8.05 Quantum Physics II, Fall 2011 FINAL EXAM You have 3 hours.

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8.05 Quantum Physics II, Fall 2011
FINAL EXAM
Thursday December 22, 9:00 am -12:00
You have 3 hours.
Answer all problems in the white books provided. Write
YOUR NAME and YOUR SECTION on your white
book(s).
There are seven questions, totalling 100 points.
None of the problems requires extensive algebra.
No books, notes, or calculators allowed.
1
Formula Sheet
• Conservation of probability
∂
∂
ρ(x, t) +
J(x, t) = 0
∂t
∂x
~
∂ ∗
2
∗ ∂
ρ(x, t) = |ψ(x, t)| ; J(x, t) =
ψ
ψ−ψ ψ
∂x
2im
∂x
• Variational principle:
Z
Z
∗
Egs ≤ dx ψ (x)Hψ(x) , for all ψ(x) satisfying
dxψ ∗ (x)ψ(x) = 1
• Spin-1/2 particle:
Stern-Gerlach :
µB =
~,
H = −~µ · B
e~
,
2me
~µ = g
~µe = −2 µB
e~ 1 ~
~
S = γS
2m ~
~
S
,
~
1
0
, |2i ≡ |z; −i = |−i =
In the basis |1i ≡ |z; +i = |+i =
0
1
~
Si = σi
2
σx =
0 1
1 0
; σy =
[σi , σj ] = 2iijk σk →
σi σj = δij I + iijk σk
0 −i
i 0
; σz =
Si , Sj = i~ ijk Sk
1 0
0 −1
(123 = +1)
→ (~σ · ~a)(~σ · ~b) = ~a · ~b I + i~σ · (~a × ~b)
eiMθ = 1 cos θ + iM sin θ , if M2 = 1
exp i~a · ~σ = 1 cos a + i~σ ·
~a a
sin a ,
a = |~a|
exp(iθσ3 ) σ1 exp(−iθσ3 ) = σ1 cos(2θ) − σ2 sin(2θ)
exp(iθσ3 ) σ2 exp(−iθσ3 ) = σ2 cos(2θ) + σ1 sin(2θ) .
~ = nx Sx + ny Sy + nz Sz = ~ ~n · ~σ .
S~n = ~n · S
2
(nx , ny , nz ) = (sin θ cos φ, sin θ sin φ, cos θ) ,
|~n; +i =
cos(θ/2)|+i +
S~n |~n; ±i = ±
sin(θ/2) exp(iφ)|−i
|~n; −i = − sin(θ/2) exp(−iφ)|+i + cos(θ/2)|−i
2
~
|n;
~ ±i
2
• Complete orthonormal basis |ii
hi|ji = δij ,
X
1=
|iihi|
i
Oij = hi|O|ji ↔ O =
X
Oij |iihj|
i,j
hi|A† |ji = hj|A|ii∗
hermitian operator: O† = O, unitary operator: U † = U −1
˜
• Position and momentum representations: ψ(x) = hx|ψi ; ψ(p)
= hp|ψi ;
Z
x̂|xi = x|xi , hx|yi = δ(x − y) , 1 = dx |xihx| , x̂† = x̂
Z
p̂|pi = p|pi ,
hx|pi = √
1
ipx exp
;
~
2π~
hx|p̂n |ψi =
hq|pi = δ(q − p) , 1 = dp |pihp| , p̂† = p̂
Z
Z
1
ipx ψ (x)
ψ̃(p) = dxhp|xihx|ψi = √
dx exp −
~
2π~
~ d n
ψ(x) ;
i dx
d n
hp|x̂n |ψi = i~
ψ̃(p) ;
dp
Z ∞
1
exp(ikx)dx = δ(k)
2π −∞
[p̂, f (x̂)] =
~ 0
f (x̂)
i
• Generalized uncertainty principle
(∆A)2 ≡ h(A − hAi)2 i = hA2 i − hAi2
2
1
(∆A)2 (∆B)2 ≥ hΨ| [A, B]|Ψi
2i
∆x ∆p ≥
~
2
1 x2 ∆
~
∆x = √ and ∆p = √
for a gaussian wavefuntion ψ ∼ exp −
2 ∆2
2
2∆
Z
+∞
dx exp −ax
2
r
=
−∞
∆H∆t ≥
~
,
2
3
π
a
∆Q
∆t ≡ dhQi dt
• Commutator identities
[A, BC] = [A, B]C + B[A, C] ,
1
1
eA Be−A = B + [A, B] + [A, [A, B]] + [A, [A, [A, B]]] + . . . ,
2
3!
eA Be−A = B + [A, B] ,
if [[A, B], A] = 0 ,
[ B , eA ] = [ B , A ]eA ,
if [[A, B], A] = 0
1
1
eA+B = eA eB e− 2 [A,B] = eB eA e 2 [A,B] ,
if [A, B] commutes with A and with B
• Time evolution
|Ψ, ti = U (t, 0)|Ψ, 0i ,
U unitary
U (t, t) = 1 , U (t2 , t1 )U (t1 , t0 ) = U (t2 , t0 ) , U (t1 , t2 ) = U † (t2 , t1 )
i~
d
d
ˆ
ˆ
|Ψ, ti = H(t)|Ψ,
ti ↔ i~ U (t, t0 ) = H(t)U
(t, t0 )
dt
dt
h i
i X i
ˆ
Time independent H: U (t, t0 ) = exp − Ĥ(t − t0 ) =
e− ~ En (t−t0 ) |nihn|
~
n
h i Z t1
i
0
0
ˆ
ˆ
[ H(t1 ) , H(t2 ) ] = 0, ∀ t1 , t2 , U (t, t0 ) = exp −
dt Ĥ(t )
~ t0
hAi = hΨ, t|AS |Ψ, ti = hΨ, 0|AH (t)|Ψ, 0i
[AS , BS ] = CS
i~
→
→
AH (t) = U † (t, 0)AS U (t, 0)
[AH (t) , BH (t) ] = CH (t)
d ˆ
ˆ H (t)] , for AS time-independent
AH (t) = [AˆH (t), H
dt
• Two state systems
~ ·B
~ → spin vector ~n precesses with Larmor frequency ω
~
H = −γ S
~ = −γB
~ = B0~ez + B1 cos ωt ~ex − sin ωt ~ey
NMR magnetic field B
i
i
~ eff · S
~ t |ψ(0)i
|ψ(t)i = exp ωtSz exp γB
~
~
~ eff = B0 1 − ω ~ez + B1 ~ex
B
ω0
4
• Harmonic Oscillator
1 2 1
1
2 2
ˆ = ↠â
ˆ+
Ĥ =
p̂ + mω x̂ = ~ω N
, N
2m
2
2
r
r
ip̂
ip̂
mω
mω
†
x̂ +
, â =
x̂ −
,
â =
2~
mω
2~
mω
r
r
~
mω~ †
†
x̂ =
(â + â ) , p̂ = i
(â − â) ,
2
2mω
[x,
ˆ p̂] = i~ , [a,
ˆ ↠] = 1 .
1
|ni = √ (a† )n |0i
n!
Ĥ|ni = En |ni = ~ω n +
1
ˆ |ni = n|ni , hm|ni = δmn
|ni , N
2
√
√
n + 1|n + 1i , â|ni = n|n − 1i .
mω 1/4
mω ψ0 (x) = hx|0i =
exp −
x2 .
π~
2~
p̂
sin ωt
xH (t) = x̂ cos ωt +
mω
pH (t) = p̂ cos ωt − mω x̂ sin ωt
↠|ni =
• Coherent states and squeezed states
i
Tx0 ≡ e− ~ p̂ x0 ,
Tx0 |xi = |x + x0 i
i
ˆ 0
|0i ,
|x̃0 i ≡ Tx0 |0i = e− ~ px
− 41
|x̃0 i = e
αa† −α∗ a
|αi ≡ D(α)|0i = e
† −α∗ a
x2
0
d2
e
x
√ 0 a†
2d
|0i ,
1
†
∗
D(α) ≡ exp αa − α a ,
2
~
mω
hx̂i
hp̂i d
α= √
+ i√
∈C
2d
2~
|0i , hx|x0 i = ψ0 (x − x0 ) ,
d2 =
†
â|αi = α|αi , |α, ti = e−iωt/2 |e−iωt αi
γ
|0γ i = S(γ)|0i ,
S(γ) = exp − (a† a† − aa) , γ ∈ R
2
1
1
|0γ i = √
exp − tanh γ a† a† |0i
2
cosh γ
|αi = eαa
|0i = e− 2 |α| eα a |0i ,
S † (γ) aS(γ) = cosh γ a − sinh γ a† ,
D† (α) aD(α) = a + α
|α, γi ≡ D(α)S(γ)|0i
5
• Orbital angular momentum operators
L̂i = ijk x̂j p̂k
1
∂2
2 ∂
+ 2
∇ = 2+
∂r
r ∂r r
2
2
L̂ = −~
2
ˆz = ~ ∂ ;
L
i ∂φ
∂2
∂
1 ∂2
+ cot θ
+
∂θ2
∂θ sin2 θ ∂φ2
∂2
1 ∂2
∂
+
+ cot θ
∂θ2
∂θ sin2 θ ∂φ2
∂
∂
±iφ
± + i cot θ
L̂± = ~e
∂θ
∂φ
• Spherical Harmonics
Y`,m (θ, φ) ≡ hθ, φ|`, mi
r
r
3
3
1
Y0,0 (θ, φ) = √
; Y1,±1 (θ, φ) = ∓
sin θ exp(±iφ) ; Y1,0 (θ, φ) =
cos θ
8π
4π
4π
r
r
15
15
2
sin θ exp(±2iφ) ; Y2,±1 (θ, φ) = ∓
sin θ cos θ exp(±iφ) ;
Y2,±2 (θ, φ) =
32π
8π
r
5
Y2,0 (θ, φ) =
3 cos2 θ − 1
16π
• Algebra of angular momentum operators J~ (orbital or spin, or sum)
[Ji , Jj ] = i~ijk Jk ;
J 2 |jmi = ~2 j(j + 1)|jmi ;
J± = Jx ± iJy ,
Jz |jmi = ~m|jmi , m = −j, . . . , j .
(J± )† = J∓
[Jz , J± ] = ±~ J± ,
[ J 2 , Ji ] = 0
1
1
Jx = (J+ + J− ) , Jy = (J+ − J− )
2
2i
[J 2 , J± ] = 0 ;
[J+ , J− ] = 2~ Jz
J 2 = J+ J− + Jz2 − ~Jz = J− J+ + Jz2 + ~Jz
J± |jmi = ~
p
j(j + 1) − m(m ± 1) |j, m ± 1i
6
• Radial equation
~2 d2
~2 `(` + 1) + V (r) +
−
uν` (r) = Eν` uν` (r) (bound states)
2m dr2
2mr2
uν` (r) ∼ r`+1 , as r → 0 .
√
2mE
π
E > 0 and lim V (r) = 0 , lim u` (r) = sin kr − ` + δ` (E) , k =
r→∞
r→∞
2
~
• Hydrogen atom
En = −
e2 1
,
2a0 n2
n = 1, 2, . . . ,
un` (r)
Y`,m (θ, φ)
r
` = 0, 1, . . . , n − 1 ,
2
a0 =
ψn,`,m (~x) =
m = −`, . . . , `
2
~
,
me2
α=
u1,0 (r) =
2r
3/2
a0
1
e
'
,
~c
137
~c ' 200 MeV–fm
exp(−r/a0 )
2r
r
u2,0 (r) =
1−
exp(−r/2a0 )
(2a0 )3/2
2a0
1
1
r2
exp(−r/2a0 )
u2,1 (r) = √
3 (2a0 )3/2 a0
• Factorization method
A=
d
+ W (x) ,
dx
A† = −
d
+ W (x)
dx
d2
+ W2 − W0
dx2
d2
H (2) = AA† = − 2 + W 2 + W 0
dx
Z x
(1)
φ0 = N exp −
W (x0 )dx0
H (1) = A† A = −
7
• Addition of Angular Momentum J~ = J~1 + J~2
Uncoupled basis : |j1 j2 ; m1 m2 i CSCO : {J12 , J22 , J1z , J2z }
Coupled basis :
|j1 j2 ; jmi
CSCO : {J12 , J22 , J 2 , Jz }
j1 ⊗ j2 = (j1 + j2 ) ⊕ (j1 + j2 − 1) ⊕ . . . ⊕ |j1 − j2 |
|j1 j2 ; jmi =
X
m1 +m2 =m
|j1 j2 ; m1 m2 i hj1 j2 ; m1 m2 |j1 j2 ; jmi
|
{z
}
Clebsch−Gordan coefficient
1
J~1 · J~2 = (J1+ J2− + J1− J2+ ) + J1z J2z
2
Combining two spin 1/2 :
| 1, 1 i = | ↑↑i ,
1
| 1, 0 i = √ (| ↑↓i + | ↓↑i) ,
2
|1, −1i = | ↓↓i .
8
1 1
⊗ =1⊕0
2 2
1
|0, 0i = √ (| ↑↓i − | ↓↑i)
2
1. True or false questions [20 points] No explanations required. Just indicate T or F
for true or false, respectively.
(1) The anti-commutator of two hermitian operators is hermitian.
(2) The Heisenberg Hamiltonian and the Schrödinger Hamiltonian are equal if the
Schrödinger Hamiltonians at different times commute.
(3) The length scale a0 · α is the classical electron radius (a0 is the Bohr radius and
α the fine structure constant).
(4) The length scale a0 · α2 is the Compton wavelength of the electron.
(5) In the factorization method either H (1) or H (2) must have a normalizable zero
energy eigenstate.
(6) If J~1 and J~2 are sets of operators each satisfying the algebra of angular momentum
the combination J~1 − J~2 also satisfies the algebra of angular momentum.
(7) Let J~1 and J~2 be angular momentum operators. The operator J~1 · J~2 commutes
with J1x + J2x .
(8) The traces of Jx , Jy , and Jz are all zero for any representation (j = 1/2, 1, 3/2, . . .).
(9) The states on the first excited level of the spherically symmetric harmonic oscillator (potential V (r) = 21 mωr2 ) fit into an ` = 1 multiplet of angular momentum.
(10) The states on the second excited level of the spherically symmetric harmonic
oscillator fit into an ` = 2 multiplet of angular momentum.
2. A short problem on Harmonic Oscillator
[10 points]
Associated with the annihilation and creation (Schrödinger) operators â and ↠there
are Heisenberg operators â(t) and ↠(t). Calculate these time-dependent Heisenberg
operators in terms of a,
ˆ ↠and other physical constants. How is the Heisenberg counˆH (t) of the number operator N
ˆ related to N
ˆ?
terpart N
9
3. Deriving and testing an inequality [15 points]
(a) Let u denote a Hermitian operator whose eigenvalues are all non-negative so that
the following expectation value is non-negative:
D1
E
2
(u − hui) ≥ 0 .
u
Use the above to prove an inequality relating h u1 i and
1
.
hui
(b) Verify your inequality for the operator u = r and the ground state of the hydrogen
atom by direct calculation of hri and h 1r i. (The second can be obtained without
doing integrals by using the virial theorem which tells you that the hT i = − 12 hV i,
where T and V are, respectively, the kinetic and potential energies).
R∞
Possibly useful integral: 0 dx xn e−x = n! for n = 1, 2, 3, . . .
4. A curious rewriting of the Hydrogen Hamiltonian
[15 points]
Consider the hydrogen atom Hamiltonian
H =
p~2
e2
− .
2m
r
We will write it as
3
1 X
x̂i x̂i H =
p̂i + iβ
p̂i − iβ
+γ,
2m i=1
r
r
where p̂i and x̂i are, respectively, the Cartesian components of the momentum and
position operators, and β and γ are real constants to be adjusted so that the two
Hamiltonians are the same.
(a) Calculate the constants β and γ.
(b) Explain carefully why for any state hHi ≥ γ.
(c) Find the wavefunction of the state for which the above energy inequality is saturated. You may assume that this wavefunction just depends on the radial coordinate.
10
5. Hamiltonian for three spin-1 particles [15 points]
Consider 3 distinguishable spin-1 particles, called 1,2, and 3, with spin operators
~1 , S
~2 and S
~3 , respectively. The spins are placed along a circle and the interactions are
S
between nearest neighbors. The hamiltonian takes the form
∆ ~ ~
~ ~
~ ~
H= 2 S
1 · S2 + S 2 · S 3 + S3 · S1 ,
~
with ∆ > 0 a constant with units of energy. For this problem it is useful to consider
~=S
~1 + S
~2 + S
~3 .
the total spin operator S
(a) What is the dimensionality of the state space of the three combined particles.
Write the Hamiltonian in terms of squares of spin operators.
(b) Determine the energy eigenvalues for H and the degeneracies of these eigenvalues.
(c) Calculate the ground state, expressing it as a superposition of states of the form
|m1 , m2 , m3 i ≡ |1, m1 i ⊗ |1, m2 i ⊗ |1, m3 i ,
where ~mi is the eigenvalue of (Sz )i and applying some suitable constraint. [Hint:
The general superposition with arbitrary coefficients has 7 candidate states. Show
that the coefficient of |0, 0, 0i is zero and determine all others. Write your answer
as a normalized state.]
6. Factorization method structure [10 points]
In solving a central potential problem we need to find the spectrum of the set of
Hamiltonians H` , with ` ≥ 0.
Assume that with some superpotential W` we suceed in constructing Hamiltonians
(1)
(2)
H` and H` such that
(1)
H` = H` + α` ,
with some constants α` assumed known and
(2)
H`
(1)
= H`+1 + β` ,
with some constants β` also assumed known. Since we know the superpotential, the
operators A` and A†` are determined as well.
(1)
Finally, assume that H`
(1)
has a zero-energy solution φ0,` , known for each `.
(a) Write expressions for the three lowest energy eigenstates φn,` of H` (n = 0, 1, 2)
and give their energies En,` . Your expressions for the states should use the zero
energy wavefunctions of H (1) and the A or A† operators. Your expressions for the
energy should use the constants α and β introduced above.
(b) Generalize to write an expression for the energy eigenstate φn,` of H` with arbitrary n ≥ 0 and find its energy En,` .
11
7. Time dependent uncertainty on a squeezed vacuum [15 points]
Recall the definition of the squeezed vacuum state
γ
|0γ i = S(γ)|0i , with S(γ) = exp − (↠↠− ââ) , γ ∈ C .
2
(a) Calculate the action of the squeezing operator S on the position and momentum
operators:
S † (γ) x̂ S(γ) = . . .
S † (γ) p̂ S(γ) = . . .
Write your answers in terms of x̂, p̂, and simple functions of γ.
(b) Let |0γ , ti denote the state that results from time evolution of the squeezed vacuum |0γ i at time zero. Determine the time dependent uncertainty (∆x(t))2 on
the state. Your answer should be of the form
(∆x(t))2 =
~
G(γ, ωt) ,
2mω
where G(γ, ωt), to be determined, is a unit-free function of γ and ωt.
[A little help with the algebra: h0|(x̂p̂ + p̂x̂)|0i = 0. ]
12
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8.05 Quantum Physics II
Fall 2013
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