Review of Linear Algebra in Quantum Mechanics (II)

Basis

A set of vectors $|\alpha_1\rangle, \cdots, |\alpha_n\rangle$ is a basis of a vector space if

• $|\alpha_1\rangle, \cdots, |\alpha_n\rangle$ are linearly independent. That is, the only solution of the equation $$\begin{equation}x_1 |\alpha_1\rangle+x_2 |\alpha_2\rangle+\cdots + x_n |\alpha_n\rangle =\mathbf{0}\end{equation}$$ is $x_1=\cdots=x_n=0$.
• Any vector $|\psi\rangle$ in the vector space is a linear combination of $|\alpha_1\rangle, \cdots, |\alpha_n\rangle$, i.e. $$\begin{equation}|\psi\rangle =c_1 |\alpha_1\rangle+c_2 |\alpha_2\rangle+\cdots + c_n |\alpha_n\rangle\end{equation}$$ for $c_1,\cdots, c_n\in \mathbb{C}$.

[Exercise 1] If there is vector $|\psi\rangle$ that can not be a linear combination of linearly independent vectors $|\alpha_1\rangle, \cdots, |\alpha_n\rangle$, then $|\alpha_1\rangle, \cdots, |\alpha_n\rangle, |\psi\rangle$ are linearly independent.

[Solution] Consider the equation $x_1 |\alpha_1\rangle+\cdots + x_n |\alpha_n\rangle +d|\psi\rangle =\mathbf{0}$, we have $d=0$. Otherwise, we have the relation $|\psi\rangle=\sum_{i=1}^n\left(-c_i/d\right)|\alpha_i\rangle$, contradicting with the assumption. The equation then reduces to $x_1 |\alpha_1\rangle+\cdots + x_n |\alpha_n\rangle =\mathbf{0}$ and thus $x_1=\cdots=x_n=0$.

[Exercise 2] Prove that the liner combination in the definition of basis is unique.

[Solution] Suppose $|\psi\rangle =\sum_{i=1}^n c_i |\alpha_i\rangle=\sum_{i=1}^nd_i |\alpha_i\rangle$, then we have $\sum_{i=1}^n (c_i-d_i) |\alpha_i\rangle=\mathbf{0}$. Since $|\alpha_1\rangle, \cdots, |\alpha_n\rangle$ are linear independent, we have $c_1=d_1,\cdots, c_n=d_n$.

Note:

• A basis is of the maximal number of linearly independent vectors.
• Given a set of linearly independent vectors, we can expand the set to be a basis by iteratively adding in a vector that can not be to linear combination of the vectors in the current set.

Linear transformation

Operators in quantum mechanics are linear transformations that map one vector to another vector satisfying the linear property $$\begin{equation}\cal{O}\Big(c_1|\psi\rangle + c_2|\phi\rangle\Big)=c_1\cal{O}|\psi\rangle + c_2 \cal{O}|\phi\rangle\,.\end{equation}$$ The only exception is the time-reversal operator which turns out be antilinear: $$\begin{equation}\cal{T}\Big(c_1|\psi\rangle + c_2|\phi\rangle\Big)=c^*_1\,\cal{T}|\psi\rangle + c^*_2\, \cal{T}|\phi\rangle\,.\end{equation}$$

A linear transformation is fully described by its transformation on the basis $\cal{O}|\alpha_j\rangle$ for $j=1,\cdots, n$. As a vector, $\cal{O}|\alpha_j\rangle$ can also be written as a linear combination of the basis: $$\begin{equation} {\cal{O}}|\alpha_j\rangle =O_{1j}|\alpha_1\rangle + O_{2j}|\alpha_2\rangle +\cdots+ O_{nj}|\alpha_n\rangle = \Big[|\alpha_1\rangle\,\,|\alpha_2\rangle\,\,\cdots\,\,|\alpha_n\rangle\Big]\begin{bmatrix} O_{1j} \\  O_{2j} \\ \vdots \\  O_{nj} \end{bmatrix}\end{equation}$$ for some coefficients $O_{1j},\cdots, O_{nj}\in \mathbb{C}$. As a result, $$\begin{equation}  \cal{O}\, \Big[|\alpha_1\rangle\,\,|\alpha_2\rangle\,\,\cdots\,\,|\alpha_n\rangle\Big]=\Big[|\alpha_1\rangle\,\,|\alpha_2\rangle\,\,\cdots\,\,|\alpha_n\rangle\Big]\underset{\text{matrix }\mathbf{O}}{\underbrace{\begin{bmatrix} O_{11} & O_{12} & \cdots & O_{1n} \\ O_{21} & O_{22} & \cdots & O_{2n} \\ \vdots  & \vdots  & \ddots & \vdots  \\ O_{n1} & O_{n2} & \cdots & O_{nn} \end{bmatrix}}}\,.\tag{1}\end{equation}$$ Remarks:

1. Under a basis, there is one-to-one correspondence between an operator and its matrix form  through Eq. (1).  We can use the words operator and matrix interchangeably.
2. There are addition and scaling operations for operators as defined by $$\begin{eqnarray}\Big({\cal{O}_1}+{\cal{O}_2}\Big)\,|\psi\rangle &:=& {\cal{O}_1}|\psi\rangle+{\cal{O}_2}|\psi\rangle\\ \Big(c\,\cal{O}\Big)\,|\psi\rangle &:=&c\Big( {\cal{O}}|\psi\rangle\Big)\end{eqnarray}$$ for any $|\psi\rangle$, which corresponds to the addition and scaling of matrices by Eq. (1).

Basis Transformation

The basis transformation $\cal{S}$ also has a matrix form $\mathbf{S}$ within the old basis $|\alpha_1\rangle, \cdots, |\alpha_n\rangle$: $$\begin{equation}\Big[|\alpha'_1\rangle\,\,\cdots\,\,|\alpha'_n\rangle\Big]\equiv\cal{S}\Big[|\alpha_1\rangle\,\,\cdots\,\,|\alpha_n\rangle\Big]=\Big[|\alpha_1\rangle\,\,\cdots\,\,|\alpha_n\rangle\Big]\mathbf{S}\,.\tag{2}\end{equation}$$

Under the new basis $|\alpha'_1\rangle, \cdots, |\alpha'_n\rangle$, the same operator $\cal{O}$ in Eq. (1) now a different matrix form: $$\begin{equation}\cal{O} \Big[|\alpha'_1\rangle\,\,|\alpha'_2\rangle\,\,\cdots\,\,|\alpha'_n\rangle\Big]=\Big[|\alpha'_1\rangle\,\,|\alpha'_2\rangle\,\,\cdots\,\,|\alpha'_n\rangle\Big]\mathbf{O'}\,.\tag{3}\end{equation}$$  We now solve the relations between matrices $\mathbf{O}$ and \mathbf{O'} as Eq. (1) and (3). Submitting Eq. (2) in Eq. (3) gives $\cal{O}\Big[|\alpha_1\rangle\,\,\cdots\,\,|\alpha_n\rangle\Big]\mathbf{S} = \Big[|\alpha_1\rangle\,\,\cdots\,\,|\alpha_n\rangle\Big] \mathbf{S}\mathbf{O'}$. Its left side is $\Big[|\alpha_1\rangle\,\,\cdots\,\,|\alpha_n\rangle\Big]\mathbf{O}\mathbf{S}$ by Eq. (1). As a result, we have the relation $$\begin{equation} \mathbf{O'}=\mathbf{S}^{-1}\,\mathbf{O}\,\mathbf{S}\,.\tag{4} \end{equation}$$ An operator's matrix forms under different basis are similar to one another. One topic in group representation theory is to find a suitable basis under which the operator's matrix form become block diagonal.