(b) Consider two plane matter waves, each with the same amplitude
ψ
0
2
/
and
traveling in opposite directions along the
x
axis. The combined wave
Ψ
is a standing
wave:
()
00
0
0
(,)
(
)
(
2 c
o
s )
.
ik
x t
i
k
x
i
k
x
it
x
te
e
e
e
e
k
x
e
ω
ωω
ψψ
−−
+
−
−
−
Ψ=
+
=
+
=
Thus, the squared amplitude of the matter wave is
(
,
)
 (
c
o
s)
(
)
,
Ψ
xt
kx e
kx
2
0
2
2
0
2
22
1
==
+
−
cos2
which is shown below.
(c) We set
Ψ
,c
o
s
b
g
b
g
2
0
2
21
2
0
=+
=
to obtain cos(2
kx
) = –1. This gives
(
)
2
2
1
0
,
1
,
2
,
3
,
kx
n
n
π
⎛⎞
+
,
=
⎜⎟
λ
⎝⎠
…
We solve for
x
:
xn
1
4
bg
λ
.
62. (a) The wave function is now given by
Ψ
(
)
.
e
e
e
e
e
i
k
x
i
k
x
=
+
+
−
−
Thus,
2
2
2
2
000
2
2
2
0
,
)

(
)
 (cos
sin
) (cos
sin
) 
4
(cos
)
2
(1 cos2
).
i
k
x
i
k
x
i
k
x
i
k
x
i
k
x
i
k
x
x
t
eee
e
ee
kx i
kx
kx i
kx
kx
kx
ψψψ
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 Spring '08
 Any
 Physics, Cos, phase velocity

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