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Unformatted text preview: PHYS851 Quantum Mechanics I, Fall 2009 HOMEWORK ASSIGNMENT 13: Solutions 1. In this problem you will derive the 2×2 matrix representations of the three spin observables from
ﬁrst principles:
(a) In the basis { ↑z ,  ↓z }, the matrix representation of Sz is of course
Sz = ↑z Sz  ↑z
↓z Sz  ↑z ↑z Sz  ↓z
↓z Sz  ↓z . (1) Use Eqs. (1) and (2) to ﬁnd the four matrix elements of Sz in the basis of its own eigenstates.
From Sz  ↑z = 2  ↑z and the orthonormality of the basis, it follows that ↑z Sz  ↑z = 2
and ↓z Sz  ↑z = 0.
From Sz  ↓z = − 2  ↓z and the orthonormality of the basis, it follows that ↑z Sz  ↓z = 0 and
↓z Sz  ↓z = − 2 .
This gives us
10
(2)
Sz =
0 −1
(b) Invert the deﬁnitions S+ = Sx + iSy and S− = Sx − iSy , to express Sx and Sy in terms of S+
and S− .
Inverting these equations gives
Sx =
Sy = 1
(S+ + Sz )
2
1
(S+ − S− )
2i (3)
(4) (c) Use the equation
S± s, ms = s(s+1) − ms (ms ±1)s, ms ±1 (5) to ﬁnd the matrix elements of S+ and S− in the basis { ↑z ,  ↓z .
This formula gives S+  ↑z = 0, S+  ↓z =  ↑z , so that orthonormality gives
S+ = 01
00 (6) Likewise, S−  ↑z =  ↓z and S−  ↓z = 0, so that
S− = 1 00
10 (7) (d) From your answers to 13.1.b and 13.1.c, derive the matrix representations of Sx and Sy for
spin1/2. 1
01
(S+ + S− ) =
2
2 10
1
01
( S+ − S− ) =
2i
2i −1 0 Sx =
Sy = (8)
= 2 0 −i
i0 (9) (e) Explicitly verify that these operators satisfy the angular momentum commutation relations. 2 [Sx , Sy ] = 4
2 = 4
2 =i 2
= i Sz
2 [Sy , Sz ] = 2 2
= i Sx
2 4
2 = 4
2 =i i0
0 −i − 2
= i Sy − 0 −i
i0 01
10 −i 0
0i 10
0 −1 0i
i0 4 =i [Sz , Sx ] = 0 −i
i0 (10) 0 −i
i0 4
2 = 01
10 10
0 −1 − 0 −i
−i 0 − 01
10 10
0 −1 0 −i
i0 (11)
10
0 −1 01
10 01
−1 0 − − 0 −1
10 01
10 10
0 −1 0 −i
i0 (12) 2 (f) Show explicitly that S 2 = 2 s(s+1)I . 2
2
2
S 2 = Sx + Sy + Sz
2 = 4
2 = 4 = 32
4 = 2 (13)
01
10 01
10
10
01 2 + 4 2 +
10
01 0 −i
i0 4 2 + 4 0 −i
i0 10
01 2 + 4 10
0 −1 2 10
01 s(s + 1)I (14) (g) Based on symmetry, write the 2×2 matrix representations of Sx , Sy , and Sz in the basis of
eigenstates of Sy .
A cyclic permutation (relabeling) of the indices leaves the commutation relations unchanged.
Thus we can relabel the indices according to x′ = y , y ′ = z m and z ′ = x. Thus the eigenstates
of y ′ are just the eigenstates of z , in terms of the primed indices we have
Sx′ = Sy = 0 −i
i0 (15) = Sz = 10
0 −1 (16) Sy ′ Sz ′ = Sx = 01
10 (17) Note that the eigenstates of Sy are only deﬁned up to a phasefactor, diﬀerent phase factor
choices will lead to diﬀerent sets of matrices. The above result, which reﬂects the permutation
symmetry of the angular momentum group, will be generated by setting the phase factors so
that
 ↑y =  ↓y = where φ is an arbitrary phase. 3 eiφ
√ ( ↑z + i ↓z )
2
iφ
e
√ ( ↑z − i ↓z )
2 (18)
(19) 2. Pauli spin matrices: The Pauli spin matrices, σx , σy , and σz are deﬁned via
S = sσ (20) (a) Use this deﬁnition and your answers to problem 13.1 to derive the 2×2 matrix representations
of the three Pauli matrices in the basis of eigenstates of Sz .
With s = 1/2, this gives
σx = 01
10 (21) σy = 0 −i
i0 (22) σz = 10
0 −1 (23) (b) For each Pauli matrix, ﬁnd its eigenvalues, and the components of its normalized eigenvectors
in the basis of eigenstates of Sz .
Each Pauli matrix has eigenvalues 1 and −1.
The eigenvectors are
 ↑z → 1
0 (24)  ↓z → 0
1 (25)  ↑x →  ↓x →  ↑y →  ↓y → 1
√
2
1
√
2
1
√
2
1
√
2 1
1 (26) 1
−1 (27) 1
i (28) 1
−i (29) (c) Use your answer to 13.2.b to obtain the eigenvalues of Sx , Sy , and Sz , as well as the components
of the corresponding normalized eigenvectors in the basis of eigenstates of Sz .
Each component of S has eigenvalues /2 and − /2.
The eigenvectors are the same as in 13.2(b). 4 3. Repeat problems 13.1.(ad) and 13.2.a for the case of a spin1 particle.
For s = 1, the eigenvalues of Sz are 1, 0, and −1.
deﬁned by Sz m = mm .
In this basis we must have 1
0
Sz =
0
We still have Sx =
From S± m = 1
2 (S+ + S− ) and Sy = 1
2i Thus we can introduce the basis {1 , 0 ,  − 1 }, 00
0 0
0 −1 (30) (S− − S+ ). s(s+1)−m(m±1)m±1 we ﬁnd
S+ m S− m = 2−m(m+1)m+1 = (31) 2 − m(m−1)m−1 (32) (33) this leads to and which gives 1S+ 0
0S+ 0
−1S+ 0 √ 0
1S+  − 1
2√
0
0S+  − 1 = 0 0
2
−1S+  − 1
00
0 1S− 0
0S− 0
−1S− 0 0
00
1S−  − 1
√
0
0S−  − 1 = 2 √ 0 −1S−  − 1
0
20 1S+ 1 0S+ 1
S+ =
−1S+ 1
1S− 1
S− = 0S− 1
−1S− 1 √ 2√
0
0
010
√
1
Sx = 2 √
1 0 1
0
2 = √
2
2
010
20
0 and (35) √ 2√
0
0 −i 0
1
− 2
Sy =
0
2 = √ i 0 −i √
2i
2
0i
0
0
−2 0
0
√ (34) The generalized Pauli matrices for spin1 would then be 010
1
σx = √ 1 0 1 2
010 0 −i 0
1
σy = √ i 0 − i 2
0i
0 10 0
0 0 0 σz =
0 0 −1 5 (36) (37) (38) (39) 4. Consider an electron whose position is held ﬁxed, so that it can be described by a simple twocomponent spinor (i.e. no r dependence). Let the initial state of the electron be spin up relative to
the zaxis,  ↑z . At time t = 0, a uniform magnetic ﬁeld is applied along the yaxis.
What is the statevector of the electron at time t > 0?
Hint: Start by writing the Hamiltonian, which should contain only the spincontribution to the
magnetic dipole energy. Then propagate the state using the energy eigenvalue representation of the
propagator, U (t) = n ωn e−iωn t ωn .
The initial state of the system is ψ (0) =  ↑z
The Hamiltonian is
e B0
gq
σy = µB B0 σy
S·B =
H = −µ · B = −
2M
2me (40) With ω0 = eB0 , the eigenvalues of H are then ± ω0 .
me
The state at time t is then
ψ (t) = e−iHt/ ψ (0) (41) = (cos(ω0 t)I − i sin(ω0 t)σy )  ↑z (43) = (cos(Ht/ ) − i sin(Ht/ )) ψ0 (42) = cos(ω0 t) ↑z + sin(ω0 t) ↓z (44) 6 5. Consider the most general normalized spin1/2 state ψ = c↑  ↑z + c↓  ↓z .
(a) Compute Sx , Sy , and Sz , with respect to this state. Sx =
Sy =
Sz = 2 01
10 c∗ , c∗
+− 2 c∗ , c∗
+− 0 −i
i0 2 c∗ , c∗
+− c+
c− 10
0 −1 = c+
c− = c+
c− 2 2 = 2 −ic−
ic+ c∗ , c∗
+− = c+
−c− c∗ , c∗
+− 2 = c−
c+ c∗ , c∗
+− (c∗ c− + c∗ c+ )
−
+ = 2i
2 (c∗ c− − c∗ c+ )
+
− (c∗ c+ − c∗ c− )
+
− (b) Compute the variances ∆Sx , ∆Sy , and ∆Sz .
We know that
2
2
2
Sx = Sy = Sz = 2 4 . So that
∆Sx = 2
Sy − Sy 2 ∆Sz = 2 2 ∆Sy = (c) Prove that ∆Sx = 2
Sx − Sx 2
Sz − Sz 2 c2 − c2 , ∆Sy =
↓
↑ 2 =
=
= 2 1 − (c∗ c− + c∗ c+ )2
+
− 2 1 + (c∗ c− − c∗ c+ )2
+
− 2 1 − (c∗ c+ − c∗ c− )2
+
− c2 + c2 , and ∆Sz = c↑ c↓ .
↓
↑ With
(c∗ c+ + c∗ c− )2 = 12 = 1
+
−
we can write ∆Sx as
2 =
= (c∗ )2 c2 − (c∗ )2 c2 − (c∗ )2 c2 + (c∗ )2 c2
+
+
+
−
−
+
−
− 2 = c+ 4 + 2c+ 2 c− 2 + c− 4 − (c∗ )2 c2 − 2c+ 2 c− 2 − (c∗ )2 c2
+
−
−
+ 2 = (c∗ c+ + c∗ c− )2 − (c∗ c− + c∗ c+ )2
+
−
+
− 2 ∆Sx = ((c∗ )2 − (c∗ )2 )(c2 − c2 )
+
−
+
− 2 c2 − c2 
+
− Similarly, we obtain
∆Sy =
=
= 2 (c∗ )2 c2 + 2c∗ c∗ c+ c− + (c∗ )2 c2 + (c∗ )2 c2 − 2c∗ c∗ c+ c− + (c∗ )2 c2
+
+
+−
−
−
+
−
+−
−
+ 2 (c∗ )2 c2 + (c∗ )2 c2 + (c∗ )2 c2 + (c∗ )2 c2
+
+
−
−
+
−
−
+ 2 c2 + c2 
+
−
7 and
∆Sz =
=
= 2 (c+ 2 + c− 2 )2 − (c+ 2 − c− 2 )2 4c+ 2 c− 2
2
c+ c−  8 6. The SternGerlach eﬀect: A SternGerlach analyzer (SGA) spatially separates the ms states,
relative to the axis of alignment, of any particle with spin sent through the device.
(a) Consider an SGA aligned along the zaxis. At the location of the beam center, the magnetic
ﬁeld inside the SGA can be written to good approximation as B (r ) = B0 (z )ez , where B0 (z )
is a monotonically increasing function of z . The operator for the spin magnetic dipole energy
is VB = −µ · B (r ). Due to conservation of energy, the electron must experience a force in the
direction of decreasing dipole energy. Show that this force is orthogonal to the Lorentz force
if the particle has charge, and that it will deﬂect the spin up and spin down states in opposite
directions.
The Lorentz force is qv × B , and so is always perpendicular to B . This means that any observed
deﬂection along the zaxis can only be attributed to spin eﬀects.
The potential for spinup relative to B , is
V↑ = ↑ VB  ↑ = µB B0 (z ) (45) V↓ = ↓ VB  ↓ = −µB B0 (z ) (46) and for spindown
This shows that the magnetic dipole energy of the spinup state increases with increasing B0 (z ),
and thus the spinup state will experience a downward force.
The magnetic dipole energy of the spindown state decreases with increasing B0 (z ), and thus
the spindown state will experience an upward force.
(b) A single electron in the  ↑z state, is directed into SGA1, which is aligned along the xaxis.
Determine the probabilities for the electron to exit SGA1 in the  ↑x and  ↓x channels.
The probabilities are
1
2
1
P1 (↓) =  ↑z  ↓x 2 =
2
P1 (↑) =  ↑z  ↑x 2 = (47)
(48) (c) The output beam from SGA1 corresponding to the  ↓x channel is then directed into SGA2,
which is aligned along the zaxis. While the output beam from SGA1 corresponding to the  ↑x
1
1
channel is directed into SGA3, which is aligned along the unit vector √2 ez + √2 ey . Determine
the probabilities for the electron to exit in each of the four output channels (i.e. two for SGA2
and two for SGA3).
The independent probabilities for SGA2 are
1
2
1
2
P2 (↓) =  ↓z  ↓z  =
2
P2 (↑) =  ↑z  ↓x 2 = (49)
(50) and for SGA3 we need to ﬁnd the eigenvectors of
1
S3 = √ ( Sz + Sy ) =
2
2
9 1
√
2
i
√
2 i
− √2
1
− √2 , (51) which must have eigenvalues ± /2. The eigenvectors are solutions to
1
i
√ a − √ b = ±a
2
2
with a = 1, we ﬁnd b = −i(1 ∓ (52) √ 2), which after normalization gives
√
 ↑z − i(1 − 2) ↓z
 ↑3 =
√
1 + (1 − 2)2
√
 ↑z − i(1 + 2) ↓z
 ↓3 =
√
1 + (1 + 2)2 (53) (54) The probabilities are then
√
√
2− 2
1
1 + (1 − 2)
√
√=
=
P3 (↑) =  ↑x  ↑3  =
2
2
1 + (1 − 2)
4−2 2
√
√
1 + (1 + 2)
1
2+ 2
√
√=
P3 (↓) =  ↑x  ↓3 2 =
=
2
1 + (1 + 2)2
4+2 2
2 (55)
(56) Computing the combined probabilities then gives
1
4
1
P (↓x ) = P1 (↓)P2 (↓) =
4
1
P (↑3 ) = P1 (↑)P3 (↑) =
4
1
P (↓3 ) = P1 (↑)P3 (↓) =
4 P (↑x ) = P1 (↓)P2 (↑) = 10 (57)
(58)
(59)
(60) 7. Work through problem 9.1 on page 990 in CohenTannoudji, transcribed below:
Consider a spin 1/2 particle. Call its spin S , and its orbital angular momentum, L, and its state
vector ψ . The two functions ψ+ (r ) and ψ− (r ) are deﬁned by
ψ± (r ) = r , ±ψ , (61) where + indicates spin up relative to the zaxis, and − indicates spin down.
Assume that:
1
ψ+ (r ) = R(r ) Y00 (θ, φ) + √ Y10 (θ, φ)
3
R(r ) 1
ψ− (r ) = √
Y1 (θ, φ) − Y10 (θ, φ)
3 (62)
(63) where r , θ , and φ are the coordinates of the particle and R(r ) is a given function of r .
(a) What condition must R(r ) satisfy for ψ to be normalized?
The normalization condition is
d3 r ψ+ (r )2 + ψ− (r)2 1= ∞ =
0 r 2 dr R(r )2 1 + ∞ =2
0 so that the condition on R(r ) is r 2 dr R(r )2
∞ 0 111
++
333
(64) r 2 dr R(r )2 = 1
2 (65) (b) Sz is measured with the particle in state ψ . What results can be found, and with what probabilities? Same question for Lz , then for Sx .
In Dirac notation, the state of the system in the basis {R, ℓ, mℓ , ms }, where R, ℓ, mℓ , ms =
1
R (r) ⊗ ℓ.mℓ (Ω) ⊗ ms (s) , and RR (r) = 2 .
1
1
1
ψ = R, 0, 0, + + √ R, 1, 0, + + √ R, 1, 1, − − √ R, 1, 0, −
3
3
3 (66) If Sz is measured, the two possible outcomes are ± /2. The probability to obtain /2 is
P+ =
=
=
= ψ + +(s) ψ 1
R, 0, 0, + + √ R, 1, 0, +
3 4
RR
3
2
3 1
R, 0, 0, + + √ R, 1, 0, +
3 (67) (68)
11 The probability to obtain − /2 is then
P− = 1 − P+ = 1
3 (69) A measurement of Lz could obtain the results 0, or . The corresponding probabilities are
P (1) =
=
=
= ψ 1 1(φ) ψ
1
R, 1, 1, −R, 1, 1, −
3
1
RR
3
1
6 (70) and 5
6
For Sx , the possible results are ± /2. The corresponding probabilities are
P ( ) = 1 − P (0) = P (+) =
=
=
=
P (−) =
= ψ  ↑x ↑x (s) ψ
1
ψ  (+ + − ) ( + + −) ψ
2
1
1
1
R, 0, 0 + √ R, 1, 1
R, 0, 0 + √ R, 1, 1
2
3
3
2
3
1 − P (+)
1
3 (71) (72) (73) (c) A measurement of L2 , with the particle in state ψ , yielded zero. What state describes the
particle just after this measurement? Same question if the measurement of L2 had given 2 2 .
A measurement of L2 can yield only two possible outcomes for this particular state. They
are 0, for ℓ = 0, and 2 2 , for ℓ = 1.
The probability to obtain ℓ = 0 is
ψ 0 0(ℓ) ψ P (0) =
=
=
= R, 0, 0, +R, 0, 0, + RR
1
2 (74) The state after the measurement is then
ψ ′ = 0 0(ℓ) ψ = √ 12 P (0) 2R, 0, 0, + (75) The probability to obtain 2 2 is
P (1) = 1 − P (0) = 1
2 (76) and the state after obtaining this result is
ψ ′ =
= 1 1(ℓ) ψ
P (1) 2
(R, 1, 0, + + R, 1, 1, − − R, 1, 0, − )
3 13 (77) ...
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This note was uploaded on 06/15/2011 for the course PH 101 taught by Professor Tressler during the Spring '08 term at Bentley.
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