Going from kronecker deltas to dirac deltas

[MaestroBach]

TL;DR Summary

Reading the intro to a book on non equilibrium green's functions, where it defines the orthonormality relation of two kets in the continuum formulation. Not sure how to understand the dirac deltas.

I'm reading Stefanucci's Nonequilibrium Many Body Theory of Quantum Systems.

In the first chapter, where it goes over basic quantum mechanics, it first defines the usual orthonormality condition I'm familiar with,

$$\langle n' | n \rangle = \delta_{n, n'} $$

where $$ | n \rangle$$ is the ket describing a particle in the interval $x_n \pm \Delta/2$

However, it then goes on to say that in the continuum formulation, the orthonormality relation becomes:

$$\langle n' | n \rangle = \lim_{\Delta\to0}\frac{\delta_{n, n'}}{\Delta} = \delta(x_{n'}-x_n) $$

I am not sure how to think about this, specifically in the case where $x_{n'} = x_n$

Obviously in all other cases, then the kronecker delta i'm used to is equivalent to the dirac delta, but in that one case, the book seems to be implying that instead of = 1, $\langle n' | n \rangle = \infty$?

 

[anuttarasammyak]

Kronecker's delta is for discrete physical quantity. Dirac's delta is for continuous physical quantity, e.g. coordinate x and momentum p.
$$<x'|x^">=\delta(x'-x^")$$
$$<x'|x'>=\delta(0)$$
infinity, so eigen vector of coordinate x whose eigenvalue is x’ cannot be normalized.

 

[MaestroBach]

Kronecker's delta is for discrete physical quantity and Dirac's delta is for continuous physical quantity, e.g. coordinate x and momentum p.
$$<x'|x^">=\delta(x'-x^")$$
$$<x'|x'>=\delta(0)$$
infinity, so eigen vector of coordinate x whose eigenvalue is x’ cannot be normalized.

I see...

Later in the same chapter, for many particles, this is given:

$$\psi(x) | y_1 ... y_N \rangle = \sum{k=1}^{N}(\pm)^{N + k} \delta(x - y_k) | y_1 ... y_{k-1} y_{k+1} ... y_N \rangle $$,

where $\psi(x)$ is the annihilation operator and x is another coordinate that CAN equal y (To be more specific, $y = r \sigma$). Does this imply that if $x = y_k$, then the entire state essentially blows up when acted on by $\psi(x)$? I guess I'm just confused because this seems different from the behavior that an annihilation operator would have in a discrete context.

My apologies if I'm not giving enough context here.

 

[anuttarasammyak]

$$\psi(x) | y_1 ... y_N \rangle = \sum_{k=1}^{N}(\pm)^{N + k} \delta(x - y_k) | y_1 ... y_{k-1} y_{k+1} ... y_N \rangle$$
Applying $\int dx$
$$\int \ dx\ \psi(x) | y_1 ... y_N \rangle = \sum_{k=1}^{N}(\pm)^{N + k} | y_1 ... y_{k-1} y_{k+1} ... y_N \rangle$$
It seems to make sense.