Hilbert space and rays:

In a very general sense, we say that quantum states of a quantum mechanical system correspond to rays in the Hilbert space $\mathcal{H}$, such that for any $c∈ℂ$ the state $\psi$ and $c\psi$ map to the same ray and hence are taken as equivalent states.

  1. How should one interpret the above in order to understand why $\psi$ and $c\psi$ are the same states? Clearly for $c= 0$ it doesn't hold, and for $c=1$ it is trivial, but why should this equivalence hold for any other $c$?

  2. Knowing Hilbert space is a complex vector space with inner product, is ray just another way of saying vectors?

  3. In the case that $c$ just corresponds to a phase factor of type $e^{i\phi}$ with $\phi \in \mathbb{R},$ then obviously $|\psi|=|e^{i\phi}\psi|,$ i.e. the norms didn't change, then what is the influence of $e^{i\phi}$ at all? In other words what does the added phase of $\phi$ to the default phase of $\psi$ change in terms of the state of the system?

Projective Hilbert space:

Furthermore, through a process of projectivization of the Hilbert space $\mathcal{H},$ it is possible to obtain a finite dimensional projective Hilbert space $P(\mathcal{H}).$ In the Projective Hilbert space, every point corresponds to a distinct state and one cannot talk in terms of rays anymore.

  1. What does such projectivization entail in a conceptual sense? I guess in other words, how are rays projected to single points in the process? and what implies the distinctness? Is such process in any way analogous to the Gram–Schmidt process used to orthonormalise a set of vectors in linear algebra?

  2. When one limits the Hilbert space to that of a certain observable of the system at hand, e.g. momentum or spin space (in order to measure the momentum and spin of a system respectively), does that mean we're talking about projective spaces already? (e.g. is the spin space spanned by up $\left|\uparrow\rangle\right.$ and down $\left|\downarrow\rangle\right.$ spins states of a system referred to as projective spin Hilbert space?)

The aim is to develop a better and clearer understanding of such fundamental concepts in quantum mechanics.


5 Answers 5


Why are states rays?

(Answer to OP's 1. and 2.)

One of the fundamental tenets of quantum mechanics is that states of a physical system correspond (not necessarily uniquely - this is what projective spaces in QM are all about!) to vectors in a Hilbert space $\mathcal{H}$, and that the Born rule gives the probability for a system in state $\lvert \psi \rangle$ to be in state $\lvert \phi \rangle$ by

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

(Note that the habit of talking about normalised state vectors is because then the denominator of the Born rule is simply unity, and the formula is simpler to evaluate. This is all there is to normalisation.)

Now, for any $c \in \mathbb{C} - \{0\}$, $P(c\psi,\phi) = P(\psi,c\phi) = P(\psi,\phi)$, as may be easily checked. Therefore, especially $P(\psi,\psi) = P(\psi,c\psi) = 1$ holds, and hence $c\lvert \psi \rangle$ is the same states as $\lvert \psi \rangle$, since that is what having probability 1 to be in a state means.

A ray is now the set of all vectors describing the same state by this logic - it is just the one-dimensional subspace spanned by any of them: For $\lvert \psi \rangle$, the associated ray is the set

$$ R_\psi := \{\lvert \phi \rangle \in \mathcal{H} \vert \exists c \in\mathbb{C}: \lvert \phi \rangle = c\lvert \psi \rangle \}$$

Any member of this set will yield the same results when we use it in the Born rule, hence they are physically indistiguishable.

Why are phases still relevant?

(Answer to OP's 3.)

For a single state, a phase $\mathrm{e}^{\mathrm{i}\alpha},\alpha \in \mathbb{R}$ has therefore no effect on the system, it stays the same. Observe, though, that "phases" are essentially the dynamics of the system, since the Schrödinger equation tells you that every energy eigenstate $\lvert E_i \rangle$ evolves with the phase $\mathrm{e}^{\mathrm{i}E_i t}$.

Obviously, this means energy eigenstates don't change, which is why they are called stationary states. The picture changes when we have sums of such states, though: $\lvert E_1 \rangle + \lvert E_2 \rangle$ will, if $E_1 \neq E_2$, evolve differently from an overall multiplication with a complex phase (or even number), and hence leave its ray in the course of the dynamics! It is worthwhile to convince yourself that the evolution does not depend on the representant of the ray we chose: For any non-zero complex $c$, $c \cdot (\lvert E_1 \rangle + \lvert E_2 \rangle)$ will visit exactly the same rays at exactly the same times as any other multiple, again showing that rays are the proper notion of state.

The projective space is the space of rays

(Answer to OP's 4. and 5. as well as some further remarks)

After noting, again and again, that the physically relevant entities are the rays, and not the vectors themselves, one is naturally led to the idea of considering the space of rays. Fortunately, it is easy to construct: "Belonging to a ray" is an equivalence relation on the Hilbert space, and hence can be divided out in the sense that we simply say two vectors are the same object in the space of rays if they lie in the same ray - the rays are the equivalence classes. Formally, we set up the relation

$$ \psi \sim \phi \Leftrightarrow \psi \in R_\phi$$

and define the space of rays or projective Hilbert space to be

$$ \mathcal{P}(\mathcal{H}) := (\mathcal{H} - \{0\}) / \sim$$

This has nothing to do with the Gram-Schmidt way of finding a new basis for a vector space! This isn't even a vector space anymore! (Note that, in particular, it has no zero) The nice thing is, though, that we can now be sure that every element of this space represents a distinct state, since every element is actually a different ray.1

(Side note (see also orbifold's answer): A direct, and important, consequence is that we need to revisit our notion of what kinds of representations we seek for symmetry groups - initially, on the Hilbert space, we would have sought unitary representations, since we want to preserve the vector structure of the space as well as the inner product structure (since the Born rule relies on it). Now, we know it is enough to seek projective representations, which are, for many Lie groups, in bijection to the linear representations of their universal cover, which is how, quantumly, $\mathrm{SU}(2)$ as the "spin group" arises from the classical rotation group $\mathrm{SO}(3)$.)

OP's fifth question

When one limits the Hilbert space to that of a certain observable of the system at hand, e.g. momentum or spin space (in order to measure the momentum and spin of a system respectively), does that mean we're talking about projective spaces already? (e.g. is the spin space spanned by up |↑⟩ and down |↓⟩ spins states of a system referred to as projective spin Hilbert space?)

is not very well posed, but strikes at the heart of what the projectivization does for us: When we talk of "momentum space" $\mathcal{H}_p$ and "spin space" $\mathcal{H}_s$, it is implicitly understood that the "total space" is the tensor product $\mathcal{H}_p \otimes \mathcal{H}_s$. That the total/combined space is the tensor product and not the ordinary product follows from the fact that the categorial notion of a product (let's call it $\times_\text{cat}$) for projective spaces is

$$ \mathcal{P}(\mathcal{H}_1) \times_\text{cat} \mathcal{P}(\mathcal{H}_2) = \mathcal{P}(\mathcal{H}_1\otimes\mathcal{H}_2)$$

For motivations why this is a sensible notion of product to consider, see some other questions/answers (e.g. this answer of mine or this question and its answers).

Let us stress again that the projective space is not a vector space, and hence not "spanned" by anything, as the fifth question seems to think.

1The inquiring reader may protest, and rightly so: If our description of the system on the Hilbert space has an additional gauge symmetry, it will occur that there are distinct rays representing the same physical state, but this shall not concern us here.

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    $\begingroup$ I would be careful with "categorial notion of a product", as long as I haven't decided on the allowed morphisms. Martin Brandenburg patiently explained this to me when I asked Is the categorical product for projective spaces essentially the tensor product?... $\endgroup$ Commented Jan 7, 2015 at 1:54
  • $\begingroup$ @user929304: Time evolution is still $\mathrm{e}^{\mathrm{i}Ht}$, and hence unitary. A space which is "spanned" is not the projective space. The projective space of a spin-1/2 system is the Bloch sphere. I cannot view the link for the projective measurements, unfortunately. $\endgroup$
    – ACuriousMind
    Commented Jan 14, 2015 at 16:55
  • $\begingroup$ @user929304: If the two operators do not commute, then choosing an eigenbasis of $A$ makes you indeed "lose" information about $B$ - this is essentially just the general form of the Heisenberg uncertainty prinicple. There's no general recipe for saying how one eigenstate translates into the other, you have to just derive the transformation using the commutation relations. $\endgroup$
    – ACuriousMind
    Commented Feb 18, 2015 at 16:40
  • $\begingroup$ @ACuriousMind thanks, and how about the case of commuting operators? if I have the eigenvectors of one of them, I know already the other will have the same set of eigenvectors just on the account of commutability? e.g. spin and position, if I represent psi in Spin basis, how to find its position eigenvectors from there? just looking for some rope, to know how to reason. Thanks again $\endgroup$
    – user929304
    Commented Feb 18, 2015 at 17:46
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    $\begingroup$ Technical nitpick: the equivalence relation is actually defined by multiplication by a nonzero complex number. So your formula should be $$ R_\psi := \{\lvert \phi \rangle \in \mathcal{H} \vert \exists c \in (\mathbb{C} \setminus \{0\}): \lvert \phi \rangle = c\lvert \psi \rangle \},$$ and a ray in projective Hilbert space is not exactly the same thing as a one-dimensional subspace: it's actually a one-dimensional subspace minus the zero vector. $\endgroup$
    – tparker
    Commented Jul 16, 2017 at 22:05

The current mathematical formulation of Quantum Mechanics is based on the theory of Operator Algebras. The fundamental Axiom is that a mechanical system is described by a C*-algebra and the set of states is given by (a restriction of) the state space of the said C*-algebra. Hilbert spaces come into play from the representation theory of C*-algebras. Given a state, that is a normalised positive linear functional on the algebra, it induces a representation of such algebra through the so-called GNS representation. The C*-algebra is now represented on a Hilbert space, and the state is now generated through a vector, viz. $$\omega(a) = (\xi_\omega,\pi_\omega(a)\xi_\omega),$$ where $\omega$ is the state and $(H_\omega,\pi_\omega,\xi_\omega)$ is the GNS triple from the GNS construction, $H_\omega$ being the Hilbert space. Normalisation of the state $\omega$ implies that $\omega(I) = 1$, which then implies $$1 = \omega(I) = (\xi_\omega,\pi_\omega(I)\xi_\omega) = (\xi_\omega,\xi_\omega)=\Vert\xi_\omega\Vert^2.$$ Hence the vector generating the state must be normalised. This is why, when considering states, the whole Hilbert space is not necessary, but it is enough to consider its projectivisation. Moreover, it is clear from the last line that, if $z\in\mathbb C$ is any phase, that is $|z|=1,$ $z\xi_\omega$ generates the same state $\omega$, and therefore the unit vector $\xi_\omega$ associated to the state $\omega$ is defined up to an unobservable phase factor.

Up to here this should have answered questions 1-3. For the remaining questions, observe that a projective Hilbert space is not always finite-dimensional (in fact it rarely is). There is no direct link to Graham-Schmidt, since you are just taking the Hilbert space and simply identifying some vectors according to an equivalence relation.

Whenever a quantum system is given to you in the plain form of a set of observables and a Hilbert space, (pure) states are considered to be normalised vectors defined up to a phase factor, that is as elements from the projective Hilbert space.

  • 1
    $\begingroup$ Thank you for posting an answer so promptly. I must admit I couldn't decipher anything from the first paragraph as it is extremely terse for me, although I'm confident it must be correct. So you can imagine that I didn't really follow the reasoning with $\omega (a)$ using the GNS representation. Any further simplification would be immensely appreciated. Regarding the "un-observable phase factors", I had already noticed that it doesn't change the norm, hence it goes un-noticed in a measurement, but what does it change? what's the point of ever multiplying a state by a phase factor then? $\endgroup$
    – user929304
    Commented Jan 6, 2015 at 11:29
  • $\begingroup$ In the expression of the state you have $\omega(a)=(\xi_\omega,\pi_\omega(a)\xi_\omega)$, so if your state is now $z\xi_\omega$, then $(z\xi_\omega,\pi_\omega(a)z\xi_\omega)=\bar z\cdot z (\xi_\omega,\pi(a)\xi_\omega)=|z|^2\omega(a)=\omega(a)$. It is because there is no point in multiplying a vector by a phase factor that you consider the projective Hilbert space. You can then exploit this equivalence to introduce arbitrary phase factors that can simplify your computations when dealing with vectors explicitly. $\endgroup$
    – Phoenix87
    Commented Jan 6, 2015 at 11:42

The notion of projective space is sensible for any vector space V (finite dimensional or not). As a set the projective space $P(V)$ is simply the set of one-dimensional sub vector spaces of the vector space. If you think briefly about it, any one-dimensional sub vector space (or ray as physicists call them) is given by a non-zero vector $v$ in $V$ and all scalar multiples $\lambda v$ of it. Obviously that description is not unique, any non-zero multiple of $v$ would do as well, so projective space is the quotient

$$P(V) = (V - \{0\})/\equiv$$

where $v \equiv w$ if there exists $\lambda \neq 0$, such that $v = \lambda w$.

In other words projective space gives you a way to talk about rays (one-dimensional sub vectorspaces) in a vectorspace. This is independent of whether the vector space is a Hilbert space.

So what do role do rays in a Hilbert space play in quantum mechanics? Some of them are pure eigenstates of an observable $Q$. Recall that observables are described by self-adjoint hermitian operators, if $v$ is an eigenvector of $Q$ for eigenvalue $\lambda$, that is $Qv = \lambda v$, so is $\mu v$ for all $\mu \neq 0$. Since it is the real eigenvalue $\lambda$ that is physically observable, it is really the one-dimensional subspace spanned by $v$ that matters. geometrically a hermitian self-adjoint operator $Q$ generates a "rotation" $e^{iQ}$ in the projective Hilbert space, whose fixed points (generally they are not really points) correspond to the so called "pure states".

Moreover given an eigenstate of an observable $Q$, $\vert \phi \rangle$ and an arbitrary state $\vert \psi\rangle$, the transition probability from $\vert \psi\rangle$ to $\vert \phi\rangle$ is given by $$P(\psi,\phi) = \frac{\langle \phi \vert \psi \rangle \langle \psi \vert \phi \rangle }{\langle \psi \vert \psi \rangle \langle \phi \vert \phi \rangle}.$$ This quotient obviously does not change if $\vert \psi\rangle$ and $\vert \phi\rangle$ are multiplied with arbitrary non zero complex numbers $\lambda, \lambda'$.

More precisely symmetries of a quantum mechanical system with hilbert space $H$ are given by a representation of the symmetry group

$$G \to PU(H)$$

in the projective unitary group of the Hilbert space, this group naturally acts on the projective space $P(H)$ and not $H$. The simplest case is that of a non-relativistic spin-1/2 particle in its rest frame. The remaining symmetry of the Gallilei group is a representation

$$SO(3) \to PU(H).$$

As it happens such projective representations are in one-to-one correspondence with representations of the double cover of $SO(3)$, $SU(2)$:

$$SU(2) \to U(H).$$

Since $SU(2)$ is compact its unitary representations are finite dimensional and its smallest irreducible representation is called the spin-1/2 representation, with $H = \mathbf{C}^2.$ So we learn that the space states of a spin-1/2 in its rest frame is

$$P(\mathbf{C}^2) = \mathbf{P}^1(\mathbf{C})$$ also known as the Riemann sphere.

Nothing mysterious is going on though, once you've chosen a set of eigenstates (say for the spin in z-direction $\vert + \rangle, \vert - \rangle$) every other state can be written as a linear combination $$\vert \psi \rangle = w \vert + \rangle + z \vert - \rangle$$ where not both $w$ and $z$ can be zero. Physically it also doesn't matter, if we scale with $\lambda \neq 0$. In mathematical language $[w:z]$ are known as homogeneous coordinates of the projective space $\mathbf{P}^1(\mathbf{C})$

This is of significance if we consider multiple spin-1/2 particles, their Hilbert space is given from general principles as

$H = H_1 \otimes \ldots \otimes H_n$

where $\otimes$ is the tensor product and the braiding $H_i \otimes H_{i+1} \to H_{i+1} \otimes H_i$ introduces a minus sign because spin 1/2 particles are fermions, that is $v_i \otimes v_{i+1} = -v_{i+1} \otimes v_i$ (Such vector spaces are also called super-vector spaces). Now again one should really look at

$$P(H) = P(H_1 \otimes \ldots \otimes H_n),$$

interestingly there only is an embedding of

$$P(H_1) \times \ldots \times P(H_n) \to P(H)$$

called the Segre embedding, tensor products of pure one-particle states don't span the multiparticle hilbert space, states that don't lie in the image are known as "entangled".

To summarize a precise separation of the Hilbert space and its projectivization help clarify lots of issues in Quantum mechanics. I haven't mentioned the study of irreducible projective representations of the Poincare group for example, Weinberg's excellent book on Quantum Field Theory covers that. Geometric ideas like the Fubini Study metric (or Bures metric) also are helpful.

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    $\begingroup$ -1 You write a lot of stuff, but don't make it clear how this relates to the original question. Also I wondered about the Segre embedding and tensor products before, and in the end didn't feel like I understood QM any better. The same it true when reading you answer. And I'm not even sure yet whether going to the projective space is a good idea in QM. Other approaches (like C* algebras) seem to be more common and offer more relevant insight, but this doesn't answer the question whether projective spaces wouldn't offer some relevant insight too. $\endgroup$ Commented Jan 6, 2015 at 14:52
  • 2
    $\begingroup$ If you read the post carefully it addresses the points raised by the author one by one, although I don't explicitly say so. Most of his questions stem from a fundamental confusion about basic concepts in quantum mechanics. $\endgroup$
    – orbifold
    Commented Jan 6, 2015 at 17:51

In addition ot Phoenix87's answer, which succintly summarizes the Hilbert space as it is now understood to emerge from the operator space structure of quantum mechanics, let me try to answer your questions more directly:

A Hilbert space is nothing but a (complete) vector space with an inner product. As you said, this provides (a version of) the state space of quantum mechanics. Any vector should thus be a state. However, all we can know is the measurement results, so states that lead to the same measurements for all measurements can be identified. Since they cannot be distinguished by any measurements, why should we think them to be different? If you have a look at how measurements work, you can then see that a global phase (i.e. $c\psi$ if $\psi$ is a vector of your Hilbert space and $c\in\mathbb{C}$ with $|c|=1$) does not change the result. If you furthermore require the measurements to be related to probabilities (Born's rule), you want your state $\psi$ to be normalized. This means that all vectors $c\psi$ with $c\in\mathbb{C}\setminus\{0\}$ will represent the same state, once you normalize.

All of this can be summarized by saying that instead of all vectors in the Hilbert space, you just consider the space that consists of the equivalence classes we have just constructed. This is then the projective Hilbert space and it is a new space - not a subset of your old space. This has nothing to do with Gram-Schmidt, because you construct a new space, while with Gram-Schmidt, you are still living in the same vector space.

In a particular example, you usually work with the Hilbert spaces and disregard global phases and always normalize your state. It's like using the projective Hilbert space, but you don't have to know about them at all.


Phoenix87 nicely summarizes the general situation. But in a pedestrian situation, with a given separable Hilbert space, switching to density operators to describe the states the whole issue disappears since they are invariant under changes of phase factors. Note that density operators are positive trace class operators with trace 1, pure states being one-dimensional projectors. Note further that more general states can exist, they are still positive functionals but are more general than density operators.

  • 2
    $\begingroup$ Could you please kindly expand on your point with density operators? because I don't directly see their relevance to the discussion at hand. Thanks $\endgroup$
    – user929304
    Commented Jan 6, 2015 at 12:39
  • $\begingroup$ -1 You are completely right with what you say, but you don't explain it in a way that the OP understands it. And there would be some more things you could explain, which would make it more clear why this answers the original question. It seems to be common on physics.se to downvote this type of answer, so I better try to adjust to this sort of culture here. $\endgroup$ Commented Jan 6, 2015 at 14:45
  • $\begingroup$ @user929304 The point is that the density operator formalism avoids all questions with phases. $\endgroup$
    – Urgje
    Commented Jan 6, 2015 at 14:55

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