Problem with normalization wave function position/momentum space

[WarDieS]

Homework Statement

We start with a pure state at t=0 of an electron is
$C e^{- a^2 x^2} \left(\begin{array}{c} 1\\ i \end{array}\right)$

Probability density of measuring momentun $p_0$ and third component of spin $- \frac{\hbar}{2}$

And probability of measuring a state with momentum between 0 and $p_0$

Homework Equations

Fourier Transform

The Attempt at a Solution

I have a wave function with the spatial and spinor (Z basis),i know the normalization for the spinor is $\frac{1}{\sqrt{2}}$ and for the spatial part i just have to solve this
$\int_{-\infty}^{\infty}\psi^{*}\psi dx = 1$

Wich gives me
$C = (\frac{2}{\pi})^{1/4} \sqrt{a}$

Now i have a spatial normalized wave function which is

$\psi = (\frac{2}{\pi})^{1/4} \sqrt {a} e^{- a^2 x^2}$

Since the momentum is not well defined in the spatial basis i can't obtain the probability of measuring $p_0$ right away, i have to use the momentum basis wave function, so i must do a Fourier transform of the function like this

$\frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{\infty}\psi e^{\frac{-i p x}{\hbar}} dx = \frac{1}{\sqrt{a \hbar}} \frac{1}{(2 \pi)^{1/4}} e^-{\frac{p^2}{4 a^2 \hbar^2}}=\phi$

Now of the probability to obtain p0 $<p_0|\phi>$

Wich i assume is the integral from p0 to p0 which is zero, and the probability between 0 and $p_0$ its the integral from 0 to $p_0$

$\int_{0}^{p_0} \phi p dp = \frac{2^{3/4} (a \hbar)^{3/2}}{\pi^{1/4}} (1-e^{\frac{-p_0^2}{4 a^2 \hbar^2}})$


But this has to be wrong, because if i take $p_0 = \infty$ it must be equal to 1 which is not, what's wrong with the normalization ? or is something else ?

Thanks !

 

[vanhees71]

The idea with the Fourier trnsform to get the momentum wave function is correct. Now just remember how to get the probability density (momentum distribution) from the momentum wave function.