31 I 3

(1) I–1 I–2 I–3 1

(2) I–1

(3) I–2 2 I–2A I–2B I–3 4

I–3A I–3B I–3C I–3D

( )

(4)

(5) 3 3

(6) 7

1

I-1 ( ) (100 )

( )

2 z = 0 z < 0

z > 0

n

~∇ · ~E = 0 (A)

~∇× ~E = −∂ ~B

∂t(B)

~∇× ~B = µ0~J + µ0ε0

∂ ~E

∂t(C)

~∇ · ~B = 0 (D)

~E, ~B ~J µ0, ε0

e m ~v

md~v

dt= e ~E

( )

SI

(1) ~J ∂ ~J/∂t ~E

∂ ~J

∂t= ~E

(2) (A)∼(D) ~B~E

~∇2 ~E = ~E + µ0ε0∂2 ~E

∂t2

n, µ0, ε0, e, m

2

(3) (z > 0) ( ω k′)

c (= 1/√ε0µ0 ) ωp

ω2 = c2k′2 + ω2p

ωp n, µ0, ε0, e, m

(z < 0) +z ~Ei = Re[E0ei(kz−ωt)]~ex

E0 ~ex x

(4) ω > ωp k′ z = 0

( ~Er, ~Et )

(5) (4) ~S⟨

~S⟩

z = +0

(6) ω < ωp z = +0

3

I-2A ( ) (50 )

Ms Mp(� Ms)

F = −GMsMp

r2

(r, θ)

ar =d2r

dt2− r

(

dθ

dt

)2

, aθ =1

r

d

dt

(

r2dθ

dt

)

2 (

=1

2r2dθ

dtaθ = 0

θ r

1

a ε πa2√1− ε2

(1) Tdθ

dtT

(2) r 1 2

dr

dt= A

d

dθ

(

1

r

)

,d2r

dt2= B

d2

dθ2

(

1

r

)

A , B T

(3) r θ r =a(1− ε2)

1 + εcosθar =

kspr2

ksp a, T

Fp = arMp =kspMp

r23

kspFs

Fs =kpsMs

r2kps C

ksp = −GMs kps = −GMp

(4) C

4

I-2B ( ) (50 )

m ω

[a, a†] = 1 a† a

|n〉 = 1√n!(a†)n |0〉 (n = 0, 1, 2, 3, · · · ) |0〉

~ 2π

(1) x =

√

~

2mω(a+ a†) 〈n′| x |n〉 〈n′| x2 |n〉

(2) 4 V1(x) = Ax V2(x) = Bx2 V3(x) =

Cx3 V4(x) = Dx4

|n〉 En

~,m, ω, n A,B,C,D

(3) V1(x) + V2(x)

5

I-3A ( ) (25 )

g M0

0

u

t = 0 t M(t)

v(t)

I-3B ( ) (25 )

k , Pλ(k) = e−λλk/k! ( k = 0, 1, 2, ...)

λ

(1) λ

(2)

1%

6

I-3C ( ) (25 )

H = − ~2

2m

(

∂2

∂x21

+∂2

∂x22

)

+ λ|x1 − x2|

1/2

x1, x2 ∈ R m

~ 2π λ

I-3D ( ) (25 )

~r = (x, y, z) f(~r) = A exp(−β|~r|2)f(~r)

∫

d3~rf(~r) = 1

A

∫ ∞

0

s2 e−as2ds =

√π

4a3/2

|~r| 〈|~r|2〉

7

31 II 3

(1) II–1 II–2 II–3 1

(2) II–1 II–2

(3) II–3 2 II–3A II–3B

( )

(4)

(5) 3 3

(6) 7

1

II-1 ( ) (100 )

m

V N a

(V/N)1/3

kB

(1) a

T

P (T, V,N) CV (T, V,N)

(2) a k

~r1 ~r2

1

2k (|~r1 − ~r2| − a)2

ka2 � kBT

T P (T, V,N) CV (T, V,N)

(2)

2π ~

(3) T0, T1, T2 T0m n T1 m a

T2 m k

~ kB

(4) T � T0

(5) T0 � T1 � T2 (i) T0 � T � T1, (ii) T1 � T � T2,

(iii) T � T2

(6) (300K )5NkB2

ω =√

k/m ωc

ωc

~ = 10−34 m2 kg s−1, kB = 10−23 m2 kg s−2 K−1

2

3

II-2 ( ) (100 )

R C m e(e > 0

) C

C

Φ

C

ψ(σ) ψ(σ+2πR) = ψ(σ)

σ = Rθ C

2π ~

6Φ

σ

θR

C

(1) Φ = 0 ψ(σ)

E

(2) Φ 6= 0 σ Aσ C

Aσ Φ R

(3) Aσ Aσ = 0

σ pσ pσ − eAσ

Aσ C

Aσ

(4) Φ Φ = 0

Φ > 0 Φ = 0

Φ Φ0

(5) Φ Φ = 0 Φ = Φ0

Φ

(

)

4

V (σ) = λ cos

(

2σ

R

)

λ

ν meν(ξ, q) q ξ

d2

dξ2ϕ(ξ) + (a− 2q cos(2ξ))ϕ(ξ) = 0 (A)

meν(ξ + π, q) = eiνπmeν(ξ, q) , meν(ξ, 0) = eiνξ

ν = ν(a, q) a, q

ν meν(ξ, q) me−ν(ξ, q) (A)

(6) ξ = σ/R, ψ(σ) = eikξϕ(ξ)

k (A)

k, a, q Φ,Φ0, λ E

(7) k, a, q (6) ψ(σ) = eikξmeν(ξ, q)

ψ(σ)

ν

(8) 0 < q � 1 q a ν

ν = 1.0 a a = 1.0

a ν 0 < q � 1

λ

(5)

(5)

ν

a

5

II-3A ( ) (50 )

1 [0, L] T (x, t)

∂T

∂t= D

∂2T

∂x2(B)

∂T (x, t)

∂x

∣

∣

∣

∣

x=0

=∂T (x, t)

∂x

∣

∣

∣

∣

x=L

= 0 (C)

(1) [0, L] T (x, t)

φn(x) (n = 0, 1, 2, · · · )

T (x, t) =∞∑

n=0

an(t)φn(x)

T (x, t) (B) (C)

φn(x) (n = 0, 1, 2, · · · ) an(t)

am (m 6= n)

φn(x) an(t)

an(t) an(0)

(2) Tn(t)

Tn(t) =

∫ L

0

dxφn(x)T (x, t)

an(0) 6= 0 Tn(t)/an(0)

(3) t = 0

T (x, 0) = A

[

1−

(

2x

L− 1

)

+1

3

(

2x

L− 1

)3]

A t∫ L

0dx T (x, t)

6

II-3B ( ) (50 )

0 ≤ y ≤ L

D B xy

e

y < 0 y = 0 D

(1) D ρ

ρ

(2) y = 0 y = L x

y = 0, L x1, x2

(3) (2) p

x1, x2 σxp σp/p x1, x2

7

31 III 3

(1) III–1 III–2 III–3 1

(2) III–1

(3) III–2 2 III–2A III–2B III–3

2 III–3A III–3B

( )

(4)

(5) 3 3

(6) 9

1

III-1 ( ) (100 )

a k

3 x

L = 4a

a+ x1 2a+ x2 3a+ x3 x1 x2 x3

M

m ω0 =√

2k/m

(1) m 2 x1 = 0 x3 = 0

t x2(t) t = 0

x2 = X dx2/dt = V

(1) 2

ωs ωc ωl

(2) xj(t)(j = 1, 2, 3) xj(t) =

Aj cosωt ω2s ω2

c ω2l ω0 β = m/M

Aj

(3) β = 1 A2 A3 A1

(4) M ω2s /ω

20 ω2

c/ω20 ω2

l /ω20

β 0 ≤ β ≤ 2 β = 0

β = 1

2

1 M m

M 0 n

x = na xn

M ω

(5) n xn = Re[Bneiωt] Bn = B0b

n B0

b m b

ω ω0 |b| 6= 1

|b| = 1 |b| < 1

M

M < m

(5) n xn = Re[Bneiωt]

(6) M xn |n|

(5) b

M x−1 x+1 b x0

M b

ω ω0 β = m/M

ω m → ∞ ω

(6) M

3

III-2A ( ) (50 )

ε0 SI

(1) e a

r

(2) (1) U0

π e π π

( )

2

I : π a π

π

II: π b 2

π 1

3e 2

3e π

1

3e −1

3e A 2

3e −2

3e

B π

A, B

b π

(3) π π I II

a/b

4

III-2B ( ) (50 )

N i σi = +1

−1 l+ σia ε+ σi∆

l, a, ε, ∆

f

E({σi}) =N∑

i=1

(ε+ σi∆)− fL (A)

L =N∑

i=1

(l + σia)

A A T

kB

(1) T σi = 1

(2) σi σi (σi − σi)2

(3) f f + δf L

L+ δL δf

δL ' kδf (B)

k

(4) (A)

E({σi}) = E0({σi})− fL

(B) k(

L− L)2

E0({σi}) f

5

III-3A ( ) (60 )

The excerpt below has been adapted from David Lee’s 1996 Nobel Prize in Physics

lecture. Read the passage carefully and answer questions [Q-a]-[Q-j] below in

clear, easy-to-read English or symbols where appropriate. The questions can be

answered without advanced physics knowledge.

The major breakthrough in our understanding a

a

a (a) the revolution a superconductivity

in field of vast a (b) congregate a

a

a

a

a

a (c)

macroscopic a (d1) BCS theory

a (d2) metal a

(d3) electrons a (d4) metal a (d5) pairs. a

a (d6)

single ground state a (d7) order parameter a

(d8) Pauli principle a

a

a (e) discrete

diatomic molecules. a

a

a

a

a

a

a

a (f) marching in lock step, a

a (g).

Why do electrons form pairs? (h) a

a

even a (i) one. a

a

a

a

a (j).

6

Answer the following questions concerning the passage above.

[Q-a] The underlined words after (a) have been scrambled. Put them in the correct

order. Change the capitalization and include punctuation as necessary.

[Q-b] Which of the following best describes the meaning of the underlined text

after (b) in this sentence? Write one of b1, b2, b3, or b4 in the answer space.

b1) distribute b2) congress

b3) accumulate b4) concrete

[Q-c] Which of the following words or phrases can replace the underlined text after

(c) without changing the meaning of the sentence? Write one of c1, c2, c3,

or c4 in the answer space.

c1) slow c2) frequent

c3) limited c4) large scale

[Q-d] Fill in the correct article (“the”, “an”, or “a”) in the underlined spaces (d1)-

(d8) in the sentences. Capitalize your answers as necessary and if no article

is needed write “φ” in the answer space.

[Q-e] Which of the following best describes the meaning of the underlined text

after (e)? Write one of e1, e2, e3, or e4 in the answer space.

e1) two atoms e2) individual pairs of atoms

e3) single correlated atoms e4) pairs of two molecules

[Q-f] Which of the following best describes the meaning of the underlined text

after (f)? Write one of f1, f2, f3, or f4 in the answer space.

f1) move together f2) move in discrete steps

f3) flow in lines f4) flow freely

[Q-g] Using several sentences and your own expressions explain why the analogy

in the section of the text ending with (g) is suitable to describe the behavior

of Cooper pairs.

[Q-h] Based on the passage after (h), explain why it is surprising that electrons

form pairs using around 20 or 30 words.

[Q-i] Complete the sentence by providing an appropriate word to fill in the space

marked by (i).

[Q-j] Complete the sentence by providing an appropriate phrase to fill in the space

marked by (j).

7

III-3B ( ) (40 )

Read the passage carefully and answer questions [Q-1]-[Q-6] below. The questions

can be answered without advanced physics knowledge.

Operational amplifiers, often called op-amps, are active linear circuit elements

with a variety of applications in experimental physics. Real op-amps have a large

internal input resistance, Rint, and a single output voltage, Vo. The output is the

product of a large number, which is called the gain G, and the voltage difference

between its two inputs, V+ and V−. An example circuit diagram showing an op-

amp’s internal structure is shown in Figure 1.

The behavior of an ideal op-amp can be summarized with the following two

golden rules. First, no current flows into its inputs. Second, when connected in

a feedback loop, such as the one connecting Cd and the terminal producing Vo in

Figure 2, the op-amp will produce whatever output reduces the voltage difference

between its input terminals to zero.

Figure 1. Op-amp circuit diagram. Figure 2. Charge sensitive amplifier.

The circuit in Figure 2 is an example of a charge sensitive amplifier. Such am-

plifiers are used to amplify the signal from a particle detector. Charge collected in

the detector accumulates in the op-amp’s feedback capacitor, Cf , inducing a volt-

age which in turn causes the op-amp’s output voltage, Vo, to increase sharply. Its

amplitude will be proportional to the integrated charge deposited in the detector

provided the circuit’s time constant, RfCf , is sufficiently longer than the duration

of the input pulse. The output will decay exponentially, eventually returning to

its nominal value if the amplifier receives no other input.

8

Answer the following questions concerning the passage above.

[Q-1] Write an expression for the output voltage Vo of the op-amp in Figure 1 in

terms of its input voltages, V+ and V−, and gain G.

[Q-2] Write an appropriate value for an ideal op-amp’s internal input resistance,

Rint.

[Q-3] Op-amps are powered externally via two additional source inputs, Vs+ and

Vs−, which are drawn as vertical lines connected to opposite sides of the

op-amp symbol between its input and output ends. Draw an example of an

op-amp symbol showing all of its inputs and outputs but without its internal

structure. The positive source, Vs+, should appear on the bottom of the

op-amp symbol.

[Q-4] Make a rough sketch of the output voltage as a function of time for a charge

sensitive amplifier that has received an input signal that is much shorter

than its time constant.

[Q-5] As in [Q-4] make a rough sketch of the output for two short input signals

with the same integrated charge but separated by a time interval about as

long as the time constant of the amplifier.

[Q-6] A voltage sensitive amplifier can be made by replacing the coupling capac-

itor, Cd, and feedback system in Figure 2 with two resistors, RA and RB,

respectively. Draw an example of such an amplifier.

9