same as evaluating
it in the interval [0,L] because of the symmetry of our system, and again, because the
particle doesn’t give a damn where we set our origin, the normalization coefficient is still
the same.
4.
(This is addressing Q5) The uncertainty principle determines that you can only know one
component of angular momentum, and the total magnitude, not the direction. The reason
is that if you knew the direction of
L
, then the particle would be bound to an orbital plane
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This note was uploaded on 02/23/2012 for the course PHYSICS 361 taught by Professor Cole during the Spring '11 term at N. Arizona.
 Spring '11
 Cole
 Physics

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