Physics 6210/Spring 2007/Lecture 5
Lecture 5
Relevant sections in text:
§
1.2–1.4
An alternate third postulate
Here is an equivalent statement of the third postulate.*
Alternate Third Postulate
Let
A
be a Hermitian operator with an ON basis of eigenvectors

i
:
A

i
=
a
i

i .
The only possible outcome of a measurement of the observable represented by
A
is one of
its eigenvalues
a
j
. The probability for getting
a
j
in a state

ψ
is
Prob
(
A
=
a
j
) =
D
i
=1

i, a
j

ψ

2
,
where the sum ranges over the
D
dimensional space of eigenvectors

i, a
j
,
i
= 1
, . . . , D
,
with the given eigenvalue
a
j
. Note that if the eigenvalue
a
k
is nondegenerate then
Prob
(
A
=
a
j
) =

j

ψ

2
.
Note also that
j

ψ
is the component of

ψ
along

j
.
Let us prove the probability formula in the case where there is no degeneracy. Allowing
for degeneracy is no problem; think about it as an exercise. Let
f
(
x
) be the characteristic
function of
a
j
. We have (exercise)
f
(
A
) =

j
j

.
Then, in the state

ψ
we have
Prob
(
A
=
a
j
) =
f
(
A
) =
ψ

(

j
j

)

ψ
=

j

ψ

2
.
Because the state vectors have unit norm, we have
1 =
ψ

ψ
=
j
ψ

j
j

ψ
=
j

j

ψ

2
,
* For now we state it in a form that is applicable to finitedimensional Hilbert spaces. We
will show how to generalize it to infinitedimensional ones (such as occur with a “particle”)
in a little while.
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Physics 6210/Spring 2007/Lecture 5
that is, the probabilities
Prob
(
A
=
a
j
) add up to unity when summed over all the eigen
values. This indicates that the probability vanishes for finding a value for
A
which is
not
one of its eigenvalues.
We can write the expectation value of an observable so that the probabilities feature
explicitly. As usual, we have
A

i
=
a
i

i ,
i

j
=
δ
ij
.
If a system is in the state

ψ
, we compute (by inserting the identity operator twice – good
exercise)
A
=
ψ

A

ψ
=
ij
ψ

i
i

A

j
j

ψ
=
i
a
i

i

ψ

2
.
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 Spring '07
 M
 Physics, mechanics, Probability, Hilbert space

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