Exercise 1.1
[latexpage]
a) Using the axioms of inner products, we are asked to prove
\[
\{<A|+<B|\}|C>\,=\, <A|C>+<B|C>
\]
Solution:
Let $<D|=\{<A|+<B|\}$. Using the second axiom of inner products
\[
\\<C|D>=<D|C>^ {\ast}
\]
Now lets substitute $|D>$
\[
\\<C|\{|A>+|B>\}=(\{<A|+<B|\}|C>)^ {\ast}
\]
Which, using the first axiom of inner products, becomes
\[
\\<C|A>+<C|B>=(\{<A|+<B|\}|C>)^ {\ast}
\]
Lastly, we compute the complex conjugates of each term
\[
\\<C|A>^ {\ast}+<C|B>^ {\ast}=(\{<A|+<B|\}|C>)^ {\ast}^ {\ast}
\]
\[
\\<A|C>+<B|C>=\{<A|+<B|\}|C>
\]
Which is the proof that we were seeking.
b) Prove $<A|A>$ is a real number
Here we will use the component notation. The bra $<A|$ and ket $|A>$ in component notation are:
\[
<A|=\sum\nolimits_{i=1} \alpha_i^{\ast}<i|
\]
\[
|A>=\sum\nolimits_{j=1} \alpha_j|j>
\]
If we take the inner product we find
\[
<A|A>=\sum\nolimits_{j=1}\sum\nolimits_{i=1} \alpha_i^{\ast}\alpha_j<i|j>
\]
Since $<i|j>=0$ when $i\neq j$ and $<i|j>=1$ when $i= j$, the previous becomes
\[
<A|A>=\sum\nolimits_{j=1} \alpha_j^{\ast}\alpha_j
\]
Now, remembering that any complex number multiplied by its complex conjugate yields a real number, we can conclude that the previous expression is a summation of real numbers. Therefore, we conclude that $<A|A>$ is a real number too
Exercise 1.2
\[
\\<B|A>=\beta_1^ {\ast}\alpha_1+\beta_2^ {\ast}\alpha_2+\beta_3^ {\ast}\alpha_3+\beta_4^ {\ast}\alpha_4+\beta_5^ {\ast}\alpha_5
\]
satisfies all the axioms of inner products.
Solution:
The first axiom is
\[
\\<C|\{|A>+|B>\}=<C|A>+<C|B>
\]
Let us introduce the bra vector $<C|$
\[
\\<C|=\left(\begin{matrix}
c^{\ast}_1 & c^{\ast}_2 & c^{\ast}_3 &c^{\ast}_4 & c^{\ast}_5
\end{matrix}\right)
\]
The left side of the axiom in explicit form is
\[
\\<C|\{|A>+|B>\}=\left(\begin{matrix}
c^{\ast}_1 & c^{\ast}_2 & c^{\ast}_3 &c^{\ast}_4 & c^{\ast}_5
\end{matrix}\right)
\left(\begin{matrix}
\alpha_1+\beta_1\\
\alpha_2+\beta_2\\
\alpha_3+\beta_3\\
\alpha_4+\beta_4\\
\alpha_5+\beta_5
\end{matrix}\right)
\]
Working out the vector multiplication and rearranging we find
\[
\\<C|\{|A>+|B>\}=
c^{\ast}_1\alpha_1+c^{\ast}_1\beta_1+
c^{\ast}_2\alpha_2+c^{\ast}_2\beta_2+
c^{\ast}_3\alpha_3+c^{\ast}_3\beta_3+
c^{\ast}_4\alpha_4+c^{\ast}_4\beta_4+
c^{\ast}_5\alpha_5+c^{\ast}_5\beta_5
\]
\[
\\<C|\{|A>+|B>\}=
(c^{\ast}_1\alpha_1+
c^{\ast}_2\alpha_2+
c^{\ast}_3\alpha_3+
c^{\ast}_4\alpha_4+
c^{\ast}_5\alpha_5)+
(c^{\ast}_1\beta_1+
c^{\ast}_2\beta_2+
c^{\ast}_3\beta_3+
c^{\ast}_4\beta_4+
c^{\ast}_5\beta_5)
\]
\[
\\<C|\{|A>+|B>\}=<C|A>+<C|B>
\]
Proving the first axiom. The second axiom is
\[
<B|A>=<A|B>^{\ast}
\]
We can start this proof computing the complex conjugate of <B|A>
\[
\\<B|A>^{\ast}=(\beta_1^ {\ast}\alpha_1+\beta_2^ {\ast}\alpha_2+\beta_3^ {\ast}\alpha_3+\beta_4^ {\ast}\alpha_4+\beta_5^ {\ast}\alpha_5)^{\ast}
\]
Computing the complex conjugate for each term in the right yields
\[
\\<B|A>^{\ast}=\beta_1\alpha_1^{\ast}+\beta_2\alpha_2^{\ast}+\beta_\alpha_3^{\ast}+\beta_4\alpha_4^{\ast}+\beta_5\alpha_5^{\ast}
\]
Since the terms in the right commute, we can write
\[
\\<B|A>^{\ast}=\alpha_1^{\ast}\beta_1+\alpha_2^{\ast}\beta_2+\alpha_3^{\ast}\beta_3+\alpha_4^{\ast}\beta_4+\alpha_5^{\ast}\beta_5=<A|B>
\]
Taking the complex conjugate of all terms we find
\[
\\(<B|A>^{\ast})^{\ast}=<A|B>^{\ast}
\]
And therefore
\[
\\<B|A>=<A|B>^{\ast}
\]