# Why does replacing bra and ket basis vectors by their row and column representations give the wrong matrix representation in a non-orthogonal basis?

I have a Hermitian operator (for a 2D Hilbert space) given by $$H=|\psi\rangle \langle \psi|+|\phi\rangle \langle \phi|$$ where $|\psi\rangle$ and $|\phi\rangle$ are normalized but not necessarily orthogonal. I know that the matrix representation of $H$ in the basis $\{|\psi\rangle,|\phi\rangle \}$ is $$\begin{pmatrix} 1 & \langle \psi|\phi \rangle \\ \langle \phi|\psi \rangle & 1 \end{pmatrix} .$$ But if I just replace $|\psi\rangle$ as \begin{pmatrix} 1 \\ 0 \end{pmatrix} and $|\phi \rangle$ as \begin{pmatrix} 0 \\ 1 \end{pmatrix} then I get matrix representation to be identity matrix. I know I have to use the fact that basis is not orthogonal to get the correct matrix representation but the replacing method would have just worked fine and given correct representation if the basis was orthonormal. What am I missing?

• You are replacing by orthogonal vectors, violating your previous supposition of them not being orthonormal. – Ignacio Vergara Kausel Apr 10 '15 at 10:22
• @IgnacioVergaraKausel that should probably be an answer – David Z Apr 10 '15 at 11:11

If $|\phi⟩$ and $|\psi⟩$ are linearly independent, then it is always possible to assign them to the column vectors $$|\phi⟩\mapsto\begin{pmatrix}1\\0\end{pmatrix} \text{ and } |\psi⟩\mapsto\begin{pmatrix}0\\1\end{pmatrix},$$ but if they're not orthogonal you're obviously going to need to work harder on the representation of the inner product in this basis.
The cleanest way to do this is to go back to thinking of the bras $⟨\phi|$ and $⟨\psi|$ as the dual basis for the dual space of $\mathcal H$. Thus they are linear functionals over $\mathbb C^2$, represented by row vectors $$⟨\phi|\mapsto\left(a\,\,\,b\right) \text{ and } ⟨\psi|\mapsto\left(c\,\,\,d\right).$$ The coefficients in these covectors are fixed by the requirement that $⟨\phi|$ and $⟨\psi|$ reproduce the old scalar product: \begin{align} \left(a\,\,\,b\right)\begin{pmatrix}1\\0\end{pmatrix}&=1,\quad& \left(a\,\,\,b\right)\begin{pmatrix}0\\1\end{pmatrix}&=⟨\phi|\psi⟩, \\ \left(c\,\,\,d\right)\begin{pmatrix}1\\0\end{pmatrix}&=⟨\psi|\phi⟩,\quad& \left(c\,\,\,d\right)\begin{pmatrix}0\\1\end{pmatrix}&=1. \end{align} This fixes the representation as $$⟨\phi|\mapsto\left(1\,\,\,⟨\phi|\psi⟩\right) \text{ and } ⟨\psi|\mapsto\left(⟨\psi|\phi⟩\,\,\,1\right).$$
Having done this, you get the correct representation for $H$: $$H=|\phi⟩⟨\phi|+|\psi⟩⟨\psi|= \begin{pmatrix}1\\0\end{pmatrix}\left(1\,\,\,⟨\phi|\psi⟩\right) + \begin{pmatrix}0\\1\end{pmatrix}\left(⟨\psi|\phi⟩\,\,\,1\right) = \begin{pmatrix} 1 & \langle \phi|\psi \rangle \\ \langle \psi|\phi \rangle & 1 \end{pmatrix}.$$