Does quantum mechanics need projective representations only due to the Born rule?

In quantum mechanics, physical states don't live in the Hilbert space, but rather on the equivalence class of rays on the Hilbert space. This is called a projective space. This is the reason why when we look for representations of symmetries of the theory, we need to abandon the classical representations and go to projective ones (that is why we replace $$SO(3)$$ with its universal cover $$SU(2)$$ to find representations of rotations and thus include spin).

The explanation for this is that states that differ by a multiplicative constant will give the same probability, given by the Born rule

$$$$P(\phi,\psi)=\frac{|\langle \phi|\psi\rangle|^2 }{|\langle \phi|\phi\rangle||\langle \psi |\psi \rangle|}$$$$

where $$P(\phi,\psi)$$ is the probability to find the state $$\psi$$ on state $$\phi$$ as we measure. The denominator becomes $$1$$ if we normalize states, but that's not necessarily required.

Given that the Born rule cannot be explained within the dynamics of quantum theories but is a separate axiom, my question is: Is the Born rule the only reason why we need projective representations? Or is there something intrinsic about quantum theories that requires projectiveness?

• In what way is the Born rule not "intrinsic about quantum theories"? Commented Sep 2, 2023 at 20:12
• How could you possibly reproduce the results of quantum experiments without projective representations? Commented Sep 2, 2023 at 20:14
• What I mean is that if I gave you a Hilbert state, all the fields in the theory, a Hamiltonian operator, etc. you could calculate their dynamics, time evolution, etc. But you could not derive Born's rule or deduce that the states in the Hilbert space give probabilities. That's what I mean by "not intrinsic". For example, the uncertainty principle IS intrinsic to the theory, given a hilbert state, commutators of unitary operators will give you a lower bound on the product of their variances. It is not an independent axiom, like the Born rule. Commented Sep 3, 2023 at 18:15

Multiplying vector by a constant is the most general transformation not changing magnitudes of all inner products only if we restrict ourselves to a single gauge of EM potentials.

But gauge is arbitrary just as the phase factor. If while changing $$\psi$$, we allow also changing EM potentials via the standard gauge transformation using some function $$\lambda (\mathbf r,t)$$:

$$\mathbf A' = \mathbf A + \nabla \lambda$$ $$\varphi' = \varphi - \partial_t \lambda$$ then function

$$\psi '(\mathbf r,t) = \psi(\mathbf r,t) e^{\frac{iq\lambda(\mathbf r,t)}{\hbar}}$$ corresponds to the same state the function $$\psi$$ did. This means that this new function obeys the same Schroeedinger equation, just with changed EM potentials. The function $$\lambda(\mathbf r,t)$$ is arbitrary.

Notice the new function $$\psi'$$ is not simply a constant times the original function, because the phase in the exponent $$\frac{q \lambda}{\hbar}$$ depends arbitrarily on position $$\mathbf r$$ (for multi-particle systems, the phase factor depend on all positions $$\mathbf r_1,\mathbf r_2,...\mathbf r_N$$ defining the classical configuration).

Thus, in the wave-mechanical description, we have even greater set of functions, not merely those related by multiplication, which all correspond to the same "physical state". Multiplying by a constant is just a special case of $$\lambda$$ which is constant for all $$\mathbf r$$.

Applying the gauge transformation, all vectors have to change this way, including the eigenvectors of the measured observable $$\phi_k$$.

This property (invariance of inner product magnitudes with respect to changing $$\psi$$ in the described way) really is due to a special property of the time-dependent Schroedinger equation, in particular due to minimal coupling: $$\psi'$$ obeys the same time-dependent equation, just with different, appropriately transformed EM potentials. The Born rule just interprets quantities that do not depend on these transformations (magnitudes of inner products).

• Thanks for your answer! However, I don't see how this extends to gauge fields. I understand that gauge theories act on matter fields on a way that generalizes the invariance under multiplication I mentioned before. But the electromagnetic field also has invariance under c-multiplication since it also enters the Born rule, and yet Gauge transformations don't affect the 4-potential in a multiplicative way... Commented Sep 3, 2023 at 18:20
• > the electromagnetic field also has invariance under c-multiplication since it also enters the Born rule Can you give a concrete example of "invariance under c-multiplication" for EM field? Commented Sep 3, 2023 at 22:42
• Well, I can imagine a state of the electromagnetic field with N photons $|k_1,k_2,\dots,k_N\rangle$. If we want to measure the probability to measure one of this photons with momentum $q$ we should calculate the inner product of the one photon state $|q\rangle$ with the afformentioned state according to the Born rule. In that case, multiplying the state of the EM field by a c-number wouldn't change the outcome of the probability. Commented Sep 4, 2023 at 3:37
• Yes, but that is a Fock state, not an EM potential operator. The inner products are invariant wrt to both changing the Fock states by a phase factor, and wrt changing EM potentials and the state via gauge transformation. The former seems to be a special case of the latter. Commented Sep 4, 2023 at 5:03
• Interesting! So how do Fock states transform under a EM gauge transformation? Commented Sep 4, 2023 at 15:01