Introduction: Same Vector, Different Coordinates
A vector exists independently of any coordinate system. When we change from one basis to another, the vector stays the same but its coordinates change. Understanding how coordinates transform is essential for quantum mechanics, where we frequently switch between position, momentum, and energy representations.
The Change of Basis Matrix
Let \(\mathcal{B} = \{\vec{b}_1, \vec{b}_2\}\) and \(\mathcal{C} = \{\vec{c}_1, \vec{c}_2\}\) be two bases for the same space. The change of basis matrix \(P\) from \(\mathcal{C}\) to \(\mathcal{B}\) has columns that are the \(\mathcal{B}\)-coordinates of each \(\vec{c}_i\).
If \([\vec{v}]_\mathcal{B}\) denotes coordinates in basis \(\mathcal{B}\), then:
\[[\vec{v}]_\mathcal{B} = P \cdot [\vec{v}]_\mathcal{C}\]Worked Examples
Example 1: From Standard to Rotated Basis
Standard basis: \(\vec{e}_1 = (1, 0)\), \(\vec{e}_2 = (0, 1)\)
Rotated basis (45°): \(\vec{b}_1 = \frac{1}{\sqrt{2}}(1, 1)\), \(\vec{b}_2 = \frac{1}{\sqrt{2}}(-1, 1)\)
To find coordinates of \(\vec{v} = (1, 0)\) in the rotated basis:
We solve \(c_1 \vec{b}_1 + c_2 \vec{b}_2 = (1, 0)\):
\[\frac{c_1}{\sqrt{2}} - \frac{c_2}{\sqrt{2}} = 1, \quad \frac{c_1}{\sqrt{2}} + \frac{c_2}{\sqrt{2}} = 0\]Result: \(c_1 = \frac{1}{\sqrt{2}}\), \(c_2 = -\frac{1}{\sqrt{2}}\)
In the rotated basis, \((1, 0)_\text{std} = (\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})_\text{rot}\)
Example 2: The Inverse Relationship
If \(P\) changes from basis \(\mathcal{C}\) to \(\mathcal{B}\), then \(P^{-1}\) changes from \(\mathcal{B}\) to \(\mathcal{C}\):
\[[\vec{v}]_\mathcal{C} = P^{-1} \cdot [\vec{v}]_\mathcal{B}\]The inverse undoes the transformation—you can always go back.
The Quantum Connection
In quantum mechanics, changing basis corresponds to changing which observable you're analyzing. The Fourier transform is precisely a change of basis from position \(|x\rangle\) to momentum \(|p\rangle\):
\[\langle p | \psi \rangle = \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{\infty} e^{-ipx/\hbar} \langle x | \psi \rangle \, dx\]The wavefunction \(\psi(x)\) and its Fourier transform \(\tilde{\psi}(p)\) describe the same quantum state in different bases.