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Physics 325 Solution to Selected Problems #1 Winter 2004
- Your textbook, Problem 4.
a.)
[Sx, Sy ] = SxSy − SySx =
¯h^2 4
[ (^0
0 −i i 0
0 −i i 0
) ]
¯h^2 4
[ (^ i 0 0 −i
−i 0 0 i
) ]
¯h^2 4
2 i 0 0 − 2 i
= i¯h
¯h/ 2 0 0 −¯h/ 2
= i¯hSz
b.) σ^2 x =
; σ^2 y =
0 −i i 0
0 −i i 0
σ^2 z =
which is consistent with σj σj = δjj I + i
l jjl^ σl^ because^ jkl^ = 0 when two of the indices are equal and^ I is the identity matrix.
σxσy =
0 −i i 0
i 0 0 −i
= iσz
σy σx =
0 −i i 0
−i 0 0 i
= −iσz
so σxσy = i
l xylσl^ and^ σy^ σx^ =^ i^
l yxlσl^ because^ xyz^ =^ −yxz^ = 1. Similarly, σy σz = −σz σy = iσx = i
l yxlσl^ and^ σz^ σx^ =^ −σxσz^ =^ iσy^ =^ i^
l zxlσl. Putting these together, σj σk = δjk I + i
l jkl^ σl
- Your textbook Prob. 4.
a.) 〈χ|χ〉 = 1 = AA∗^ ( − 3 i 4 )
3 i 4
= |A|^2 (9 + 16) = 25 |A|^2 => A =
eiα 5
where α is an arbitrary real phase angle.
b.) 〈Sx〉 = 〈χ|Sx|χ〉 =
( − 3 i 4 )
3 i 4
¯h 2
( − 3 i 4 )
3 i
¯h 2
〈Sy 〉 = 〈χ|Sy |χ〉 =
( − 3 i 4 )
0 −i i 0
3 i 4
¯h 2
( − 3 i 4 )
− 4 i − 3
¯h 2
×
¯h 2
〈Sz 〉 = 〈χ|Sz |χ〉 =
( − 3 i 4 )
3 i 4
¯h 2
( − 3 i 4 )
3 i − 4
¯h 2
×
¯h 2
c.) Because σSj =
〈S^2 j 〉 − 〈Sj 〉^2 and S^2 j = I¯h^2 /4 where I is the identity matrix, we have:
〈S j^2 〉 =
( − 3 i 4 )
3 i 4
¯h^2 4
¯h^2 4
and σSx =
¯h^2 /4 = ¯h/2; σSy = ¯h/ 2
1 − (24/25)^2 = 7¯h/50; and
σSz = ¯h/ 2
1 − (7/25)^2 = 24¯h/50.
d.) σSx σSy = 7¯h^2 100
¯h 2
×
7¯h 50
¯h 2
|〈Sz 〉|; σSx σSz = 24¯h^2 100
¯h 2
×
24¯h 50
¯h 2
|〈Sy 〉|
σSy σSz =
7 × 24¯h^2 (50)^2
¯h 2
× 0 =
¯h 2
|〈Sx〉|
.
- Your textbook Prob. 4. a.) We must solve Sˆy χ(y)^ = λχ(y):
det
−λ −i¯h/ 2 i¯h/ 2 −λ
= 0 => λ^2 − ¯h^2 /4 = 0 => λ± = ±¯h/ 2
.
Letting χ+ =
a+ b+
and χ− =
a− b−
, we have:
Sˆy χ(+y ) = λ+χ(+y ) => ¯h 2
0 −i i 0
a+ b+
¯h 2
a+ b+
−ib+ ia+
a+ b+
=> b+ = ia+
So we have χ (y)
i
after making the norm equal to 1, with an eigenvalue of +¯h/2.
Sˆy χ( −y ) = λ−χ( −y ) => ¯h 2
0 −i i 0
a− b−
¯h 2
a− b−
−ib− ia−
−a− −b−
=> b− = −ia−
So we have χ( −y )= √^12
−i
after making the norm equal to 1, with an eigenvalue of −¯h/2.
b.)
χ =
a b
= αχ (y)
(y) − =^
α √ 2
i
β √ 2
−i
α + β i(α − β)
Therefore α + β = a
2 and α − β = −ib
2 giving us α = (a − ib)/
2 and β = (a + ib)/
- If we measure
Sy for state χ, we will get a result +¯h/2 with a probability of |α|^2 = |a − ib|^2 /2 or a result −¯h/2 with a probability of |β|^2 = |a + ib|^2 /2.
1 2 |a − ib|^2 +
|a + ib|^2 =
[
(a∗^ + ib∗)(a − ib) + (a∗^ − ib∗)(a + ib)
]
[
a∗a + b∗b + ib∗a − ia∗b + a∗a + b∗b − ib∗a + ia∗b
]
(2a∗a + 2b∗b) = 1
because χ is normalized (|a|^2 + |b|^2 = 1).
c.) Because S^2 y = I¯h^2 /4 where I is the identity matrix, all states χ are eigenstates of S^2 y with eigenvalue
¯h^2 /4: you will measure ¯h^2 /4 with a probability of 1.
