(f) YES.
The function in figure 5.16(f) appears to satisfy all of the conditions, so it can be a wave
function, although it is not clear how the amplitude can decrease with increasing x without the
wavelength changing.
(e) NO.
The function is not single valued.
(d) YES.
(c) NO.
The first derivative is discontinuous in the middle.
Beiser also probably
intends to imply that the function goes to infinity at the middle.
(b) YES if you assume the "interval shown" is only where the function is defined.
NO if
you assume the "interval shown" means the entire shown part of the positive xaxis.
(a) YES.
"YES" means allowed, "NO" means not allowed.
Which of the wave functions in Fig. 5.16 cannot have physical significance in the interval shown? Why not?
Problem 5.2
Physics 107
The function in figure 5.15(f) is not continuous, so it cannot be a wave function.
This is tricky,
because it is possible that the derivative could be continuous and finite.
The function in figure 5.15(e) appears to satisfy all of the conditions, so it can be a wave
function.
The function in figure 5.15(d) does not satisfy the condition for a continuous first derivative, so it
cannot be a wave function.
In addition, the function is not continuous.
Beiser in his answer implies the
function goes to infinity, which is not obvious from the figure as it is drawn.
The function in figure 5.15(c) does not satisfy the condition for a continuous first derivative, so it
cannot be a wave function.
The function in figure 5.15(b) is not singlevalued, so it cannot be a wave function.
The function in figure 5.15(a) appears to satisfy all of the conditions over the interval shown, so it can be
a wave function.
If you want to be picky, its zero slope everywhere means the momentum is zero
everywhere, which is not allowed by the uncertainty principle.
I will not be that tricky on the exam.
must be finite, and that
Φ
must be singlevalued and continuous with finite, singlevalued, and
continuous first derivatives.
∞
−
∞
V
Φ
()
2
⌠
⌡
d
Recall that
Which of the wave functions in Fig. 5.15 cannot have physical significance in the interval shown? Why not?
Problem 5.1
Physics 107
1
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View Full DocumentThis function has
singularities at
π
/2, 3
π
/2,
etc.
.
Φ
x
()
x
x2
−π
⋅
2
⋅
2
π
⋅
100
+
,
2
π
⋅
..
:=
Φ
x
At
a
nx
⋅
:=
(b)
There are singularities wherever cos(x)=0, so
this is not an allowed wave function if its range
includes an xvalue where there is a singularity.
Φ
x
x
We can see the function better if we choose the plot scale ourselves.
There are singularities here, which seem to
scale differently.
Because of our choice of
values of x, the function does not always
"hit" the singularity at exactly the same
position.
Φ
x
x
First plot
Φ
(x), letting Mathcad choose the plot scale.
⋅
2
⋅
2
π
⋅
100
+
,
2
π
⋅
..
:=
Φ
x
1
cos x
:=
(a)
The secant is 1/cosine.
Let's plot the secant function for small values of x.
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 Fall '09
 Abra
 pH, Energy, wave function, dx, k2

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