Question

Consider a three-dimensional vector space spanned by an Orthonormal basis $\overline{)\mathbf{1}\mathbf{>}\mathbf{,}\overline{)\mathbf{2}\mathbf{>}\mathbf{,}\overline{)\mathbf{3}\mathbf{>}}}}$. Kets $\overline{)\mathbf{\alpha }\mathbf{>}}$and $\overline{)\mathit{\beta }\mathbf{>}}$are given by

$\mathbf{|}\mathbf{}\mathit{\alpha }\mathbf{⟩}\mathbf{=}\mathit{i}\mathbf{|}\mathbf{}\mathbf{1}\mathbf{⟩}\mathbf{-}\mathbf{2}\mathbf{|}\mathbf{}\mathbf{2}\mathbf{⟩}\mathbf{-}\mathit{i}\mathbf{|}\mathbf{}\mathbf{3}\mathbf{⟩}\mathbf{,}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{|}\mathit{\beta }\mathbf{>}\mathbf{=}\mathit{i}\mathbf{|}\mathbf{}\mathbf{1}\mathbf{⟩}\mathbf{+}\mathbf{2}\mathbf{|}\mathbf{}\mathbf{3}\mathbf{⟩}\mathbf{.}$

(a)Construct$\mathbf{<}\overline{)\mathbf{\alpha }}$and $\mathbf{<}\overline{)\mathbf{\beta }}$(in terms of the dual basis

$\mathbf{}\mathbf{⟨}\mathbf{1}\mathbf{|}\mathbf{,}\mathbf{⟨}\mathbf{2}\mathbf{|}\mathbf{,}\mathbf{⟨}\mathbf{3}\mathbf{\left|}\mathbf{\right)}\mathbf{.}$
(b) Find $\mathbf{⟨}\mathit{\alpha }\mathbf{\mid }\mathbf{}\mathit{\beta }\mathbf{⟩}$and$\mathbf{}\mathbf{⟨}\mathit{\beta }\mathbf{\mid }\mathbf{}\mathit{\alpha }\mathbf{⟩}\mathbf{,}$and confirm that

$\mathbf{⟨}\mathit{\beta }\mathbf{\mid }\mathit{\alpha }\mathbf{⟩}\mathbf{=}\mathbf{⟨}\mathit{\alpha }\mathbf{\mid }\mathit{\beta }{\mathbf{⟩}}^{\mathbf{*}}\mathbf{.}$
(c)Find all nine matrix elements of the operator$\stackrel{\mathbf{⏜}}{\mathbf{A}}\mathbf{}\mathbf{\equiv }\mathbf{|}\mathbf{}\mathit{\alpha }\mathbf{⟩}\mathbf{⟨}\mathit{\beta }\mathbf{|}$, in this basis, and construct the matrix A. Is it hermitian?

Short answer

(a) $$\begin{gathered} ⟨\alpha |=-i⟨1|-2⟨2|+i⟨3| \\ ⟨\beta |=-i⟨1|+2⟨3| \end{gathered}$$ .

(b) $$\begin{gathered} ⟨\alpha \mid \beta ⟩=1+2i \\ ⟨\beta \mid \alpha ⟩=1-2i \end{gathered}$$ .

(c) $A=\left(\begin{array}{ccc}1& 0& 2i\\ 2i& 0& -4\\ -1& 0& -2i\end{array}\right)$And it is not a hermitan matrix.

Step-by-step solution

Given data

Consider a three-dimensional vector space spanned by an Orthonormality basis $|1⟩,|2⟩,|3⟩$.

Kets localid="1658312431665" $|\alpha ⟩$and $|\beta ⟩$are given by

$$\begin{gathered} |\alpha ⟩=i|1⟩-2|2⟩-i|3⟩, \\ |\beta ⟩=i|1⟩+2|3⟩. \end{gathered}$$

(a) Construct ⟨α| and ⟨β| 

Here we consider the following two kets:

$$\begin{gathered} |\alpha ⟩=i|1⟩-2|2⟩-i|3⟩ \\ |\beta ⟩=i|1⟩+2|3⟩ \end{gathered}$$

In order to construct bras $⟨\alpha |$and $⟨\beta |$we need to take a complex conjugate of each factor and substitute kets for bras.

Following the above recipe we have for $⟨\alpha |$and $⟨\beta |$:

$$\begin{gathered} ⟨\alpha |=-i⟨1|-2⟨2|+i⟨3| \\ ⟨\beta |=-i⟨1|+2⟨3|. \end{gathered}$$

(b) To find ⟨α∣ β⟩ and ⟨β∣ α⟩.

The inner product$⟨\alpha \mid \beta ⟩$ is now given by

$⟨\alpha \mid \beta ⟩=(-i⟨1|-2⟨2|+i⟨3|\left)\right(i|1⟩+2|3⟩)$.

Term by term multiplication yields

$⟨\alpha \mid \beta ⟩=-i^2⟨1\mid 1⟩-2i⟨1\mid 3⟩-2i⟨2\mid 1⟩-4⟨2\mid 3⟩+i^2⟨3\mid 1⟩+2i⟨3\mid 3⟩$

This is an Orthonormal basis which means that

$⟨i\mid j⟩=\left\{\begin{array}{l}1,i=j\\ 0,i\ne j\end{array}\right.$

Applying this to the above equation we find

$$\begin{gathered} ⟨\alpha \mid \beta ⟩=-i^2\cdot 1-2i\cdot 0-2i\cdot 0-4\cdot 0+i^2\cdot 0+2i\cdot 1 \\ ⟨\alpha \mid \beta ⟩=1+2i \end{gathered}$$

Similarly, we have

$$\begin{gathered} ⟨\beta \mid \alpha ⟩=(-i⟨1|+2⟨3\left|\right)\left(i\right|1⟩-2|2⟩-i|3⟩) \\ ⟨\beta \mid \alpha ⟩=-i^2\cdot 1+2i\cdot 0+i^2\cdot 0+2i\cdot 0-4\cdot 0-2i\cdot 1 \\ ⟨\beta \mid \alpha ⟩=1-2i \end{gathered}$$

We now also see that

$⟨\beta \mid \alpha ⟩=⟨\alpha \mid \beta {⟩}^{*}$.

(c) To find all nine matrix elements.

We now consider the operator$\stackrel{⏜}{A}=|\alpha ⟩⟨\beta |$.

Its matrix elements are given by

$A_{ij}=⟨i\mid \alpha ⟩⟨\beta \mid j⟩$
Let us now compute the first row of the matrix A:

$$\begin{gathered} A_{11}=⟨1\left|\right(i|1⟩-2|2⟩-i|3⟩)(-i⟨1|+2⟨3\left|\right)\mid 1⟩ \\ {A}_{11}=(i+0+0)(-i+0) \\ {A}_{11}=1 \\ {A}_{12}=⟨1\left|\right(i|1⟩-2|2⟩-i|3⟩)(-i⟨1|+2⟨3\left|\right)\mid 2⟩ \\ {A}_{12}=(i+0+0)(0+0) \\ {A}_{12}=0 \end{gathered}$$

$$\begin{gathered} A_{13}=⟨1\left|\right(i|1⟩-2|2⟩-i|3⟩)(-i⟨1|+2⟨3\left|\right)\mid 3⟩ \\ {A}_{13}=(i+0+0)(0+2) \\ {A}_{13}=2i \end{gathered}$$

Similarly, other elements are

$$\begin{gathered} A_{21}=⟨2\left|\right(i|1⟩-2|2⟩-i|3⟩)(-i⟨1|+2⟨3\left|\right)\mid 1⟩ \\ {A}_{21}=2i \end{gathered}$$

$$\begin{gathered} A_{22}=⟨2\left|\right(i|1⟩-2|2⟩-i|3⟩)(-i⟨1|+2⟨3\left|\right)\mid 2⟩ \\ {A}_{22}=0 \end{gathered}$$

$$\begin{gathered} A_{23}=⟨2\left|\right(i|1⟩-2|2⟩-i|3⟩)(-i⟨1|+2⟨3\left|\right)\mid 3⟩ \\ {A}_{23}=-4 \end{gathered}$$

$$\begin{gathered} A_{31}=⟨3\left|\right(i|1⟩-2|2⟩-i|3⟩)(-i⟨1|+2⟨3\left|\right)\mid 1⟩ \\ {A}_{31}=-1 \end{gathered}$$

$$\begin{gathered} A_{32}=⟨3\left|\right(i|1⟩-2|2⟩-i|3⟩)(-i⟨1|+2⟨3\left|\right)\mid 2⟩ \\ {A}_{32}=0 \end{gathered}$$

$$\begin{gathered} A_{33}=⟨3\left|\right(i|1⟩-2|2⟩-i|3⟩)(-i⟨1|+2⟨3\left|\right)\mid 3⟩ \\ {A}_{33}=-2i. \end{gathered}$$

Determining that A is hermitian or not

We can now write out the full matrix explicitly:

$A=\left(\begin{array}{ccc}1& 0& 2i\\ 2i& 0& -4\\ -1& 0& -2i\end{array}\right)$

This is not a hermitan matrix (note, for example, that $A_{21}\ne A_{21}^{*}$).