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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?

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$\begingroup$ "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" - what do you think the "potential level" is if not the energy? $\endgroup$

ACuriousMind
– ACuriousMind ♦

2021-08-04 17:17:56 +00:00
Commented Aug 4, 2021 at 17:17
$\begingroup$ You might find this question, and the associated answers interesting $\endgroup$

By Symmetry
– By Symmetry

2021-08-04 18:32:03 +00:00
Commented Aug 4, 2021 at 18:32
1

$\begingroup$ @ACuriousMind To put my question in a better context let me quote the part from an answer given here by Sahand: physics.stackexchange.com/q/614186 $\endgroup$

Lost
– Lost

2021-08-05 09:18:23 +00:00
Commented Aug 5, 2021 at 9:18
$\begingroup$ "I think what is typically meant by "zero-point energy" in this context is actually zero-point kinetic energy. Exactly as you say, a constant offset $E_0$ in the Hamiltonian $H=H_0+E_{0}$ has no consequences on the dynamics and classically speaking, it corresponds to picking a different reference point for your potential energy function." $\endgroup$

Lost
– Lost

2021-08-05 09:19:22 +00:00
Commented Aug 5, 2021 at 9:19
3

$\begingroup$ Please edit information that clarifies your question into the question instead of just leaving it in comments $\endgroup$

ACuriousMind
– ACuriousMind ♦

2021-08-07 11:21:40 +00:00
Commented Aug 7, 2021 at 11:21

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