As we discussed before, quantum mechanics is deterministic in the sense that if

Question

As we discussed before, quantum mechanics is deterministic in the sense that if we leave the system alone (i.e., don't change the Hamiltonian) then the state of the system for all future times and for all past times is determined. |psi (t)) = U(t)|psi(0)) |psi(-t)) = U (t)|osu(0)) = U(-t)|psi(0)) We can run the state either forward or backward in time. If we reverse time the wave packet at time t will become narrower until at time t = 0 we will recover the initial wave packet. |psi(0)) = U(-t)|psi(t)) But sigma(t) that we found above depends quadratically on t. So, why does the wave packet shrink in width when Delta t is negative? sigma(t) = sigma(1 + h^2 t^2/4m^2 sigma^2)^1/2

Explanation / Answer

According to the uncertainty principle, the lesser the time is the higher is the energy of the wave packet. So at negative time, i.e. when the time is negative, the energy of the wavepacket shrink or peaked in mathematical sense.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.