How Does a Harmonic Oscillator Behave When Traveling at Constant Velocity?

[tamiry]

hi,

i'm trying to see how does an HO, traveling with constant speed v looks like. suppose a unitless system
$$H = P^2+(X-vt)^2$$
define
$$Y = X-vt$$
then
$$H = P^2+Y^2$$

i can see that [P,Y] = -i (unitless - no h-bar) so i guess it means that P and Y are conjugate space/momentum operators. therefore the solution for this is, using ehrenfest theorem
$$<Y> = Y(0)cos(t)+P(0)sin(t)$$
$$<P> = P(0)cos(t)-Y(0)sin(t)$$
where Y(0) is the expectation value of Y at t = 0 and the same for P(0).
now, going back to X, assuming the state is a square-integrable one
$$Y(0) = X(0) - v*0 = X(0)$$
$$<Y> = <X-vt> = <X> - vt<state|state> = <X>-vt$$
$$<X> = X(0)cos(t)+P(0)sin(t)+vt$$
$$<P> = P(0)cos(t)-X(0)sin(t)$$

now this makes some sense, <X> really oscillates around a value increasing with rate v, but P seems unchanged. I'd expect P to have a constant part as well, with size v since that is the constant velocity.
i've written the ehrenfest theorem equations for the original P,X and I've noticed that if I set P_new = P-v is solves those equations. so where did I get it wrong?

thanks a lot

 

[azntoon]

for the help!You are on the right track, but you made a mistake. The P that you defined in the Hamiltonian H=P^2+Y^2 is the momentum operator associated to the Y coordinate, and not the original momentum operator P. To get the solution for the original momentum operator, you need to use the Ehrenfest theorem for the original P,X coordinates: <P> = P(0)cos(t)-X(0)sin(t) + v. This equation has the expected form, with a constant term v that represents the constant velocity of the HO.