# Time free evolution of the harmonic ground state (Quantum mechanics)

I have an atom in the ground state for a harmonic potential. At time $$t = 0$$ the parabolic potential is switched off. How can I derive the time evolution of the wave function $$\psi(t; x)$$ during the free evolution?

I know that, in order to calculate the time evolution, I need to consider the eigenvalues of the new Hamiltonian. Furthermore I should express the ground state as a linear combination of the eigenstates of the new Hamiltonian. I don't know how to do so, because the eigenstates of a free particles are planar waves.

How can I compute the time evolution of the ground state? Thank you very much.

## 2 Answers

What you are describing is known as a quantum quench, where an initial quantum state belonging to some Hamiltonian $$H_{0}$$ is suddenly evolved by a different Hamiltonian $$H'$$ after some switching time $$t=0$$. The equation governing the quench is the Schrödinger equation for the state ($$\hbar=1$$): $$\begin{eqnarray} |\Psi(t)\rangle = U(t,t_{0})|\Psi(t_{0})\rangle \hspace{10pt}U(t,t_{0})=\mathcal{T}e^{-i\int_{t_{0}}^{t}dt'H'(t')} \end{eqnarray}$$ where $$\mathcal{T}$$ is the time-ordering operator and $$U(t,t_{0})$$ the unitary time-evolution operator. In the simpler case of $$H'$$ being time-independent, time-ordering is not needed and one has: $$\begin{eqnarray} |\Psi(t)\rangle = e^{-iH'(t-t_{0})}|\Psi(t_{0})\rangle. \end{eqnarray}$$ Now, to compute the ground evolution in an exact way, you might expand the RHS using the completeness identity for the eigenstates of $$H'$$: $$\begin{eqnarray} H'|\lambda\rangle = \lambda|\lambda\rangle\to \sum_{k}|\lambda_{k}\rangle\langle \lambda_{k}|=1. \end{eqnarray}$$ Using that, the time-evolved state can be expressed as: $$\begin{eqnarray} |\Psi(t)\rangle = \sum_{k}e^{-iH'(t-t_{0})}|\lambda_{k}\rangle\langle\lambda_{k}|\Psi(t_{0})\rangle=\sum_{k}\phi_{k}(t)|\lambda_{k}\rangle,\hspace{10pt}\phi_{k}(t)=\langle \lambda_{k}|\Psi(t_{0}) \rangle e^{-i\lambda_{k}(t-t_{0})}. \end{eqnarray}$$ You are left with calculating the overlaps of the initial state with the eigenbasis of the $$H'$$ operator.

• The result for a time independent Hamiltonian switched on at $t_0$ is also known as the "sudden approximation". (Since in reality the switching between the Hamiltonians will not be instantaneous). Dec 14, 2021 at 19:57

Your difficulty is that the parabolic potential has discrete eigenstates, and in particular your atom is in the ground state, which has a positive energy eigenvalue. When the potential is removed, the eigenstates are planar waves.

But the ground state eigenstate (the square modulus of which is integrable) can be expanded in an integral of plane waves (rather than as a sum, in contradistinction from Zarathustra's answer which is essentially correct, but fails to take into account the difference between a sum, even infinite, of discrete terms, and a continuous integral). In fact, what you have to do is a Fourier transform of the ground state eigenstate at time $$t_{0}$$. This would be the relevant, continuous, equivalent of what Zarathustra calls the "the overlaps of the initial state with the eigenbasis of the $$H'$$ operator", $$\langle \lambda_{k}|\Psi(t_{0}) \rangle$$.

Then you have to evolve this Fourier transform in time, which is easy because the Hamiltonian $$H'$$ consists in the kinetic energy only.

Finally do an inverse Fourier transform on the time advanced expression you find in the previous step.

Because the ground state eigenstate in a parabolic potential is a Gaussian function of $$x$$, its Fourier transform is also a Gaussian function of the continuous wavevector $$k$$ (rather than the discrete $$k$$ of Zarathustra). Because $$H'$$ is just the kinetic energy, the evolution in time is just the multiplication by one more, time dependent, Gaussian function of $$k$$, which preserves the Gaussian character during the entire time evolution.

The inverse Fourier transform will therefore just be, again, a Gaussian function of $$x$$, but with a width that keeps increasing in time. The initially localized wave-function will remain a localized Gaussian at all times, but keeps spreading, the width increasing with time.

I leave the details of the calculations to you, but I'll give you a rough sketch, ignoring all normalisations and with hbar=1 to simplify. Also I write just $$t$$ for $$t-t_0$$.

Initial ground state of a parabolic potential $$e^{(-x^2/2d^2)}$$ for some width $$d$$ related to the potentiel.

Fourier transform $$e^{(-d^2k^2/2)}$$ at the removal of the potential

Time-advanced expression, with kinetic energy $$E_k=k^2/2m$$

$$e^{(-d^2k^2/2)}e^{(-ik^2t/2m)}=e^{-k^2(d^2+it/m)/2}$$

A "complex" Gaussian, but still a Gaussian

Final wavefunction, after inverse FT

$$e^{-x^2/2(d^2+it/m)}=e^{-x^2(d^2-it/m)/2(d^4+t^2/m^2)}$$

This describes a very dense oscillation with time of the phase along $$x$$, within a Gaussian envelope

$$e^{-x^2d^2/2(d^4+t^2/m^2)}$$

of squared width $$(d^4+t^2/m^2)/d^2=d^2+t^2/d^2m^2$$

The fine details, I leave to you.

• Thank you very much Alfred. My difficulty was precisely in dealing with the continuous eigenstates for the free particle and the discrete eigenstates of the parabolic potential. Your explanation is clear and you have helped me a lot.
– stik
Dec 15, 2021 at 9:54