Why is the momentum-space wavefunction for a free particle not a function of time? Suppose the initial wave function of a free particle is given by $\psi(x,0)$. Now to find how the wave function evolves with time we generally do the Fourier transform of the wave function at $t=0$.
The Fourier transform is given by $\phi(k)$ and the time dependence of the wave function is given by
$$\int_{-\infty}^{\infty} \phi(k)\exp\biggl(ikx-\frac{ihk^2t}{4m\pi}\biggr)\mathrm{d}k$$
where $k$ is given by
$$k=\frac{\sqrt{2mE}}{h/2\pi}$$
This is the procedure as given in Griffiths' quantum mechanics 2nd edition.
Now my question is why the amplitude corresponding to $k$, $\phi(k)$, is only a function of $k$, not time, i.e, $\phi(k,t)$?
The statement given in the book is,

$$\Psi(x, t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\phi(k) e^{i\bigl(kx - \frac{\hbar k^2}{2m}t\bigr)}\mathrm{d}k$$

Why it is not $\phi(k,t)$?
 A: 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 $\phi(k,\omega)$. 
A: 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 $\phi(k)$
A: It is not a function of time because time is a separate dimension unrelated to particles in free space:  particles don't necessarily travel in time, as there is no absolute or even relative reference frame for them to be measured against (unless you put it in there, like a particle accelerator, for example).
