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?
It depends what you're doing, and indeed most of the quantum optics literature dismisses the term as it does not contribute to the dynamics. However, it is important that beginning students form an intuition for how and where zero-point energies come in, and why they are necessary.
Take a look at the eigenfunctions of the harmonic oscillator, in position space:
Notice, in particular, the behaviour at the classical turning points, where the baselines cross the potential. These are the inflection points of the wavefunctions, where the oscillatory behaviour turns into exponential decay. Even for the ground state, these two points must be spatially separated, to allow the exponential decay on the left to turn round into a decreasing function and match into exponential decay on the right, and for these two points to be separated the energy of the ground state needs to be separated from the bottom of the well. This is the essence of the zero-point energy, and until you internalize all the implications of 'classically allowed' and 'classically forbidden' on the wavefunction, it's best to be explicitly reminded that it exists.
On the other hand, once you've done that, there is little point in lugging that term around. If you dig a little deeper into the literature, you'll see people start to drop the term in settings where it is not important. Some examples:
The Dicke model (e.g. equation 6),
and many, many others. For a good look at what people actually use in the literature, I would recommend searching for 'quantum harmonic oscillator' on the arXiv. This will turn up many papers you won't understand, but it is not that complicated to discard the ones that don't have QHO hamiltonians in them, and distinguish the ones that use hamiltonians of the form $\tfrac1{2m}p^2+\tfrac12 m\omega^2 x^2$ from the ones that use the form $\hbar\omega a^\dagger a$.
It's also worth mentioning that you can't always drop the term. In quantum field theory in particular, you are often faced with a system that is an infinite collection of harmonic oscillators, for which vacuum energy must be treated carefully. On another branch of that, zero-point energies can have measurable effects, for example through the Casimir effect, in which case you obviously can't neglect it.
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.
I disagree that :
Since energy can always be shifted by a constant value without changing anything,
You are maybe thinking of classical potential energy , but the mass of a proton is fixed, for example, it cannot be shifted by a constant value, and at rest $E=mc^2$. This statement is not general and can only be true for the solutions of non relativistic equations.
Edit after comments:
After the comments I realized the question is about a change of the zero of energy, that would not affect the energy levels but the y axis of the potential, which would acquire a negative lower point, so that the first energy level is at 0.
This change would only introduce an overall phase factor ( see answer by dextercioby) in the time dependent solutions.
The harmonic oscillator is a very useful quantum mechanical solution because all symmetric potentials have as a first term in their series expansion the x**2. Thus it is extensively used in most many body problems in chemistry, and not only, to model the different collective potentials arising in lattices.
The reason then is for simplicity and esthetics, not to introduce extra complexity in the form of the potential so that the generic equation is described by the simplest functional form of the potential, x**2.
