Normalization and the probability amplitude

[redtree]

Given two probability amplitude wavefunctions, one in position space $\psi(r,k)$ and one in wavenumber space $\phi(r,k)$, where $r$ and $k$ are Fourier conjugates, how is it possible for the modulus squared, i.e., probability density, of BOTH wavefunctions to be normalized? It seems that only one of the two probability densities can be normalized, and one must choose to normalize in either position OR wavenumber space. Is my understanding correct?

 

[stevendaryl]

Given two probability amplitude wavefunctions, one in position space $\psi(r,k)$ and one in wavenumber space $\phi(r,k)$, where $r$ and $k$ are Fourier conjugates, how is it possible for the modulus squared, i.e., probability density, of BOTH wavefunctions to be normalized? It seems that only one of the two probability densities can be normalized, and one must choose to normalize in either position OR wavenumber space. Is my understanding correct?

On the contrary, the norm of a wavefunction is the same whether you normalize in position space or momentum (wavenumber) space.

I don't know what you mean by $\psi(r,k)$. Let's simplify to one spatial dimension, so a wave function is just a function $\psi(x)$ in position space. Then the corresponding representation in momentum space, $\tilde{\psi}(k)$ is defined by:

$\tilde{\psi}(k) = \frac{1}{\sqrt{2\pi}} \int e^{-ikx} \psi(x) dx$. Then the norms are the same:

$\int (\psi(x))^* \psi(x) dx = \int (\tilde{\psi}(k))^* \tilde{\psi}(k) dk$

 

[redtree]

Assume the probability density is the normalized Gaussian, such that:
\begin{equation}
\begin{split}
\left|\phi(x) \right|^2&=\frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{x^2}{2 \sigma^2}}
\end{split}
\end{equation}

Thus:
\begin{equation}
\begin{split}
\int_{-\infty}^{\infty} dx\frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{x^2}{2 \sigma^2}}&=1
\end{split}
\end{equation}Where:
\begin{equation}
\begin{split}
\mathcal{F}\left[\frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{x^2}{2 \sigma^2}}\right]&=e^{-2 \pi^2 \sigma^2 k^2}
\end{split}
\end{equation}

However:
\begin{equation}
\begin{split}
\int_{-\infty}^{\infty} dk
e^{-2 \pi^2 \sigma^2 k^2}&=\frac{1}{\sqrt{2 \pi \sigma^2}}
\\
&\text{i.e., } <>1
\end{split}
\end{equation}

So what am I missing?

 

[stevendaryl]

You don't take the Fourier transform of the square of the function, you take the Fourier transform of the function, and then square it.

So your function is $\phi(x) = (2\pi \sigma^2)^{-\frac{1}{4}} e^{-\frac{x^2}{4\sigma^2}}$

Then $\tilde{\phi}(k) = (2\pi \sigma^2)^{-\frac{1}{4}} (2\pi)^{-\frac{1}{2}} \int e^{-\frac{x^2}{4\sigma^2} } e^{-ikx} dx = (2\pi)^{-\frac{3}{4}} (\sigma)^{-\frac{1}{2}} \int e^{-\frac{x^2}{4\sigma^2} } e^{-ikx} dx$

To do the integral, let $x = 2\sigma y + i \sigma k$. Then $e^{-\frac{x^2}{4\sigma^2} } e^{-ikx} = e^{-(y+ik \sigma)^2 - \sigma^2 k^2}$

So $(2\pi)^{-\frac{3}{4}} (\sigma)^{-\frac{1}{2}} \int e^{-\frac{x^2}{4\sigma^2} } e^{-ikx} dx = (2\pi)^{-\frac{3}{4}} (\sigma)^{-\frac{1}{2}} 2 \sigma e^{-\sigma^2 k^2} \sqrt{\pi}$

So

$\tilde{\phi}(k) = (\frac{2\sigma^2}{\pi})^{\frac{1}{4}} e^{-\sigma^2 k^2}$

Square it to get the probability density:

$|\tilde{\phi}(k)|^2 = (\frac{2\sigma^2}{\pi})^{\frac{1}{2}} e^{-2\sigma^2 k^2}$

You'll find that $\int |\tilde{\phi}(k)|^2 dk = 1$.

 

[stevendaryl]

$|\tilde{\phi}(k)|^2 = (\frac{2\sigma^2}{\pi})^{\frac{1}{2}} e^{-2\sigma^2 k^2}$

It occurs to me that "wave number" might not be the same thing as my $k$. It might be defined so that $k \equiv 2\pi k'$ where $k'$ is the wave number. If that's the case, then the density as a function of $k'$ must be adjusted.

 

[redtree]

Got it; I figured I was making a simple mistake. I use $e^{-2 \pi i k x}$ instead of $e^{-i k x}$ for the Fourier transform so my answer a little different but normalizes just the same. Thanks!