Review of Linear Algebra in Quantum Mechanics (III)

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:

  1. \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.
  2. \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).
  3. 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.
  4. 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.
  5. \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).
  6. \langle 1\,f|\psi\rangle=1\,\langle f|\psi\rangle=\langle f|\psi\rangle, and thus 1\,f=f.
  7. \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.
  8. \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.

Location: 美国加利福尼亚帕罗奥图

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