Linear Equations & Quantum Mechanics

[StevieTNZ]

Say we can find a solution to the Schrödinger equation, which gives a superposition of two states x1 and x2. Say this occurs at time0.

The equation evolves over time, say to time1.

At time1, are x1 and x2 (the same states found by solving the Schrödinger equation at time0) still superposition states in a solution to the Schrödinger equation at time1?

 

[_PJ_]

Say we can find a solution to the Schrödinger equation, which gives a superposition of two states x1 and x2. Say this occurs at time0.

The equation evolves over time, say to time1.

At time1, are x1 and x2 (the same states found by solving the Schrödinger equation at time0) still superposition states in a solution to the Schrödinger equation at time1?

Yes, since as soon as the superposition is reconciled as a definite result, the wavefunction no longer applies.

 

[G01]

If we can write our initial state as

$$|\psi(t=0)>=A|1>+B|2>$$

where |1> & |2> are our orthogonal states.

If we want the solution at a later time, we then apply the time evolution operator to the system:

$$|\psi(t)>=e^{iHt/\hbar}|\psi(0)>$$

So, if |1> and |2> are both eigenfunctions of H, the Hamiltonian, the solution at time t will remain a superposition of |1> and |2>, but can be a different superposition:

i.e.

$$|\psi(t)>=A'|1>+B'|2>=Ae^{iE_1t/\hbar}|1>+Be^{iE_2t/\hbar}|2>$$

Notice however:

If |1> and |2> are not energy eigenstates, then we need to expand them in terms of energy eigenstates before we apply the time evolution operator. The resulting state at time t will still be able to be written as a superposition of energy eigenstates, but not necessarily as a superposition of |1> and |2>.

Yes, since as soon as the superposition is reconciled as a definite result, the wavefunction no longer applies.

I am not sure what you are saying.

 

[StevieTNZ]

If we can write our initial state as

$$|\psi(t=0)>=A|1>+B|2>$$

where |1> & |2> are our orthogonal states.

If we want the solution at a later time, we then apply the time evolution operator to the system:

$$|\psi(t)>=e^{iHt/\hbar}|\psi(0)>$$

So, if |1> and |2> are both eigenfunctions of H, the Hamiltonian, the solution at time t will remain a superposition of |1> and |2>, but can be a different superposition:

i.e.

$$|\psi(t)>=A'|1>+B'|2>=Ae^{iE_1t/\hbar}|1>+Be^{iE_2t/\hbar}|2>$$

Just so I follow, the states |1> and |2> are the same at t=0 and t=1, just that the solution to the equation at t=1 will look different? i.e you add $$e^{iHt/\hbar}$$ to the front of each state in the solution at t=1.

 

[G01]

Just so I follow, the states |1> and |2> are the same at t=0 and t=1, just that the solution to the equation at t=1 will look different? i.e you add $$e^{iHt/\hbar}$$ to the front of each state in the solution at t=1.

Yes,

If the state |a> is an energy eigenstate, at t=t1, The state |a> will be the same (up to the phase factor $e^{iE_at/\hbar}$ )

However, if the state |a> is not an energy eigenstate, then the time evolution operator $e^{iHt/\hbar}$ does not give just a phase factor like above.