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Showing that a momentum space wave function is normalized

  1. Oct 15, 2015 #1
    1. The problem statement, all variables and given/known data

    Using the following expression for the Dirac delta function: $$\delta(k-k')=\frac{1}{2\pi} \int_{-\infty}^{\infty}e^{i(k-k')x} \mathrm{d}x$$

    Show that if a position space wave function $$\Psi(x,t)$$ is normalized at time t=0, then it is also true that the corresponding momentum space $$\Phi(p_x,t)$$ is normalized at t=0.

    2. Relevant equations

    $$\Phi(p_x,0)=\frac{1}{\sqrt{2\pi \hbar}} \int_{-\infty}^{\infty} e^{-i\frac{p_xx}{\hbar}}\Psi(x,0)\mathrm{d}x$$

    From which it follows that $$\Phi^*(p_x,0)=\frac{1}{\sqrt{2\pi \hbar}} \int_{-\infty}^{\infty} e^{i\frac{p_xx}{\hbar}}\Psi^*(x,0)\mathrm{d}x$$

    3. The attempt at a solution

    We need to show that $$\int_{-\infty}^{\infty} \Phi(p_x,0)\Phi^*(p_x,0)\mathrm{d}p_x=1$$

    To get started, we look at the modulus squared of the momentum space wave function, which can be written as a double integral instead of a product of integrals: $$\Phi^*(p_x,0)\Phi(p_x,0)=\frac{1}{2\pi \hbar} \left( \int_{-\infty}^{\infty} e^{i\frac{p_xx}{\hbar}}\Psi^*(x,0)\mathrm{d}x \right)\left( \int_{-\infty}^{\infty} e^{-i\frac{p_xx}{\hbar}}\Psi(x,0)\mathrm{d}x \right)=\frac{1}{2\pi \hbar} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{i(\frac{p_xy}{\hbar}-\frac{p_xx}{\hbar})}\Psi(x,0)\Psi^*(y,0)\mathrm{d}y \mathrm{d}x$$

    From here, I'm pretty sure I'm meant to manipulate the double integral until it looks like I'm integrating a delta function, using the form of the Dirac delta given to me, to simplify, and then simplify further using the fact that the position wave function is normalized. I just can't see how to manipulate the integrand to get that far.
     
  2. jcsd
  3. Oct 15, 2015 #2

    TSny

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    OK. This looks good to me.

    You won't be able to do that. Try using your result to set up ##\int_{-\infty}^{\infty} \Phi(p_x,0)\Phi^*(p_x,0)\mathrm{d}p_x## and then see if you can manipulate to get the Dirac delta function.
     
  4. Oct 15, 2015 #3

    blue_leaf77

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    Integrate the last expression w.r.t. to ##p_x## and use the Dirac function definition.
     
  5. Oct 15, 2015 #4
    Thank you both for the suggestion. So I have ##\Phi^*(p_x,0)\Phi(p_x,0)=\frac{1}{2\pi \hbar} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{i(\frac{p_xy}{\hbar}-\frac{p_xx}{\hbar})}\Psi(x,0)\Psi^*(y,0)\mathrm{d}y \mathrm{d}x##, and so ##\int_{-\infty}^{\infty} \Phi^*(p_x,0)\Phi(p_x,0)\mathrm{d}p_x=\frac{1}{\hbar}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\Psi(x,0)\Psi^*(y,0) \left( \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i(\frac{y}{\hbar}-\frac{x}{\hbar})p_x} \mathrm{d}p_x\right)\mathrm{d}y\mathrm{d}x##
    ##=\frac{1}{\hbar}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\Psi(x,0)\Psi^*(y,0) \delta(\frac{y}{\hbar}-\frac{x}{\hbar})\mathrm{d}y\mathrm{d}x##

    With the change of variables ##y'=\frac{y}{\hbar},x'=\frac{x}{\hbar}##, we get ##\hbar\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\Psi(x'\hbar,0)\Psi^*(y'\hbar,0) \delta(y'-x')\mathrm{d}y'\mathrm{d}x'=\hbar\int_{-\infty}^{\infty}\Psi(x'\hbar,0)\Psi^*(x'\hbar,0) \mathrm{d}x'=\int_{-\infty}^{\infty}\Psi(x,0)\Psi^*(x,0) \mathrm{d}x=1##

    And so ##\Phi## is normalized at time ##t=0##. I'm unsure if I used the property of the delta function correctly, however, since the argument of ##\Psi^*## was ##y'\hbar## and not simply ##y'##...
     
  6. Oct 15, 2015 #5

    TSny

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    Your work looks good to me. You could have made the change of variable back in the px integral by letting k = px/ħ.
     
    Last edited: Oct 16, 2015
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