Why not drop $hbaromega/2$ from the quantum harmonic oscillator energy?

Since energy can always be shifted by a constant value without changing anything, why do books on quantum mechanics bother carrying the term $\hbar\omega/2$ around?

To be precise, why do we write $H = \hbar\omega(n + \frac{1}{2})$ instead of simply $H = \hbar\omega n$.

Is there any motivation for not immediately dropping the term?

Answer

Consider a potential, which approximately can be described by two harmonic oscillators with different base frequencies, for example (working in dimensionless units) $$U=1-e^{-(x-4)^2}-e^{-\left(\frac{x+4}2\right)^2}$$

It will look like

Now let's look at two lowest energy states of the Hamiltonian

$$H=-\frac1m \frac{\partial^2}{\partial x^2}+U,$$

taking for definiteness $m=50$, so that the lowest energy states are sufficiently deep. Now, at the origin of oscillator at left it can be shown to be

$$U_L=\frac14(x+4)^2+O((x+4)^4),$$ and the for right one we'll have $$U_R=(x-4)^2+O((x-4)^4)$$

If two lowest levels are sufficiently deep that their wavefunction don't overlap, then we can approximate them as eigenstates of each of the harmonic oscillators $U_L$ and $U_R$. See how these two states look:

You wanted to remove zero of the total energy by shifting the potential. Of course, you could do this for a single oscillator. But now you have to select, which one to use. And if you select some, you'll still get zero-point energy for another.

Thus, this trick isn't really useful. It tries to just hide an essential feature of quantum harmonic oscillator and quantum states in general: in bound states there is lowest bound on energy, which can't be overcome by the quantum system, although classically the energy could be lower.

Zero-point energy is the difference between minimum total energy and infimum of potential energy. It can't be "dropped" by shifting the potential energy.