# What is the probability of measuring $p$ in the momentum space?

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

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: $$\boxed{ \Phi(p,t)= \frac{1}{\sqrt{2\pi\hbar}} \int _{-\infty}^{\infty}dx e^{-i px/\hbar} \Psi(x,t)}$$