Harmonic oscillator with potential shifted by a constant

Question

I've been thinking a lot about changes to the harmonic oscillator potential, and I was looking into the problem where

$$V(x) = \frac{1}{2}m\omega ^2 x^2 + C$$

where $C$ is some positive real constant. I was able to convince myself that, in this case, our eigenenergies would take the form $E_n = \hbar \omega (n+\frac{1}{2}) + C$, which is basically our "standard" harmonic oscillator energies shifted by $C$.

What I'm having trouble with are the eigenstates. Since $[H_\text{new}, H_\text{SHO}] =0$, they should share the same set of eigenstates. This means that we can use $|n\rangle$ as are eigenstates.

However, how can this make sense? Suppose we start with $$H = \frac{p^2}{2m} + \frac{1}{2}m\omega ^2 x^2 + C$$ and our particle is in the state $|0\rangle$. As some time, we snap our fingers that $H$ immediately changes to $$H = \frac{p^2}{2m} + \frac{1}{2}m\omega ^2 x^2 \, ,$$ our normal harmonic oscillator. If I check the probability of measuring the energy $E_0 = \hbar \omega /2$, I would get

$$|\langle 0|0 \rangle|^2 = 1 \, . $$

This seems weird to me because we originally had an energy of $\hbar \omega /2 + C$ ... this would imply that we are measuring in the same basis, but our measurement depends on the Hamiltonian we are using.

My conclusion is that the eigenstates are not the same, but I can't see why!

Answer 1

Your "snap of the fingers" really means that your Hamiltonian is

$$H(t) = \frac{p^2}{2m} + \frac{1}{2}m \omega^2 - \theta(t)C$$

where $\theta(t)$ is the Heaviside step function.

The eigenstates will only be equivalent asymptotically for $t \rightarrow\pm\infty$ since you are now dealing with a time-dependent Hamiltonian.

Answer 2

It should actually be noted that you cannot measure absolute energy. What you would actually have experimental access to is the difference between energy levels, in the form (for instance) of a photon emitted following a transition.

Hence, the $n$ in $|n\rangle$ for a harmonic oscillator does not label the absolute energy of the state, but rather the number of quanta (excitations).

The absolute energy of the $n^{\mathrm{th}}$ state is: $$ E_{\mathrm{zp}} + n\hbar \omega,$$ where $E_{\mathrm{zp}}$ is the zero-point energy, whose experimental measurements so far do not make sense.

Depending on which convention for operator ordering you use, i.e. $H = \hbar\omega a^\dagger a $ or $H = \hbar\omega a a^\dagger$, the exact value of the zero-point energy varies. The energy spacing, however, does not.

In fact, adding a constant shift to the Hamiltonian only amounts to a pure phase factor in the wavefunction, which has hence no physical (observable) consequence:

$$ \mathrm{i}\hbar\frac{\partial \psi}{\partial{t}} = (H+C) \psi \quad \Leftrightarrow \quad \mathrm{i}\hbar\frac{\partial (\psi\,e^{\mathrm{i}Ct/\hbar})}{\partial{t}} = H (\psi\,e^{\mathrm{i}Ct/\hbar}).$$

In your specific case, the time-dependent Schroedinger equation can still be solved analytically by separation of variables.

From

$$H(x,t) \psi(x,t) = \mathrm{i}\hbar\, \frac{\partial}{\partial t}\psi(x,t)$$

with $ H(x,t) = -\partial_x^2 + V(x) + C(t)$ and $C(t) = C\theta(t)$, we assume separation of variables: $\psi(x,t) = X(x)T(t)$.

This reduced the equation to:

$$ -\frac{1}{X}\frac{\mathrm{d}}{\mathrm{d}x^2}X(x)+V(x)=\mathrm{i}\hbar\,\frac{1}{T}\frac{\mathrm{d}}{\mathrm{d}t} T(t) -C(t). $$

Both sides equal to a constant $E$ as they depend on different independent variables.

So the spatial part, i.e. the Time-indepenent Schroedinger equation, comes out exactly the same. The time part, $T(t)$, can be solved analytically to give:

$$ T(t) = e^{-\frac{i E t}{\hbar }}+ \left [e^{-\frac{i E t}{\hbar }-\frac{\mathrm{i} t}{\hbar }}-e^{-\frac{\mathrm{i} E t}{\hbar }}\right ]\theta(t) ,$$

where I used an initial condition $T(t) \bigg|_{t=0} =1$. Note that $|T(t)|^2 =1$ so it's still a pure phase factor.

As expected from the earlier discussion, then, the energy eigenstates are the same, for the shift in the energy just becomes a phase factor.