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L.J. Sham April 3, 2009
Physics 130A Problem Set 1
- Discrete variable In a benzene ring (C 6 H 6 ), φj (x), where j = 1 − 6 , are six identical (normalized) wave functions of an electron localized around each C site with no overlap with one another, where x is the distance around a circle through the C atoms with the origin to be specified by the user. An electron wave function is given by,
ψ(x) =
∑^6
j=
φj (x)cj , (1)
where cj is a complex coefficient.
(a) Find the condition so that ψ(x) is normalized. (b) If c 3 = 1, c 4 = 1 and the rest of the coefficients are zero, normalize the wave function. What are the expectation value of x and its uncertainty? What do they mean? (c) If c 1 = 2, c 2 = 1, c 6 = 1 and the rest of the coefficients are zero, what are the most meaningful expectation value and its uncertainty? (d) If cj = eijπ/^3 , what are the expectation value of x and its uncertainty? What do they mean?
- Griffiths, Problem 1.3, p.12.
- Given the particle wave function is a plane wave, Ψ(x, t) = Aeikx−iωt, for 0 ≤ x ≤ A−^2 , and zero elsewhere, where A, k, ω are real constants. Please answer all questions backed up by reasons and work.
(a) What is the probability density function? (b) What is the probability of finding the particle in the left half of the interval? (c) Does the probability everywhere add up to 1? (d) Is there any relation between the state of this particle with the classical example of a free particle motion?
- Griffiths, Problem 1.5, p.14.
- Griffiths, Problem 1.7, p.18.
References
[1] M. Abramowitz, M. and I. A. Stegun, Handbook of Mathematical Functions (Dover, New York, 1965).
L.J. Sham April 3, 2009
Physics 130A Problem Set 1
- Discrete variable In a benzene ring (C 6 H 6 ), φj (x), where j = 1 − 6 , are six identical (normalized) wave functions of an electron localized around each C site with no overlap with one another, where x is the distance around a circle through the C atoms with the origin to be specified by the user. An electron wave function is given by,
ψ(x) =
∑^6
j=
φj (x)cj , (1)
where cj is a complex coefficient.
(a) Find the condition so that ψ(x) is normalized. (b) If c 3 = 1, c 4 = 1 and the rest of the coefficients are zero, normalize the wave function. What are the expectation value of x and its uncertainty? What do they mean? (c) If c 1 = 2, c 2 = 1, c 6 = 1 and the rest of the coefficients are zero, what are the most meaningful expectation value and its uncertainty? (d) If cj = eijπ/^3 , what are the expectation value of x and its uncertainty? What do they mean?
Solution –
(a) Denote the integral of two wave functions as,
〈j|k〉 :=
dx φ∗ j (x)φk(x). (2)
Since the two local wave functions on different carbon sited do not overlap and each is normalized,
〈j|k〉 = δj,k. (3)
Then, ∫ dx φ∗ j (x)φk(x) =
∑^6
j,k=
c∗ j 〈j|k〉ck =
j,k
c∗ j δj,kck =
k
|ck|^2. (4)
The normalized state is,
ψ(x) =
∑^6
j=
φj (x)˜cj , where ˜cj =
cj √∑ k |ck|
2.^ (5)
around the sites are uniform but the actual value of the mean position is meaningless beaucse of the arbitrary assignment of the distance modulo 6. For example, x 6 = 0 , x 1 = a, x 2 = 2a, x 3 0 = ± 3 a, x 4 = − 2 a, x 5 = −a. Note that the value would depend on the two possible assigned values at j = 3. The uncertainty, is
∆x =
a =
a = 1. 7 a. (14)
The uncertainty is independent of the origin but the value does not reflect the uniform distribution around the atom sites.
- Griffiths, Problem 1.3, p.12.
Solution – We will use the Gamma function for the integral [1], ∫ (^) ∞
−∞
dy y^2 ne−y
2
0
dt tn−^
(^12) e−t^ = Γ(n + 12 ); (15)
Γ(n + 12 ) = (n − 12 )Γ(n − 12 ); Γ(^12 ) =
π. (16)
(a) The integral is evaluated by changing variable y =
λ(x − a), ∫ (^) ∞
−∞
dx e−λ(x−a) 2 =
λ
−∞
dy e−y 2 =
π λ
Hence the normalization constant,
A =
λ π
(b) Note that by symmetry about x = a, ∫ (^) ∞
−∞
dx (x − a)e−λ(x−a)
2 = 0, and, 〈x〉 = a. (19)
From Eq. ( ?? ),
(∆x)^2 = 〈(x − 〈x〉)^2 〉 =
λ π
−∞
dx (x − a)^2 e−λ(x−a)
2
2 λ
the uncertainty ∆x is given by,
∆x =
2 λ
Note that sometimes it is easier to use 〈(x − 〈x〉)^2 〉 and sometimes 〈x^2 〉 − 〈x〉^2. It is definitely the former in this case. If you really need 〈x^2 〉,
〈x^2 〉 = (∆x)^2 + 〈x〉^2 =
2 λ
-4 -2 2 4
1
Figure 1: Graph of ρ(x)/A vs x.
(c) Plot of ρ(x)/A = e−(x−^0 .5) (^2) / 2 .
- Given the particle wave function is a plane wave, Ψ(x, t) = Aeikx−iωt, for 0 ≤ x ≤ A−^2 , and zero elsewhere, where A, k, ω are real constants. Please answer all questions backed up by reasons and work.
(a) What is the probability density function? (b) What is the probability of finding the particle in the left half of the interval? (c) Does the probability everywhere add up to 1? (d) Is there any relation between the state of this particle with the classical example of a free particle motion?
Solution –
(a) The probability density function is
ρ(x, t) = |Ψ(x, t)|^2 = A^2. (23)
(b) The length of the interval is L = A−^2. The probability of finding the particle in the left half of the interval is 1/2 for the uniform probability distribution. (c) Yes, ∫ (^) L
0
dx ρ(x, t) = LA^2 = 1. (24)
- Griffiths, Problem 1.7, p.18.
Solution –
d〈p〉 dt
dx
∂t
[
Ψ∗(x, t)
i
∂x
Ψ(x, t)
]
dx
[
iℏ
∂t
∂x
+ Ψ∗^
∂x
iℏ
∂t
)]
The first and second integrals in the last line are, by using the Schr¨odinger equation, respec- tively,
(I) = −
dx
ℏ^2
2 m
∂^2 Ψ∗
∂x^2
∂x
dx V Ψ∗^
∂x
(II) =
dx
∂x
iℏ
∂t
integration by parts
dx
∂x
ℏ^2
2 m
∂^2 Ψ
∂x^2
dx
∂x
V Ψ. (32)
Their sum is
(I) + (II) = −
ℏ^2
2 m
dx
[
∂^2 Ψ∗
∂x^2
∂x
∂x
∂^2 Ψ
∂x^2
]
dx V
[
Ψ∗^
∂x
∂x
]
ℏ^2
2 m
dx
∂x
[
∂x
∂x
]
dx V
∂x
[Ψ∗Ψ] (34)
dx
∂V
∂x
[Ψ∗Ψ] =
dx Ψ∗
[
∂V
∂x
]
Ψ, integration by parts (35)
provided that, [ ∂Ψ∗ ∂x
∂x
]∞
−∞
= 0, [V Ψ∗Ψ]∞−∞ = 0. (36)
References
[1] M. Abramowitz, M. and I. A. Stegun, Handbook of Mathematical Functions (Dover, New York, 1965).
