Ψ(x) = 1 √ 2π ˆ k 0 +∆k/2 k 0 −∆k/2 1 √ ∆k ikx dk = 1 √ 2π ik0x ix √ ∆k ( i∆kx/2 − i∆kx/2 ) = ik 0 x √ 2π sin( ∆kx 2 ) x √ ∆k 2 |Ψ(x)| 2 = ∆k 2π sin 2 ( ∆kx 2 ) ( ∆kx 2 ) 2 , ∆ p∆x = ∆k( 4π ∆k ) = 4π L x = y p z − zp y , L y = zp x − xp z , L z = xp y − yp x [L x , L y ] = [(yp z − zp y ), (zp x − xp z )] = [yp z , zp x ] − [yp z , xp z ] − [zp y , zp x ] + [ zp y , xp z ] = y[ p z , z] p x + x[z, p z ] p y = i (−yp x + xp y ) = i L z .
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8/19/2019 Quantum Physics Mid-term Exam Question
http://slidepdf.com/reader/full/quantum-physics-mid-term-exam-question 1/3
Ψ(x) = 1√
2π
ˆ k0+∆k/2
k0−∆k/2
1√ ∆k
ikxdk
= 1√
2π
ik0x
ix√
∆k(
i∆kx/2 −
i∆kx/2)
=
ik0x
√ 2π
sin(∆kx2 )
x√ ∆k2
|Ψ(x)|2 = ∆k
2π
sin2(∆kx2 )
(∆kx2 )2
,
∆ p∆x = ∆k( 4π∆k ) = 4π
Lx = ypz−zpy, Ly = zpx−xpz, Lz = xpy−ypx
[Lx, Ly] = [(ypz − zpy), (zpx − xpz)]
= [ypz, zpx] − [ypz, xpz] − [zpy, zpx] + [zpy, xpz]
= y[ pz, z] px + x[z, pz] py
= i (−ypx + xpy) = i Lz.
8/19/2019 Quantum Physics Mid-term Exam Question
http://slidepdf.com/reader/full/quantum-physics-mid-term-exam-question 2/3
[Lx, Ly] = i xyzLz
[Li, Lj ] = i ijkLk; {i,j,k} = {x, y, z}.
∆Lx∆Ly ≥ 1
2|[Lx, Ly]|
≥
2 Lz
.
[a, a†] = [(
mω
2 x + i
p√ 2mω
), (
mω
2 x − i
p√ 2mω
)]
= − i
2 [x, p] +
i
2 [ p, x]
= − i(2i )
2 = 1,
a a† H = ω(a†a + 12)
[a, H ] = [a, ω(a†a + 1
2)]
= ω[a, a†a]
= ω([a, a†]a + a†[a, a]) = ωa
[a†, H ] = ω[a†, a†a]
= ω([a†, a†]a + a†[a†, a]) = − ωa†.
H |n = E |n,
[a, H ]|n = ωa|n = aH |n − Ha|n [a†, H ]|n = − ωa†|n = a†H |n − Ha†|n⇔ Ha|n = aE |n − ωa|n ⇔ Ha†|n = a†E |n + ωa†|n
= (E − ω)(a|n) H (a†|n) = (E + ω)(a†|n).
a|n E ω a
a : |n → |n − 1 a† : |n → |n + 1
H = ω(a†a + 12) n|n ≥ 0
ωn|a†a|n = n|(H − 1
2 ω)|n
= (E − 1
2 ω)n|n ≥ 0
E min = 12 ω
E = (n + 12) ω
p = −i ddx x|0 ≡ ψ0(x)
x|a|n = 0 =
mω
2 xx|0 + i
(−i )√ 2mω
d
dxx|0 = 0
⇔ dψ0(x)
ψ0(x) = −mωxdx,
ψo(x) = A
−mωx2/2
8/19/2019 Quantum Physics Mid-term Exam Question
http://slidepdf.com/reader/full/quantum-physics-mid-term-exam-question 3/3
k =
2mE 2
κ =
2m(V 0−E )2
E = 0.5V 0
β = kκ = 1.
T =
1 +
1 + 12
214
12
1 − 1
2
sinh2
2m(0.5V 0)
2 (14 × 10−9)
−1
≈ 0.0098 = 0.98%
T = 50%
1
2 =
1 + 64 sinh2
2m(0.5V 0)
2 L
−1
L =
2
2m(0.5V 0) sinh−1(
1
8) ≈ 1.66
T > 50%)
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