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763312A QUANTUM MECHANICS I - Solution set 1 - Autumn 2016
a) Show that
1 iθ
e + e−iθ
2
1 iθ
e − e−iθ .
sin θ =
2i
cos θ =
(1.1)
(1.2)
b) Use the Euler’s formula and express sin(3φ) and cos(3φ) in terms of sin(φ)
and cos(φ).
Solution
a) From Euler’s formula,
eiφ = cos φ + i sin φ
(1.3)
we get that
eiθ + e−iθ
cos θ
=
cos(θ) + i sin(θ) + cos(−θ) + i sin(−θ)
=
cos(θ) + i sin(θ) + cos(θ) − i sin(θ)
=
2 cos θ. ⇔
1 iθ
e + e−iθ
2
=
(1.4)
Similarly
eiθ − e−iθ = 2i sin θ ⇔
1 iθ
sin θ =
e − e−iθ
2i
b) Using the result from a), we get that
1 3iφ
sin(3φ) =
e − e−3iφ
2i
3 i
1 h iφ 3
=
e
− e−iφ
.
2i
Now we use Euler’s formula and simplify
sin(3φ)
=
1
(cos φ + i sin φ)3 − (cos φ − i sin φ)3 )
2i
3 sin φ cos2 φ − sin3 φ
=
3 sin φ − 4 sin3 φ.
=
(1.5)
(1.6)
(1.7)
By using the same recipe for cos(3φ) we obtain
cos(3φ)
=
=
cos3 φ − 3 cos φ sin2 φ
3
4 cos φ − 3 cos φ.
1
(1.8)
s1
λ
ΔlP
a
θP
θP
s2
xP
P
d
Figure 1: Setup
2. Show that in the double-slit experiment, the spacing ∆x between two
adjacent maxima is
λd
∆x =
,
a
where λ is the wave length, a is the separation between the slits, and d is
the distance between the slits and the detector. Assume that you conduct
the double-slit experiment for bullets with m = 7 g and v = 400 m/s. Can
you observe an interference pattern for any reasonable values of a and d?
Solution
Let us choose a maxima on the detector (any maxima will do), say it’s
located at point P. Because P is a maxima, the difference in distance from
the slits s1 and s2 to the detector has to be an integer multiple of λ,
∆lP = nλ,
n ∈ Z.
(2.1)
From Figure 1 we get that
∆lP
xP
= nλ = a sin θ ≈
=
nλd
a
2
xP
⇔
d
(2.2)
(2.3)
Now the next maxima is located at point Q where
∆lQ = (n + 1)λ.
(2.4)
Like for ∆lP , we obtain for ∆lQ that
∆lQ
=
(n + 1)λ = a sin θ ≈
⇔ xQ
=
(n + 1)λd
.
a
xQ
d
(2.5)
(2.6)
So the spacing between P and Q is
∆x = xQ − xP =
λd
(n + 1)λd nλd
−
=
.
a
a
a
(2.7)
The de Broglie wave length for the bullet is
λ=
h
≈ 2.37 · 1034 m.
p
So d/a would have to be about 1031 , which is not practical.
3
(2.8)
3. Consider the Gaussian distribution
2
ρ(x) = Ae−λ(x−a) ,
where A, a, and λ are positive real constants. (Look up any integrals you
need.)
a) Determine A so that ρ(x) can be interpreted as a probability density.
b) Find hxi , hx2 i , and σ.
Solution
a) For ρ(x) to be a probability density its integral over all x must be unity.
So
Z ∞
Z ∞
2
1 =
ρ(x)dx = A
e−λ(x−a) dx
(3.1)
−∞
−∞
Z ∞
2
1
e−y dy
(3.2)
= A√
λ −∞
√ 1
= A π√ .
(3.3)
λ
√
Where we used change of variables, y = λ(x − a), and Gauss’ integral.
Thus
r
λ
A =
.
(3.4)
π
b) The expectation value of a quantity O(x) is defined as
Z ∞
hO(x)i =
O(x)ρ(x)dx.
(3.5)
−∞
Using the definition gives
r Z ∞
r Z ∞
2
y
a
λ
λ
−λ(x−a)2
hxi =
xe
dx =
+√
e−y dy
π −∞
π −∞ λ
λ
r
Z ∞
Z ∞
2
2
1
a
π
−y
−y
=√
ye dy + √
e dy = 0 +
a=a
π
π −∞
πλ −∞
(3.6)
where we used the same change of variables as in a) and Gauss’ integral.
4
For the square of x we obtain
r Z ∞
r Z ∞
2
λ
λ
2 −λ(x−a)2
2
x e
dx =
(x − a)2 e−λ(x−a) dx
hx i =
π −∞
π −∞
r Z ∞
r Z ∞
2
λ
λ
−λ(x−a)2
2
+ 2a
xe
dx − a
e−λ(x−a) dx,
π −∞
π −∞
(3.7)
where we have added and substracted terms in order to complete the
square (x − a)2 in the first term on the right-hand side. This can now be
solved using integration by parts:
Z ∞
Z ∞
Z
2
2
2
1 ∞ −y2
1 ∞
y 2 e−y dy =
y ∗ ye−y dy = − y ∗ e−y +
e dy
2 −∞
2 −∞
−∞
−∞
√
π
,
(3.8)
=0+
2
where we have used Gauss’ integral on the last step. The asterisks (*)
signify multiplication. With an appropriate change of variables (the same
one as before), we can use the above to calculate the first integral on
the right-hand side of 3.7. Furthermore, we already calculated the second
integral when calculating hxi and the last integral is simply Gauss’ integral
once again. Collecting our results we get:
r r
Z
1 1 ∞ 2 −y2
λ π
1 + 2λa2
2
2
2
hx i = √
=
.
(3.9)
y e dy + 2a − a
π λ
2λ
π λ −∞
p
Standard deviation σ is defined as σ = hx2 i − hxi2 . In our case we have
r
1
1 + 2λa2
− a2 = √ .
(3.10)
σ=
2λ
2λ
As an aside, sometimes integrals such as the ones in this exersice can
be evaluated by taking derivatives of a known integral with respect to a
parameter that is not the integration variable, in our case λ:
r
Z ∞
d
π
−λ(x−a)2
e
dx =
(3.11)
λ
dλ
−∞
√
Z ∞
2
1 π
−
(x − a)2 e−λ(x−a) dx = − 3/2 .
(3.12)
2λ
−∞
This process can be repeated to get higher powers of (x − a). Instead
of integration by parts, we could have used this method to calculate the
first integral on the right-hand side of 3.7 and arrived at the same result.
This method is sometimes called differentiating under the integral sign or
Feynman’s Trick, after Richard Feynman obviously.
5
4. Study Griffiths’s Example 1.1. (on page 10 in the older editions, discussed
at the lecture).
a) Find the standard deviation of the distribution in Example 1.1.
b) What is the probability that a photograph, selected at random, would
show a distance x more than one standard deviation away from the
average?
Solution
a) We are given the following probability density
1
ρ(x) = √ .
2 hx
(4.1)
Using this we can compute the expectation values hxi and hx2 i and the
stantard deviation σ:
Z h
Z h
√
h
1
1 h 2
1
(4.2)
hxi =
xdx = √ x3/2 = ,
x √ dx = √
3
2 hx
2 h 0
2 h 03
0
2
Z
hx i =
0
h
1
1
x √ dx = √
2 hx
2 h
2
Z
h
0
1 h 2
h2
x3/2 dx = √ x5/2 =
,
5
2 h 05
r
q
h2
h2
2
2
σ = hx2 i − hxi =
−
= √ h ≈ 0.298h.
5
9
45
(4.3)
(4.4)
b) We may employ the concept of a complementary event:
P (|x − hxi| > σ) = 1 − P (|x − hxi| < σ).
(4.5)
The probability of an event is calculated by integrating the probability
density over the desired range:
Z
hxi+σ
Z
h
3 +σ
1
√ dx
2 hx
hxi−σ
s
s
h
1 3 +σ √
1
2
1
2
= 1 − √ h 2 x = 1 −
+√ +
−√
3
3
45
45
2 h 3 −σ
P =1−
ρ(x)dx = 1 −
(4.6)
h
3 −σ
≈ 0.393.
(4.7)
(4.8)
6
5. Solve the differential equation
y 00 (x) − y 0 (x) − 2y(x) = −20 sin 2x.
(i)
Solution
In author’s opinion, Fundamentals of Differential Equations and Boundary
Value Problems, Nagle, second edition, is a good introductory level book.
Note that the method used here is described in that book.
We immediately see that Eq. (i) is a second order inhomogeneous differential equation with constant coefficients. Furthermore, we easily see that
the corresponding homogeneous equation reads
y 00 (x) − y 0 (x) − 2y(x) = 0.
(5.1)
The auxiliary equation associated with Eq. (5.1) reads
r2 − r − 2 = 0.
(5.2)
The roots of the auxiliary equation are
r1
= −1,
(5.3)
r2
=
(5.4)
2.
Thus the general solution of Eq. (5.1) is
y0 (x) = Ae−x + Be2x ,
(5.5)
where A and B are arbitrary constants.
Let us now construct the particular solution yp (x) by using the method
of undetermined coefficients. Substituting
yp (x) = a sin 2x + b cos 2x
(5.6)
into Eq. (i) yields
−4a sin 2x−4b cos 2x−2a cos 2x+2b sin 2x−2a sin 2x−2b cos 2x = −20 cos 2x.
(5.7)
This equation is solved for all x if the coefficients a and b satisfy the pair
of equations
−6a + 2b = −20
(5.8)
−6a − 2b = 0.
The solution to this pair of equations is
a = 3
b = −1.
7
(5.9)
Now we have constructed the particular solution yp (x). The general solution
y(x)
= y0 (x) + yp (x)
= Ae−x + Be2x + 3 sin 2x − cos 2x.
8
(5.10)