Simple Harmonic Oscillator State Evolution

This question asks for the minimum time required for a simple harmonic oscillator, initially in a superposition of its ground state and first excited state, to return to its initial state.

The simple harmonic potential is given by \(V(x) = \frac{1}{2} m\omega^2 x^2\). The energy eigenstates of this potential are denoted by \(|n\rangle\), where \(n = 0, 1, 2, \dots\), and the corresponding energy eigenvalues are \(E_n = (n + \frac{1}{2})\hbar\omega\).

The initial state of the particle at time \(t=0\) is given as:

\(|\psi(t=0)\rangle = \sqrt{\frac{1}{2}} (|0\rangle + |1\rangle)\)

Here, \(|0\rangle\) is the ground state with energy \(E_0 = \frac{1}{2}\hbar\omega\), and \(|1\rangle\) is the first excited state with energy \(E_1 = \frac{3}{2}\hbar\omega\).

Time Evolution of the State

The time evolution of a quantum state \(|\psi(t)\rangle\) is governed by the time-dependent Schrödinger equation. If the initial state is a superposition of energy eigenstates, the state at a later time \(t\) is given by:

\(|\psi(t)\rangle = e^{-iHt/\hbar} |\psi(t=0)\rangle\)

Since \(|n\rangle\) are energy eigenstates, \(e^{-iHt/\hbar} |n\rangle = e^{-iE_n t/\hbar} |n\rangle\). Applying this to the initial state:

\(|\psi(t)\rangle = e^{-iHt/\hbar} \left( \sqrt{\frac{1}{2}} |0\rangle + \sqrt{\frac{1}{2}} |1\rangle \right)\)

\(|\psi(t)\rangle = \sqrt{\frac{1}{2}} e^{-iE_0 t/\hbar} |0\rangle + \sqrt{\frac{1}{2}} e^{-iE_1 t/\hbar} |1\rangle\)

Substituting the energy eigenvalues \(E_0\) and \(E_1\):

\(|\psi(t)\rangle = \sqrt{\frac{1}{2}} e^{-i(\frac{1}{2}\hbar\omega) t/\hbar} |0\rangle + \sqrt{\frac{1}{2}} e^{-i(\frac{3}{2}\hbar\omega) t/\hbar} |1\rangle\)

\(|\psi(t)\rangle = \sqrt{\frac{1}{2}} e^{-i\frac{1}{2}\omega t} |0\rangle + \sqrt{\frac{1}{2}} e^{-i\frac{3}{2}\omega t} |1\rangle\)

Minimum Time for State Return

The oscillator will be found in the "same state" when the state vector \(|\psi(t)\rangle\) is proportional to the initial state vector \(|\psi(t=0)\rangle\). This means \(|\psi(t)\rangle = e^{i\phi} |\psi(t=0)\rangle\) for some overall phase factor \(e^{i\phi}\).

So, we require:

\(\sqrt{\frac{1}{2}} e^{-i\frac{1}{2}\omega t} |0\rangle + \sqrt{\frac{1}{2}} e^{-i\frac{3}{2}\omega t} |1\rangle = e^{i\phi} \left( \sqrt{\frac{1}{2}} |0\rangle + \sqrt{\frac{1}{2}} |1\rangle \right)\)

\(\sqrt{\frac{1}{2}} e^{-i\frac{1}{2}\omega t} |0\rangle + \sqrt{\frac{1}{2}} e^{-i\frac{3}{2}\omega t} |1\rangle = \sqrt{\frac{1}{2}} e^{i\phi} |0\rangle + \sqrt{\frac{1}{2}} e^{i\phi} |1\rangle\)

Comparing the coefficients of the orthogonal states \(|0\rangle\) and \(|1\rangle\):

• Coefficient of \(|0\rangle\): \(e^{-i\frac{1}{2}\omega t} = e^{i\phi}\)
• Coefficient of \(|1\rangle\): \(e^{-i\frac{3}{2}\omega t} = e^{i\phi}\)

From these two equations, we must have:

\(e^{-i\frac{1}{2}\omega t} = e^{-i\frac{3}{2}\omega t}\)

Dividing both sides by \(e^{-i\frac{1}{2}\omega t}\) (which is non-zero):

\(1 = \frac{e^{-i\frac{3}{2}\omega t}}{e^{-i\frac{1}{2}\omega t}}\)

\(1 = e^{-i\frac{3}{2}\omega t + i\frac{1}{2}\omega t}\)

\(1 = e^{i(-\frac{3}{2} + \frac{1}{2})\omega t}\)

\(1 = e^{-i\omega t}\)

For \(e^{-i\omega t}\) to be equal to 1, the exponent \(-i\omega t\) must be an integer multiple of \(2\pi i\).

\(-i\omega t = i 2\pi N\), where \(N\) is an integer (\(N = 0, \pm 1, \pm 2, \dots\)).

\(\omega t = -2\pi N\)

Since we are looking for a positive time \(t\) and \(\omega > 0\), we must have \(N\) as a negative integer. Let \(N' = -N\), where \(N'\) is a positive integer (\(N' = 1, 2, 3, \dots\)).

\(\omega t = 2\pi N'\)

\(t = \frac{2\pi N'}{\omega}\)

We are looking for the minimum positive time, which occurs when \(N' = 1\).

\(t_{min} = \frac{2\pi}{\omega}\)

At this time, \(\omega t = 2\pi\). Let's check the phase \(e^{i\phi}\):

\(e^{i\phi} = e^{-i\frac{1}{2}\omega t} = e^{-i\frac{1}{2}(2\pi)} = e^{-i\pi} = -1\)

So, at \(t = \frac{2\pi}{\omega}\), the state becomes \(|\psi(t)\rangle = -|\psi(t=0)\rangle\), which is proportional to the initial state.

Calculation Summary

The steps to find the minimum time are:

1. Write the initial state as a superposition of energy eigenstates.
2. Determine the energy eigenvalues for the involved states.
3. Apply the time evolution operator \(e^{-iHt/\hbar}\) to get the state at time \(t\).
4. Set the state at time \(t\) equal to the initial state multiplied by an overall phase factor \(e^{i\phi}\).
5. Compare coefficients of the basis states to find the condition on \(t\).
6. Solve for \(t\) and find the minimum positive value.

The condition \(e^{-i\omega t} = 1\) implies \(\omega t\) is a multiple of \(2\pi\). The minimum positive time is when \(\omega t = 2\pi\), leading to \(t = \frac{2\pi}{\omega}\).

Final Answer

The minimum time after which the oscillator will be found in the same state is \(\frac{2\pi}{\omega}\).