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}<br /> 1\\<br /> i<br /> \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

which 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 &lt;p_0|\phi&gt;

which 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.