Dual Vector Space
Given a vector space $V$, its dual $V^*$ is a set of linear maps $f:\,V\rightarrow \mathbb{C}$. Following the Dirac notation, we denote such map $f$ by $\langle f|$. When acting on a vector $|\psi\rangle$, it leads to a complex number denoted by $\langle f|\psi\rangle$. The map is linear such that \begin{equation}\langle f|\,\Big(c_1|\psi\rangle + c_2|\phi\rangle\Big)=c_1\,\langle f|\psi\rangle+c_2\langle f|\phi\rangle\,.\end{equation} $V^*$ becomes a vector space when the linear maps are equipped with the addition and scaling operations: \begin{eqnarray}\langle f+g| \psi\rangle&=& \langle f| \psi\rangle + \langle g| \psi\rangle\,,\\ \langle c\,f|\psi\rangle&=& c \langle f|\psi\rangle\,,\end{eqnarray} for all $f, g \in V^*$, $|\psi\rangle\in V$ and $c\in\mathbb{C}$.
Note:
- $\langle f|\psi\rangle$ is bilinear in each of the two arguments.
- We will use Greek letters for vectors in the vector space and use English letters for maps in the dual space.
[Exercise 1] Verify that $V^*$ satisfies the eight axioms of vector space.
[Solution] The definition of $V^*$ fully leverages the properties of complex numbers:
- $\langle f+g| \psi\rangle= \langle f| \psi\rangle + \langle g| \psi\rangle= \langle g| \psi\rangle + \langle f| \psi\rangle= \langle g+f| \psi\rangle$, and thus $f+g = g+f$.
- $\langle (f+g) + h| \psi\rangle= \langle f + g| \psi\rangle + \langle h| \psi\rangle= \langle f| \psi\rangle + \langle g| \psi\rangle + \langle h| \psi\rangle=\langle f| \psi\rangle + \langle g+h | \psi\rangle= \langle f+(g+h)| \psi\rangle$ , and thus $(f+g) + h=f+(g+h)$.
- The zero element in $V^*$ is the map $i$ such that $\langle i|\psi\rangle=0$ for any $|\psi\rangle\in V$. Now, for any $f\in V^*$, $\langle f+i |\psi\rangle=\langle f| \psi\rangle + \langle i| \psi\rangle=\langle f| \psi\rangle+0=\langle f| \psi\rangle$, and thus $f+i=f$.
- For any $f\in V^*$, $-f$ is the map such that $\langle -f|\psi\rangle=-\langle f|\psi\rangle$. $\langle f+(-f)|\psi\rangle=\langle f|\psi\rangle+\langle -f|\psi\rangle=\langle f|\psi\rangle-\langle f|\psi\rangle=0=\langle i|\psi\rangle$, and thus $f+(-f)=i$.
- $\langle c_1(c_2 f)| \psi\rangle= c_1 \langle c_2 f| \psi\rangle=c_1c_2\langle f| \psi\rangle=\langle (c_1c_2) f| \psi\rangle$, and thus $ c_1(c_2 f) = (c_1c_2 f)$.
- $\langle 1\,f|\psi\rangle=1\,\langle f|\psi\rangle=\langle f|\psi\rangle$, and thus $1\,f=f$.
- $\langle c\,(f+g)|\psi\rangle=c\,\langle f+g|\psi\rangle=c\langle f|\psi\rangle+c\langle g|\psi\rangle=\langle cf|\psi\rangle+\langle cg|\psi\rangle=\langle cf+cg|\psi\rangle$, and thus $c\,(f+g)=cf + cg$.
- $\langle (c_1+c_2)f|\psi\rangle=(c_1+c_2)\,\langle f|\psi\rangle=c_1\langle f|\psi\rangle+c_2\langle f|\psi\rangle=\langle c_1f|\psi\rangle+\langle c_2 f|\psi\rangle=\langle c_1f+c_2f|\psi\rangle$, and thus $(c_1+c_2)f=c_1f+c_2f$.
Dual Basis
Given a
basis $|\alpha_1\rangle, \cdots, |\alpha_n\rangle$ in $V$, there is a specific basis $\langle b_1|, \cdots, \langle b_n|$ in $V^*$ satisfying the relations \begin{equation}\langle b_i| \alpha_j\rangle=\delta_{ij}\tag{1}\end{equation} for $i, j=1, \cdots, n$. In the following two exercises, we prove that such $\langle b_1|, \cdots, \langle b_n|$ as defined in Eq. (1) indeed form a basis in $V^*$.
[Exercise 2] Prove that $\langle b_1|, \cdots, \langle b_n|$ are linearly independent.
[Solution] Consider the equation $x_1\langle b_1|+\cdots+x_n\langle b_n|=\langle i|$. We iteratively act both side of the equation on the basis vector $|\alpha_j\rangle$ in $V$ and obtain $x_j=0$ for $j=1,\cdots, n$.
[Exercise 3] Prove that any linear map $\langle f|$ is a linear combination of $\langle b_1|, \cdots, \langle b_n|$.
[Solution] Indeed, for any $f$, we have $\langle f|=\sum_{j=1}^n \langle f|\alpha_j\rangle\,\langle b_j|$.
Remarks:
We can expand any linear map $\langle f|=\sum_{i=1}^n f_i\langle b_i|$ under the dual basis and expand any vector $|\psi\rangle=\sum_{i=1}^n c_i|\alpha_i\rangle$ under the basis. With the relation (1), we have \begin{equation}\langle f| \psi\rangle=\sum_{i,j=1}^n f_ic_j\langle b_i|\alpha_j\rangle=\sum_{i,j=1}^n f_ic_j\delta_{ij}=\sum_{i=1}^n f_ic_i\,.\end{equation} We can view the linear map and the vector as a row and column \begin{equation}\langle f| = \left[f_1, f_2, \cdots, f_n\right]\,,\quad\quad\quad |\psi\rangle =\left[\begin{array}{c}c_1\\ c_2\\ \vdots \\ c_n\end{array}\right]\,, \end{equation}$\langle f| \psi\rangle$ is just the ordinary matrix multiplication between a row and a column.
Unnatural Isomorphism
A vector space $V$ is unnaturally isomorphic to its dual space $V^*$. That is, given a fixed basis $|\alpha_1\rangle, \cdots, |\alpha_n\rangle$ in $V$ and the corresponding dual basis $\langle b_1|, \cdots, \langle b_n|$ in $V^*$, there is one-to-one correspondence between the vector space and dual space through the same coefficients: \begin{equation}|\psi\rangle=\sum_{i=1}^n \color{red}{c_i}|\alpha_i\rangle\in V\quad\longleftrightarrow\quad\langle f|=\sum_{i=1}^n \color{red}{c_i}\,\langle b_i|\in V^*\,.\tag{2}\end{equation}
However, the one-to-one correspondence (2) is NOT natural since its validation depends on a pre-selected basis and dual basis. If we
change the basis by some matrix $\mathbf{S}$, \begin{equation}\Big[|\alpha'_1\rangle\,\,\cdots\,\,|\alpha'_n\rangle\Big]=\Big[|\alpha_1\rangle\,\,\cdots\,\,|\alpha_n\rangle\Big] \mathbf{S}\,,\end{equation} the dual basis changes in the opposite way to maintain the relation (1): \begin{equation}\left[\begin{array}{c}\langle b''_1|\\ \vdots \\ \langle b''_n|\end{array}\right]=\mathbf{S}^{-1}\left[\begin{array}{c}\langle b_1|\\ \vdots \\ \langle b_n|\end{array}\right]\,. \end{equation} Consequently, for the same $|\psi\rangle$ and $\langle f|$ in Eq. (2), the coefficients of $|\psi\rangle$ in Eq. (2) become \begin{equation}\left[\begin{array}{c}c'_1\\ \vdots \\ c'_n\end{array}\right]=\mathbf{S}^{-1}\left[\begin{array}{c}c_1\\ \vdots \\ c_n\end{array}\right]\,,\end{equation} while the coefficients of $\langle f|$ in Eq. (2) become \begin{equation}\left[c''_1, \cdots, c''_n\right]= \left[c_1, \cdots, c_n\right]\mathbf{S}\,,\end{equation} which are not equal anymore after a basis transformation.
Note: a vector space $V$ becomes naturally isomorphic to its dual space $V^*$ once equipped with
an inner product.
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