# 1 Coordinate Transformations

Given complex vector space $$V$$, the set of all linear functions $$L:V\to \mathbb{C}$$ forms the dual space to $$V$$, which is also a vector space denoted by $$V^*$$. When $$V$$ is finite dimensional, $$V^*$$ has the same dimension as $$V$$. The pairing of vector spaces to their duals requires disambiguation of elements in $$V$$, called contravariant vectors or kets, and operatorsin $$V^*$$, called dual vectors, covectors, covariant vectors, or bras.

Covariant components transform synergistically (“co”) in the same way as the basis vectors, whereas contravariant components transform anti-synergistically (“contra”) in the opposite way to basis vectors. Transformation of basis vectors refers to mapping of the basis vectors in the original coordinate system into new basis vectors. Transformation of vector components refers to the change in the vector components (coefficients) relative to two different sets of coordinate axes.

$\underbrace{(New\ basis\ vectors,\ ν')=(Transform\ matrix,\ T)\times (Original\ basis\ vectors,\ ν)}_{base\ vectors\ covary}\ ,$

$\underbrace{(Vector\ components\ in\ the\ new\ coordinate\ system,\ u_{ν'})=\\ (Inverse\ Transpose\ Transform\ matrix,\ (T^t)^{-1})\times (vector\ components\ in\ the\ original\ coordinate\ system,\ u_ν}_{base\ vectors\ covary}\ ,$

Denote the new basis vector $$v' = \begin{pmatrix} v_{1}' \\ v_{2}' \end{pmatrix}$$, and the original basis vectors $$v = \begin{pmatrix} v_{1} \\ v_{2} \end{pmatrix}$$. Denote the vector components in the new coordinate system $$u_{v^\prime} = \begin{pmatrix} u^{1\prime} \\ u^{2\prime} \end{pmatrix}$$, and the vector in the old coordinate system $$u_v = \begin{pmatrix} u^{1} \\ u^{2} \end{pmatrix}$$.

Any given (fixed) vector $$\alpha\in \mathbb{R}^2$$ would have unique representations in either the new ($$v'$$) or the old ($$v$$) coordinate systems

$\alpha \equiv \left \langle \begin{pmatrix} u^{1\prime}\\u^{2\prime} \end{pmatrix} | \begin{pmatrix} v_{1}^\prime \\ v_{2}^\prime \end{pmatrix} \right \rangle = \underbrace{\begin{pmatrix} u^{1\prime} & u^{2\prime} \end{pmatrix}}_{new\ components} \begin{pmatrix} v_{1}^\prime \\ v_{2}^\prime \end{pmatrix} \equiv \begin{pmatrix} u^{1} & u^{2} \end{pmatrix} \underbrace{\begin{pmatrix} v_{1} \\ v_{2} \end{pmatrix}}_{old\ basis}\ .$

Then, there exists a linear mapping $$T$$ transforming the old basis into the new basis, $$T:v\to v'$$, $\underbrace{\begin{pmatrix} v_1' \\ v_2' \end{pmatrix}}_{v'} = T \underbrace{\begin{pmatrix} v_1 \\ v_2 \end{pmatrix}}_{v}\ .$

$\alpha = \left \langle \begin{pmatrix} u^{1\prime}\\u^{2\prime} \end{pmatrix} | \begin{pmatrix} v_{1}^\prime \\ v_{2}^\prime \end{pmatrix} \right \rangle = \left \langle\begin{pmatrix} u^{1\prime}\\u^{2\prime} \end{pmatrix} |\ T \begin{pmatrix} v_{1} \\ v_{2} \end{pmatrix}\right \rangle = \left \langle T^t \begin{pmatrix} u^{1\prime}\\u^{2\prime} \end{pmatrix} | \begin{pmatrix} v_{1} \\ v_{2} \end{pmatrix}\right \rangle = \left \langle\begin{pmatrix} u^{1}\\u^{2} \end{pmatrix} | \begin{pmatrix} v_{1} \\ v_{2} \end{pmatrix}\right \rangle\ .$

Due to the uniqueness of the coordinate components, $$u_{v} = \begin{pmatrix} u^{1} \\ u^{2} \end{pmatrix}$$ and $$u_{v^\prime} = \begin{pmatrix} u^{1\prime} \\ u^{2\prime} \end{pmatrix}$$,

$\underbrace{\begin{pmatrix} u^{1}\\u^{2} \end{pmatrix}}_{u_{v}} = T^t \underbrace{\begin{pmatrix} u^{1\prime}\\u^{2\prime} \end{pmatrix}}_{u_{v'}}\ .$

The transposed basis transformation matrix $T^t = \begin{bmatrix} T_{1}^{1} & T_{2}^{1} \\ T_{1}^{2} & T_{2}^{2} \end{bmatrix}$ may be used to map the coefficient components of the vector $$\alpha = u^1v_1 + u^2v_2$$ from the old coordinate system, $$v$$, to components in the new basis, $$v^\prime$$:

$u_v = T^tu_{v\prime}\ \ \Longleftrightarrow \ \ u_{v^\prime} = \left(T^t\right)^{-1}u_v\ .$

# 2 Example

Here is one example of a change of basis transformation $\left| \begin{array}{cc} v_1' = v_1 + 2 v_2 \\ v_2' = v_1 + 3 v_2 \end{array} \right.\ .$

$T\equiv \begin{pmatrix} T_1^1 & T_1^2 \\ T_2^1 & T_2^2 \end{pmatrix}= \begin{pmatrix} 1 & 2 \\ 1 & 3 \end{pmatrix}\ , \ \ T^{-1} = \begin{pmatrix} 3 & -2 \\ -1 & 1 \end{pmatrix}\ ,$

$T^t\equiv \begin{pmatrix} T_1^1 & T_2^1 \\ T_1^2 & T_2^2 \end{pmatrix}= \begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix}\ , \ \ (T^t)^{-1} = \begin{pmatrix} 3 & -1 \\ -2 & 1 \end{pmatrix}\ ,$

$ν'=Tν\ \ , \ \ v=T^{-1}v' \ .$

Suppose the vector $$u_ν=u^1 ν_1+u^2 ν_2$$ in the old system and in the new coordinate system, $$u_{ν'} = u_{ν'}^1 ν_1' + u_{ν'}^2 ν_2'$$. then,

$u_ν=u^1 ν_1+u^2 ν_2 = \left\langle \begin{pmatrix} u^1 \\ u^2 \end{pmatrix} | \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} \right\rangle= u^1 (3ν_1' - 2ν_2') + u^2 (-ν_1' + ν_2')=\cdots =\\ \underbrace{(3u^1-u^2)}_{u_{ν'}^1} ν_1' + \underbrace{(-2u^1 + u^2)}_{u_{ν'}^2} v_2' = u_{ν'}\ .$

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