Physics 214 - Quantum I - Solutions #11
Griffiths 4.1) a) Any commutator between different coordinate operators is zero since multiplying by or
differentiating by one coordinate on a function of another independent coordinate
[ uj , uk ] = 0
[ u j , k ] =
Physics 214 - Quantum I - Solutions #5
a) The answer is the quwstion!
d
Starting with
2
dx
(
2
(2 )
= x K
)
2
d
d
( x ) = u ( x ) then
u ( x ) = x K ( x ), and the two first order equations are equivalent to the original
dx
dx
second order equation. The
Physics 214 - Quantum I - Solutions #3
Griffiths 2.1) The hints in all of the parts lead the way.
a) Following the first hint, the wavefunction
i
( x , t) = A ( x ) e
E0+ i
t
h
(
)
2
=
(
A
)
2
e
2
t
h
would have an exponentially growing probability di
Physics 214 - Quantum I - Solutions #4
2
2
n x
n h2
with En =
. Shifting the position of the well will
sin
2
a
a
2 m a
2
1) The situation with the well 0<x<a is n ( x ) =
only change the form of the wavefunction by a displacement of the position varia
Physics 214 - Quantum I - Solutions #2
Note - Complex conjugation in this document is annotated by a bar over the expression, i.e.
( a + i b ) = a i b . This is a quirk of Mathcad, the software used to wright up these solutions.
Griffiths 1.7) This proof
Physics 214 - Quantum I - Solutions #6
1) The classically allowned region associated with a state n is where
m x > n +
1
2 2
2
.The probability of finding the particle inside of this non-classical region is
Pnc = 2
2 n+1
Changing variables of the integ
Physics 214 - Quantum I - Quiz #3 - 03/05/10 Name_
1) Show how the free particle Schrodinger's equation gives waves with a group velocity equivalent to the classical particle
velocity.
Answer) The free particle wavefunction = A e
i ( k x t)
solves Schrodi
Physics 214 - Quantum I - Solutions #12
Griffiths 4.8) a) The function j1 (x) is
j1 ( x) =
sin( x )
x
so the radial function is
For l=2, we have
2
cos( x )
x = k r
x
and must solve
u ( r) = A r j1 ( k r)
d
2
2
u
dr
2
2
2
u = k u
d
2
dx
Calculate the sec
Physics 214 - Quantum I - Solutions #10
Griffiths 3.27) a) Following the measurement of observable A, the state is in the eigenstate associated
with the eigenvalue obtained by the measurement, so the system is in the state 1 .
b) The system, being in supe
Physics 214 - Quantum I - Solutions #8
1) With Schrodinger's equation
E0 =
h2
d
2
2 m dx 2
h2
( x ) = E the bound state solution was
2
=
2 m
e
x
=
m
h2
a) Both <x> and <p> are zero by the left-right symmetry of the potential. For the spatial varian
Physics 214 - Quantum I - Solutions #9
Griffiths 3.3) Following the hint, take h = f + g and expand both sides of < h | Qh > = < Qh |h >.
< f | Qf > + < f | Qg > + < g | Qf > + < g | Qg > = < Qf | f > + < Qf | g > + < Qg | f > + < Qg | g >
Cancel out the
Physics 214 - Quantum I - Solutions #1
1) The result for the fringe spacing s was essentially the wavelength expanded by the ratio of the
distance to the screen with the slit spacing, D/d. The piece of this not specified directly is the
wavelength, which
