# Is there a consensus on the meaning of bras in non-orthogonal bases?

For a basis $\mathcal B$ of some vector space (be it Euclidean or Hilbert) $\mathcal H$ there always exists a dual basis $\mathcal B^*\subset \mathcal H^*$ s.t. $\phi^j(\phi_i)=\delta_i^j\ \forall \phi^j\in\mathcal B^*,\ \phi_i\in\mathcal B$. Furthermore, for an orthonormal basis, the map $\mathcal I: \mathcal H\to\mathcal H^*,\ \phi\to\langle \phi,\cdot\rangle$ is an isomorphism which establishes a one-to-one relationship between a basis and its dual counterpart, i.e. $\mathcal I(\phi_i)(\phi_j)=\langle\phi_i,\phi_j\rangle=\delta_{ij}$ or $\mathcal I(\phi_i)\equiv\phi^i$, essentially Riesz representation theorem. I always understood this as the motivation behind writing $\langle\phi_i|\equiv \phi^j=\mathcal I(\phi_i)$ in the Dirac notation. For orthonormal bases, the scalar product is the "natural" way to relate a basis to its dual basis.

Usually when doing quantum mechanics, this is actually the case, as bases are some sort of eigenbases of self-adjoint operators. However in quantum chemistry, bases quite often are non-orthogonal, i.e. $\langle \phi_i, \phi_j\rangle\neq\delta_{ij}$ for two basis vectors $\phi_{i,j}$. Note that I explicitely did not use Dirac-notation here, but the actual Hilbert space scalar product $\langle\cdot,\cdot\rangle$.

Is it true that the meaning of a bra $\langle\phi_i|$ in the sense stated above stays unchanged in this case? On the one hand I tend to believe that, as objects like an "overlap matrix" with components $S_{ij}=\langle\phi_i|\phi_j\rangle$ wouldn't make much sense otherwise, for that $\langle\phi_i|\phi_j\rangle=\delta_{ij}$ by definition of the dual basis and the Dirac notation (as I understand it). Thus the $\langle\phi_i|$-basis can no more be dual to $|\phi_i\rangle$ but instead must be dual to the basis $\mathcal B'=\{S^{-1}|\phi_i\rangle\}$ which in turn implies that the $S_{ij}$ are the components of a matrix that also transforms between $\mathcal B$ and $\mathcal B'$ - this makes me suspicious.

Or is that $\langle\phi_i|$ and $|\phi_j\rangle$ are still dual to oneanother and the definition of the matrix components w.r.t. $\mathcal B$ actually should be $S_{ij}=\langle\phi_i,\phi_j\rangle\neq\langle\phi_i|\phi_j\rangle=\delta_{ij}$, or in other words $\langle\phi_i|\equiv \mathcal I(\phi_i)\circ S^{-1}$? This would make appear Dirac's notation look like a bad abuse of notation as it hides away the $S^{-1}$ and still implies that $\langle\phi_i,\cdot\rangle$ is dual to $\phi_j$ which is only right for orthonormal bases.

A third possibility would be that the scalar product is actually changed like $\langle\cdot,\cdot\rangle\to\langle\cdot,S^{-1}\cdot\rangle$, but I doubt that.

In the end, all approaches are equivalent, but is there a consensus in which one is used? If yes, which one?

• I have never seen calculations in quantum mechanics that used a non-othonormal basis, so somebody may correct me here, but I would be amazed if the convention was not that bras were defined by the isomorphism in the Riesz representation theorem. If $\langle\phi|$ is defined as an element of the duel basis, then it is basis dependant and then so are expressions like $\langle\phi|A|\phi\rangle$. This seems incredibly unnatural to me. Quantum mechanics has a privileged inner product but the choice of basis is arbitrary, so it makes sense to define things in terms of the scalar product. – By Symmetry May 14 '18 at 16:05

I disagree with your statement that the inner product provides an isomorphism between bases of $V$ and of $V^*$; in every treatment I've seen, it provides an isomorphism of vectors and covectors. For any $|\psi\rangle \in V$, $\langle \psi |$ is defined by $\langle \psi | \phi \rangle \equiv \langle \psi | (| \phi \rangle) = (|\psi\rangle , |\phi\rangle)$, where I've used function notation $\langle \psi | (| \phi \rangle)$ to emphasize that $\langle \psi |$ is a functional.
Using this definition, the duals of the elements of a basis $\{|\phi_i\rangle \}$ are given by the inner product just like for any other vector, and if the basis is not orthonormal then we will have $\langle \phi_i | \phi_j \rangle = S_{ij}$, where $S_{ij}$ is not the identity matrix anymore; this is the point of defining the overlap matrix.
• "I disagree with your statement that the inner product provides an isomorphism between bases" What I was trying to emphasize was that for orthonormal bases, this isomorhpism yields the actual dual basis. If however the basis is not orthonormal, this remains an isomorhphism, but applied to a basis, this will not be the corresponding dual basis, but another one, as the dual basis always fulfills $\phi^i(\phi_j)=\delta^i_j$, irrespective of it being orthonormal or not. – Jodocus May 14 '18 at 16:41