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?

  • $\begingroup$ 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. $\endgroup$ – 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.

  • $\begingroup$ "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. $\endgroup$ – Jodocus May 14 '18 at 16:41
  • 1
    $\begingroup$ @Jodocus: OK, in that case you're correct, and the answer is that we don't use the dual basis, we use the basis formed by the duals of our basis vectors. Be careful, though, because (for example) in relativity it's the other way around: we do use the dual basis. It's definitely confusing. $\endgroup$ – Javier May 14 '18 at 16:51
  • $\begingroup$ I see, so you vote for option #1? Things really start to get confusing when operators even stop being complex symmetric (which happens at my case) rendering left and right eigenvectors of such an operator different, so I want to fix one standard for myself once and forall and follow it stoically $\endgroup$ – Jodocus May 14 '18 at 16:58
  • $\begingroup$ @Jodocus yes, at first I thought you said something different, but now that I read it more carefully I agree with option #1. I also agree that it can get quite confusing! Maybe you should use component notation instead of Dirac's like relativists do. $\endgroup$ – Javier May 14 '18 at 17:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.