Why can't two different quantum states evolve into the same final state? Is it true that two different states cannot evolve into the same final state? Can they achieve this state at different times? If yes, what is the proof?
 A: Let $\Psi$ and $\Phi$ be two states that evolve into the same state after some time $t$. Time evolution after time $t$ is given by a unitary operator $U(t)$. In particular, this means that $U(t)$ is invertible, so we have $U(t)^{-1}U(t) = 1$. Now we have by assumption that $U(t)\Psi = U(t)\Phi$. Multiplying both sides of this equation by $U(t)^{-1}$ from the left we obtain $\Psi = \Phi$. So if two states evolve into the same state after some time $t$, they were the same to begin with.
Time evolution furthermore has the property that $U(t)U(s) = U(t+s)$. Let $t\neq 0$. Now suppose that we have some state $\Phi$ with the property that $U(s)\Phi \neq \Phi$. Set $\Psi = U(s)\Phi$. We then get
\begin{equation}
 U(t)\Psi = U(t)(U(s)\Phi) = U(t+s)\Phi.
\end{equation}
We then have that the states $\Psi$ and $\Phi$, (which are different), evolve into the same state $U(t+s)\Phi$, but after different times.
In fact, this is the only way this can happen. That is, if there are two different states $\Phi \neq \Psi$ such that $U(t)\Phi = U(t')\Psi$, then there exists a time $s$ such that $\Psi = U(s) \Phi$. Maybe finding a proof for this claim is a good exercise.
A: $\def\ket#1{|#1\rangle}$
More simply. As to first question: let $\ket{a,0}$ be a state vector
taken at time 0, $\ket{a,t}$ the same state evolved at time $t$.
Note 1: Schrödinger picture is used, where states evolve in time, observables don't.
Note 2: I'm using a rather different ket notation. I don't write things like $\ket{\Psi(t)}$ because I think this is a misinterpretation of Dirac's notation.
We have 
$$\ket{a,t} = U(t)\,\ket{a,0}$$ 
where $U(t)$ is unitary. Assume now that another $\ket{b,0}$ exists, such that also 
$$\ket{a,t} = U(t)\,\ket{b,0}.$$
Then
$$\ket{b,0} = U^{-1}(t)\,\ket{a,t} = U^{-1}(t)\,U(t)\,\ket{a,0} = \ket{a,0}.$$
Now for the second question. You're asking if
$$\ket{a,t} = \ket{b,t'} \tag1$$
could happen, for $t'\ne t$. Let's expand eq. (1), using $U$:
$$U(t)\,\ket{a,0} = U(t')\,\ket{b,0}$$
$$\ket{b,0} = U^{-1}(t')\,U(t)\,\ket{a,0} = U(-t')\,U(t)\,\ket{a,0} = 
U(t-t')\,\ket{a,0}.$$
This is the condition $\ket{a,0}$ and $\ket{b,0}$ are to obey, in order that $\ket{a,t}$ and $\ket{b,t'}$ be equal.
A: Excluding wave function collapse, wave functions evolve deterministically, and this determinism goes both ways in time. So if you take  $\Psi$ and $\Phi$ such that there is some $t_0$ for which $\Psi(t_0)=\Phi(t_0)$, then as long as they evolve under the same transformation, you have  $\Psi(t)=\Phi(t)$ for all $t$. Thus, you can have neither two wavefunctions that are the same at one time evolve into different wavefunctions at a different time, nor two wavefunctions that are different at one time evolve into the same wavefunction at another time.
A state can evolve into what another state was at another time, i.e. $\Psi(t_1)=\Phi(t_2)$. But if they evolve under a time-constant transformation, then if we define $\Delta t= t_2-t_1$, then $\Psi(t)=\Phi(t+\Delta t)$ for all $t$; the two states are simply time-shifted version of each other.
A: 
Is it true that two different states cannot evolve into the same final state?

That depends on exactly what you mean.
If we consider the total state of a closed system, then two different states will never simultaneously evolve into the same state at any later time.
You may have learned that quantum states evolve with a unitary transformation, i.e.
$$ \lvert \Psi(t) \rangle = U(t) \lvert \Psi(0) \rangle$$
where $U(t)$ is unitary, which means that $U(t)^\dagger = U(t)^{-1}$.
That being the case,
\begin{align}
\langle \Phi(t)|\Psi(t) \rangle 
&= \langle U(t) \Phi(0) | U(t) \Psi(0) \rangle \\
\text{(definition of Heritian conjugate)} \quad &= \langle \Phi(0) | U(t)^\dagger U(t) | \Psi (0) \rangle \\
\text{(unitarity)} \quad &= \langle \Phi(0) | U(t)^{-1} U(t) | \Psi (0) \rangle \\
&= \langle \Phi(0) | \Psi (0) \rangle \, .
\end{align}
So you can see, the inner product between two states does not change as time evolves.
Two states that are the same have inner product of 1, but states that are not the same have inner product not 1.
Therefore, two states that are not initially the same cannot become the same later under unitary evolution.
On the other hand, if we allow measurement, it is possible for two initially different states to wind up being the same.
For example, if we have a two level system starting in state $(\lvert 0 \rangle + e^{i \phi} \lvert 1 \rangle)/\sqrt 2$ for any value of $\phi$, it could collapse to $\lvert 0 \rangle$ after a measurement.
Note, however, that in this case there is randomness, i.e. we cannot make a situation where two intitially different states deterministically evolve to the same final state.
If you could do that, I'm pretty sure you could control the future, communicate faster than light, and destroy the entire universe.

Can they achieve this state at different times?

Yeah, sure.
Consider a two level system with Hamiltonian
$$ H = \hbar \frac{\omega}{2} \sigma_x \, .$$
The propagator for this system is
$$U(t) = \cos(\omega t / 2) \mathbb{I} - i \sin(\omega t / 2) \sigma_x =
\left(
  \begin{array}{cc}
  \cos(\omega t / 2) && -i \sin(\omega t / 2) \\
  -i \sin(\omega t / 2) && \cos(\omega t / 2)
  \end{array}
\right)$$
where $\mathbb{I}$ means the identity.
If we start with state $\lvert 0 \rangle$, then the state at time $t$ is
$$
U(t) \lvert 0 \rangle = \cos(\omega t / 2) \lvert 0 \rangle - i \sin(\omega t / 2) \lvert 1 \rangle
$$
Similarly, if we had started with $i \lvert 1 \rangle$, we'd get
$$U(t) i \lvert 1 \rangle = \sin(\omega t / 2) \lvert 0 \rangle + i \cos(\omega t / 2) \lvert 1 \rangle \, .
$$
Now look at two particular times:
$$U(t = \pi / 2 \omega) \lvert 0 \rangle
= \cos(\pi / 4) \lvert 0 \rangle - i \sin(\pi / 4) \lvert 1 \rangle
= \frac{1}{\sqrt 2}(\lvert 0 \rangle - i \lvert 1 \rangle)$$
and
$$U(t = 3 \pi / 2 \omega) i \lvert 1 \rangle
= \sin(3\pi/4)\lvert 0 \rangle  + i \cos(3 \pi / 4) \lvert 1 \rangle
= \frac{1}{\sqrt 2}(\lvert 0 \rangle - i \lvert 1 \rangle) \, .$$
So we can see that two initially different states evolve to the same state, but at different times.
