I recently read that a quantum state is actually defined by a ray and not a vector. That is it is possible to multiply a state $\psi$ by any complex number $c\in \mathbb{C}$ and you won't be changing the physics in any way. I understand this mathematically, but I don't understand what the physical meaning of such an "equivalent state" would be since the new state need not be normalised if $c$ is not of the form $e^{i\phi}$.


There is no particularly interesting new physical significance to such a state vector. As you already stated, it represents exactly the same physical state. The only difference is that, on taking the modulus squared, the new state gives an unnormalised probability distribution over possible measurement outcomes. You can easily extract the probability of obtaining a measurement outcome corresponding to the (possibly unnormalised) state $|\phi\rangle$ from an unnormalised state $|\psi\rangle$ by using the Born rule: $$ \mathrm{Pr}(\phi) = \frac{|\langle \phi | \psi \rangle |^2}{\langle \phi | \phi \rangle \langle \psi | \psi \rangle }. $$ Clearly, using normalised states is just a handy convention that avoids any worries about calculating the denominator above. There is nothing wrong with formulating quantum mechanics without normalising state vectors to unity. Indeed, most people avoid bothering with the overall normalisation factors until they are needed at the very end of the calculation, since they just add unsightly clutter to the mathematics and have no physical significance.


Let me first consider a finite level $N$-dimensional system whose pure states are $N$ dimensional vectors. All quantum mechanical results can be equivalently obtained by considering normalized state vectors spanning the unit $2N-1$-sphere or the state space of rays spanning the $N-1$ dimensional complex projective space $CP^{N-1}$. However, in many aspects the second option (rays) is simpler and natural,(even though not taught in introductory quantum mechanics courses) , because the complex projective space is a complex manifold and the complex field is algebraically closed. Thus, by working with rays we have the powerful tools of complex analysis and geometry at our disposal.

Moreover, the complex projective space is a symplectic manifold, which makes the concepts of quantization and the quantum classical transition more natural. Thus in the ray picture, every quantum system has a natural classical counterpart (a symplectic manifold), which is not easy to describe in real picture.

A prominent example is the spin which is not easy to describe classically in the first (real) picture while in the complex picture has a very simple description as a system having the two dimensional sphere as its phase space, please see for example section on the quantization of $S^2=CP^1$ in the following work by Vathsan. This quantization generalizes to more complex systems in which the complex picture treatment is more favorable, please see for example the following work by Borthwick.

The quantization in the complex picture is especially favorable when the phase space possesses a Kähler structure. Kähler manifolds became relevant to many very modern treatments in physics for example many moduli spaces such as (Calabi-Yau spaces) are Kähler, please see this review by Martin Schlichenmaier. Compact Kähler manifolds describe the classical mechanics of internal degrees of freedom (again generalizing the spin case).

Kähler quantization also works when the number of states becomes infinite, such as in the case of the Harmonic oscillator. In this case the quantization space becomes the Bargmann space, please see section 3.1 in the following review by Todorov.

Kähler quantization was also used to quantize infinite dimensional manifolds relevant in quantum field theory and string theory. One of the most known examples is the Kähler quantization of $Diff(S^1)/S^1$ by Bowick and Rajeev

  • $\begingroup$ Thanks so much for the links to the various papers and reviews! $\endgroup$ – user34801 Oct 11 '13 at 2:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.