Kashmir said:
This implies that a delta function wavefunction in momentum space would give only one value on measurement. So the probability of measuring that particular eigenvalue should be ##1##.
If ##P## is the probability density then ##P d p=|\varphi(p)|^{2} d p## will give me the probability of getting a momentum value in a range ##dp## around ##p##
If Dirac delta function represents a state that results in only one measured value ##p##, we should then have ##P d p=|\varphi(p)|^{2} d p =1 ## but ##\varphi(p)=\delta\left(p-p^{\prime}\right)## so ##\delta^{2}(p-p) d p \neq 1##
So how does a Dirac delta function correspond to a single measurement of ##p## ?
@Nugatory ,
@vanhees71 ,
@PeroK , can you help me here?
It does not correspond to a single measurement of ##p## nor does it represent a state of a particle. The (pure) states of a particle can only be represented by a wave function that is square integrable, i.e., which can be normalized to 1:
$$\int_{\mathbb{R}} \mathrm{d} p |\varphi(p)|^2=1.$$
There is no proper square-integrable eigenfunction of the momentum operator ##\hat{p}##, but it has only a continuous spectrum ##p \in \mathbb{R}##. The corresponding "eigenfunctions" of such "generalized eigenvalues" are "generalized functions" or "distributions". They are not square-integrable.
That's easy to see in the position representation, where ##\hat{p}=-\mathrm{i} \hbar \partial_x##. The eigenvalue equation reads
$$-\mathrm{i} \hbar \partial_x u_p(x)=p u_p(x).$$
The solutions are
$$u_p(x)=N \exp(\mathrm{i} p x/\hbar).$$
That's for sure not square integrable, because ##|u_p(x)|^2=|N|^2=\text{const}##. Nevertheless all ##p \in \mathbb{R}## are "generalized eigenvalues" of the momentum operator and the eigenfunctions can be "normalized" to a Dirac-##\delta## distribution,
$$\langle u_p|u_{p'} \rangle=\int_{\mathbb{R}} u_p^*(x) u_{p'}(x)=\delta(p-p').$$
As is known from the theory of Fourier transformations for that we simply have to set the normalization constant to
$$N=\frac{1}{\sqrt{2 \pi \hbar}} \; \Rightarrow \; u_p(x)=\langle x|p \rangle=\frac{1}{\sqrt{2 \pi \hbar}} \exp(\mathrm{i} p x/\hbar).$$
They build a complete set of orthonormal functions in the sense that you can represent a dense subset of square integrable functions as a Fourier transform,
$$\psi(x)=\langle x|\psi \rangle = \int_{\mathbb{R}} \mathrm{d} p \langle x|p \rangle \langle p|\psi \rangle = \int_{\mathbb{R}} \mathrm{d} p \frac{1}{\sqrt{2 \pi \hbar}} \exp(\mathrm{i} p x/\hbar) \tilde{\psi}(p)$$
and in the other way
$$\tilde{\psi}(p) = \langle p|\psi \rangle=\int_{\mathbb{R}} \mathrm{d} x \langle p|x \rangle \langle x|\psi \rangle = \int_{\mathbb{R}} \mathrm{d} x \frac{1}{\sqrt{2 \pi \hbar}} \exp(-\mathrm{i} p x/\hbar) \psi(x).$$
It's also clear that the transformation from the "position-space wave function" to the "momentum-space wave function" is a unitary transformation since
$$\langle \psi_1|\psi_2 \rangle=\int_{\mathbb{R}} \mathrm{d} x \psi_1^*(x) \psi_2(x) = \int_{\mathbb{R}} \mathrm{d} p \tilde{\psi}_1^*(p) \tilde{\psi}_2(p).$$
