# Showing that a momentum space wave function is normalized

## 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.

TSny
Homework Helper
Gold Member
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.

blue_leaf77
Homework Helper
Integrate the last expression w.r.t. to ##p_x## and use the Dirac function definition.

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'##...

TSny
Homework Helper
Gold Member
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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