Consider the following Hamiltonian: $$\hat H=\frac{\hbar\omega}{2}(\hat x^2+\hat p^2)-\frac{\hbar\omega}{2}\hat 1 =\frac{\hbar\omega}{2}(\hat x^2+\hat p^2-\hat 1 )$$
After defining annihilation and creation operators in the usual way, this can be written as:$$\hat H=\left(\hat a^{\dagger}\hat a +\frac{\hat 1}{2}-\frac{\hat 1}{2}\right)\hbar\omega$$
Thus,
$$E_n=n\hbar\omega$$
Now, in the case of a harmonic oscillator we know from the normalization condition of number states that $n\geq0$ is satisfied and hence $n$ can be taken zero for the ground state. For a HO this gives the ground state energy as $\hbar\omega/2$ but here it gives zero. I tried to see if there was something that constraints $n$ to be greater than $0$ but I couldn't find it.
Also, since the ground state for this system can be built explicitly from the definitions of $\hat a$ and $\hat a ^{\dagger}$ so I think that the ground wave function for this case will also not go to zero if $n=0$. Therefore, this cannot be a reason to constraint $n$ to be greater than zero.
Does that mean that ground state energy in this case can be zero?
But this conclusion seems wrong to me since I have just added a constant factor to the Hamiltonian of a Harmonic oscillator which is nothing but shifting the potential level and I don't think that should change the zero-point energy.
So, what's going on here?
EDIT:
Maybe I have not indicated my doubt clearly.
I know that a constant offset in the hamiltonian does not change the dynamics of the system.
I also understand that for many systems we generally make the ground state energy as zero reference assigning it a value of zero as we are interested only in the excitations or difference in energy levels.
The problem that I have is that I think (I might be wrong) that zero-point energy represents real fluctuations arising due to the uncertainty principle and this can be seen in a $0$K scenario where even at that temperature there is still kinetic energy.
But as can be seen in the above example, this energy seems to come out to be zero. What does this mean? How is it consistent with the fact that the lowest state should have some energy?
Best Answer
I'll start with things you already understand. The question considers a model with Hamiltonian $$ H \propto x^2+p^2-1 \tag{1} $$ with a positive overall factor, where $x$ and $p$ are operators satisfying $$ [x,p]=i. \tag{2} $$ As noted in the question, the Hamiltonian can also be written $$ H\propto a^\dagger a \tag{3} $$ with $$ a\propto x-ip. \tag{4} $$ The eigenstate of $H$ with the lowest eigenvalue is the state $|0\rangle$ that satisfies $$ a|0\rangle = 0. \tag{5} $$ This is the ground state. The eigenvalue (which is sometimes called "zero point energy") is irrelevant, because shifting $H$ by a constant term does not change the fact that (5) has the lowest eigenvalue among all eigenstates of $H$. You already understand this.
I'm not sure what you want to call "kinetic energy" in this context, but suppose it's the operator $$ K \propto p^2. \tag{6} $$ The ground state an eigenstate of the total energy operator $H$, so measuring $H$ in the ground state would always give the same outcome. But the ground state is not an eigenstate of the kinetic energy operator $K$. If we measure $K$ in the state $|0\rangle$, and if we repeat this experiment a jillion times, we'll get a jillion different outcomes. This is an example of the so-called uncertainty principle: if you measure an observable in a state that is not one of the observable's eigenstates, you get diverse outcomes. Sometimes people describe this using words like "vacuum fluctuations," but I don't know why. Maybe they're imagining Hidden Variables.
When we say that a system is at absolute zero, we mean that it is in its ground state — the state of lowest total energy. In this example, that's the state satisfying (5). In some systems, like an ideal gas, total energy and kinetic energy are the same thing. The harmonic oscillator is not one of those systems, at least not if "kinetic energy" means (6). In general, zero absolute temperature does not mean zero kinetic energy. Zero absolute temperature does mean time-independent, and the ground state — the ray represented by the vector $|0\rangle$ — is indeed independent of time.