Creation and annihilation operators in complex projective space

Question

Note 1: The Hilbert space $\mathcal{H}$ of a quantum system cannot possibly be the genuine space of states in quantum mechanics. In particular, $\vec{0} \in \mathcal{H}$, which is not a valid state. Hence, to call $\mathcal{H}$ the space of states would be to call $\vec{0}$ a state, which as stated above is not a valid state.

Note 2: Even in "conventional" quantum mechanics, the space of states is not a vector space. It is the set of trace-class, positive-definite, trace-one operators over Hilbert space, denoted $\overline{\mathcal{B}}^+(\mathcal{H})$. Also known as density operators.

(At the least) Finite dimensional quantum mechanics formally has $\mathbb{C}P^{n-1}$ as the raw space of states. To be explicit, $\mathbb{C}P^{n-1}$ corresponds to the rays in $\mathbb{C}^n$. There does not exist an additive identity element (a zero element) of $\mathbb{C}P^{n-1}$. However, naively creation and annihilation operators require the notion of an additive identity element as evidenced by the expression $$\hat{a} \lvert 0 \rangle = 0 \text{ (the zero vector)}$$ where $\hat{a}: \mathcal{H} \to \mathcal{H}$. Without access to such an additive identity element, how does one talk about creation and annihilation operators $\hat{a}: \mathbb{C}P^{n-1} \to \mathbb{C}P^{n-1}$ in $\mathbb{C}P^{n-1}$?

Answer 1

The postulates of Quantum Mechanics demand that there be a Hilbert space of states ${\cal H}$. States are then unit rays on ${\cal H}$, so we may take $$\mathscr{S} = \{[\Psi] : \Psi\in \mathcal{H}, \Psi \sim \lambda \Psi, \lambda\in \mathbb{C}\setminus\{0\}, |\Psi|=1\}.$$

For ${\cal H}=\mathbb{C}^n$ this is indeed a subspace of $\mathbb{CP}^{n-1}$ (note that there is the unit ray condition: states should be normalized due to the probabilistic interpretation).

Now, the equation $a_p|0\rangle=0$ is not an equation in $\mathscr{S}$, rather it's an equation in ${\cal H}$. So while it is true that the states are really unit rays and therefore elements of ${\mathscr{S}}$, you cannot simply dispense with ${\cal H}$ and work just with $\mathscr{S}$. In fact, most of the time we just work in ${\cal H}$ with normalized states $|\Psi|=1$ and recall that they are defined up to a global phase because of the equivalence relation.

Answer 2

Remember that state vectors on the Hilbert space are defined modulo a phase and a norm. The normalization postulate can be dropped as long as you define $$ \langle \hat{O} \rangle = \frac{\langle \psi | \hat{O} | \psi \rangle}{\langle \psi|\psi \rangle} $$ (this implies that the whole distribution is normalized since is given by the expectation values on the projectors). So the phase space is by definition necessarily $\mathcal{P(H)}$, whether we like it or not.

Now, regarding your objection, you are implicitly assuming that the operators must be defined on $\mathcal{P(H)}$, which is not necessary. As long as you can compute the distributions, you're good. In order to find a distribution, you don't need to actually apply the operator, you just need the expectation values of projectors, which are well-defined scalar functions on our phase space, being representative-independent.

Wait, but what about the Schrodinger equation? You need to apply the Hamiltonian there!
Well, not necessarily. $\mathcal{P(H)}$ has a natural Kahler structure, which is essentially the complex version of symplectic geometry (classical Hamiltonian mechanics). Turns out that you can use the same machinery you use in the classical case, i.e. defining a Hamiltonian vector field s.t. $$ \Omega(\cdot, X_H) = dH$$ where $\Omega$ is the natural symplectic form and $H$ is the expectation value of the Hamiltonian operator (again a well-defined scalar on $\mathcal{P(H)}$).
You can then explicitly show, with some basic calculation, that the Schrodinger equation is a particular case where you choose your representatives to be normalized. This is called the geometrical formulation of quantum mechanics.