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.
If you are only looking at the dynamics of a quantum harmonic oscillator, you can go ahead and drop this term. However, the universe isn't a single particle trapped in a $1$ dimensional harmonic oscillator. Thus we very, very frequently are looking at systems like these:
In such more complicated situations, often the groundstate energy of the harmonic oscillator will have physical consequences. Consider the following real-world examples:
Adiabatic release from optical lattices
Using counterpropagating laser beams detuned from atomic transitions, we can create potentials that trap atoms proportional to $\cos(2kx)$, where $k$ is the angular wavenumber of the laser. Then near the bottoms of these potentials, they can be approximated as a harmonic oscillator. (picture from the NIST website)
Many atomic physics experiments will use something like RSC or EIT cooling too cool atoms to the groundstate of these localized potentials. Those experiments typically want the atoms to be as cold as possible. If you suddenly turn off the laser beams, the atoms in free space are left with roughly the groundstate energy of the harmonic oscillator they were trapped in (there's one physical consequence of the groundstate energy). However, if you slowly turn off the laser beams, the resulting atoms are colder. This is because you're effectively turning down $c$ over time in the potential $V=c(t)x^2/2$. Because of the adiabatic theorem, the atoms stay in the groundstate, and that groundstate energy decreases over time. Here are two papers where this is done:
Molecular energy levels
Consider this diagram of molecular energy levels from hyperphysics. For very simple diatomic molecules, we can calculate an effective potential as a function of distance between the two nuclei, and the relative motion of the two atoms is governed by this effective potential (via the Born-Oppenheimer approximation). This effective potential can then be approximated as a quadratic near the equilibrium point. And the picture shows an important idea - that quadratic coefficient is different for different electronic energy levels. Sometimes in atomic physics experiments we want to drive electronic transitions in molecules or photoionize molecules. In those circumstances, the groundstate energy of the harmonic oscillator actually modifies the wavelength of light needed to drive these transitions.
So yeah, if you stayed in one harmonic oscillator, you could redefine energy to remove the groundstate energy. But if you have transitions between different harmonic oscillators or excite to a regime where the harmonic oscillator approximation is no longer valid, the groundstate energy will have physical consequences.
The $\nabla \mathbf{B}$ force in a nonuniform magnetic field
In a lot of quantum mechanics classes, shortly after learning about the harmonic oscillator you will learn about Landau levels, the quantum mechanical solution to a charged particle moving in a magnetic field. You will find that in this situation, the energy levels are the same as those arising in a 2 dimensional harmonic oscillator. Now what happens if the magnetic field varies in magnitude with position? Perhaps lets say the length scale over which the magnetic field varies is small compared to the orbital radius of the groundstate of the charged particle. In classical mechanics, we find that this charged particle doing a "cyclotron orbit" also feels an "effective force" pointing in the direction of the gradient of the magnetic field $\nabla \mathbf{B}$ (described in the "guiding center" wiki). In quantum mechanics, we will find that this effective force still occurs for atoms in the groundstate of the potential created by the local magnetic field. This is because the quadratic coefficient of the harmonic oscillator will vary with position, and because of the adiabatic theorem and the Born-Oppenheimer approximation, the particle will feel an effective force pushing it toward places where it has a lower groundstate energy.
As a less realistic but simpler toy model, consider a system which is a harmonic oscillator in $x$, but the coefficient varies with $y$: $V=c(y)x^2/2$. Let's say that the length scale of variation in $y$ is much smaller than the size of the groundstate in $x$. And maybe we assume the particle is in the groundstate in $x$. You will find that in $y$, the particle moves in an effective potential given by $\hbar\omega(y)/2$, where $\omega(y)=\sqrt{c(y)/m}$.
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.
