I have a wave function $\Psi (x,t)$. According to the Max Born postulate, $\lvert\Psi (x,t)\rvert ^2$ is the probability density. This quantity specifies the probability, per length of the $x$ axis, of finding the particle near the coordinate $x$ at time $t$.
If at $t=0$ I make a Fourier transform for the momentum space, $$\phi(p)=\frac{1}{\sqrt{2 \pi \hbar}} \int _{-\infty} ^{+\infty} \psi(x)e^{-ipx/\hbar} dx$$ does $\vert\phi(p)\rvert ^2$ specifies the probability of finding the particle near the momentum $p$ at time $t=0 \hspace{1mm}$?
In this sense, given $\Psi(x,t)$, how could I write $\phi(p)$ at any time $t$, i.e. $\Phi(p,t)\hspace{1mm}$?


1 Answer 1


You’re exactly right: $|\phi(p)|^2$ gives the probability of measuring momentum $p$ at time $t=0$. An analogous relation holds for the time-dependent case: $$\Phi(p,t)=\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}dx e^{-i px/\hbar}\Psi(x,t)$$ This is simply due to the fact that one independently transforms between position and momentum space and between time and frequency space.

For a “proof” that this is the case, consider: $$\Phi(p,t)=\langle p|\Psi(t)\rangle=\int _{-\infty}^{\infty}dx \langle p|x\rangle\langle x|\Psi(t)\rangle= \int _{-\infty}^{\infty}dx \langle p|x\rangle \Psi(x,t)$$ Now, note that a free particle momentum eigenstate in the position basis is a plane wave $\langle x|p\rangle= \frac{1}{\sqrt{2\pi\hbar}} e^{i px/\hbar} $, so $\langle p|x\rangle=\langle x|p\rangle^*= \frac{1}{\sqrt{2\pi\hbar}} e^{-i px/\hbar}$. Finally then, we arrive at: $$ \boxed{ \Phi(p,t)= \frac{1}{\sqrt{2\pi\hbar}} \int _{-\infty}^{\infty}dx e^{-i px/\hbar} \Psi(x,t)}$$


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