Problem with normalization wave function position/momentum space

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Homework Statement


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

Probability density of measuring momentun [itex]p_0[/itex] and third component of spin [itex]- \frac{\hbar}{2}[/itex]

And probability of measuring a state with momentum between 0 and [itex]p_0[/itex]


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 [itex]\frac{1}{\sqrt{2}}[/itex] and for the spatial part i just have to solve this
[itex] \int_{-\infty}^{\infty}\psi^{*}\psi dx = 1[/itex]

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

Now i have a spatial normalized wave function wich is

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

Since the momentum is not well defined in the spatial basis i cant obtain the probability of measuring [itex]p_0[/itex] right away, i have to use the momentum basis wave function, so i must do a fourier transform of the function like this

[itex] \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 [/itex]

Now of the probability to obtain p0 [itex] <p_0|\phi> [/itex]

Wich i assume is the integral from p0 to p0 wich is zero, and the probability between 0 and [itex] p_0[/itex] its the integral from 0 to [itex] p_0[/itex]

[itex] \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}}) [/itex]


But this has to be wrong, because if i take [itex] p_0 = \infty [/itex] it must be equal to 1 wich is not, whats wrong with the normalization ? or is something else ?

Thanks !
 

Answers and Replies

  • #2
vanhees71
Science Advisor
Insights Author
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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.
 

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