How does $U(x,t;x',0) to delta(x-x')$ as $t to 0$ preserve the norm?

Question

If the propagator approaches the dirac delta as the difference in time decreases then we would expect
$$lim_{t to 0} langle Psi(x,t) | Psi(x',0) rangle$$
$$int_{- infty}^{+ infty} |Psi(x',0|^2 delta(x-x')^2 dx$$
But how is that ever equal to 1?

yuggib · Accepted Answer

The integral kernel of the evolution operator converges (in a strong sense, i.e. when applied to wavefunctions) to the integral kernel of the identity operator, that is the delta distribution, as the time interval approaches zero.
However, the action of operators described by integral kernels on wavefunctions is not the one the OP writes.
In particular
$$langle psi (0),psi(t)rangle=langle psi (0),U(t)psi(0)rangle=int mathrm{d}xmathrm{d}x' psi(x)U(t;x,x')psi(x')$$
The limit $tto 0$ of the scalar product can be taken because of the strong continuity property of evolution operators (in particular it can be roughly speaking exchanged with the integral), yielding
$$lim_{tto 0}langle psi (0),psi(t)rangle=int mathrm{d}xmathrm{d}x' psi(x)delta(x-x')psi(x')=langle psi(0),psi(0)rangle$$

How does $U(x,t;x',0) to delta(x-x')$ as $t to 0$ preserve the norm?

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