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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.