28 Sep 2009 ... PHYS851 Quantum Mechanics I, Fall 2009. HOMEWORK ASSIGNMENT 3:
Solutions. Fundamentals of Quantum Mechanics. 1. [10pts] The ...
PHYS851 Quantum Mechanics I, Fall 2009
HOMEWORK ASSIGNMENT 3: Solutions Fundamentals of Quantum Mechanics 1. [10pts] The trace of an operator is defined as T r{A} = set.
P
m hm|A|mi,
where {|mi} is a suitable basis
(a) Prove that the trace is independent of the choice of basis. Answer: Let {|mi} and {|em i} Pbe two independent P basis sets for our Hilbert space. We must show that m hem |A|em i = m hm|A|mi. Proof: X hem |A|em i = m
=
X
mm′ m′′
X
mm′ m′′
=
X
m′ m′′
=
X
m′ m′′
=
X m′
=
X m
hem |m′ ihm′ |A|m′′ ihm′′ |em i
(1)
hm′′ |em ihem |m′ ihm′ |A|m′′ i
(2)
hm′′ |m′ ihm′ |A|m′′ i
(3)
δm′ m′′ hm′ |A|m′′ i
(4)
hm′ |A|m′ i
(5)
hm|A|mi
(6)
(b) Prove the linearity of the trace operation by proving T r{aA + bB} = aT r{A} + bT r{B}. Answer:
T r{aA + bB} =
X hm|aA + bB|mi
(7)
m
=
X
(ahm|A|mi + bhm|B|mi)
(8)
X X hm|A|mi + b hm|B|mi
(9)
m
= a
m
m
= a T r{A} + b T r{B}
1
(10)
(c) Prove the cyclic property of the trace by proving T r{ABC} = T r{BCA} = T r{CAB}. Answer: First, if T r{ABC} = T r{BCA} then it follows that T r{BCA} = T r{CAB}, so we need only prove the first identity. T r{ABC} =
X hm|ABC|mi
(11)
m
=
X
mm′ m′′
=
X
mm′ m′′
hm|A|m′ ihm′ |B|m′′ ihm′′ |C|mi
(12)
hm′′ |C|mihm|A|m′ ihm′ |B|m′′ i
(13)
= T r{CAB}
2
(14)
2. Consider the system with three physical states {|1i, |2i, |3i}. In this basis, the Hamiltonian matrix is: 1 2i 1 (15) H = −2i 2 −2i 1 2i 1 Find the eigenvalues {ω1 , ω2 , ω3 } and eigenvectors {|ω1 i, |ω2 i, |ω3 i} of H. Assume that the initial state of the system is |ψ(0)i = |1i. Find the three components h1|ψ(t)i, h2|ψ(t)i, and h3|ψ(t)i. Give all of your answers in proper Dirac notation. Answer: The eigenvalues are solutions to det |H − ~ωI| = 0
(16)
Taking the determinate in Mathematica gives 4ω + 4ω 2 − ω 3 = 0
(17)
ω(ω 2 − 4ω − 4) = 0
(18)
which factorizes as which has as its solutions ω1 = 2(1 −
√ 2)
ω2 = 0
√ ω3 = 2(1 + 2)
(19) (20) (21)
the corresponding eigenvectors are √ 1 |ω1 i = (|1i + 2i|2i + |3i) 2 1 |ω2 i = √ (−|1i + |3i) 2 √ 1 |ω3 i = (|1i − 2i|2i + |3i) 2
(22) (23) (24)
The components of |ψ(t)i are found via |ψ(t)i = e−iHt |ψ(0)|i, giving √ √ 1 2 + e−i2(1− 2)t + e−i2(1+ 2)t 4 √ √ i h2|ψ(t)i = √ e−i2(1− 2)t − e−i2(1+ 2)t 2 2 √ √ 1 h3|ψ(t)i = −2 + e−i2(1− 2)t + e−i2(1+ 2)t 4
h1|ψ(t)i =
The mathematic script I used to work this problem is on the following page:
3
(25) (26) (27)
4
3. Cohen-Tannoudji: pp 203-206: problems 2.2, 2.6, 2.7 2.2 (a) The operator σy is hermitian: † ∗ 0 −i 0 i 0 −i † σy = = = = σy i 0 −i 0 i 0 We find the eigenvalues via det |σy − ωI| = 0: −ω −i det i −ω
= ω2 − 1 = 0
(28)
(29)
The solutions are ω = 1 and ω = −1. Let the corresponding eigenvectors be |+i and |−i, so that σy |±i = ±|±i.
(30)
Hit this equation with the bra h1| and insert the projector onto the {|1i, |2i} basis: (−h1|σy |1i ± 1) h1|±i − h1|σy |2ih2|±i = 0
(31)
inserting the values of the matrix elements of σy then gives: ±h1|±i + ih2|±i = 0
(32)
a non-normalized solution is then h1|±i = i
h2|±i = ∓1
(33) (34)
the normalized eigenvectors are then given, up to arbitrary overall phase-factors, by |+i = |−i =
1 √ (i|1i − |2i) 2 1 √ (i|1i + |2i) 2
(35) (36)
(b) The projectors are given by I± = |±ih±|. In matrix form, in the {|1i, |2i} basis, these are h1|±ih±|1i h1|±ih±|2i I± = (37) h2|±ih±|1i h2|±ih±|2i ! i ∓1 i −i =
=
√ √ √ √ 2 2 2 2 ∓1 −i ∓1 √ ∓1 √ √ √ 2 2 2 2 1 i ∓ 2 2 1 ± 2i 2
2 = I and I + I = I: we need to show that I± ± + − 1 ∓i 1 ∓i 2 2 2 2 2 I± = ±i ±i 1 1 2 2 21 12 i i + ∓ ∓ 4 4 4 4 = 1 1 + ± 4i ± 4i 4 4 1 i ∓ 2 2 = 1 ± 2i 2 = I±
5
(38)
(39)
(40) (41) (42) (43)
I+ + I− =
1 2
+ 2i
− 2i 1 2
+ 21 − 2i 1 0 = 0 1 = I =
1 2
+ 2i
(c) The results for M and Ly are attached.
6
+
1 2
− 2i i
− 2i + 2 1 1 2 + 2
+ 2i 1 2
(44) (45) (46) (47)
2_2M.nb
1 H* Enter the matrix M *L M = 882, I Sqrt@2D
