02 Quick Notes on Linear Transformation

Author

Guest_1013

Published

2026-08-07

Transformation between Coordinate Systems

We have 2 sets of coordinates in space, namely (\hat{x}_i) and (\hat{x}^{\prime}_i), i = 1, 2, ... N.

Suppose there is a linear transformation between these 2 coordinates:

(..., \hat{x}^{\prime}_i, ...) = (...,\hat{x}_j, ...) M \tag{1}

Transformation of Coordinate Components

Now, consider a vector \bm{A}. It apparently has 2 different representations in 2 coordinate systems:

\bm{A} = (...,\hat{x}_j, ...) \cdot (..., A_i, ...)^\text{T} = (...,\hat{x}^{\prime}_j, ...) \cdot (..., A^{\prime}_i, ...)^\text{T} \tag{2}

With Equation 1 , we have:

\begin{aligned} \bm{A}&=(...,\hat{x}_{j},...)M\cdot(...,A_{i}^{\prime},...)^{\text{T}} \\ &=\hat{x}_{j}M_{ji}A_{i}^{\prime}=(...,\hat{x}_{j},...)\cdot M(...,A_{i}^{\prime},...)^{\text{T}} \\ &=(...,\hat{x}_{j},...)\cdot(...,A_{i},...)^{\text{T}} \\ \end{aligned}

which means

(...,A_{i},...)^{\text{T}}=M(...,A_{i}^{\prime},...)^{\text{T}}

or

(...,A^{\prime}_{i},...)^{\text{T}}=M^{-1}(...,A_{i},...)^{\text{T}} \tag{3}

Also, for orthogonal coordinate systems, M^\text{T} = M^{-1}.

Dirac Notation

For real coordinates that are orthonormal (which is often the case),

\ket{\hat{x}^{\prime}_j} = \sum_i \ket{\hat{x}_i} \braket{\hat{x}_i | \hat{x}^{\prime}_j} = \ket{\hat{x}_i}M_{ij} \\ \Rightarrow M_{ij} = \braket{\hat{x}_i | \hat{x}^{\prime}_j}

For vectors,

\ket{A} = \ket{\hat{x}_i}A_i \Rightarrow A_i = \braket{\hat{x}_i|A}

Then,

\begin{aligned} A_i &= \braket{\hat{x}_i|A} \\ &= \braket{\hat{x}_i|\hat{x}^{\prime}_j}\braket{\hat{x}^{\prime}_j|A} \\ &= \braket{\hat{x}_i|\hat{x}^{\prime}_j}A^{\prime}_j\\ &= M_{ij} A^{\prime}_j \end{aligned}

or

A^{\prime}_j = \braket{\hat{x}^{\prime}_j|\hat{x}_i}A_i = M_{ij}^* A_i\\ \Rightarrow M^\prime = M^\dagger = M^\text{T}