PHYS 507
Answers to Homework I (Fall ’05)
1.
(a)
We know that
u
2
has to be orthogonal to
u
1
, i.e.,
h
u
1

u
2
i
= 0. Using this, we can
completely determine
u
2
. Let
u
2
=
•
a
b
‚
.
Then
h
u
1

u
2
i
=
1
√
6
£
2
1 +
i
/
•
a
b
‚
=
2
a
+ (1 +
i
)
b
√
6
= 0
.
From here we find
a
=

(1 +
i
)
b/
2 and therefore
u
2
=
•
a
b
‚
=
b
•

(1 +
i
)
/
2
1
‚
.
Here,
b
can be any complex number, i.e., the equation
h
u
1

u
2
i
= 0 is satisfied by
any complex value of
b
. However, in applications, especially in the computation
of probabilities, it is advantageous to have
u
2
to be normalized to unity, i.e.,
h
u
2

u
2
i
= 1. This condition reads,
h
u
2

u
2
i
=

b

2
1
2
+ 1
¶
=
3

b

2
2
= 1
.
As a result, we have

b

=
p
2
/
3. Normalization gives only the value of modulus
of
b
and not its phase. There is no way to determine its phase and moreover we
don’t need to because it is just an overall phase factor for
u
2
. For this reason, we
choose the phase angle of
b
to be anything we like. For example,
b
=
p
2
/
3 is one
such possible choice. In that case we have
u
2
=
1
√
6
•

(1 +
i
)
2
‚
.
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 Spring '11
 starg
 u2, Orthogonal matrix, cos cos

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