# Why is the momentum-space wavefunction for a free particle not a function of time?

The Fourier transform $\phi(k)$ is a function only of $k$ and not of time because it indicates the amplitude of each plane wave that compose the wave function.

The amplitudes are conserved in time, because the plane waves linear superpose among them and don't interact.

The evolution in time so is not in the amplitudes $\phi(k)$, but you can observe how evolves each plane wave and summed again the evolved ones with the previous amplitudes $\phi(k)$

Time evolution in quantum mechanics is usually done with the so-called time evolution operator, $$U(t_f,t_i)=\text{e}^{-\text{i}H(t_f-t_i)/\hbar},$$ such that, when applied on a wave function at initial time $t=t_i$, $$U(t_f,t_i)\, \psi(x,t_i) = \psi(x, t_f),$$ the wave function at the final time $t=t_f$ follows. For a free particle, the Hamiltonian reads, $$H=\frac{p^2}{2m},$$ so we get the wave function for $t=t_f$ if we let $H$ act on $\psi(x)$. Before we can do that, we have to express $p$ as a derivative acting on $x$, $$p=\hbar k \to -\text{i}\hbar \partial_x\quad\Rightarrow\quad H=-\frac{\hbar^2}{2m}\partial_x^2$$ (note $\partial_x \equiv \tfrac{\partial}{\partial x}$) so that we write, $$\psi(x, t_f) = \text{e}^{\text{i}\frac{\hbar}{2m}\partial_x^2(t_f-t_i)}\, \psi(x,t_i).$$ According to your question, we will use $t_i=0$ and $t_f=t$: $$\psi(x, t) = \text{e}^{\text{i}\frac{\hbar}{2m}\partial_x^2t}\, \psi(x, 0).$$ I'm not exactly sure why, but in order to match your image, we now do a Fourier transform on the initial wave function, $$\psi(x,0) = \frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^\infty \text{d}k\, \text{e}^{ikx}\, \phi(k).$$ With this, we arrive at the equation \begin{align*} \psi(x, t) &= \text{e}^{\text{i}\frac{\hbar}{2m}\partial_x^2 t} \frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^\infty \text{d}k\, \text{e}^{ikx}\, \phi(k)\\ &= \frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^\infty \text{d}k\, \text{e}^{\text{i}\frac{\hbar}{2m}\partial_x^2 t} \text{e}^{ikx}\, \phi(k)\\ &= \frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^\infty \text{d}k\, \text{e}^{\text{i}\frac{\hbar}{2m}(\text ik)^2 t} \text{e}^{ikx}\, \phi(k)\\ &= \frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^\infty \text{d}k\, \text{e}^{-\text{i}\frac{\hbar k^2}{2m} t} \text{e}^{ikx}\, \phi(k)\\ &= \frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^\infty \text{d}k\, \text{e}^{i\left( kx -\frac{\hbar k^2}{2m}t\right)}\, \phi(k). \end{align*} The third equality follows by an implicit Taylor expansion and letting every power of $\partial_x$ act on $\text e^{\text ikx}$, see e.g. here or here for details.

So, to answer your question: $\phi(k)$ is only a function of $k$, because we did the Fourier transform only on $\psi(x)$. If we were to do a Fourier transform on $\psi(x,t)$, we would get a $\phi(k,\omega)$.