# Is the zero vector necessary to do quantum mechanics?

Textbook quantum mechanics describes systems as Hilbert spaces $$\mathcal{H}$$, states as unit vectors $$\psi \in \mathcal{H}$$, and observables as operators $$O: \mathcal{H} \to \mathcal{H}$$. Ultimately, we only end up computing expectation values or probability amplitudes. Neither of these computations seem to necessitate use of the zero vector $$0 \in \mathcal{H}$$.

Can we remove the zero vector and reproduce the predictions of the above described theory of quantum mechanics (having modified whatever structures as necessary)?

• The zero vector is required, for example, in the formulation of the Quantum Harmonic Oscillator. The annihilation operator acting on the ground state results in the zero vector, avoiding states with indefinite negative energy. Commented Jun 29 at 11:45
• In addition to the above, physicists also love to pretend that quantum theory is a lot about matrices. In linear algebra, zero is privileged. You absolutely need it. Commented Jun 29 at 12:26
• Without the zero vector, you lose the vector space structure of the Hilbert space. So even if it does not appear in the final step of calculating an expectation value or a probability amplitude, you absolutely need it to justify the calculations leading up to them, or to even make the theory mathematically consistent. Commented Jun 29 at 13:17
• My question includes the possibility of reformulating the mathematics to describe quantum mechanics @paulina Commented Jun 30 at 4:49
• @ToddWilcox: No, if we remove the zero vector and forget that there was an origin, then we have an affine space, which is a weaker concept but still capable of hosting linear algebra. Commented Jun 30 at 15:26

Quantum theory can be completely stated in the complex projective space of a complex Hilbert space. The basic object from this perspective is the transition probability of a couple of rays (pure states). That is computed as a suitable distance on the said projective space which makes it complete as a metric space.

So, in principle, the zero vector does not play any explicit role as it does not determine an element of the projective space.

However, the above approach turns out to be mathematically cumbersome and awkward when performing explicit computations. In summary, it is practically convenient to work in the associated Hilbert space. This is practical on the one hand, and cause of difficult issues on the other hand, for instance when dealing with groups of symmetries and their unitary projective representations. The use of $$SU(2)$$ in place of $$SO(3)$$ to describe the representations of the physical rotations on states is an effect of that choice. In the projective space the relevant group is the physical one ($$SO(3)$$), but it is replaced by its universal covering $$SU(2)$$ when dealing in the Hilbert space.

Varadarajan’s book “The geometry of quantum theory” develops (also) these ideas.

• Following a comment above, I just realized I don't understand how creation/annihilation operators work in projective Hilbert space. As you said, the zero vector is not an element of projective Hilbert space, so it would seem that $a$ would send the vacuum out of the projective Hilbert space, and so isn't an operator in that case. That seems wrong. Any thoughts? Commented Jun 29 at 15:30
• Creation and annihilation operators are not in the projective space, all that construction is done in the Hilbert space, sure. If you use vectors you are already outside the projective space. I never tried to deal with the harmonic oscillator in the projective space (I suspect that it would be like to commit suicide). However, it must be possible as all physical notions like observables and all that can be defined in the projective space. Evidently observables are not selfadjoint operators in the projective space, but mathematical objects with the same physical information. Commented Jun 29 at 16:44
• @paulina mixed states are classical mixtures of pure states, so they can be described also in the projective space, using classical probability densities thereon. I repeat: I never tackled the difficult task to pass all the theory to the projective space (maybe there is some attempt in Varadarajan's book), but in principle it must be possible, since all physical notions in terms of states, symmetries, and elementary observables can be defined therein... Commented Jun 29 at 16:47
• As far as my opinion is concerned that is based on a misleading statement: if you rotate an electron of 2$\pi$ you have done nothing as physics requires. There is at least one discussion on this fact in PSE. I do not wish to come back on this discussion, sorry. Commented Jun 29 at 22:15
• @Tobias Funke Thanks, I was just referring to that answer of mine. Commented Jun 30 at 7:27

This is largely a philosophical question, but I would argue that yes, the zero vector is necessary for QM to make sense conceptually. A few examples (off the top of my head) of places where the zero vector comes up:

1. The result of acting the annihilation operator on the ground state of the quantum harmonic oscillator. (This is important for ensuring that the spectrum of the QHO is only one-sided infinite rather than two-sided infinite, in which case it wouldn't be bounded below and would therefore be thermodynamically unstable.)
2. If there were no zero vector then the zero operator would not be defined either, since it's defined as the operator that takes every vector to the zero vector. So it wouldn't be meaningful to talk about any two operators commuting, i.e. their commumutator being the zero operator. Nor would you be able to say things like $$a^2 = 0$$ for a fermionic annihilation operator, which is one way of encoding the Pauli exclusion principle.
3. In the presence of symmetries that block-diagonalize the Hamiltonian (or the time-evolution operator), we can decompose the Hilbert space into an inner direct sum of one symmetry sector $$S$$ and the other sectors (the subspace orthogonal to $$S$$). Then if $$|\psi_S \rangle$$ lies in the symmetry sector $$S$$, then $$U|\psi_S \rangle = |\psi'_S\rangle \oplus 0_{\perp S}$$, which requires a notion of the zero vector of the orthogonal subspace of the Hilbert space.
4. Most fundamentally: without a zero vector, the Hilbert space wouldn't be a vector space, and all kinds of mathematical machinery would immediately break (or at least need to become immensely more complicated in order to "manually" exclude the possibility of the zero vector coming out of any calculation).

The squared absolute value of a wave function, integrated to the whole space, must be 1. Thus, the zero vector is not a wave function. But it still can happen anywhere in various QM formulas, and it is even needed to describe deeper math things (groups must have a zero-like element).