# Ground state of harmonic oscillator moving with constant velocity

This post primarily concerns my confusion regarding time-dependent Hamiltonian.

Let's think about a ball resting at the center of a harmonic oscillator potential. Now view the problem in a reference frame moving with constant velocity $v$. In classical mechanics, the ball and the center of the harmonic oscillator move together with velocity $v$ in this reference frame, and the ball still rests on the (moving) center. Conversely, if we have a ball sitting in the moving center of the harmonic oscillator, we can move to the reference frame where they both stand still. Nothing surprising here.

Now if I try to describe the moving harmonic oscillator in quantum mechanical Hamiltonian, it looks like the following:

$$\hat{H} = \frac{\hat{p}^2}{2m} + \frac{1}{2}m\omega^2 (\hat{x}-vt)^2$$

What is the ground state of this Hamiltonian? My first naive guess would be that it is simply proportional to $\exp\left[-(m\omega/2\hbar) (x-vt)^2\right]$, and it seems right in that it satisfies $\hat{H}|\psi\rangle = (\hbar\omega/2) |\psi\rangle$, but this is not correct because 1) the previous equation applies for time-independent Hamiltonian, and 2) the wavefunction (times $e^{-i Et/\hbar}$) does not satisfy $i\hbar \frac{d}{dt} |\psi\rangle = \hat{H} |\psi\rangle$.

We can do a unitary transformation of $\hat{H}$ so that we end up with a time-independent Hamiltonian. If you choose $\hat{U}(t) = e^{i\hat{p}vt/\hbar}$, then $\hat{U}\hat{p}\hat{U}^\dagger = \hat{p}$, $\hat{U}\hat{x}\hat{U}^\dagger = \hat{x}+vt$. The unitary transformation takes us to the frame where the Schrodinger equation is $i\hbar \frac{d|\psi'\rangle}{dt} = \hat{H'}|\psi'\rangle$, where $|\psi\rangle = \hat{U}^\dagger |\psi'\rangle$ and $\hat{H'} = \hat{U}\hat{H}\hat{U}^\dagger - i\hbar \hat{U} \frac{d \hat{U}^\dagger}{dt}$. So in this frame, the Hamiltonian is

$$\hat{H'} = \frac{\hat{p}^2}{2m} + \frac{1}{2}m\omega^2 \hat{x}^2 - \hat{p} v$$

The Hamiltonian $\hat{H'}$ is now time-independent. Note the unitary transformation we did is not really Galilean transformation, since $\hat{p}$ stayed the same, but if you try $\hat{U} = \exp\left[i/\hbar (\hat{p} vt + \hat{x} mv)\right]$ you get what you expect and addition of constant $\frac{1}{2}mv^2$ to $\hat{H'}$. Anyway, what is notable about $\hat{H'}$ is that it describes a harmonic oscillator with coupling between vibrational states.

As I said in the beginning of the post, in classical mechanics, analysis of a ball resting on the moving center of the harmonic oscillator is simple - just do Galilean transformation. The ground state of the static harmonic oscillator and harmonic oscillator moving with constant velocity shouldn't be much different. But somehow I cannot show this starting from Schrodinger's equation, and this is where I need much help.

P.S: $\hat{H'}$ is useful in the sense that it lets you calculate the excitation probability of particle in a harmonic oscillator doing linear transport. In a perfect linear transport, there is a square pulse of velocity. Now if we restrict ourselves to the ground and the first excited state, $\hat{H'}$ looks like a Rabi problem, and you can obtain the excitation probability using the Rabi formula. This has been experimentally tested in Phys. Rev. Lett. 91, 010407.

• Questions asking for help solving a problem are off topic on this site. If you have a conceptual question here, I don't see what it is. – sammy gerbil Feb 5 '18 at 23:55
• You can perform a consistent treatment of Galilean-boosted states. Besides the shift in the position dependence of the wave function, states also acquire a time dependent phase. In particular, for systems invariant under Galilean transformations, you can check that boosting a solution of Schrodinger's equation produces a solution. You can find the details in section 7.3 of Gottfried and Yan's book, Quantum Mechanics, 2$^{\mathrm{nd}}$ edition. – secavara Feb 6 '18 at 1:34
• I guess my question was really a conceptual question about the consistency of Galilean boost in quantum mechanics, which @secavara helped answer (thank you for the reference). It seems I was too casual in ignoring the (+1/2 mv^2) term just because it is a constant term to the Hamiltonian. – wcc Feb 6 '18 at 3:08