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Problem with normalization wave function position/momentum space
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[QUOTE="WarDieS, post: 4637805, member: 231907"] [h2]Homework Statement [/h2] 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] [h2]Homework Equations[/h2] Fourier Transform [h2]The Attempt at a Solution[/h2] 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 which 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 can't 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 which 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 which is not, what's wrong with the normalization ? or is something else ? Thanks ! [/QUOTE]
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Problem with normalization wave function position/momentum space
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