Hi Everyone,

Attached are the solutions for the first three problems only (to not spoil the fun of the 7 problems that are due this week).

Update(9/15/2015): The full set of solution is posted below.

Solutions 6 (full)

Hi all,

I’ve attached the solutions for Chapter 5. (We’re about halfway done with this book!)

Chapter5Solutions

Hi all,

I’ve attached the exercise solutions for Chapters 3 and 4 as separate pdf’s. Please let me know if there are any mistakes or errors in logic.

Chapter3 Solutions

Chapter4 Solutions

Thanks!

Exercise 2.1

Assume that  and $\ket{d}$ are normalized basis vectors. And let:

$\ket{l} = \frac{1}{\sqrt{2}}\ket{u} - \frac{1}{\sqrt{2}}\ket{d}$

 Prove that $\ket{r}$ is orthogonal to $\ket{l}$

 pf/ $\ket{r}$ and $\ket{l}$ are orthogonal iff $\braket{r}{l} = 0$

\begin{tabular}{p{13cm}}

 Now $\bra{r} = \overline{\ket{r}}$

 And $\overline{\ket{r}} = \overline{\dfrac{1}{\sqrt{2}}\ket{u} + \dfrac{1}{\sqrt{2}}\ket{d}}$

$= \overline{\dfrac{1}{\sqrt{2}}\ket{u}} + \overline{\dfrac{1}{\sqrt{2}}\ket{d}} $   \hfill $(\overline{a+b} = \overline{a}+\overline{b})$

$= \overline{\dfrac{1}{\sqrt{2}}}\overline{\ket{u}} + \overline{\dfrac{1}{\sqrt{2}}}\overline{\ket{d}} $   \hfill $(\overline{ab} = \overline{a}\overline{b})$

So $ \bra{r} = \dfrac{1}{\sqrt{2}}\bra{u} + \dfrac{1}{\sqrt{2}}\bra{d} $   \hfill $(\overline{\ket{a}} = \bra{a})$

\end{tabular}

 Now,  $\braket{r}{l} = (\dfrac{1}{\sqrt{2}}\bra{u} + \dfrac{1}{\sqrt{2}}\bra{d}) \cdot  (\dfrac{1}{\sqrt{2}}\ket{u} - \dfrac{1}{\sqrt{2}}\ket{d})$

 \begin{tabular}{p{13cm}}

 = $\dfrac{1}{2}\braket{u}{u} - \dfrac{1}{2} \braket{l}{l}$

 = $ \dfrac{1}{2} *1 - \dfrac{1}{2}*1 = 0$ \hfill (Since $\bra{u}$ is normalized, $\braket{u}{u} = 1$)

 So therefore $\braket{r}{l} = 0$

 \end{tabular}

Exercise 2.2

 Exercise 2.3

 Let $\ket{i} = a\ket{u} + b\ket{d}$ and $\ket{o} =  c\ket{u} + e\ket{d}$

 \begin{tabular}{p{13cm}}

 a) $a^*a = \braket{u}{i}\braket{i}{u} = 0.5$  \hfill (b/c when the spin is set to in, the prob of up is 0.5)

 b)

 c)

 Say that $a^*b$ was real. Then $a^*b = (a^* b)^* = ba^*$. (real number are equal to their complex conjugates.) But then $a^*b + ba^* = 0 = 2a^*b$. But then either $a^*$ or $b$ must be 0. But then either $a^*a$ or $b^*b$ must be equal to 0 which contradicts.

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