Note that I wrote this quickly, so try to find mistakes and correct them:

Okay so I'm assuming this is a 1D square well, since there was no ambiguity stated as to which degenerate state the n=2 wave function is in. Let's use the convention that the potential is zero from 0 to $\pi$ without a loss of generality (our answer will be generally invariant to rescaling).

At t=0 we can immediately say that the probability of finding a particle int he left half of ensemble (1) is 1/2. That's because $|\psi_1|^2$ and $|\psi_2|^2$ are both even about the point $\pi/2$ so the mod-square of the wave function of every particle integrates to 1/2 over the left part of the region.

For ensemble (2) (at t=0) we have a different result: $$\frac{1}{\pi}\int_0^{\pi/2} dx \left|c_1 \sin x + c_2 \sin(2 x) \right|^2 $$ where $c_1$ and $c_2$ are just complex numbers of magnitude 1 (which are legitimate degrees of freedom for the respective wave functions). Actually, since the answer is clearly invariant under global U(1) rotation, the only phase degree of freedom we care about in the brackets is the phase difference, so we can rotate to a phase where $c_1 = 1$ and $c_2 = e^{i \theta}; \, \theta \in [0,2 \pi)$. We can then expand it as such:

$$\frac{1}{\pi}\int_0^{\pi/2} dx \left( \sin^2x + 2 \sin\theta\sin x \sin 2x + \sin^22x \right) $$

That's easy to do. The answer is $$\frac{3 \pi + 8 \sin\theta}{6 \pi} $$

Notice that the answer would simply be 1/2 if $\psi_1$ and $\psi_2$ were in phase. Alas, you didn't include that restraint in your question, so there's the general answer. Thus if $\psi_2$ is ahead of $\psi_1$ by some phase in $(0,\pi)$ then we're more likely to find the particle in the left half. If it's behind by some phase in that same range, we're more likely to find it in the right half. Otherwise, if it's on the boundaries of that range, the chance is 1/2 again. So the answer to the question ("which is more likely?") depends on the relative phase of $\psi_1$ and $\psi_2$.

Now let's look at situation (2):

As time evolves, we have the following: $$\psi \rightarrow e^{-i H t} \psi $$

you'll forgive me for using units of $\hbar = 1$ here.

For the first ensemble, it's easy to see that this should have no effect. Half of the systems are just in state $e^{-i E_1 t} \psi_1$ and the other half are in state $e^{-i E_2 t} \psi_2$. In both cases, their wave function is just multiplied by an overall phase factor that gets cancelled out when you mod-square it, so the integral over the left half of the box won't change.

For system 2 we get: $$(\psi_1 + \psi_2)/\sqrt{2} \rightarrow (e^{-i E_1 t} \psi_1 + e^{-i E_2 t} \psi_2)/\sqrt{2}, $$ which is easy to show by simply acting the unitary time operator on the whole thing. So assuming (like we did before) that $\psi_2$ was offset by phase $\theta$ from $\psi_1$ at $t=0$, we now have $\psi_2$ offset by phase $\theta - (E_2 - E_1) t.$

But this just makes our lives simple, since we already did most of the work before. This is just the same problem with a different phase factor. For ensemble 2 at time t, the probability, when taking a random system from the ensemble, of finding the particle in the left side of the box is simply $$p = \frac{3 \pi + 8 \sin \left[\theta - (E_2 - E_1)t\right]}{6 \pi}. $$

So now you have an answer that changes in time. Similar to the analysis I did for $\theta$ at time 0, you can figure out at which times you're more likely to find the particle in the left side for ensemble 1, and at which times it's more likely for ensemble 2. (Remember ensemble 1 is still always $1/2$; ensemble 2 oscillates between $(3 \pi- 8)/6 \pi$ and $(3 \pi + 8)/6 \pi.$)

Note that I wrote this quickly, so try to find mistakes and correct them:

Okay so I'm assuming this is a 1D square well, since there was no ambiguity stated as to which degenerate state the n=2 wave function is in. Let's use the convention that the potential is zero from 0 to $\pi$ without a loss of generality (our answer will be generally invariant to rescaling).

At t=0 we can immediately say that the probability of finding a particle int he left half of ensemble (1) is 1/2. That's because $|\psi_1|^2$ and $|\psi_2|^2$ are both even about the point $\pi/2$ so the mod-square of the wave function of every particle integrates to 1/2 over the left part of the region.

For ensemble (2) (at t=0) we have a different result: $$\frac{1}{\pi}\int_0^{\pi/2} dx \left|c_1 \sin x + c_2 \sin(2 x) \right|^2 $$ where $c_1$ and $c_2$ are just complex numbers of magnitude 1 (which are legitimate degrees of freedom for the respective wave functions). Actually, since the answer is clearly invariant under global U(1) rotation, the only phase degree of freedom we care about in the brackets is the phase difference, so we can rotate to a phase where $c_1 = 1$ and $c_2 = e^{i \theta}; \, \theta \in [0,2 \pi)$. We can then expand it as such:

$$\frac{1}{\pi}\int_0^{\pi/2} dx \left( \sin^2x + 2 \cos\theta\sin x \sin 2x + \sin^22x \right) $$

That's easy to do. The answer is $$\frac{3 \pi + 8 \cos\theta}{6 \pi} $$

Notice that the answer would simply be 1/2 if $\psi_1$ and $\psi_2$ were in phase. Alas, you didn't include that restraint in your question, so there's the general answer. Thus if $\psi_2$ shifted from $\psi_1$ by some phase in $(-\pi/2,\pi/2)$ then we're more likely to find the particle in the left half. And you can work out the consequence of different values of $\theta.$ So the answer to the question ("which is more likely?") depends on the relative phase of $\psi_1$ and $\psi_2$.

Now let's look at situation (2):

As time evolves, we have the following: $$\psi \rightarrow e^{-i H t} \psi $$

you'll forgive me for using units of $\hbar = 1$ here.

For the first ensemble, it's easy to see that this should have no effect. Half of the systems are just in state $e^{-i E_1 t} \psi_1$ and the other half are in state $e^{-i E_2 t} \psi_2$. In both cases, their wave function is just multiplied by an overall phase factor that gets cancelled out when you mod-square it, so the integral over the left half of the box won't change.

For system 2 we get: $$(\psi_1 + \psi_2)/\sqrt{2} \rightarrow (e^{-i E_1 t} \psi_1 + e^{-i E_2 t} \psi_2)/\sqrt{2}, $$ which is easy to show by simply acting the unitary time operator on the whole thing. So assuming (like we did before) that $\psi_2$ was offset by phase $\theta$ from $\psi_1$ at $t=0$, we now have $\psi_2$ offset by phase $\theta - (E_2 - E_1) t.$

But this just makes our lives simple, since we already did most of the work before. This is just the same problem with a different phase factor. For ensemble 2 at time t, the probability, when taking a random system from the ensemble, of finding the particle in the left side of the box is simply $$p = \frac{3 \pi + 8 \cos \left[\theta - (E_2 - E_1)t\right]}{6 \pi}. $$

So now you have an answer that changes in time. Similar to the analysis I did for $\theta$ at time 0, you can figure out at which times you're more likely to find the particle in the left side for ensemble 1, and at which times it's more likely for ensemble 2. (Remember ensemble 1 is still always $1/2$; ensemble 2 oscillates between $(3 \pi- 8)/6 \pi$ and $(3 \pi + 8)/6 \pi.$)

Note that I wrote this quickly, so try to find mistakes and correct them:

Okay so I'm assuming this is a 1D square well, since there was no ambiguity stated as to which degenerate state the n=2 wave function is in. Let's use the convention that the potential is zero from 0 to $\pi$ without a loss of generality (our answer will be generally invariant to rescaling).

At t=0 we can immediately say that the probability of finding a particle int he left half of ensemble (1) is 1/2. That's because $|\psi_1|^2$ and $|\psi_2|^2$ are both even about the point $\pi/2$ so the mod-square of the wave function of every particle integrates to 1/2 over the left part of the region.

For ensemble (2) (at t=0) we have a different result: $$\frac{1}{\pi}\int_0^{\pi/2} dx \left|c_1 \sin x + c_2 \sin(2 x) \right|^2 $$ where $c_1$ and $c_2$ are just complex numbers of magnitude 1 (which are legitimate degrees of freedom for the respective wave functions). Actually, since the answer is clearly invariant under global U(1) rotation, the only phase degree of freedom we care about in the brackets is the phase difference, so we can rotate to a phase where $c_1 = 1$ and $c_2 = e^{i \theta}; \, \theta \in [0,2 \pi)$. We can then expand it as such:

$$\frac{1}{\pi}\int_0^{\pi/2} dx \left( \sin^2x + 2 \sin\theta\sin x \sin 2x + \sin^22x \right) $$

That's easy to do. The answer is $$\frac{3 \pi + 8 \sin\theta}{6 \pi} $$

Notice that the answer would simply be 1/2 if $\psi_1$ and $\psi_2$ were in phase. Alas, you didn't include that restraint in your question, so there's the general answer. Thus if $\psi_2$ is ahead of $\psi_1$ by some phase in $(0,\pi)$ then we're more likely to find the particle in the left side. If it's behind by some phase in that same range, we're more likely to find it in the left side. Otherwise, if it's on the boundaries of that range, the chance is 1/2 again. So the answer to the question ("which is more likely?") depends on the relative phase of $\psi_1$ and $\psi_2$.

Now let's look at situation (2):

As time evolves, we have the following: $$\psi \rightarrow e^{-i H t} \psi $$

you'll forgive me for using units of $\hbar = 1$ here.

For the first ensemble, it's easy to see that this should have no effect. Half of the systems are just in state $e^{-i E_1 t} \psi_1$ and the other half are in state $e^{-i E_2 t} \psi_2$. In both cases, their wave function is just multiplied by an overall phase factor that gets cancelled out when you mod-square it, so the integral over the left half of the box won't change.

For system 2 we get: $$(\psi_1 + \psi_2)/\sqrt{2} \rightarrow (e^{-i E_1 t} \psi_1 + e^{-i E_2 t} \psi_2)/\sqrt{2}, $$ which is easy to show by simply acting the unitary time operator on the whole thing. So assuming (like we did before) that $\psi_2$ was offset by phase $\theta$ from $\psi_1$ at $t=0$, we now have $\psi_2$ offset by phase $\theta - (E_2 - E_1) t.$

But this just makes our lives simple, since we already did most of the work before. This is just the same problem with a different phase factor. For ensemble 2 at time t, the probability, when taking a random system from the ensemble, of finding the particle in the left side of the box is simply $$p = \frac{3 \pi + 8 \sin \left[\theta - (E_2 - E_1)t\right]}{6 \pi}. $$

So now you have an answer that changes in time. Similar to the analysis I did for $\theta$ at time 0, you can figure out at which times you're more likely to find the particle in the left side for ensemble 1, and at which times it's more likely for ensemble 2. (Remember ensemble 1 is still always $1/2$; ensemble 2 oscillates between $(3 \pi- 8)/6 \pi$ and $(3 \pi + 8)/6 \pi.$)

Note that I wrote this quickly, so try to find mistakes and correct them:

Okay so I'm assuming this is a 1D square well, since there was no ambiguity stated as to which degenerate state the n=2 wave function is in. Let's use the convention that the potential is zero from 0 to $\pi$ without a loss of generality (our answer will be generally invariant to rescaling).

At t=0 we can immediately say that the probability of finding a particle int he left half of ensemble (1) is 1/2. That's because $|\psi_1|^2$ and $|\psi_2|^2$ are both even about the point $\pi/2$ so the mod-square of the wave function of every particle integrates to 1/2 over the left part of the region.

For ensemble (2) (at t=0) we have a different result: $$\frac{1}{\pi}\int_0^{\pi/2} dx \left|c_1 \sin x + c_2 \sin(2 x) \right|^2 $$ where $c_1$ and $c_2$ are just complex numbers of magnitude 1 (which are legitimate degrees of freedom for the respective wave functions). Actually, since the answer is clearly invariant under global U(1) rotation, the only phase degree of freedom we care about in the brackets is the phase difference, so we can rotate to a phase where $c_1 = 1$ and $c_2 = e^{i \theta}; \, \theta \in [0,2 \pi)$. We can then expand it as such:

$$\frac{1}{\pi}\int_0^{\pi/2} dx \left( \sin^2x + 2 \sin\theta\sin x \sin 2x + \sin^22x \right) $$

That's easy to do. The answer is $$\frac{3 \pi + 8 \sin\theta}{6 \pi} $$

Notice that the answer would simply be 1/2 if $\psi_1$ and $\psi_2$ were in phase. Alas, you didn't include that restraint in your question, so there's the general answer. Thus if $\psi_2$ is ahead of $\psi_1$ by some phase in $(0,\pi)$ then we're more likely to find the particle in the left half. If it's behind by some phase in that same range, we're more likely to find it in the right half. Otherwise, if it's on the boundaries of that range, the chance is 1/2 again. So the answer to the question ("which is more likely?") depends on the relative phase of $\psi_1$ and $\psi_2$.

Now let's look at situation (2):

As time evolves, we have the following: $$\psi \rightarrow e^{-i H t} \psi $$

you'll forgive me for using units of $\hbar = 1$ here.

For the first ensemble, it's easy to see that this should have no effect. Half of the systems are just in state $e^{-i E_1 t} \psi_1$ and the other half are in state $e^{-i E_2 t} \psi_2$. In both cases, their wave function is just multiplied by an overall phase factor that gets cancelled out when you mod-square it, so the integral over the left half of the box won't change.

For system 2 we get: $$(\psi_1 + \psi_2)/\sqrt{2} \rightarrow (e^{-i E_1 t} \psi_1 + e^{-i E_2 t} \psi_2)/\sqrt{2}, $$ which is easy to show by simply acting the unitary time operator on the whole thing. So assuming (like we did before) that $\psi_2$ was offset by phase $\theta$ from $\psi_1$ at $t=0$, we now have $\psi_2$ offset by phase $\theta - (E_2 - E_1) t.$

But this just makes our lives simple, since we already did most of the work before. This is just the same problem with a different phase factor. For ensemble 2 at time t, the probability, when taking a random system from the ensemble, of finding the particle in the left side of the box is simply $$p = \frac{3 \pi + 8 \sin \left[\theta - (E_2 - E_1)t\right]}{6 \pi}. $$

So now you have an answer that changes in time. Similar to the analysis I did for $\theta$ at time 0, you can figure out at which times you're more likely to find the particle in the left side for ensemble 1, and at which times it's more likely for ensemble 2. (Remember ensemble 1 is still always $1/2$; ensemble 2 oscillates between $(3 \pi- 8)/6 \pi$ and $(3 \pi + 8)/6 \pi.$)