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8/9/2019 Joseph Son

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8/9/2019 Joseph Son


Using ψi = √n i eiθ i we nd

ihn 1

2√n 1eiθ 1 + iθ1 √n 1 eiθ 1 = −

ne 2

C √n 1 eiθ 1 −

E J

n 0

√n 2 eiθ 2 , (3)

ihn 2

2√n 2

eiθ 2 + iθ2 √n 2 eiθ 2 =ne 2

C

√n 2 eiθ 2

−

E J

n 0

√n 1 eiθ 1 , (4)

or

ihn 1

2 −hn 1 θ1 = −ne 2

C n 1 −

E J

n 0

√n 1 n 2 eiφ , (5)

ihn 2

2 −hn 2 θ2 =ne 2

C n 2 −

E J

n 0

√n 1 n 2 e− iφ , (6)

Taking real and imaginary parts,

θ1 =ne 2

hC +

E J

hn 0

n 2

n 1

cos φ, (7)

n 1 = −E J

hn 0

√n 1 n 2 sin φ, (8)

θ2 = −ne 2

hC +

E J

hn 0 n 1

n 2cos φ, (9)

n 2 =E J

hn 0

√n 1 n 2 sin φ. (10)

Taking differences, we nd the equations for n and φ,

φ = θ2 −θ1 = −2ne 2

hC −E J

hn 0 n 2

n 1 − n 1

n 2 cos φ ≈ −2ne 2

hC , (11)

n = n 2 −n 1 = +2E J

hn 0

√n 1 n 2 sin φ ≈E J

hsin φ, (12)

noting that n 1 ≈ n 2 ≈ n 0 / 2. Taking the sums, we nd that n 0 = constant, and anamusing but not very relevant equation for the time dependence of θ1 + θ2 .

2. We identify a pair (electrical) current from side 1 to side 2 as

J s = ( −2e)n2

= −eE J

hsin φ ≡J 0 sin φ, (13)

using eq. (12), where the maximum current is

J 0 =eE J

h=

2πeE J

h=

πE J

ϕ0, (14)

where ϕ0 = h/ 2e is the ux quantum.

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8/9/2019 Joseph Son


3. To exhibit oscillatory behavior we use eq. (12) in the derivative of eq. (11) to nd

φ ≈ −2e2 E J

h2 C sin φ, (15)

If E J is positive, then there are oscillations about φ = 0 whose angular frequency is

given byωJ = 2e2 E J

h2 C (16)

for small amplitudes.If E J is negative, then there are oscillations about φ = π, since sin(π −φ) = sin φ whiled2 (π −φ)/dt 2 = −φ. The frequency of oscillation is again given by eq. (16), now using

|E J |.The form of eq. (15) suggests that −E J sin φ be considered as a generalized force withrespect to coordinate φ. That is

F φ ∝−E J sin φ = −∂U ∂φ , (17)

so thatU ∝−E J cos φ + ... (18)

This further suggests that we reconsider the system in terms of coordinates n and φ.A Hamiltonian in terms of these coordinates would include the electrostatic energy(ne)2 / 2C as well as the tunneling energy −E J cos φ. We recall that E J /n 0 was theamplitude for one out of the total of n 0 pairs to tunnel across the junction, so E J isthe normalized tunneling energy for the whole system. Then, a suitable Hamiltonianis

H = (ne)2

2C −E J cos φ. (19)

We could then deduce the equations of motion for n and φ from H via

n =ih

[H, n ] =ih

∂H ∂φ

[φ, n], φ =ih

[H, φ] =ih

∂H ∂n

[n, φ], (20)

provided we know the commutation relation [ φ, n]. Working backwards (or otherwise?)we nd that we need

[n, φ] = i. (21)

Thus,

n = ih

(E J sin φ)(−i) = E J sin φh

, (22)

which agrees with eq. (12), and

φ =ih

ne 2

C (i) = −

ne 2

hC (23)

which more or less agrees with eq. (11).

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8/9/2019 Joseph Son


4. If a voltage V = V 1 −V 2 is applied across the junction, we expect charge Q1 = V C =(−2e)(−n/ 2) = en to be held on side 1, and the negative of this on side 2. Then,eq. (1) becomes

φ ≈ −2eV

h ≡ −ω , (24)

whereω = 2eV

h. (25)

Equation (24) integrates to φ = −ωt.

The battery holds the charge difference across the junction xed at en = V C , but canbe a source of sink of charge such that a current can ow in the circuit. The claim isthat in the present case, the current is given by eq. (13), so

J s = −J 0 sin ωt. (26)

A possible argument (Feynman Red Book III, p. 21-15) is that as φ moves off zero,

eq. (13) describes the resulting current across the junction in an isolated system. Fora system hooked to a battery, this current ows through the entire system, while thenumbers of pairs n 1 and n 2 remain xed at n/ 2 = V C/ 2e.

Accepting this argument, the DC voltage of the battery results in an AC current inthe circuit of angular frequency (25).

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