- #1
Rubber Ducky
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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.