# Question about equivalence of inner product in Dirac notation and the overlap integral

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.

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. $$\langle x \vert \psi \rangle = \psi(x)$$.

• Ok but what about ${\bf 1} = \sum_n | n \rangle \langle n |$? – Andrew Steane Nov 23 '19 at 13:29
• 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. – Max Stammer Nov 23 '19 at 13:35
• 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. – Andrew Steane Nov 23 '19 at 14:47