This is a deceptively simple question, and I would like to answer it in the context of Condensed Matter Theory. We will start with the question: why do objects in our everyday life appear to be localized?
The Model:
Let's consider a simple model for a quantum crystal. The Hamiltonian for a collection of atoms held together by harmonic forces between neighbors is given by:
$$\hat{H}=\sum_{\vec{x},\vec{\delta}} \frac{\hat{P}^2(\vec{x})}{2m} + \frac{1}{2}m\omega_0 (\hat{X}(\vec{x})-\hat{X}(\vec{x}+\hat{\delta}))^2.$$
The position of the atoms $\vec{x}$ may be one, two, or three-dimensional, and the connection between interacting atoms $\vec{\delta}$ may cover nearest-neighbours, next-nearest-neighbours, or any other interatomic distance. We are thus not just considering an oversimplified model for a hypothetical piece of material, but also the family of Hamiltonians which in principle describe all solids.
The crucial point to notice, is that this Hamiltonian commutes with the operator of total (or centre-of-mass) momentum:
$$\hat{P}_{tot}=\sum_{\vec{x}}\hat{P}(\vec{x})$$
$$\implies [\hat{H},\hat{P}_{tot}]=0.$$
The vanishing of this commutator is a consequence of the homogeneity of space: Any translationally invariant Hamiltonian commutes with the total momentum operator!
That aside, if two quantum mechanical operators commute, this implies that it is possible to find a set of states which are simultaneous eigenstates of both operators. The fact that the Hamiltonian for a crystal commutes with the total momentum operator thus implies that all eigenstates of the crystal are total momentum eigenstates. Because of Heisenberg’s uncertainty principle, states in which the value of the total momentum can be known with certainty correspond to states in which the centre-of-mass position is entirely unpredictable. In other words, the total momentum eigenstates, and hence the eigenstates of the crystal Hamiltonian, are all completely spread out over all of space!
This is not the end of the story however. We can write the Hamiltonian in reciprocal space, and separate it into two independent parts:
$$\hat{H}=\hat{H}_{coll} + \sum_{\vec{k} \ne 0} \hat{H}_{int}(\vec{k}).$$
The part of the Hamiltonian for non-zero values of $\vec{k}$ describes the internal dynamics of the crystal, in terms of all of its phonon excitations and their interactions. The part at $\vec{k} = 0$ on the other hand, describes the collective dynamics of the crystal as a whole. It can be straightforwardly shown that in the case of the crystal, the collective part of the Hamiltonian is given by:
$$\hat{H}_{coll}=\frac{\hat{P}^2_{tot}}{2mN}.$$
From now on, we will only consider the properties of the collective part of the Hamiltonian, and completely ignore the internal, phonon-related part (this is possible because the two parts of the Hamiltonian commute; and at extremely low temperatures, the collective part of the Hamiltonian is the only part that matters).
As expected, the eigenstates of the collective Hamiltonian are the eigenstates $|P \rangle$ of the total momentum operator, and the corresponding energies are given by the kinetic energy $E_P=\frac{P^2}{2mN}$ of the crystal as a whole. The ground state of the crystal is the state with total momentum $P = 0$. Its wavefunction is thus completely and evenly spread out over all of space, with equal amplitude and even equal phase at every possible positions. Additionally, the energy separating the ground state from the collective excitations with non-zero total momentum, is inversely proportional to the total mass $mN$ of the whole crystal. This means that as we consider a larger and larger crystal, it becomes easier and easier to make excitations of the crystal. In fact, in the limit $N\rightarrow \infty$ of an infinitely large crystal, it would cost no energy at all to make collective excitations, and all states in the collective part of the spectrum become degenerate with the ground state. In that limit, a wave packet of total momentum states with a well-defined centre-of-mass position would have the same energy expectation value as the zero-momentum state.
Of course, real crystals are not infinitely large, and superpositions of momentum states do cost energy to create. But the pieces of matter we are interested in (like you) contain a very large, albeit finite, number of atoms. How difficult would it then be to make a superposition of crystal eigenstates which has a well-defined centre-of-mass position? In terms of the collective Hamiltonian, this amounts to adding a perturbation which tends to localize the crystal:
$$\hat{H}'_{coll}=\frac{\hat{P}^2}{2mN}+V\hat{X}^2_{com}.$$
Here V is the strength of a potential which tends to localize the centre of mass position $\hat{X}_{com}$ at the origin of our coordinate system. Since $\hat{P}_{tot}$ and $\hat{X}_{com}$ obey canonical commutation relations, the perturbed Hamiltonian $\hat{H}'_{coll}$ is that of a harmonic oscillator, with the well known energies and eigenstates:
$$\hat{H}'_{coll}=\hbar \omega (\hat{n}+1/2),$$
$$\langle x|n=0 \rangle = (\frac{2mNV}{\pi^2 \hbar^2})^{\frac{1}{8}} exp(-\sqrt{\frac{mNV}{2\hbar^2}}x^2).$$
The ground state in the presence of a perturbation corresponds to a Gaussian wave packet of width $\sigma = \frac{\hbar}{\sqrt{2mNV}}$. The effect of spontaneous symmetry breaking emerges when we consider two non-commuting limits:
$$\lim_{N\rightarrow \infty} \lim_{V \rightarrow 0} \langle x|n=0 \rangle = constant,$$
$$\lim_{V \rightarrow 0} \lim_{N\rightarrow \infty} \langle x|n=0 \rangle = \delta(x).$$
Conclusion:
For a system of any finite size, no matter how large, the perturbation can be made small enough for the ground state wavefunction to be essentially spread out over the entire universe. But for an infinitely large system, any perturbation, no matter how weak, is enough to completely localize the wavefunction into a single position. In the thermodynamic limit, the localization happens even in the presence of only an infinitesimal potential, which in effect means that the wavefunction can spontaneously localize and break the translational symmetry of the unperturbed Hamiltonian.
But even if the material is not infinitely large, the important message of the last equation is that even if N is not yet truly infinite, the approach towards the thermodynamic limit is singular. This implies in particular that as you consider larger and larger pieces of matter, a weaker and weaker perturbation suffices to make its ground state a localized wave packet. Therefore, in a realistic piece of material, even entirely immeasurable perturbations are easily strong enough to completely localize the matter into a single position.
Finally, it's important to note that if we find a crystal (or a person like you) in a localized state, that thus implies not only that the crystal picks out a preferred location in space, but also that it exists in a state that is not the ground state of its Hamiltonian. In fact, it is not even an eigenstate of the Hamiltonian. Spontaneous symmetry breaking explains both how an infinitesimal perturbation can suffice to pick a location and break the symmetry, and how that same infinitesimal perturbation suffices to make the localized state the ground state of the perturbed system.
Therefore, to finally answer your question:
You are indeed localized into a single position.
Source: Lectures notes from a course on Condensed Matter Theory by Jasper van Wezel from University of Amsterdam.
Update:
The comment by OP caught me off guard:
When some perturbation causes "me" to become localized in a single position, does this mean "I" am completely spread out in total momentum space? Or is my total momentum also localized to a single value?
I didn't realize at the time, but my answer only explained why objects appear to have a well-defined position - it did not provide an answer as to why they also appear to have a well-defined momentum.
Therefore, I Fourier transformed the ground state wave function $\psi_0(x)=\langle x|n=0 \rangle$, to momentum space $\tilde{\psi_0}(p)$ and applied the two limits. The results were:
$$\lim_{N\rightarrow \infty} \lim_{V \rightarrow 0} \tilde{\psi_0}(p)=\delta(p)$$
$$\lim_{V \rightarrow 0} \lim_{N\rightarrow \infty} \tilde{\psi_0}=constant.$$
First thing to note is that this is the exact opposite of the results we got for the wavefunction in position space. Secondly, as before, we are interested in the equation where we first take the $N\rightarrow \infty$ limit, and then the $V \rightarrow 0$ limit (this corresponds to having a very large object and then applying an infinitesimal perturbation). Thus, the answer is that the wavefunction is momentum space is completely spread out, i.e. it's a constant. This is not what I expected, even though it makes total sense due to Heisenberg's uncertainty principle.
Triggered by this result, I consulted the professor himself.
This is basically what he said (paraphrased to make more sense in our context):
The result at which we arrived is indeed, a very good point. The example of the quantum crystal is unfortunately a not well-behaved mathematical object, unlike antiferromagnets and Bose-Einstein condensates which are much more mathematically clean, and subtleties do not come up. The crystal was given merely as an example to illustrate that everyday objects can display strange collective quantum behavior.
Some problems with the quantum crystal are that
- its order parameter is hard to define,
- the required symmetry-breaking field $V$ for the crystal is not obvious.
In reality, the symmetry-broken ground state is a Gaussian wavefunction whose width does not depend on $N$, does not get completely localized in the thermodynamic limit, and does not have undefined momentum. This is because in order to have an extensive energy for the symmetry-broken Hamiltonian, the symmetry-breaking field V needs to scale with the inverse system size, i.e. $V$ should be replaced by $V/N$ in all equations.
This raises the question:
Does that mean that translational symmetry was not broken after all?
The answer is simply no, and this is because we've been looking at it the wrong way all along. Rather than considering the wavefunction itself, and saying whether or not it is absolutely localized, we should consider how localized it is, compared to some natural measure for localization. Since we're talking about a single crystal without any other objects around, the only sensible measure is to ask whether or not the crystal is localized to within its own size. In other words, whether fluctuations in position, $\sqrt{<x^2>}$, are small compared to the linear size of the object, $L$. You can then see that, by taking the right order of limits, the fluctuations are always large in absolute size (for small $V$) but that $\sqrt{<x^2>} / L$ disappears in the thermodynamic limit.
So an object (or a person) is localized in real space in units of its own size (which are the only relevant units!).
Similarly, for the fluctuations in momentum, we can show that although they almost disappear in absolute size ($\sim V$), they are still precisely large enough to allow the object not to fluctuate compared to its own size.
(Hopefully) Final Conclusion:
An everyday object is still a wave packet in the thermodynamic limit, with uncertainty in both position and momentum. But the fluctuations in position and momentum relative to any natural scale for measuring them, do disappear, and the object thus appears localized in both position and momentum on human scales.
My apologies if my original answer was somewhat misleading!
All credit goes to Jasper van Wezel.
