Mathematical Methods for Physicists: A concise introduction - Site Map

LINEAR VECTOR SPACES
Step 3.
Clearly
he 1 jIIi ˆ he 1 j2i he 1 je 1 ihe 1 j2i ˆ 0; i:e:; …IIj?je 1 i:
Dividing jIIi by its norm (length), we now have the second basis vector
and it is orthogonal to the ®rst base vector je 1 i and of unit length.
To construct the third member of the orthonormal basis, consider
jIIIi ˆj3ije 1 ihe 1 jIIIije 2 i 2 jIIIi
which is orthogonal to both je 1 i and je 2 i. Dividing by its norm we get
je 3 i.
Continuing in this way, we will obtain an orthonormal basis je 1 i; je 2 i; ...; je n i.
The Cauchy±Schwarz inequality
If A and B are non-zero vectors in E 3 , then the dot product gives A B ˆ AB cos ,
where is the angle between the vectors. If we square both sides and use the fact
that cos 2 1, we obtain the inequality
…A B† 2 A 2 B 2 or jA Bj AB:
This is known as the Cauchy±Schwarz inequality. There is an inequality corresponding
to the Cauchy±Schwarz inequality in any inner-product space that
obeys Axioms B.1±B.4, which can be stated as
p
jhUjWij jUjjWj; jUj ˆ
hUjUi
etc:;
…5:13†
where jUi and jWi are two non-zero vectors in an inner-product space.
This can be proved as follows. We ®rst note that, for any scalar , the following
inequality holds
0 jhU ‡ WjU ‡ Wij 2 ˆ hU ‡ WjU ‡ Wi
ˆ hU
ˆ jU
jUi ‡ hW
j 2 ‡* hV
jUi ‡ hU
jWi ‡ hWjWi
jUi ‡ hUjWi ‡ jj 2 jWj 2 :
Now let ˆ hUjWi*=jhUjWij, with real. This is possible if jWi 6ˆ0, but if
hUjWi ˆ0, then Cauchy±Schwarz inequality is trivial. Making this substitution
in the above, we have
0 jU
j 2 ‡ 2jhUjWij‡ 2 jWj 2 :
This is a quadratic expression in the real variable with real coecients.
Therefore, the discriminant must be less than or equal to zero:
4jhUjWij 2 4jUj 2 jWj 2 0
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