Showing that a momentum space wave function is normalized

  • #1

Homework Statement



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.

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

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.
 

Answers and Replies

  • #2
TSny
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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: $$\Phi^*(p_x,0)\Phi(p_x,0)=\frac{1}{2\pi \hbar} \left( \int_{-\infty}^{\infty} e^{i\frac{p_xx}{\hbar}}\Psi^*(x,0)\mathrm{d}x \right)\left( \int_{-\infty}^{\infty} e^{-i\frac{p_xx}{\hbar}}\Psi(x,0)\mathrm{d}x \right)=\frac{1}{2\pi \hbar} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{i(\frac{p_xy}{\hbar}-\frac{p_xx}{\hbar})}\Psi(x,0)\Psi^*(y,0)\mathrm{d}y \mathrm{d}x$$
OK. This looks good to me.

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, ....
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.
 
  • #3
blue_leaf77
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Integrate the last expression w.r.t. to ##p_x## and use the Dirac function definition.
 
  • #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'##...
 
  • #5
TSny
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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'##...
Your work looks good to me. You could have made the change of variable back in the px integral by letting k = px/ħ.
 
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