I have a question regarding how the the stationary states (eigenstates) are arrived at in Schrodinger's wavefunction, please.

In the graph below, taken from https://www.youtube.com/watch?v=2V0Xmc0ow80&list=PLdCdV2GBGyXM0j66zrpDy2aMXr6cgrBJA&index=4:

we see the resulting position wave shape distribution over a long time period using the potential constraint V(x) = x^2 and the initial wave shape W0(x) = [(2x^3-3x)*exp(-x^2/2)/sqrt(1.25323), 0]. The array indexing refers to [Re_part, Im_part].

The resulting probability function shape (in green) given this initial probability distribution, which we deliberately chose to be W0(x), remains the same over time regardless how the first R part and the second I part is intra-changing individually.

This is referred to as the stationary (or eigen) state, if we only consider the "real" physical green function above (which is the shape of the probability distribution of finding the particle at location x), and ignore the red and blue internal components which are not individually discernable by us.

My question is, this stable state is achieved by first inputting the initial probability shape of W0(x) = [(2x^3-3x)*Math.exp(-x^2/2)/Math.sqrt(1.25323), 0], and letting the shape evolve over time.

If we start with a random "incorrect" initial shape for W0(X) to start with, the resulting evolution over time will probably not automatically evolve into the steady/stationary state by itself.

So, by what process, supposing we just throw an electron into this V(x)=x^2 constraint, for which the electron will probably not find itself in this magical Hermite state to begin with, by what process will the particle discover and settle over time into the stationary (eigen) state like the green wave distribution above?

If the Schrodinger equation claims complete description over the time evolution of the probability wave, then shouldn't the starting condition makes no difference at all, and the particle will always end up in a stationary ("stable") state no matter what the starting W0(x) looks like, just by following the rules of the Schrodinger function?

Thank you for any help/pointers on this.

I have a question regarding how the the stationary states (eigenstates) are arrived at in Schrodinger's wavefunction, please.

In the graph below, taken from https://www.youtube.com/watch?v=2V0Xmc0ow80&list=PLdCdV2GBGyXM0j66zrpDy2aMXr6cgrBJA&index=4:

we see the resulting position wave shape distribution over a long time period using the potential constraint V(x) = x^2 and the initial wave shape W0(x) = [(2x^3-3x)*exp(-x^2/2)/sqrt(1.25323), 0]. The array indexing refers to [Re_part, Im_part].

The resulting probability function shape (in green) given this initial probability distribution, which we deliberately chose to be W0(x), remains the same over time regardless how the first R part and the second I part is intra-changing individually.

This is referred to as the stationary (or eigen) state, if we only consider the "real" physical green function above (which is the shape of the probability distribution of finding the particle at location x), and ignore the red and blue internal components which are not individually discernable by us.

My question is, this stable state is achieved by first inputting the initial probability shape of W0(x) = [(2x^3-3x)*Math.exp(-x^2/2)/Math.sqrt(1.25323), 0], and letting the shape evolve over time.

If we start with a random "incorrect" initial shape for W0(X) to start with, the resulting evolution over time will probably not automatically evolve into the steady/stationary state by itself.

So, by what process, supposing we just throw an electron into this V(x)=x^2 constraint, for which the electron will probably not find itself in this magical Hermite state to begin with, by what process will the particle discover and settle over time into the stationary (eigen) state like the green wave distribution above?

If the Schrodinger equation claims complete description over the time evolution of the probability wave, then shouldn't the starting condition makes no difference at all, and the particle will always end up in a stationary ("stable") state no matter what the starting W0(x) looks like, just by following the rules of the Schrodinger wavefunction?

I have a question regarding how the the stationary states (eigenstates) are arrived at in Schrodinger's wavefunction, please.

In the graph below, taken from https://www.youtube.com/watch?v=2V0Xmc0ow80&list=PLdCdV2GBGyXM0j66zrpDy2aMXr6cgrBJA&index=4:

we see the resulting position wave shape distribution over a long time period using the potential constraint V(x) = x^2 and the initial wave shape W0(x) = [(2x^3-3x)*Math.exp(-x^2/2)/Math.sqrt(1.25323), 0]. The array indexing refers to [Re_part, Im_part].

The resulting probability function shape (in green) given this initial probability distribution, which we deliberately chose to be W0(x), remains the same over time regardless how the first R part and the second I part is intra-changing individually.

This is referred to as the stationary (or eigen) state, if we only consider the "real" physical green function above (which is the shape of the probability distribution of finding the particle at location x), and ignore the red and blue internal components which are not individually discernable by us.

My question is, this stable state is achieved by first inputting the initial probability shape of W0(x) = [(2x^3-3x)*Math.exp(-x^2/2)/Math.sqrt(1.25323), 0], and letting the shape evolve over time.

If we start with a random "incorrect" initial shape for W0(X) to start with, the resulting evolution over time will probably not automatically evolve into the steady/stationary state by itself.

So, by what process, supposing we just throw an electron into this V(x)=x^2 constraint, which will probably not end up with this magical Hermite state to begin with, by what process will the particle discover and settle over time into the stationary (eigen) state for example the green wave above?

If the Schrodinger equation claims complete description over the time evolution of the probability wave as long (as the particle is not observed), then shouldn't any starting condition makes no difference, and the particle will always end up in a stationary ("stable") state no matter what the starting W0(x) looks like, just by following the rules of the Schrodinger function?

Thank you for any help/pointers on this.

I have a question regarding how the the stationary states (eigenstates) are arrived at in Schrodinger's wavefunction, please.

In the graph below, taken from https://www.youtube.com/watch?v=2V0Xmc0ow80&list=PLdCdV2GBGyXM0j66zrpDy2aMXr6cgrBJA&index=4:

we see the resulting position wave shape distribution over a long time period using the potential constraint V(x) = x^2 and the initial wave shape W0(x) = [(2x^3-3x)*exp(-x^2/2)/sqrt(1.25323), 0]. The array indexing refers to [Re_part, Im_part].

The resulting probability function shape (in green) given this initial probability distribution, which we deliberately chose to be W0(x), remains the same over time regardless how the first R part and the second I part is intra-changing individually.

This is referred to as the stationary (or eigen) state, if we only consider the "real" physical green function above (which is the shape of the probability distribution of finding the particle at location x), and ignore the red and blue internal components which are not individually discernable by us.

My question is, this stable state is achieved by first inputting the initial probability shape of W0(x) = [(2x^3-3x)*Math.exp(-x^2/2)/Math.sqrt(1.25323), 0], and letting the shape evolve over time.

If we start with a random "incorrect" initial shape for W0(X) to start with, the resulting evolution over time will probably not automatically evolve into the steady/stationary state by itself.

So, by what process, supposing we just throw an electron into this V(x)=x^2 constraint, for which the electron will probably not find itself in this magical Hermite state to begin with, by what process will the particle discover and settle over time into the stationary (eigen) state like the green wave distribution above?

If the Schrodinger equation claims complete description over the time evolution of the probability wave, then shouldn't the starting condition makes no difference at all, and the particle will always end up in a stationary ("stable") state no matter what the starting W0(x) looks like, just by following the rules of the Schrodinger function?

Thank you for any help/pointers on this.