## Weird dreams

I woke up today having a dream still in my head from the night, but it was a strange one. I was expanding out the Dirac notation representation of an operator in matrix form, but the symbols in the kets were elaborate pictures of Disney princesses that I was drawing with forestry scenery in the background, including little bears. At the point that I woke up from the dream, I noticed that I’d gotten the proportion of the bears wrong in one of the pictures, and they looked like they were ready to eat one of the princess characters.

## Guts

As a side effect of this weird dream I actually started thinking about matrix element representation of operators.

When forming the matrix element of an operator using Dirac notation the elements are of the form $$\bra{\textrm{row}} A \ket{\textrm{column}}$$. I’ve gotten that mixed up a couple of times, so I thought it would be helpful to write this out explicitly for a $$2 \times 2$$ operator representation for clarity.

To start, consider a change of basis for a single matrix element from basis $$\setlr{\ket{q}, \ket{r} }$$, to basis $$\setlr{\ket{a}, \ket{b} }$$

\label{eqn:operatorMatrixElement:20}
\begin{aligned}
\bra{q} A \ket{r}
&=
\braket{q}{a} \bra{a} A \ket{r}
+
\braket{q}{b} \bra{b} A \ket{r} \\
&=
\braket{q}{a} \bra{a} A \ket{a}\braket{a}{r}
+ \braket{q}{a} \bra{a} A \ket{b}\braket{b}{r} \\
&+ \braket{q}{b} \bra{b} A \ket{a}\braket{a}{r}
+ \braket{q}{b} \bra{b} A \ket{b}\braket{b}{r} \\
&=
\braket{q}{a}
\begin{bmatrix}
\bra{a} A \ket{a} & \bra{a} A \ket{b}
\end{bmatrix}
\begin{bmatrix}
\braket{a}{r} \\
\braket{b}{r}
\end{bmatrix}
+
\braket{q}{b}
\begin{bmatrix}
\bra{b} A \ket{a} & \bra{b} A \ket{b}
\end{bmatrix}
\begin{bmatrix}
\braket{a}{r} \\
\braket{b}{r}
\end{bmatrix} \\
&=
\begin{bmatrix}
\braket{q}{a} &
\braket{q}{b}
\end{bmatrix}
\begin{bmatrix}
\bra{a} A \ket{a} & \bra{a} A \ket{b} \\
\bra{b} A \ket{a} & \bra{b} A \ket{b}
\end{bmatrix}
\begin{bmatrix}
\braket{a}{r} \\
\braket{b}{r}
\end{bmatrix}.
\end{aligned}

Suppose the matrix representation of $$\ket{q}, \ket{r}$$ are respectively

\label{eqn:operatorMatrixElement:40}
\begin{aligned}
\ket{q} &\sim
\begin{bmatrix}
\braket{a}{q} \\
\braket{b}{q} \\
\end{bmatrix} \\
\ket{r} &\sim
\begin{bmatrix}
\braket{a}{r} \\
\braket{b}{r} \\
\end{bmatrix} \\
\end{aligned},

then

\label{eqn:operatorMatrixElement:60}
\bra{q} \sim
{\begin{bmatrix}
\braket{a}{q} \\
\braket{b}{q} \\
\end{bmatrix}}^\dagger
=
\begin{bmatrix}
\braket{q}{a} &
\braket{q}{b}
\end{bmatrix}.

The matrix element is then

\label{eqn:operatorMatrixElement:80}
\bra{q} A \ket{r}
\sim
\bra{q}
\begin{bmatrix}
\bra{a} A \ket{a} & \bra{a} A \ket{b} \\
\bra{b} A \ket{a} & \bra{b} A \ket{b}
\end{bmatrix}
\ket{r},

and the corresponding matrix representation of the operator is

\label{eqn:operatorMatrixElement:100}
A \sim
\begin{bmatrix}
\bra{a} A \ket{a} & \bra{a} A \ket{b} \\
\bra{b} A \ket{a} & \bra{b} A \ket{b}
\end{bmatrix}.

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

Using braket logic expand

## (a)

\label{eqn:braketManip:20}
\textrm{tr}{X Y}

## (b)

\label{eqn:braketManip:40}
(X Y)^\dagger

## (c)

\label{eqn:braketManip:60}
e^{i f(A)},

where $$A$$ is Hermitian with a complete set of eigenvalues.

## (d)

\label{eqn:braketManip:80}
\sum_{a’} \Psi_{a’}(\Bx’)^\conj \Psi_{a’}(\Bx”),

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

## (a)

\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}

## (b)

\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}

so $$\lr{ X Y }^\dagger = Y^\dagger X^\dagger$$.

## (c)

Let’s presume that the function $$f$$ has a Taylor series representation

\label{eqn:braketManip:140}
f(A) = \sum_r b_r A^r.

If the eigenvalues of $$A$$ are given by

\label{eqn:braketManip:160}
A \ket{a_s} = a_s \ket{a_s},

this operator can be expanded like

\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}

To compute powers of this operator, consider first the square

\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}

The pattern for higher powers will clearly just be

\label{eqn:braketManip:220}
A^k =
\sum_{a_s} a_s^k \ket{a_s} \bra{a_s},

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

\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}

The exponential expansion is

\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}

## (d)

\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}

# References

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