Last lecture we started learning Dirac notation. In my lecture notes it says that $\langle\psi_i| \psi_j\rangle$ is defined as the inner product between the two wave functions, which is equal to the overlap integral $\int \psi ^* \psi dV$. Based on what I understand, $\langle\psi_i| \psi_j\rangle$ is just a sum of the $n^{th}$ components of $\psi_i$ and $\psi_j$ multiplied together. Which is what $\psi_j ^* \psi_i$ is. Why is it equal to the overlap integral? I am probably getting confused with something basic here.


1 Answer 1


You can understand the relationship between these two expressions when using a common trick performed in quantum mechanical calculations, which is multiplication by one.

The unit matrix in quantum mechanics can be expressed as the sum of the outer product over a complete basis, i.e. $ 1 = \sum_n \vert n \rangle \langle n \vert$, where the states $ \vert n \rangle$ form a complete basis in your Hilbert space. The same completness relation can be used for the eigenstates of your coordinate operator, which is given by

$$ 1 = \int \text dx \vert x \rangle \langle x \vert, $$

Inserting this into your inner product you obtain

$$ \langle \psi_i \vert \psi_j \rangle = \langle \psi_i \vert \int \text dx \vert x \rangle \langle x \vert \vert \psi_j \rangle = \int \text dx \langle \psi_i \vert x \rangle \langle x \vert \psi_j \rangle = \int \text dx\, \psi_i^*(x) \psi_j(x), $$

where we have used that the projection of your state vector on an eigenstate of the momentum operator is the wavefunction in coordinate representation, i.e. $\langle x \vert \psi \rangle = \psi(x)$.

  • $\begingroup$ Ok but what about ${\bf 1} = \sum_n | n \rangle \langle n |$? $\endgroup$ Commented Nov 23, 2019 at 13:29
  • $\begingroup$ I am not sure what you mean by this, but if you insert this discrete basis you would simply get $\langle \psi_i \vert \psi_j \rangle = \sum_n \langle \psi_i \vert n \rangle \langle n \vert \psi_j \rangle$. Thus you have expressed your inner product in terms of a new basis. $\endgroup$
    – MST
    Commented Nov 23, 2019 at 13:35
  • 1
    $\begingroup$ Yes. My point is that the original question is all about the distinction between a discrete and a continuous basis, and I felt that your answer, although it makes a useful contribution, does not really address that distinction or give insight into the sense in which a wavefunction is like a vector with an infinite number of components. $\endgroup$ Commented Nov 23, 2019 at 14:47

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