 ᖺᗘ≀⌮ᕤᏛᑓᨷ኱Ꮫ㝔ධᏛヨ㦂ၥ㢟≀⌮Ꮫ๓༙㸦➨  ၥ㸪➨  ၥ㸧ၥ㢟␒ྕ$3$3ゎ⟅᫬㛫ศὀ ព ஦ 㡯ヨ㦂㛤ጞࡢྜᅗࡀ࠶ࡿࡲ࡛㸪ၥ㢟ᩥࢆぢ࡞࠸ࡇ࡜ࠋ ➨  ၥࡣゎ⟅⏝⣬  ᯛ࠾ࡼࡧୗ᭩⏝⣬  ᯛࠊ➨  ၥࡣゎ⟅⏝⣬  ᯛ࠾ࡼࡧୗ᭩⏝⣬ ᯛࢆ౑⏝ࡍࡿࡇ࡜ࠋ
ゎ⟅⏝⣬࠾ࡼࡧୗ᭩⏝⣬ࡢ⿬㠃ࡢ౑⏝ࡣ⚗Ṇࡍࡿࠋࡍ࡭࡚ࡢゎ⟅⏝⣬࠾ࡼࡧୗ᭩⏝⣬ࡢୖ᪉ࡢᣦᐃࡉࢀࡓ⟠ᡤ࡟㸪ཷ㦂␒ྕࢆᛀࢀࡎ࡟
グධࡍࡿࡇ࡜ࠋ
᪥ᮏㄒࡲࡓࡣⱥㄒ࡛ゎ⟅ࡍࡿࡇ࡜ࠋ
ゎ⟅ࡣゎ⟅⏝⣬ࡢᐇ⥺ࡢᯟෆ࡟グධࡍࡿࡇ࡜ࠋ
ゎ⟅࡟㛵ಀࡢ࡞࠸グྕ㸪➢ྕ࡞࡝ࢆグධࡋࡓ⟅᱌ࡣ↓ຠ࡜ࡍࡿࠋ
᪥ᮏㄒࡢၥ㢟ᩥࡣ  ࣮࣌ࢪ㸪ⱥㄒࡢၥ㢟ᩥࡣ  ࣮࣌ࢪ࡟᭩࠿ࢀ࡚࠸ࡿࠋ
ၥ㢟ᩥࡢࢫࢡ࣮ࣟࣝ㸪ᣑ኱࠾ࡼࡧ⦰ᑠࡣࡋ࡚ࡼ࠸ࠋ࣮࣮࢟࣎ࢻ᧯సࡣ⚗Ṇࡍࡿࠋ࣭ゎ⟅࡟ࡣ⤖ᯝࡔࡅ࡛࡞ࡃᑟฟ㐣⛬ࡶྵࡵࡿࡇ࡜ࠋ
࣭ࢿࢵࢺ࣮࣡ࢡࢺࣛࣈࣝࡀ⏕ࡌࡓሙྜ࡛ࡶゎ⟅ࢆ⥆ࡅࡿࡇ࡜ࠋ

1

2021
Department of Applied Physics
The Graduate School Entrance Examination

Physics, first half (Problem 1, Problem 2)
Problem Number AP1, AP2Answer Time 120 minutes
GENERAL INSTRUCTIONS

1. Do not look at the Problems until the start of the examination has been announced.
2. Use 3 Answer Sheets and 4 Draft Sheets for the Problem 1. Use 4 Answer Sheets and 4 Draft
Sheets for the Problem 2.
3. Do not use the back faces of the Answer Sheets or the Draft Sheets.
4. Fill in your examinee number in the designated places at the top of all the Answer Sheets
and the Draft Sheets.
5. Answers must be written in Japanese or English.
6. Answers must be marked within the solid frame on the Answer Sheets.
7. Any Answer Sheet with marks or symbols irrelevant to your answers is considered to beinvalid.
8. The Problems are described in Japanese on page 5-9 and in English on page 11-15.
9. Scrolling, expansion and reduction of the Problems are permitted by use of a mouse or otherpointing device only. Keyboard operation is prohibited.

• Show the derivation processes as well as the results.
• Continue the answer even if network trouble occurs.

2

3

4

第1問
近年、量子コンピューターが注目を浴びているが、その基礎原理は量子力学である。量子系のエネ
ルギーはハミルトニアン Ĥ で定義され、時刻 t の状態ベクトル |ψ (t) に対するシュレーディンガー
方程式は
di |ψ (t) = Ĥ |ψ (t)(1)dtで与えられる。ここで i は虚数単位、d/dt は時間に関する微分、Ĥ は系を記述するハミルトニアン、 はプランク定数を 2π で割ったものである。時刻 t0 における初期条件を|ψ (t0 ) = |ψ0 とする。また、パウリ行列を


0 1
σ̂x =
,
1 0


σ̂y =

0 −i
i 0

(2)




,

σ̂z =

1 0
0 −1


(3)

とする。

[1] 時刻 t における状態ベクトルは、時間発展演算子 Û (t − t0 ) を用いて
|ψ (t) = Û (t − t0 ) |ψ0 (4)と書くことができる。ハミルトニアンが時間に依存しない場合に Û (t − t0 ) を Ĥ を用いて表せ。

[2] ハミルトニアン Ĥ がエルミートであることから時間発展演算子 Û (t − t0 ) がユニタリー演算子
であることを示せ。
[3] 設問 [2] の結果を用いて状態ベクトルの内積 ψ (t) |ψ (t) は時間に依存しないことを示せ。

まず、1 量子ビット系について考える。1 量子ビットは 2 準位系で記述され、その演算は 2 行 2 列の行列で記述できる。以下では、ある時刻 t = τ で考え、t0 = 0 とおく。

[4] ハミルトニアンがĤ = aσ̂z(5)で与えられるとき、Û (τ ) を行列表示で求めよ。ここに、a はエネルギーの次元を持つ実数である。

[5] 量子コンピューターにおける基本的な演算として、位相シフト変換がある。これは φ を実数と
して


1 0
(6)
Ûφ =
0 eiφ
で与えられる。Û (τ ) = Ûφ を与えるハミルトニアン Ĥ を行列表示で一つ求めよ。

5[6] NOT 変換は ÛNOT = σ̂x で与えられる。Û (τ ) = ÛNOT を与えるハミルトニアン Ĥ を行列表示で一つ求めよ。
[7] ハミルトニアンがĤ = a [(sin θ cos φ) σ̂x + (sin θ sin φ) σ̂y + (cos θ) σ̂z ] + bIˆ2

(7)で与えられるとき、Û (τ ) を行列表示で求めよ。ここに、θ と φ は実数であり、a と b はエネル
ギーの次元を持つ実数である。また、Iˆ2 は 2 行 2 列の単位行列である。

[8] アダマール（Hadamard）変換はパウリ行列をÛH† σ̂z ÛH = σ̂x ,ÛH† σ̂x ÛH = σ̂z(8)の様に変換する演算であり、具体的には

1
ÛH = √
2



1 1
1 −1


(9)

で与えられる。Û (τ ) = ÛH となるハミルトニアン Ĥ を行列表示で一つ求めよ。

次に、2 量子ビット系について考える。その演算は 4 行 4 列の行列で記述できる。

[9] 制御位相シフト変換は φ を実数として
⎛

1
⎜ 0
⎜
ÛC-φ = ⎜
⎝ 0
0

⎞
0 0
0 0 ⎟
⎟
⎟
1 0 ⎠
0 eiφ

0
1
0
0

(10)

で与えられる。Û (τ ) = ÛC-φ を与えるハミルトニアン Ĥ を行列表示で一つ求めよ。4 行 4 列の行列であるが、対角行列であることから設問 [5] との類推を用いて計算せよ。

[10] 制御 NOT 変換は

⎛

1
⎜ 0
⎜
ÛCNOT = ⎜
⎝ 0
0

0
1
0
0

0
0
0
1

⎞
0
0 ⎟
⎟
⎟
1 ⎠
0

で与えられる。Û (τ ) = ÛCNOT を与えるハミルトニアン Ĥ を行列表示で一つ求めよ。

6

(11)

第2問
適当な外力を弾性体に印加すると、弾性体は変形し、変形前の弾性体中における任意の点 x = (x1 , x2 , x3 )は変位する。変位を十分微小とし、かつ弾性体を連続体と見なす近似では、変位ベクトル u を x の関数とした変位ベクトル場 u(x, t) によって時刻 t における変形を表すことができる。変形前の弾性体において、点 x を中心に含む、十分微小な仮想的な直方体を考える。簡単のため、微小直方体の各辺を直交座標軸 x1 , x2 , x3 軸に沿ってとり、その長さをそれぞれ Δx1 , Δx2 , Δx3 とする（図１）。この体積素の変位ベクトル u は下記のニュートンの運動方程式にしたがって時間発展する。

ρΔx1 Δx2 Δx3∂ 2 u(x, t)= F (x, t)Δx1 Δx2 Δx3
∂t2

(1)

ここで、ρ は弾性体の単位体積あたりの質量、F (x, t) はこの体積素に働く単位体積あたりの力である。
この体積素に働く力 F (x, t)Δx1 Δx2 Δx3 が、体積素を取り囲む弾性媒質からこの体積素の各表面に加わる力の合力である場合を考えよう。ここで、図１に示した体積素の「上面」を、「x3 軸の正方向に法線ベクトルを持ち、位置 x = (x1 , x2 , x3 + Δx3 /2) を中心とする表面」としよう（すなわち、法線ベクトルの向きを着目する体積素から見て外側に向かって定義するものとする）。この微小な面素に、隣接する弾性媒質が及ぼす単位面積当たりの力（応力）を、応力ベクトル p(3) (x , t) と表す。同様に、
x1 軸、x2 軸の正方向に法線ベクトルを持つ体積素表面に対する応力ベクトルは、それぞれ p(1) , p(2)と表される。

p(3)(x', t)

ୖ㠃

x'
x3
x1

x2

'x3

x = (x1, x2, x3)x' = (x1, x2, x3 + 'x3/2)
x'' = (x1, x2, x3 − 'x3/2)

x
x''
'x2

'x1
ୗ㠃
図１

[1] 式 (1) の左辺と右辺の次元が一致していることを示せ。
[2] x を中心に含むこの体積素において、 x3 軸の負方向の法線ベクトルを持ち、
位置 x = (x1 , x2 , x3 −Δx3 /2) を中心とする面素（図１中の「下面」）に働く応力は、−p(3) (x , t)となることを数行以内で説明せよ。
(k)[3] p(k) , F の xj 軸方向の成分（j, k = 1, 2, 3）を pj , Fj とすると、これらは以下の関係を満たす。
(k)∂pj (x, t)

Fj (x, t) =k=1,2,3∂xk

j = 1 の場合について、これを示せ。
変位ベクトルが時刻 t に依存する場合の例として、無限に広がる弾性体中を弾性波が伝搬する場合を考える。簡単のため、弾性的な性質は等方的とすると、応力ベクトル p(k) (x, t) と変位ベクトル7

u(x, t) の成分の間には以下の関係が成り立つ。ここで、以下の式における um は u の xm 軸方向の成分 (m = 1, 2, 3) を表す。

(k)pj (x, t) =

1
cjkmn
2
m,n

∂um (x, t) ∂un (x, t)
+
∂xn
∂xm

(j, k, m, n = 1, 2, 3)(2)

cjkmn = λ(δjk δmn ) + μ(δjm δkn + δjn δkm )(3)

ここで δjk はクロネッカーのデルタを表す。すなわち、

cjjjj = λ + 2μ, cjjkk = λ (j = k), cjkjk = cjkkj = μ (j = k), cjkmn = 0（左記以外の j, k, m, n）であり、λ, μ は x, t によらない弾性率（正の定数）を表す。

[4] 変位ベクトル u(x, t) が満たす運動方程式は下記のように表すことができる。 A を λ, μ を用いて表せ。計算過程も示すこと。
ρ∂2u= μ∇2 u + A∇(∇ · u)∂t2[5] u(x, t) が x1 軸方向に伝播する平面波の弾性波を表している場合を考え、この等方弾性体における縦波平面波と横波平面波の位相速度をそれぞれ定数 ρ, λ, μ を用いて表せ。
半無限の弾性体において、横波平面波の弾性波が境界で反射する場合を考える。境界の位置を x2 = 0とする。簡単のため、弾性波は x1 -x2 面内にのみ変位しており（u3 (x, t) = 0）、変位は x3 方向には一様（∂u1 /∂x3 = ∂u2 /∂x3 = 0）であるものとする。なお、弾性体の表面は自由端であり、表面上の任意の位置、任意の時刻において、 p(2) = 0 を満たす。表面付近に局在した弾性波が励起される可能性は考えない。

x2
x3

x1
,
,

,

,

,

図２

[6] ある特別な角度 αt (0 ≤ αt < π/2) で、角振動数 ω t 、波数ベクトル kt の横波平面波の弾性波を入射したところ、反射波は横波平面波のみであり、その反射角、角振動数、波数ベクトルはそ(2)
れぞれ αt  , ω t  , kt  であった（図２左）。境界条件 p1 = 0 を考えることにより、 αt  , ωt  を αt ,
ωt を用いてそれぞれ表せ。ここで、横波入射波を ut (x, t) = Re[ut0 exp(i(k t ·x − ωt t))]、横波
反射波を ut  (x, t) = Re[ut0  exp(i(k t  ·x − ωt  t))] と置くと、界面を含む任意の位置における変位は u(x, t) = ut (x, t) + ut  (x, t) と表せることに留意せよ。なお、Re[· · ·] は複素数の実部、 iは虚数単位を表す。
(2)[7] さらに境界条件 p2 = 0 を考え、設問 [6] の状況が起こる入射角 αt を全て求めよ。

8[8] 設問 [7] で求めた角度以外の入射角で横波平面波の弾性波を入射すると、反射波は横波平面波の
みならず、縦波平面波の弾性波も伴う。その反射角、角振動数、波数ベクトルをそれぞれ αl  ,ω l  , kl  とすると、縦波反射波は ul  (x, t) = Re[ul0  exp(i(kl  ·x − ωl  t))] と表される（図２右）。
このとき、前述の境界条件 p(2) = 0 は縦波の反射弾性波も含めて満たされることに留意し、
kt sin αt = kl  sin αl を示せ。ここで kt , kl  はそれぞれ kt , kl  の絶対値である。

910Problem 1
Recently, quantum computation based on quantum mechanics attracts much attention. Theenergy of quantum systems is deﬁned by the Hamiltonian Ĥ. The evolution of the state vector|ψ (t) with time t is described by the Schrödinger equation. It is given byid|ψ (t) = Ĥ |ψ (t) ,dt(1)

where i is the imaginary number unit, d/dt is a diﬀerential operator with respect to time, and  is
Planck’s constant divided by 2π. We assume an initial condition at time t0 as|ψ (t0 ) = |ψ0  .
The Pauli matrices are given by


0 1
,
σ̂x =
1 0


σ̂y =

0 −i
i 0

(2)




,

σ̂z =

1 0
0 −1


.

(3)

[1] The state vector at time t is written with the use of the time-evolution operator Û (t − t0 ),|ψ (t) = Û (t − t0 ) |ψ0  .

(4)Express Û (t − t0 ) in terms of Ĥ when the Hamiltonian is independent of time.
[2] Show that the time-evolution operator Û (t − t0 ) is a unitary operator by using the fact thatthe Hamiltonian Ĥ is Hermitian.
[3] Show that the inner product of the state vector ψ (t) |ψ (t) is independent of time based on
the result of Question [2].

First, let us consider one-qubit systems. The one-qubit is described by a two-level system, where
its operation is described by a two-by-two matrix. We take a certain time t = τ and set t0 = 0 in
the following questions.

[4] Express a matrix representation of Û (τ ) when the Hamiltonian is given byĤ = aσ̂z ,(5)where a is a real number with the dimension of energy.
[5] The phase-shift transformation is a basic operation in quantum computation. It is deﬁned by


1 0
Ûφ =
,
(6)
0 eiφ
where φ is a real number. Express one matrix representation of the Hamiltonian Ĥ thatsatisﬁes Û (τ ) = Ûφ .

11

[6] The NOT transformation is deﬁned by ÛNOT = σ̂x . Express one matrix representation of the
Hamiltonian Ĥ that satisﬁes Û (τ ) = ÛNOT .
[7] Express the matrix representation of Û (τ ) when the Hamiltonian is given byĤ = a [(sin θ cos φ) σ̂x + (sin θ sin φ) σ̂y + (cos θ) σ̂z ] + bIˆ2 ,(7)

where θ and φ are real numbers, a and b are real numbers with the dimension of energy, andIˆ2 is the two-by-two identity matrix.
[8] The Hadamard transformation is an operator which transforms the Pauli matrices asÛH† σ̂z ÛH = σ̂x ,and is explicitly given by1ÛH = √
2

ÛH† σ̂x ÛH = σ̂z ,


1 1
1 −1

(8)


.

(9)

Express one matrix representation of the Hamiltonian Ĥ that satisﬁes Û (τ ) = ÛH .

Next, let us consider two-qubit systems, whose operations are described by four-by-four matrices.

[9] The controlled-phase-shift transformation is deﬁned by
⎛

1
⎜ 0
⎜
ÛC-φ = ⎜
⎝ 0
0

⎞
0 0
0 0 ⎟
⎟
⎟,
1 0 ⎠
0 eiφ

0
1
0
0

(10)

where φ is a real number. Express one matrix representation of the Hamiltonian Ĥ thatsatisﬁes Û (τ ) = ÛC-φ . This question is analogous to Question [5], where the matrix isdiagonal although it is a four-by-four matrix.
[10] The controlled-NOT transformation is deﬁned by
⎛

1
⎜ 0
⎜
ÛCNOT = ⎜
⎝ 0
0

0
1
0
0

0
0
0
1

⎞
0
0 ⎟
⎟
⎟.
1 ⎠
0

Express one matrix representation of the Hamiltonian Ĥ that satisﬁes Û (τ ) = ÛCNOT .

12

(11)

Problem 2
The application of an external force to an elastic body induces deformation. Let x = (x1 , x2 , x3 )be an arbitrary position in the elastic body before the deformation. Under the approximationthat the deformation is suﬃciently small and the elastic body is viewed as a continuous medium,the deformation at time t can be expressed by the displacement vector ﬁeld u(x, t), in which
the displacement vector u is a function of x. In the elastic medium before the deformation, let
us consider a hypothetical inﬁnitesimal volume element of rectangular shape centered at x. For
simplicity, suppose that the sides are parallel to the x1 -, x2 -, and x3 -axes of the Cartesian coordinate
system, and let Δx1 , Δx2 , and Δx3 be the length of each side of the rectangular volume element,
as shown in Fig. 1. Then, the time-evolution of the displacement vector u of the volume element
obeys the following Newtonian equation of motion:
ρΔx1 Δx2 Δx3∂ 2 u(x, t)= F (x, t)Δx1 Δx2 Δx3 ,
∂t2

(1)

where ρ is the mass of the unit volume of the elastic medium, and F (x, t) is the force per unit
volume acting on the volume element.
Consider the case that the force acting on the volume element, F (x, t)Δx1 Δx2 Δx3 , is the sum of
the forces that the surrounding elastic medium exerts on each surface of the volume element. Let us
represent “the upper surface” indicated in Fig. 1 as “the surface centered at x = (x1 , x2 , x3 +Δx3 /2)
with a normal vector toward the positive direction of the x3 -axis”; that is, we deﬁne a normal vector
toward the direction from the center of a considered volume element to the outside. Then, we deﬁne
the stress vector p(3) (x , t) as a force per unit area that the neighboring elastic medium exerts on
this surface element. Similarly, p(1) and p(2) denote the stress vectors acting on the surfaces ofthe volume element with normal vectors toward the positive directions of the x1 - and x2 -axes,
respectively.

p(3)(x', t)Upper surface

x'
x3
x1

x2

'x3

x = (x1, x2, x3)x' = (x1, x2, x3 + 'x3/2)
x'' = (x1, x2, x3 − 'x3/2)

x
x''
'x2

'x1
Lower surface
Figure 1[1] Show that the left- and right-hand sides of Eq. (1) have the same dimension.
[2] Regarding the volume element centered at x, show within a few sentences that −p(3) (x , t)gives the stress vector at “the lower surface” centered at x = (x1 , x2 , x3 − Δx3 /2) with anormal vector toward the negative direction of the x3 -axis (see Fig. 1).

13

(k)

[3] Let pj and Fj (j, k = 1, 2, 3) be the components along the xj -axis of p(k) and F , respectively.
These quantities satisfy the following relation:(k)∂pj (x, t)

Fj (x, t) =k=1,2,3∂xk.

Derive the equation for the case j = 1.
Next, we consider an elastic plane wave propagating in an inﬁnite elastic medium, as an example
for the case that the displacement vector is time t dependent. For simplicity, the elastic properties
are assumed to be isotropic, and thus, the components of the stress vector p(k) (x, t) and thedisplacement vector u(x, t) satisfy the following relations, where um denotes the xm -axis componentof u (m = 1, 2, 3):(k)pj (x, t) =

1
cjkmn
2
m,n

∂um (x, t) ∂un (x, t)
+
∂xn
∂xm

(j, k, m, n = 1, 2, 3),cjkmn = λ(δjk δmn ) + μ(δjm δkn + δjn δkm ),

(2)
(3)

where δjk represents the Kronecker delta and hence
cjjjj = λ + 2μ, cjjkk = λ (j = k), cjkjk = cjkkj = μ (j = k), cjkmn = 0 (all other cases).
λ and μ are positive elastic moduli of the elastic medium, which are independent of x and t.
[4] The equation of motion for u(x, t) can also be expressed as follows:∂2u= μ∇2 u + A∇(∇ · u).
∂t2
Express A in terms of μ and λ while showing the derivation.
ρ[5] Consider the case that u(x, t) describes an elastic plane wave propagating along the x1 -axis.
Obtain the phase velocities of the longitudinal and transverse waves in the isotropic elastic
medium by using ρ, λ, and μ.
Consider a semi-inﬁnite elastic medium and suppose that a transverse elastic plane wave is reﬂectedby the surface. We deﬁne the position of the surface as x2 = 0. For simplicity, let us assume that theelastic wave is accompanied only by a displacement vector within the x1 -x2 -plane (i.e., u3 (x, t) = 0)
and the displacement is uniform along the x3 -direction (i.e., ∂u1 /∂x3 = ∂u2 /∂x3 = 0). Note thatthe surface of the elastic medium is a free boundary and hence p(2) = 0 is satisﬁed at any arbitrary
position on the surface and at any time. Neglect the possibility that a localized elastic wave near
the surface is excited.
[6] When a transverse elastic plane wave with an angular frequency ωt and a wave vector kt
arrives at the surface at a certain incident angle αt (0 ≤ αt < π/2), we observe only atransverse elastic plane wave as the reﬂected wave, whose reﬂection angle, angular frequency,
and wave vector are αt  , ω t  , and kt  , respectively (see the left panel of Fig. 2). By considering(2)
the boundary condition, p1 = 0, express αt  and ωt  in terms of αt and ωt . When we express
the incident and reﬂected transverse plane waves as ut (x, t) = Re[ut0 exp(i(k t ·x − ωt t))] andut  (x, t) = Re[ut0  exp(i(kt  ·x−ωt t))], respectively, note that the displacement at an arbitrary
position including the surface is given by u(x, t) = ut (x, t) + ut  (x, t). Re[· · ·] denotes thereal part of a complex number and i represents the imaginary number unit.

14

x2
x3

x1
,

,

,

,

,

Figure 2(2)[7] By further considering the boundary condition, p2 = 0, ﬁnd all possible values of the incident
angle αt for which the situation supposed in Question [6] occurs.
[8] When a transverse elastic plane wave arrives at the surface at an arbitrary incident angle
other than the angles derived in Question [7], the reﬂected waves consist of not only transverse
but also longitudinal elastic plane waves. Let αl  , ω l  , and kl  be the reﬂection angle, angularfrequency and wave vector of the longitudinal reﬂected wave, respectively. Then, the reﬂectedtransverse plane wave is expressed as ul  (x, t) = Re[ul0  exp(i(kl  ·x−ωl  t))] (see the right panelof Fig. 2). Given that the boundary condition, p(2) = 0, holds when taking into account also
the longitudinal reﬂected wave, derive the following relation:
kt sin αt = kl  sin αl  ,where kt and kl  denote the magnitude of kt and kl  , respectively.

15 ᖺᗘ≀⌮ᕤᏛᑓᨷ኱Ꮫ㝔ධᏛヨ㦂ၥ㢟≀⌮Ꮫᚋ༙㸦➨  ၥ㸪➨  ၥ㸧ၥ㢟␒ྕ$3$3ゎ⟅᫬㛫ศὀ ព ஦ 㡯ヨ㦂㛤ጞࡢྜᅗࡀ࠶ࡿࡲ࡛㸪ၥ㢟ᩥࢆぢ࡞࠸ࡇ࡜ࠋ ➨  ၥࡣゎ⟅⏝⣬  ᯛ࠾ࡼࡧୗ᭩⏝⣬  ᯛࠊ➨  ၥࡣゎ⟅⏝⣬  ᯛ࠾ࡼࡧୗ᭩⏝⣬ ᯛࢆ౑⏝ࡍࡿࡇ࡜ࠋ
ゎ⟅⏝⣬࠾ࡼࡧୗ᭩⏝⣬ࡢ⿬㠃ࡢ౑⏝ࡣ⚗Ṇࡍࡿࠋࡍ࡭࡚ࡢゎ⟅⏝⣬࠾ࡼࡧୗ᭩⏝⣬ࡢୖ᪉ࡢᣦᐃࡉࢀࡓ⟠ᡤ࡟㸪ཷ㦂␒ྕࢆᛀࢀࡎ࡟
グධࡍࡿࡇ࡜ࠋ
᪥ᮏㄒࡲࡓࡣⱥㄒ࡛ゎ⟅ࡍࡿࡇ࡜ࠋ
ゎ⟅ࡣゎ⟅⏝⣬ࡢᐇ⥺ࡢᯟෆ࡟グධࡍࡿࡇ࡜ࠋ
ゎ⟅࡟㛵ಀࡢ࡞࠸グྕ㸪➢ྕ࡞࡝ࢆグධࡋࡓ⟅᱌ࡣ↓ຠ࡜ࡍࡿࠋ
᪥ᮏㄒࡢၥ㢟ᩥࡣ  ࣮࣌ࢪ㸪ⱥㄒࡢၥ㢟ᩥࡣ  ࣮࣌ࢪ࡟᭩࠿ࢀ࡚࠸ࡿࠋ
ၥ㢟ᩥࡢࢫࢡ࣮ࣟࣝ㸪ᣑ኱࠾ࡼࡧ⦰ᑠࡣࡋ࡚ࡼ࠸ࠋ࣮࣮࢟࣎ࢻ᧯సࡣ⚗Ṇࡍࡿࠋ࣭ゎ⟅࡟ࡣ⤖ᯝࡔࡅ࡛࡞ࡃᑟฟ㐣⛬ࡶྵࡵࡿࡇ࡜ࠋ
࣭ࢿࢵࢺ࣮࣡ࢡࢺࣛࣈࣝࡀ⏕ࡌࡓሙྜ࡛ࡶゎ⟅ࢆ⥆ࡅࡿࡇ࡜ࠋ

1

2021
Department of Applied Physics
The Graduate School Entrance Examination

Physics, second half (Problem 3, Problem 4)
Problem Number AP3, AP4
Answer Time 120 minutes
GENERAL INSTRUCTIONS

1. Do not look at the Problems until the start of the examination has been announced.
2. Use 2 Answer Sheets and 4 Draft Sheets for the Problem 3. Use 4 Answer Sheets and 4 Draft
Sheets for the Problem 4.
3. Do not use the back faces of the Answer Sheets or the Draft Sheets.
4. Fill in your examinee number in the designated places at the top of all the Answer Sheets
and the Draft Sheets.
5. Answers must be written in Japanese or English.
6. Answers must be marked within the solid frame on the Answer Sheets.
7. Any Answer Sheet with marks or symbols irrelevant to your answers is considered to beinvalid.
8. The Problems are described in Japanese on page 5-8 and in English on page 9-13.
9. Scrolling, expansion and reduction of the Problems are permitted by use of a mouse or otherpointing device only. Keyboard operation is prohibited.

• Show the derivation processes as well as the results.
• Continue the answer even if network trouble occurs.

2

3

4

ୈ3໰
ಉछϘʔεཻࢠ͔ΒͳΔཧ૝‫ؾ‬ମʹ͓͚ΔϘʔεɾΞΠϯγϡλΠϯ‫ڽ‬ॖʹ͍ͭͯߟ͑Α͏ɻ1 ཻࢠঢ়ଶͷΤωϧΪʔΛ i ʢi = 0, 1, 2, · · · ʣͱͯ͠ɺ0 (= 0) ≤ 1 ≤ 2 ≤ · · · ͕੒Γཱͭͱ͢Δɻ·ͨɺΤωϧΪʔ͕  ͔Β  + d ͷؒʹ͋Δ 1 ཻࢠঢ়ଶ਺Λ D()d ͱ͢Δʢͳ͓ D() ͸ঢ়ଶ਺ີ౓ɺd ͸ਖ਼ͷඍখྔʣɻ‫ܥ‬ͷԹ౓Λ T > 0 ͱͯ͠ɺϘϧπϚϯఆ਺͸؆୯ͷͨΊ kB = 1 ͱ͢Δɻͳ͓ɺεϐϯࣗ༝౓͸ߟ͑ͳ͍΋ͷͱ͢Δɻ·ͨɺཻࢠͷ࣭ྔΛ m ͱͯ͠ɺϓϥϯΫఆ਺Λ 2π ͰΘͬͨఆ਺Λ ͱॻ͘ɻԽֶϙςϯγϟϧΛ μ ≤ 0 ͱ͢Δͱɺ1 ཻࢠঢ়ଶ i Λ઎ΊΔཻࢠ਺ͷ‫ظ‬଴஋Λද͢Ϙʔε෼෍͸ɺμ = 0 ͔ͭ i = 0 ͷ৔߹Λআ͍ͯf (i ) =1e(i −μ)/T − 1Ͱ༩͑ΒΕΔɻҎԼͷઃ໰ʹ౴͑Αɻͨͩ͠ɺඞཁͳΒ͹ੵ෼ެࣜ√  
 ∞ 1/2
 ∞
π
x
3
π2
x
dx
=
ζ
,
dx
=
ex − 12
2
ex − 1
6
0
0

(1)

(2)

Λ༻͍Αɻ͜͜Ͱ ζ(·) ͸θʔλؔ਺Ͱ͋Δɻ

[1] ·ͣɺN ( 1) ‫ݸ‬ͷཻࢠ͕ 3 ࣍‫ݩ‬ͷശʢ͢ͳΘͪҰลͷ௕͕͞ L ͷཱํମͷྖҬʣʹೖ͍ͬͯΔঢ়‫گ‬Λߟ͑Α͏ɻ͜ͷശ಺ͷ 1 ཻࢠ೾ಈؔ਺ʹ͸प‫ظ‬త‫ڥ‬ք৚͕݅՝͞Ε͓ͯΓɺL ͸े෼େ͖͍ͱ͢Δɻ
[1.1] ӡಈྔۭؒͷ୯Ґମੵ͋ͨΓͷ 1 ཻࢠΤωϧΪʔ‫ݻ‬༗ঢ়ଶͷ਺Λ‫ٻ‬ΊɺL ͳͲͰදͤɻ
[1.2] D() Λ‫ٻ‬ΊΑɻʢಋग़աఔΛ໌‫ͤه‬Αɻʣ
[1.3] શཻࢠ਺͕ N Ͱ͋Δͱ͍͏৚݅͸ੵ෼Ͱॻ͘͜ͱ͕Ͱ͖ɺμ < 0 ͷͱ͖ ∞D()f ()d = N(3)0Ͱ༩͑ΒΕΔɻ͔͋͠͠Δ Tc ΑΓ௿Թͩͱɺ͜ͷ৚݅Λຬͨ͢Խֶϙςϯγϟϧ͕ μ ≤ 0ͷൣғʹଘࡏ͠ͳ͘ͳΔɻμ = 0 ͷ৔߹ʹ (3) ࣜͷࠨลͷੵ෼Λߟ͑Δ͜ͱʹΑΓɺ͜ͷΑ͏ͳ Tc Λ‫ٻ‬ΊɺL ͱ N ͳͲͰදͤɻ

[1.4] ‫ܥ‬ͷԹ౓ T ͕ Tc ΑΓ௿Թͷͱ͖ɺ‫ج‬ఈঢ়ଶΛ઎ΊΔཻࢠ਺ N0 Λ‫ٻ‬ΊɺN ɺT ɺTc Ͱදͤɻͳ͓͜͜Ͱɺ‫ج‬ఈঢ়ଶʹ‫ࢹڊ‬తͳ਺ͷཻࢠ͕‫ڽ‬ॖ͠ N0 ͕ N ͷΦʔμʔʹͳΔ͜ͱΛɺϘʔεɾΞΠϯγϡλΠϯ‫ڽ‬ॖͱ‫Ϳݺ‬ɻ
[1.5] લ໰·ͰͷઃఆͰɺۭ͕ؒ 2 ࣍‫ݩ‬ͷ৔߹ɺϘʔεɾΞΠϯγϡλΠϯ‫ڽ‬ॖ͕‫͜ى‬Βͳ͍͜ͱΛ؆ܿʹٞ࿦ͤΑɻʢ2 ࣍‫ݩ‬ͷঢ়ଶ਺ີ౓ͳͲΛ܎਺·Ͱਖ਼֬ʹಋ͘ඞཁ͸ͳ͍ɻʣ[2] ۭ͕ؒ 2 ࣍‫ݩ‬ͷ৔߹Λߟ͑Δɻલ໰·Ͱͷശ͸ଘࡏͤͣɺ୅ΘΓʹ N ( 1) ‫ݸ‬ͷཻࢠ͕ 2 ࣍‫ݩ‬ͷௐ࿨ϙςϯγϟϧʹଋറ͞Ε͍ͯΔঢ়‫گ‬Λߟ͑Α͏ɻ͜ͷϙςϯγϟϧ͸౳ํతͰ͋Γɺ1 ཻࢠΤωϧΪʔ͸ 2 ͭͷ 0 Ҏ্ͷ੔਺ (nx , ny ) Λ༻͍ͯ  = ω(nx + ny ) ͱॻ͚Δͱ͢Δɻ͜͜Ͱω > 0 ͸֯प೾਺Ͱ͋Γɺω ͸ T ʹൺ΂ͯे෼খ͍͞ͱ͢Δɻ·ͨɺθϩ఺ৼಈ͸ແࢹͨ͠ɻ
[2.1] D() Λ‫ٻ‬ΊΑɻͨͩ͠   ω ͱͯ͠ɺ ͷ࠷ߴ࣍ͷ߲͚ͩΛ‫ٻ‬ΊΑɻ
[2.2] શཻࢠ਺͕ N Ͱ͋Δͱ͍͏৚݅͸ɺ(3) ࣜʹલ໰ [2.1] Ͱ‫ٻ‬Ίͨ D() Λ୅ೖ͢ΔͱಘΒΕΔɻ͔͋͠͠Δ Tc ΑΓ௿Թͩͱɺ͜ͷ৚݅Λຬͨ͢Խֶϙςϯγϟϧ͕ μ ≤ 0 ͷൣғʹଘࡏ͠ͳ͘ͳΔɻ͜ͷΑ͏ͳ Tc Λ‫ٻ‬Ίɺω ͱ N ͳͲͰදͤɻ

5

[2.3] લ໰ [2.2] ͰɺN → ∞ ͷ೤ྗֶ‫ݶۃ‬Λͱͬͯ΋ Tc ͕ൃࢄͤͣɺ·ͨθϩʹ΋ͳΒͣ༗‫ݶ‬஋ʹͱͲ·ΔͨΊʹ͸ɺϙςϯγϟϧΛͲͷΑ͏ʹௐઅ͢Ε͹ྑ͍͔ߟ͑Α͏ɻ1 ཻࢠͷௐ࿨ϙςϯγϟϧ V̂ ͕ɺҐஔԋࢉࢠ (x̂, ŷ) ͱ௕͞ͷ࣍‫ݩ‬Λ΋ͭ਺ L0 Λ༻͍ͯ  2 ŷx̂ 2V̂ = V0+(4)
L0
L0ͱॻ͚Δͱͯ͠ɺN ͱ L0 ͷຬͨ͢΂͖ؔ܎Λٞ࿦ͤΑɻ͜͜Ͱ V0 ͸ L0 ΍ N ʹ͸ґଘ͠ͳ͍ͱ͢Δɻ

[2.4] ‫ܥ‬ͷԹ౓ T ͕ Tc ΑΓ௿Թͷͱ͖ɺ‫ج‬ఈঢ়ଶΛ઎ΊΔཻࢠ਺ N0 Λ‫ٻ‬ΊɺN ɺT ɺTc Ͱදͤɻ6ୈ4໰
ਅۭதʹஔ͔Εͨి‫ؾ‬త෼‫ ཰ۃ‬αʢ> 0ʣΛ࣋ͭதੑཻࢠʹΑΔి࣓೾ͷࢄཚʹ͍ͭͯߟ͑Δɻதੑཻࢠ͸ೖࣹ͢Δి࣓೾ͷ೾௕ΑΓ΋े෼খ͘͞ɺͦͷத৺Λ‫ ఺ݪ‬O ʹͱΔɻೖࣹ೾͸ z ํ޲ʹͷΈి৔Λ΋ͭฏ໘೾Ͱ͋Γɺ֯ৼಈ਺ ω0 Ͱৼಈ͢Δɻೖࣹ೾ʹΑΓੜͨ࣌ؒ͡มԽ͢Δཻࢠͷి‫ؾ‬૒‫ۃ‬ࢠϞʔϝϯτ͸ z ࣠ʹฏߦͰɺͦͷ z ੒෼͸ p(t) = αE0 cos ω0 t Ͱද͞ΕΔɻ͜͜Ͱ E0 (> 0) ͸ి৔ͷৼ෯Ͱ͋Δɻೖࣹ೾ʹΑΓ༠‫͞ى‬Εͨৼಈ͢Δి‫ؾ‬૒‫ࢠۃ‬Ϟʔϝϯτ͔Βͷ์ࣹΛࢄཚ೾ͱ͢Δɻ͜͜Ͱ ε0 , μ0 , c ͸ͦΕͧΕਅۭͷ༠ి཰ɺಁ࣓཰ɺޫ଎Ͱ͋ΔɻҎԼͰ͸ɺೖࣹ೾ʹΑΓཻࢠʹੜͨ͡ి‫ؾ‬૒‫ࢠۃ‬Ϟʔϝϯτ͕ͭ͘Δɺԕํͷ఺ X(࠲ඪϕΫτϧr = (r sin θ cos φ, r sin θ sin φ, r cos θ)) ʹ͓͚ΔϙςϯγϟϧΛ‫ٻ‬ΊΑ͏ɻ͜͜Ͱ‫࠲ۃ‬ඪΛਤ̍ͷΑ
1 ∂r͏ʹఆٛ͠ɺ఺ X ʹ͓͚Δ୯ҐϕΫτϧΛ er = ∂∂rr ɺeθ = 1r ∂∂θr ɺeφ = r sinθ ∂φ ͱ͢Δɻ‫ ఺ݪ‬O ͱ
2πc఺̭ͷ‫ ཭ڑ‬r ͸ɺೖࣹ೾ͷ೾௕ ω0 ΑΓ΋े෼େ͖͍ͱ͢Δɻ͜͜Ͱ͸தੑཻࢠͷి‫ؾ‬૒‫ࢠۃ‬ϞʔϝϯτΛɺਤ̎ʹࣔ͢‫ݻ‬ఆ͞Εͨ఺̍ɺ఺্̎ʹͦΕͧΕ +q(t)ɺ−q(t) ͷిՙྔΛஔ͘͜ͱʹΑΓද‫͢ݱ‬Δɻ఺̍ɺ̎ؒͷ‫཭ڑ‬Λఆ਺ l ͱ͢Δͱɺి‫ؾ‬૒‫ࢠۃ‬Ϟʔϝϯτͷ z ੒෼͸ q(t) Λ༻͍ͯp(t) = q(t)l = αE0 cos ω0 t ͱදͤΔʢఆ਺ l ͸ೖࣹ೾ͷ೾௕ΑΓे෼খ͍͞ʣɻ

z

z
X

θ

ࢋ௥
ࢋథ

ା

ࢋఏ
‫ݎ‬

O
x

X

଴



y

Ⅼ㸯


/2

߶



ି

θ
O

y

/2
x

ᅗ



Ⅼ㸰


୰ᛶ⢏Ꮚ

ᅗ

ਤ̎ͷΑ͏ʹ఺̭ͱ఺̍ɺ఺̎ͷ‫཭ڑ‬ΛͦΕͧΕ r+ ɺr− ͱͨ͠ͱ͖ɺి‫ؾ‬૒‫ࢠۃ‬Ϟʔϝϯτ͕ͭ͘Δ఺ X Ͱͷ࣌ؒ஗ԆΛ‫ؚ‬ΉεΧϥʔϙςϯγϟϧ ϕ ͸ϩʔϨϯπήʔδͰ−q(t − rc− ) 1+q(t − rc+ ) 1+
ϕ(r, t) =
4πε0
r+
4πε0
r−

(1)

ͷΑ͏ʹॻ͖දͤΔɻઃ໰ [1]-[6] ʹ͓͍ͯ͸ి‫ؾ‬త෼‫ʹ͕ؒ࣌཰ۃ‬ґΒͳ͍ఆ਺ α = α0 Ͱ͋Δͱͯ͠ҎԼͷ໰͍ʹ౴͑Αɻ

[1] ·ͣɺి৔͕੩ి৔ E0 (ω0 = 0) Ͱ͋Γ࣌ؒ஗Ԇ͕ແࢹͰ͖Δ৔߹Λߟ͑Δɻ͜ͷͱ͖ɺl  rͰ͋Δ͜ͱΛ౿·͑Δͱɺࣜ (1) ͷεΧϥʔϙςϯγϟϧ͸
1
1
α0 E0 cos θ
C· +D· 2
ϕ(r) =
4πε0rr
ͷΑ͏ʹۙࣅͰ͖Δɻ rl ͷ 2 ࣍Ҏ্ͷ߲Λແࢹ͢Δ͜ͱͰ CɺD Λ‫ٻ‬ΊΑɻ

7

[2] ҎԼͷઃ໰ [2]-[7] Ͱ͸ɺω0 l  c Λຬͨ͢༗‫ݶ‬ͷ֯ৼಈ਺ ω0 = 0 Λ࣋ͭి࣓೾ͷཻࢠʹΑΔࢄ
ཚΛߟ͑Δɻ͜ͷͱ͖࣌ؒ஗ԆΛ‫ؚ‬Ήࣜ (1) ͷεΧϥʔϙςϯγϟϧ͸
1
1
α0 E0 cos θ
F· +G· 2ϕ(r, t) =4πε0rr
ͷΑ͏ʹۙࣅͰ͖Δɻ rl , ωc0 l ͷ 2 ࣍Ҏ্ͷ߲Λແࢹ͢Δ͜ͱͰ FɺG ʹೖΔࣜΛ r, c, ω0 , t Λ༻͍ͯॻ͖දͤɻ

[3] ఺ X Ͱͷ࣌ؒ஗ԆΛ‫ؚ‬ΉϕΫτϧϙςϯγϟϧ A(r, t) ͸ҎԼͷΑ͏ʹॻ͖දͤΔɻdp(t − rc )1ez
4πε0 c2 r
dt

A(r, t) =ཻࢠʹੜͨ͡ৼಈ͢Δి‫ؾ‬૒‫ࢠۃ‬Ϟʔϝϯτ͕࡞Δ఺ X ʹ͓͚ΔϕΫτϧϙςϯγϟϧ A(r, t)Λ‫࠲ۃ‬ඪͷ୯ҐϕΫτϧͱ α0 , r ͳͲΛ༻͍ͯॻ͖දͤɻ͜͜Ͱɺez ͸ z ํ޲ͷ୯ҐϕΫτϧͰ͋Δɻ

఺ X ʹ͓͚Δి৔ͱ࣓ଋີ౓Λ‫ٻ‬ΊΑ͏ɻ͜͜Ͱ͸ԕํ΁఻ൖ͢Δ੒෼ʹண໨͠ɺઃ໰ [2] Ͱ‫ٻ‬ΊͨϙςϯγϟϧதͰ 1r ͷ߲ͷΈΛߟ͑ɺ r12 ͷ߲Λແࢹ͢Δɻ·ͨɺઃ໰
[4][5] Ͱ͸‫ࢉܭ‬աఔͰ͋ΒΘΕ:::::::::
Δ r12 ͷ߲΋ແࢹͤΑɻ:::::::::[4] ઃ໰ [2] Ͱ‫ٻ‬Ίͨ ϕ(r, t) ʹؔͯ͠ɺ఺ X ʹ͓͚Δ ∇ϕ(r, t) Λ‫࠲ۃ‬ඪͷ୯ҐϕΫτϧΛ༻͍ͯॻ͖දͤɻ͜͜Ͱɺ೚ҙͷؔ਺ f ʹରͯ͠ɺҎԼͷ͕ࣜ੒Γཱͭ͜ͱΛ༻͍ͯΑ͍ɻ
∇f =

∂f
1 ∂f
1 ∂f
er +
eθ +
eφ
∂r
r ∂θ
r sin θ ∂φ[5] ఺ X ʹ͓͚Δࢄཚ೾ͷి৔ E ͱ࣓ଋີ౓ B Λ‫࠲ۃ‬ඪͷ୯ҐϕΫτϧΛ༻͍ͯॻ͖දͤɻ͜͜
Ͱɺ࣌ؒมԽ͢Δి৔ E ͱ࣓ଋີ౓ B ͸࣍ͷࣜͰ༩͑ΒΕΔɻ
E = −∇ϕ −∂A,∂t

B =∇×AඞཁͰ͋Ε͹೚ҙͷϕΫτϧ F = Fr er + Fθ eθ + Fφ eφ ʹରͯ͠੒Γཱͭ࣍ͷࣜΛ༻͍ͯΑ͍ɻ



1 ∂
1 ∂Fθ
(sin θ Fφ ) −r sin θ ∂θr sin θ ∂φ





∇×F =




1 ∂Fr
1 ∂
1 ∂Fr
1 ∂
er +
−
(rFφ ) eθ +(rFθ ) −eφr sin θ ∂φ
r ∂r
r ∂r
r ∂θ

[6] ཻࢠʹΑΓࢄཚ͞Εͨి࣓೾ͷΤωϧΪʔྲྀີ౓Λද͢఺ X ʹ͓͚ΔϙΠϯςΟϯάϕΫτϧ
S Λ‫࠲ۃ‬ඪͷ୯ҐϕΫτϧΛ༻͍ͯॻ͖දͤɻ͜͜ͰϙΠϯςΟϯάϕΫτϧ S ͸ҎԼͷࣜͰ༩͑ΒΕΔɻ
1
S=
E×B
μ0·ͨɺS ͷେ͖͞ |S| Λ θ, φ ΛͦΕͧΕԣ࣠ʹͱΓਤࣔͤΑɻ[7] ࣍ʹి‫ؾ‬త෼‫ ཰ۃ‬α ͕࣌ؒతʹมԽ͢Δ৔߹Λߟ͑Δɻ͜ͷΑ͏ͳ෼‫཰ۃ‬ͷมௐ͸෼ࢠ΍֨ࢠͷৼಈʹΑͬͯੜ͡Δ͜ͱ͕͋Δɻి‫ؾ‬త෼‫ ཰ۃ‬α ͷ࣌ؒมԽͷ֯ৼಈ਺ ωv ͸ೖࣹ೾ͷ֯ৼಈ਺ ω0 ʹൺ΂ͯখ͍͞ͱ͢Δɻ͜ͷͱ͖ి‫ؾ‬త෼‫཰ۃ‬͸ α(t) = α0 + α1 cos ωv t Ͱද͞ΕΔɻ͜͜Ͱ α0 , α1 ͸ཻࢠʹґଘͨ͠ఆ਺Ͱ͋Δɻཻࢠʹੜͨ͡ి‫ؾ‬૒‫ࢠۃ‬Ϟʔϝϯτͷ z ੒෼͸p(t) = α(t)E0 cos ω0 t Ͱ͋Δͱͯ͠ɺཻࢠʹΑΓࢄཚ͞Εͨి࣓೾ͷ఺ X ʹ͓͚Δి৔Λॻ͖
දͤɻ·ͨɺࢄཚ͞Εͨి࣓೾ͷ֯ৼಈ਺Λ͢΂ͯॻ͖දͤɻ

8

Problem 3
Consider Bose-Einstein condensation of an ideal Bose gas composed of identical particles. Let ibe the 1-particle energy levels (i = 0, 1, 2, · · · ) with 0 (= 0) ≤ 1 ≤ 2 ≤ · · · , and let D()d be thenumber of 1-particle states between the energies  and  + d, where D() is the density of statesand d is a small energy increment. Let T > 0 be the temperature of the system. For simplicity,
set the Boltzmann constant to unity (i.e., kB = 1). We ignore the spin degrees of freedom. Letm be the mass of the particles and  be the Planck constant divided by 2π. Denote the chemical
potential by μ ≤ 0. Then, the Bose distribution describing the expectation value of the number ofparticles that occupy the 1-particle state i is given byf (i ) =1e(i −μ)/T − 1,(1)except for the case where μ = 0 and i = 0. Answer the following Questions. If needed, use theintegral formulas√  
 ∞ 1/2
 ∞
x
π
x
3
π2
dx
=
ζ
,
dx
=
,
(2)
ex − 12
2
ex − 1600where ζ(·) is the zeta function.
[1] Consider the situation that N ( 1) particles are conﬁned in a three-dimensional cubic box
with side length L. Impose periodic boundary conditions on the 1-particle wavefunction inthis box and assume that L is suﬃciently large.
[1.1] Express the number of 1-particle energy eigenstates per unit volume of momentum spacein terms of L.
[1.2] Find D(). (Write down the derivation.)[1.3] For μ < 0, the condition that the total number of particles is N is given by the following
integral equation:
∞D()f ()d = N.

0(3)

If the temperature of the system is lower than a critical temperature Tc , a chemical
potential that satisﬁes the above condition (3) and μ ≤ 0 does not exist. By calculating
the integral on the left-hand side of Eq. (3) for μ = 0, express Tc in terms of L and N .
[1.4] Suppose that the temperature T of the system is lower than Tc . Express the number
N0 of particles that occupy the ground state in terms of N , T , and Tc . Note that BoseEinstein condensation means that a macroscopic number of particles condense in the
ground state and N0 becomes of the order of N .
[1.5] Consider the previous setup but now suppose that the spatial dimension is two. Describe
concisely the reason why Bose-Einstein condensation does not occur in this case. (Youdo not need to derive the complete formula of the density of states in two dimensions.)[2] Consider the case that the spatial dimension is two. We remove the box of the previous
setup and instead assume that N ( 1) particles are trapped by a two-dimensional harmonic
potential. This potential is isotropic and the 1-particle energy is given by  = ω(nx + ny ),where nx and ny are integers equal to or greater than zero. Here, ω > 0 is the angularfrequency and suppose that ω is suﬃciently smaller than T . Note that the zero-point energy
is ignored.

9[2.1] Find D(). Suppose that   ω and only write down the highest order term withrespect to .
[2.2] The condition that the total particle number is N is obtained by substituting D()(obtained in Question [2.1]) to Eq. (3). If the temperature of the system is lower thana critical temperature Tc , a chemical potential that satisﬁes the above condition andμ ≤ 0 does not exist. Express Tc in terms of ω and N .
[2.3] Consider the adjustment of the potential with which Tc obtained in Question [2.2] does
not diverge and does not go to zero when taking the thermodynamic limit N → ∞.
Suppose that the 1-particle potential V̂ is written as  2 ŷx̂ 2V̂ = V0
+
,
(4)
L0
L0
where x̂ and ŷ are the position operators, L0 has the dimension of length, and V0 does
not depend on L0 and N . Find the relationship between N and L0 to meet the abovementioned condition of the potential adjustment.
[2.4] Suppose that the temperature T of the system is lower than Tc . Express the number N0of particles that occupy the ground state in terms of N , T , and Tc .

10

Problem 4
We consider the scattering of an electromagnetic wave by a neutral particle with an electric polarizability α(> 0) in vacuum. The particle is suﬃciently smaller than the wavelength of the incident
electromagnetic wave. The center of the particle is located at the origin O. The incident plane
wave with the angular frequency ω0 has an electric ﬁeld component only parallel to the z axis.
The time-dependent z-component of the electric dipole moment of the particle induced by the incident wave is expressed as p(t) = αE0 cos ω0 t. E0 (> 0) is the amplitude of the electric ﬁeld of theincident wave. The radiation from the oscillating electric dipole moment induced by the incident
wave is deﬁned as the scattering wave. Here, ε0 , μ0 and c are the dielectric constant and magnetic
permeability in vacuum, and the speed of light, respectively.
In the following, consider the potential caused by the electric dipole moment of the particle inducedby the incident wave at the point X (position vector r = (r sin θ cos φ, r sin θ sin φ, r cos θ)), which
is located at a large distance from the particle. The polar coordinates are deﬁned as in Fig. 1. The
1 ∂runit vectors at the point X are er = ∂∂rr , eθ = 1r ∂∂θr , eφ = r sinθ ∂φ . The distance r between theorigin O and point X is suﬃciently larger than the wavelength of the incident wave 2πcω0 . Supposethat the electric dipole moment of the particle is expressed by the electric charges +q(t) and −q(t)located at the ﬁxed points 1 and 2 in Fig. 2, respectively. In terms of q(t), the z-component of theelectric dipole moment is described as p(t) = q(t)l = αE0 cos ω0 t, where the distance between theﬁxed points 1 and 2 is l. Here, l is suﬃciently smaller than the wavelength of the incident wave.

z

z

ࢋ௥
ࢋథ

X

θ

ା

ࢋఏ
‫ݎ‬

Point 1
 

଴

O
x

X

y

/2

߶

/2
x

Figure 1



ି

θ

O

y

Point 2
Neutral particle
Figure 2

As shown in Fig. 2, the distances from the points 1 and 2 to point X are r+ and r− , respectively.
The retarded scalar potential ϕ in Lorentz gauge condition is described asϕ(r, t) =−q(t − rc− ) 1+q(t − rc+ ) 1
+
.
4πε0
r+
4πε0
r−

(1)

In the following Questions [1]-[6], assume that the electric polarizability is a time-independent
constant given as α = α0 .
[1] First, consider the particle in the static ﬁeld E0 (ω0 = 0) without retardation. On the condi-

11

tions that l  r, the scalar potential in Eq. (1) is approximated by
α0 E0 cos θ1
1
ϕ(r) =
C· +D· 2 .
4πε0rr
Ignoring the second and higher order terms of rl , calculate C and D.
[2] In Questions [2]-[7], consider the scattering of the electromagnetic ﬁeld with ﬁnite angularfrequency ω0 = 0 satisfying ω0 l  c. The scalar potential with retardation in Eq. (1) isapproximated by11
α0 E0 cos θ
F· +G· 2 .
ϕ(r, t) =4πε0rr
Ignoring the second and higher order terms of rl and ωc0 l , express F and G by using r, c, ω0 ,and t.
[3] At the point X, the vector potential A(r, t) including the retardation is described as follows,A(r, t) =dp(t − rc )1ez .
4πε0 c2 r
dt

Describe the vector potential A(r, t) at point X induced by the oscillating electric dipole
moment of the particle, by using α0 , r and the unit vectors of the polar coordinates. The ez
is a unit vector along the z direction.

Consider the electric ﬁeld and magnetic ﬂux density at point X. Considering the propagation of
the wave to large distances from the particle, we take into account the terms with 1r and ::::::
ignore theignore the terms
terms with r12 in the potential obtained in Question [2]. In Questions [4] and [5], :::::::
of r12 in the calculation.
[4] Describe ∇ϕ(r, t), in which ϕ(r, t) is the scalar potential obtained in Question [2], at point
X by using the unit vectors of the polar coordinates. You may use the following equation for
an arbitrary function f ,
∇f =

∂f
1 ∂f
1 ∂f
er +
eθ +
eφ .
∂r
r ∂θ
r sin θ ∂φ[5] Describe the electric ﬁeld E and magnetic ﬂux density B at point X by using the unit vectorsof the polar coordinates. Here, the time-dependent electric ﬁeld E and magnetic ﬂux density
B are expressed by∂A
E = −∇ϕ −,B = ∇ × A.
∂t
You may use the following formula for an arbitrary vector F = Fr er + Fθ eθ + Fφ eφ ,


1 ∂Fθ
1 ∂
(sin θ Fφ ) −r sin θ ∂θr sin θ ∂φ



∇×F =




1 ∂
1 ∂Fr
1 ∂
1 ∂Fr
er +
−
(rFφ ) eθ +(rFθ ) −eφ .
r sin θ ∂φ
r ∂r
r ∂r
r ∂θ


12

[6] Describe the Poynting vector S, which expresses the energy ﬂux of the electromagnetic wave
scattered from the particle, at the point X by using the unit vectors of the polar coordinates.
Here, the Poynting vector is expressed as
S=

1
E × B.
μ0

Draw the shapes of the amplitude of the Poynting vector |S| with respect to the horizontal
axes θ and φ.
[7] Next, we assume that the electric polarizability α is time-dependent. The oscillations of theelectric polarizability can be induced by vibrations of molecules and lattices. The angularfrequency of the electric polarizability (ωv ) is smaller than that of the incident wave (ω0 ).
The electric polarizability is described as α(t) = α0 + α1 cos ωv t, where α0 , α1 are constants
speciﬁc for the particle. The z-component of the electric dipole moment on the particle isexpressed as p(t) = α(t)E0 cos ω0 t. Describe the electric ﬁeld, which is scattered from theparticle, at point X. Write all angular frequencies of the scattered electromagnetic wave.

13