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# Some bra-ket manipulation problems.([1] pr. 1.4)

Using braket logic expand

## (a)

\begin{equation}\label{eqn:braketManip:20}

\textrm{tr}{X Y}

\end{equation}

## (b)

\begin{equation}\label{eqn:braketManip:40}

(X Y)^\dagger

\end{equation}

## (c)

\begin{equation}\label{eqn:braketManip:60}

e^{i f(A)},

\end{equation}

where \( A \) is Hermitian with a complete set of eigenvalues.

## (d)

\begin{equation}\label{eqn:braketManip:80}

\sum_{a’} \Psi_{a’}(\Bx’)^\conj \Psi_{a’}(\Bx”),

\end{equation}

where \( \Psi_{a’}(\Bx”) = \braket{\Bx’}{a’} \).

# Answers

## (a)

\begin{equation}\label{eqn:braketManip:100}

\begin{aligned}

\textrm{tr}{X Y}

&= \sum_a \bra{a} X Y \ket{a} \\

&= \sum_{a,b} \bra{a} X \ket{b}\bra{b} Y \ket{a} \\

&= \sum_{a,b}

\bra{b} Y \ket{a}

\bra{a} X \ket{b} \\

&= \sum_{a,b}

\bra{b} Y

X \ket{b} \\

&= \textrm{tr}{ Y X }.

\end{aligned}

\end{equation}

## (b)

\begin{equation}\label{eqn:braketManip:120}

\begin{aligned}

\bra{a} \lr{ X Y}^\dagger \ket{b}

&=

\lr{ \bra{b} X Y \ket{a} }^\conj \\

&=

\sum_c \lr{ \bra{b} X \ket{c}\bra{c} Y \ket{a} }^\conj \\

&=

\sum_c \lr{ \bra{b} X \ket{c} }^\conj \lr{ \bra{c} Y \ket{a} }^\conj \\

&=

\sum_c

\lr{ \bra{c} Y \ket{a} }^\conj

\lr{ \bra{b} X \ket{c} }^\conj \\

&=

\sum_c

\bra{a} Y^\dagger \ket{c}

\bra{c} X^\dagger \ket{b} \\

&=

\bra{a} Y^\dagger

X^\dagger \ket{b},

\end{aligned}

\end{equation}

so \( \lr{ X Y }^\dagger = Y^\dagger X^\dagger \).

## (c)

Let’s presume that the function \( f \) has a Taylor series representation

\begin{equation}\label{eqn:braketManip:140}

f(A) = \sum_r b_r A^r.

\end{equation}

If the eigenvalues of \( A \) are given by

\begin{equation}\label{eqn:braketManip:160}

A \ket{a_s} = a_s \ket{a_s},

\end{equation}

this operator can be expanded like

\begin{equation}\label{eqn:braketManip:180}

\begin{aligned}

A

&= \sum_{a_s} A \ket{a_s} \bra{a_s} \\

&= \sum_{a_s} a_s \ket{a_s} \bra{a_s},

\end{aligned}

\end{equation}

To compute powers of this operator, consider first the square

\begin{equation}\label{eqn:braketManip:200}

\begin{aligned}

A^2 =

&=

\sum_{a_s} a_s \ket{a_s} \bra{a_s}

\sum_{a_r} a_r \ket{a_r} \bra{a_r} \\

&=

\sum_{a_s, a_r} a_s a_r \ket{a_s} \bra{a_s} \ket{a_r} \bra{a_r} \\

&=

\sum_{a_s, a_r} a_s a_r \ket{a_s} \delta_{s r} \bra{a_r} \\

&=

\sum_{a_s} a_s^2 \ket{a_s} \bra{a_s}.

\end{aligned}

\end{equation}

The pattern for higher powers will clearly just be

\begin{equation}\label{eqn:braketManip:220}

A^k =

\sum_{a_s} a_s^k \ket{a_s} \bra{a_s},

\end{equation}

so the expansion of \( f(A) \) will be

\begin{equation}\label{eqn:braketManip:240}

\begin{aligned}

f(A)

&= \sum_r b_r A^r \\

&= \sum_r b_r

\sum_{a_s} a_s^r \ket{a_s} \bra{a_s} \\

&=

\sum_{a_s} \lr{ \sum_r b_r a_s^r } \ket{a_s} \bra{a_s} \\

&=

\sum_{a_s} f(a_s) \ket{a_s} \bra{a_s}.

\end{aligned}

\end{equation}

The exponential expansion is

\begin{equation}\label{eqn:braketManip:260}

\begin{aligned}

e^{i f(A)}

&=

\sum_t \frac{i^t}{t!} f^t(A) \\

&=

\sum_t \frac{i^t}{t!}

\lr{ \sum_{a_s} f(a_s) \ket{a_s} \bra{a_s} }^t \\

&=

\sum_t \frac{i^t}{t!}

\sum_{a_s} f^t(a_s) \ket{a_s} \bra{a_s} \\

&=

\sum_{a_s}

e^{i f(a_s) }

\ket{a_s} \bra{a_s}.

\end{aligned}

\end{equation}

## (d)

\begin{equation}\label{eqn:braketManip:n}

\begin{aligned}

\sum_{a’} \Psi_{a’}(\Bx’)^\conj \Psi_{a’}(\Bx”)

&=

\sum_{a’}

\braket{\Bx’}{a’}^\conj

\braket{\Bx”}{a’} \\

&=

\sum_{a’}

\braket{a’}{\Bx’}

\braket{\Bx”}{a’} \\

&=

\sum_{a’}

\braket{\Bx”}{a’}

\braket{a’}{\Bx’} \\

&=

\braket{\Bx”}{\Bx’} \\

&= \delta_{\Bx” – \Bx’}.

\end{aligned}

\end{equation}

# References

[1] Jun John Sakurai and Jim J Napolitano. *Modern quantum mechanics*. Pearson Higher Ed, 2014.