Calculating the Wavefunction Associated with any Ket Vector

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We can express any ket vector \(|\psi⟩\) (representing all the different possible states) in its “component form” as \(|\psi⟩=\sum_{i=1}^{N}\psi_i|i⟩\) where \(|i⟩\) are all the different possible basis vectors one could decompose \(|\psi⟩\) with respect to, \(\psi_i\) are the different components (which, in general, can be complex numbers), and the value of \(N\) is simply the number of dimensions in the space. The basis vectors \(|i⟩\) are by definition orthogonal vectors whose magnitudes equal one. This is analogous to the unit vectors \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\) which are, by definition, perpendicular vectors whose magnitudes equal 1.

We shall now derive an equation which will allow us to find the components \(\psi_i\) of any complex vector . Let’s start by taking the inner product between \(|\psi⟩\) and any basis vector \(|j⟩\) to get

$$⟨j|\psi⟩=\sum_{i=1}^{N}\psi_i⟨j|i⟩.$$

If \(i≠j\), then we are considering the inner product between two different basis vectors which, by definition, are orthogonal. Therefore all the terms \(⟨j|i⟩\) in the sum in which \(i≠j\) are zero. When \(i=j\), we are taking the inner product between the same two basis vectors which have equal magnitudes of one and point in the same direction; thus \(⟨j|i⟩\)=1\) when \(i=j\). This means that the inner product \(⟨j|i⟩\) simply is just the Kronecker delta \(𝛿_{ij}\); thus \(⟨j|i⟩=𝛿_{ij}\) and

$$⟨j|\psi⟩=\sum_{i=1}^{N}\psi_i𝛿_{ij}.$$ 

In the sum, all of the terms become zero except for the \(\psi_j𝛿_{jj}=α_j\) term. Thus, the equation simplifies to

$$⟨j|\psi⟩=\psi_j.\tag{23}$$

This result indicates that we're at the halfway point towards our goal of deducing that \(\psi_i=⟨L_i|\psi⟩\)—something I said, earlier, that I'd eventually prove. The only discrepancy between this equation and Equation (23) is the dummy variable (which doesn't matter) and the fact that there's the bra \(⟨j|\) instead of \(⟨L_i|\). But once we prove that the eigenvectors \(|L_i⟩\) of any observable \(\hat{L}\) form a complete orthonormal basis, then you'll be able to breath a sigh of relief. Since this analysis applies to any set of basis vectors \(|i⟩\), the generality of this analysis permits us to substitute \(|L_i⟩\) for \(|i⟩\) in order to obtain our desired result.