8.05, Quantum Physics II, Fall 2012
TEST
Wednesday October 24, 12:00-1:30pm
You have 90 minutes.
Answer all problems in the white books provided. Write
YOUR NAME and YOUR SECTION on your white
book(s).
There are five questions, totalling 100 points.
None of the problems requires extensive algebra.
No books, notes, or calculators allowed.
TIME MANAGEMENT: I suggest 40 minutes for the
first three questions and 25 minutes for each of the remaining two.
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
2im
∂x
∂x
• Variational principle:
• Spin-1/2 particle:
dx ψ ∗ (x)Hψ(x)
Egs ≤ R
, for all ψ(x)
dxψ ∗ (x)ψ(x)
R
Stern-Gerlach :
µB =
~,
H = −~µ · B
e~
,
2me
~µ = g
~µe = −2 µB
e~ 1 ~
~
S = γS
2m ~
~
S
,
~
1
0
In the basis |1i ≡ |z; +i = |+i =
, |2i ≡ |z; −i = |−i =
0
1
~
Si = σi
2
σx =
0 1
1 0
[σi , σj ] = 2iǫijk σk →
σi σj = δij I + iǫijk σk
; σy =
0 −i
i 0
; σz =
Si , Sj = i~ ǫijk Sk
1 0
0 −1
(ǫ123 = +1)
→ (~σ · ~a)(~σ · ~b) = ~a · ~b I + i~σ · (~a × ~b)
eiMθ = 1 cos θ + iM sin θ , if M2 = 1
~a sin a ,
exp i~a · ~σ = 1 cos a + i~σ ·
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
• Bras and kets: For an operator Ω and a vector v, we write |Ωvi ≡ Ω|vi
Adjoint: hu|Ω† vi = hΩu|vi
|α1 v1 + α2 v2 i = α1 |v1 i + α2 |v2 i ←→ hα1 v1 + α2 v2 | = α1∗ hv1 | + α2∗ hv2 |
• Complete orthonormal basis |ii
hi|ji = δij ,
1=
X
|iihi|
X
Ωij |iihj |
i
Ωij = hi|Ω|ji ↔ Ω =
i,j
hi|Ω† |ji = hj|Ω|ii∗
Ω hermitian: Ω† = Ω,
• Matrix M is normal ([M, M † ] = 0) ←→
U unitary: U † = U −1
unitarily diagonalizable.
˜
• 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
p̂|pi = p|pi ,
hx|pi = √
ipx 1
;
exp
~
2π~
hq|pi = δ(q − p) , 1 = dp |pihp| , p̂† = p̂
Z
Z
1
ipx ψ̃(p) = dxhp|xihx|ψi = √
dx exp −
ψ (x)
~
2π~
~ d n
ψ(x) ;
hx|p̂ |ψi =
i dx
n
Z
d n
hp|x̂ |ψi = i~
ψ̃(p) ;
dp
Z ∞
1
exp(ik x)dx = δ(k)
2π −∞
n
[p̂, f (x̂)] =
~ ′
f (x̂)
i
• Generalized uncertainty principle
(∆A)2 ≡ h(A − hAi)2 i = hA2 i − hAi2
2
1
(∆A) (∆B) ≥ hΨ| [A, B]|Ψi
2i
2
2
∆x ∆p ≥
~
2
1 x2 ~
∆
∆x = √ and ∆p = √
for a gaussian wavefuntion ψ ∼ exp −
2 ∆2
2
2∆
r
Z +∞
π
2
dx exp −ax =
a
−∞
3
Time independent operator Q :
∆H∆t ≥
~
,
2
• Commutator identities
d
i
hQi = h [H, Q]i
dt
~
∆Q
∆t ≡ dhQi dt
[A, BC] = [A, B]C + B[A, C] ,
1
1
eA Be−A = B + [A, B] + [A, [A, B]] + [A, [A, [A, B]]] + . . . ,
3!
2
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
• Harmonic Oscillator
1
1 2 1
2 2
, N̂ = ↠â
Ĥ =
p̂ + mω x̂ = ~ω N̂ +
2
2
2m
r
r
mω
ip̂
mω
ip̂
†
,
x̂ +
, â =
x̂ −
â =
mω
2~
mω
2~
r
r
~
mω~ †
x̂ =
(â + ↠) , p̂ = i
(â − â) ,
2mω
2
ˆ â ] = −â ,
[N,
[x̂, p̂] = i~ , [â, ↠] = 1 ,
ˆ , ↠] = ↠.
[N
1
|ni = √ (a† )n |0i
n!
Ĥ|ni = En |ni = ~ω n +
1
|ni , N̂ |ni = n|ni , hm|ni = δmn
2
√
√
n + 1|n + 1i , â|ni = n|n − 1i .
mω 1/4
mω ψ0 (x) = hx|0i =
exp −
x2 .
π~
2~
↠|ni =
4
1. True or false questions [20 points] No explanations required. Just indicate T or F
for true or false, respectively.
(1) The parity operator, Pˆ |xi = |−xi, is both hermitian and unitary.
(2) For position eigenstates: |−xi = −|xi
(3) Every N × N matrix has N eigenvectors.
(4) For a finite dimensional Hilbert space with N energy levels E1 ≤ E2 ≤ . . . ≤ EN ,
hψ|H|ψi ≤ EN for all normalized trial wavefunctions |ψi.
The following three questions refer to a one-dimensional x ∈ (∞, ∞) Schrödinger
problem with a real, continuous and finite potential V (x) = V (−x) that admits
five bound states.
(5) The highest energy bound state has four nodes.
(6) Let ψ(x) denote a bound state wavefunction. There may exist points x0 (6= ∞)
where both ψ and is spatial derivative ψ ′ vanish.
(7) An excited even bound state can have ψ(x = 0) = 0.
The following three questions refer to the operator
R = exp
i
~
πŜx
where Sˆx is the spin operator in the x direction.
(8) The operator R is unitary.
(9) The operator R is diagonalizable.
(10) R =
2i
Ŝ
~ x
.
2. Expectation values and uncertainty [15 points]
A useful equation relates the action of a Hermitian operator Ω on a normalized state
|Ψi to its expectation value hΩi and its non-zero uncertainty ∆Ω in the state:
Ω|Ψi = hΩi|Ψi + (∆Ω)|Ψ⊥ i .
(1)
Here |Ψ⊥ i is (i) a normalized state, and (ii) is orthogonal to |Ψi. Prove the above
equation by showing that the state |Ψ⊥ i defined by (1) has the two properties we
claim it has.
5
3. A property of complex vector spaces [15 points]
Consider two vectors |ui and |vi in a complex vector space V as well as the linear
operator T : V → V.
(a) Simplify the following expression
i
1h
hu + v|T |u + vi − hu − v |T |u − vi − i hu + iv|T |u + ivi +i hu − iv|T |u − ivi
4
and write your answer in terms of (some or all of) the overlaps hu|T |ui, hu|T |vi, hv|T |ui,
and hv|T |vi.
(b) Use your result in (a) to prove that if hw|T |wi = 0 for all w ∈ V then T is the
zero operator.
[Comment: This is a remarkable property of complex vector spaces that is not true in
real vector spaces. For a two-dimensional real vector space let T be the linear operator
that rotates vectors by 90◦ . Then T is non-zero even though hv|T |vi = 0.]
4. An Anharmonic Oscillator [25 points]
Consider a particle of mass m moving in one-dimension under the influence of an x2n
potential
~2 x2n
,
2m L2n+2
with L a constant with units of length.
V (x) =
n ≥ 2,
(a) Use dimensional analysis to estimate the ground state energy of the system up to
an undetermined dimensionless number.
(b) Consider the trial wavefunction ψ(x) = exp −b2 x2 for a variational analysis of
the ground with b a parameter to be adjusted to obtain the best bound. Consider
also the following integrals
Z ∞
2
dy e−2y y 2k = ck , k = 0, 1, 2, . . .
−∞
and assume the constants ck known. Determine the function F (b) that bounds
the ground state energy as
Egs ≤ F (b)
You do not have to minimize over b, but simplify your result for F , which also
depends on other constants of the problem.
(c) Sketch the potential in the limit of n → ∞. What do you expect the ground state
energy to be? Explain.
(d) One can show (don’t try!) that the value of b for best variational estimate in
√
part (b) goes like bL ∼ n for large n. Sketch the expected ψ and explain if this
is becoming a better or worse representation of the expected ground state as n
becomes larger and larger.
6
5. A Three Dimensional Harmonic Oscillator [25 points]
Consider a particle of mass m confined to a three-dimensional harmonic potential with
rotational symmetry:
3 2
X
p̂i
1
Ĥ =
+ mω 2 x̂2i ,
2m 2
i=1
where [x̂i , p̂j ] = i~δij . The corresponding raising and lowering operators are denoted
ˆi = ↠âi , also with
as â†i and âi , with i = 1, 2, 3, and we have number operators N
i
i = 1, 2, 3. (You are welcome to drop the “hats” from the a’s and a† to save time!)
(a) Using the â†i operators and a common vacuum state |0i for the three âi ’s, explicitly
construct all states in (i) the ground state, (ii) the first excited level, and (iii) the
second excited level. Give their energies and state the degeneracy (The states
need not be normalized.)
We introduce the useful linear combinations
1
âL ≡ √ (â1 + iâ2 ),
2
1
âR ≡ √ (â1 − iâ2 ).
2
ˆ in terms of the number operators
(b) Express H
†
ˆR = ↠a
N̂L = â†L âL , N
R ˆR , N̂3 = â3 â3 .
ˆL , N
ˆR , N
ˆ3 } form a complete set of commuting observables. What
Explain why {N
is the energy of the state |nL , nR , n3 i?
(c) Make the list of states of the ground state and first two excited levels (as in (a))
but using the â†L , â†R , and â†3 operators to build the states.
(d) Consider the angular momentum operator
ˆR − N
ˆL ) .
L̂z ≡ ~ (â†R aˆR − â†L âL ) = ~(N
ˆ and L
ˆ z form a complete set of commuting observShow that [Ĥ , L̂z ] = 0. Do H
ables? If yes, give an argument. If not, give an example of two degenerate states
that are not distinguished.
ˆ+ ≡ L
ˆ x + iL
ˆ y and given by
(e) Now consider the angular momentum operator L
√
L̂+ ≡ 2 ~ ( â†3 âL − â†R â3 ).
ˆ + Hermitian? From the second excited states of the oscillator, one can form
Is L
ˆ z and by L
ˆ + (i.e. L
ˆ z |ψi = 0 and
a unique state |ψi that is killed both by L
L̂+ |ψi = 0). Find it. This is the unique state of the second excited level that has
no angular momentum whatsoever!
7
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8.05 Quantum Physics II
Fall 2013
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