Confusion related to momentum eigenstates

[amjad-sh]

When we deal with an infinite-dimensional basis,the normalization condition of this basis becomes <x|x'>=δ(x-x')(here for example the position basis).Same thing for momentum eigenstates <p|p'>=δ(p-p').
Lets look now on the eigenvalue problem of the momentum operator:
$\hat p | p \rangle =p | p\rangle$
projecting $\langle x |$ on both sides this will yield to a differential equation of the form :

$-iħ\frac{d\psi_{p}(x)}{dx}=p\psi_{p}(x)$

where $\psi_{p}(x)=\langle x | p\rangle$

this will finally yield that $\psi_{p}(x)=\frac{1}{√(2πħ)}e^{ipx/ħ}$

so $| p \rangle \leftrightarrow \frac{1}{√2πħ} (e^{ ipx_{1} } | x_{1} \rangle +e^{ ipx_{2} } |x_{2} \rangle+ . . . . . +e^{ ipx_{n} } | x_{n} \rangle)$ where n goes to infinity and $x_{n}=x_{n-1}+dx$
but doesn't this violate<x|x'>=δ(x-x')? since if we project,for example,<x1| on |p> and we take by consideration δ(x-x')=<x|x'>, the result will be <x|p>=∞ and not $\frac{1}{√2πħ}e^{ ipx_{1} }$ ?

 

[vanhees71]

I do not understand what you mean with the last paragraph. The consistency between the generalized position and momentum eigen bases is given by a well-known theorem in Fourier analysis:
$$\langle x|x' \rangle=\int_{\mathbb{R}} \mathrm{d} p \langle x|p \rangle \langle p|x' \rangle=\int_{\mathbb{R}} \mathrm{d} p \frac{1}{2 \pi \hbar} \exp[\mathrm{i} p(x-x')/\hbar]=\delta(x-x').$$

 

[amjad-sh]

I do not understand what you mean with the last paragraph.

I mean multiply <x1| by |p> , the result will be 1/√(2πħ) *exp(ipx1)*<x1|x1> and as <x1|x1>=∞ then <x1|p>=∞ and not 1/√(2πħ)*exp(ipx1).

 

[blue_leaf77]

$| p \rangle \leftrightarrow \frac{1}{√2πħ} (e^{ ipx_{1} } | x_{1} \rangle +e^{ ipx_{2} } |x_{2} \rangle+ . . . . . +e^{ ipx_{n} } | x_{n} \rangle)$

That's where the mistake lies because that's not the proper way to expand a state into position basis. You should instead operate the identity operator $1 = \int |x\rangle \langle x| dx$ to the ket $|p\rangle$.
$$
|p\rangle = \int |x\rangle \langle x|p\rangle dx
$$
You can easily see if you project this onto certain position basis $|x'\rangle$, you will obtain what you want.

 

[vanhees71]

I mean multiply <x1| by |p> , the result will be 1/√(2πħ) *exp(ipx1)*<x1|x1> and as <x1|x1>=∞ then <x1|p>=∞ and not 1/√(2πħ)*exp(ipx1).

I don't understand your math at all. As you have derived correctly yourself, the generalized momentum eigenstates in the position representation read
$$u_p(x)=\langle x|p \rangle=\frac{1}{\sqrt{2 \pi \hbar}} \exp(\mathrm{i} p x)$$
without any undetermined (diverging) factors.