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Gleason's Theorem famously asserts that (appropriately defined) measures on the lattice of a complex Hilbert space can be implemented by density operators via the trace operation, except in the case where the dimension of the Hilbert space is equal to 2.

My question is what to make of the exceptional case. More specifically, in the 2-dimensional case, does a Gleason measure which cannot be implemented by a density operator correspond to a real physical quantum state?

To put the question yet one more way, should we take the fundamental mathematical definition of a quantum state to be a Gleason measure or a density operator, since these are not quite the same thing?

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  • $\begingroup$ Is this question about the ontology of quantum states in the quantum logic approach? According to Pitowsky ("Quantum Mechanics as a Theory of Probability", 2008), "In our scheme quantum states are just assignments of probabilities to possible events, that is, possible measurement outcomes". In that scheme I don't know if talking about a "real physical quantum state" makes any sense at all. As for the limitation on dimension size, it does not seem of a very interesting nature according to the counter-example in ncatlab.org/nlab/show/… $\endgroup$ – Stéphane Rollandin Jun 18 '16 at 18:07
  • $\begingroup$ No, it isn't about the ontology of quantum states. The entire point of the Hilbert space framework is to decide which probability assignments are admissible and which ones aren't. So you could replace my qualifier "real physical" with "admissible" if you like. However, what 'admissible' really amounts to here is that it is possible in principle to set up an experiment which achieves the probabilistic predictions of the states in question. In short, please don't dismiss this as a question of interpretation - it definitely has nothing to do with that. $\endgroup$ – Physics Footnotes Jun 18 '16 at 20:00
  • $\begingroup$ Maybe of interest, the amazing exposition of the Hilbert space formalism in physics.stackexchange.com/a/52253/109928. $\endgroup$ – Stéphane Rollandin Jun 18 '16 at 21:02
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I think the particular statement of Gleason's theorem is really important in elucidating this perceived ambiguity. Wikipedia states it as:

Theorem. Suppose H is a separable Hilbert space of complex dimension at least 3. Then for any quantum probability measure on the lattice Q of self-adjoint projection operators on H there exists a unique trace class operator W such that P(E) = Tr(W E) for any self-adjoint projection E in Q.

I would bet this is the statement most of us are familiar with. However I think its statement for C* algebras clears up your question:

Definition. Let $\rho: \mathcal{P} \to [0, 1]$ such that for every finite family $\{ P_1, ..., P_n: P_i \in \mathcal{P} \}$ of pairwise orthogonal projections we have $\rho(\sum_{i=1}^n P_i) = \sum_{i=1}^n \rho(P_i)$ , then $\rho$ is a finitely additive measure on $\mathcal{P}$.

If the family is not finite, but countable, then $\rho$ is a sigma-finite measure.

Theorem. If $\dim(\mathcal{H}) \neq 2$ then each finitely additive measure on $\mathcal{P}$ can be uniquely extended to a state on $\mathcal{B}(\mathcal{H})$. Conversely the restriction of every state to $\mathcal{P}$ is a finitely additive measure on $\mathcal{P}$.

The same holds for sigma-finite measures and normal states: Every sigma-finite measure can be extended to a normal state and every normal state restricts to a sigma-finite measure.

Clearly if the dimension of your space is 2, then your measure can not be uniquely extended to a state on the set of bounded operators of your Hilbert space. Here a state is considered as a density matrix and thus is a bounded operator on your Hilbert space. This means a measure either has multiple states associated with it or there is no such extension. In the first case this means the set of orthogonal projections onto your Hilbert space is not enough to determine a unique state. The first case cannot occur for a projective Hilbert space since the only difference would be a phase. Thus the second case must be the problem. This means there exist probability measures on your projections which do not correspond to a state.

However I believe the converse does hold, for any given state restricted to $\mathcal{P}$ there is a corresponding finitely additive measure. Considering the second statement of the theorem, the answer to your question is that "states" are a priori independent objects from probability measures. This theorem allows us to interpret states as a probability measure but apparently we cannot take the probability measure as the definition of a state in two dimensions. If you want to label states by probability measures, you're out of luck in two dimensions.

For a clear counter example check out this reference which I based my answer on.


Update June 26th

Another way to consider how the measure must be relate to a state is to start from the set of all possible general 2 dimensional Hilbert Space states and see what the measure has to look like. We can represent an arbitrary vector in terms of the basis $\{|u_1\rangle, |u_2\rangle\}$:

$$|v\rangle = \cos(\beta)|u_1\rangle + e^{i\alpha}\sin(\beta)|u_2\rangle$$

We recognize that the above is really an equivalence class of vectors related by an overall phase. We can find the density operator for the pure state above:

$$\rho_v = |v\rangle \langle v| = \left( \begin{array}{ccc} \cos^2\beta & e^{i\alpha}\sin\beta\cos\beta \\ e^{-i\alpha}\sin\beta\cos\beta & \sin^2\beta \end{array}\right)$$

We can write our measure as a continuous function $f_v(|u\rangle) = \langle u|\rho_v |u \rangle$ where $|u\rangle$ is an arbitrary element of our 2 dimensional projective Hilbert Space. That means it can be expanded just the same as the vector which determined our state:

$$|u\rangle = \cos\theta|u_1\rangle + e^{i\phi}\sin\theta|u_2\rangle$$

Our function is written as:

$$f_v(|u\rangle) = \langle u| \rho_v |u \rangle = f_v(\theta,\phi)$$

After some calculation we get that our measure is given by:

$$\begin{align*}f_v(\theta,\phi) &= \cos^2\beta\cos^2\theta + \sin^2\beta\sin^2\theta \\ &+ 2\cos(\alpha-\phi)\sin\beta\cos(\beta)\sin\theta\cos\theta \end{align*}$$

For mixed states we have:

$$\rho_m = \sum_{v} p_v \rho_v$$

$$\begin{align*}f_m(\theta,\phi) &= \sum_{v} p_v\big[ \cos^2\beta_v\cos^2\theta + \sin^2\beta_v\sin^2\theta \\ &+ 2\cos(\alpha_v-\phi)\sin\beta_v \cos\beta_v\sin\theta \cos\theta\ \big] \end{align*}$$

Without any loss of generality we can simplify the measure for any state as:

$$f(\theta,\phi) = c_{11} \cos^2\theta + c_{22}\sin^2\theta + \big(c_{12}e^{i\phi} + c_{21}e^{-i\phi}\big)\sin\theta \cos\theta$$

What we get is the spherical harmonics, $Y_{0}^{0}$ and $Y_{2}^{m}$ for all allowable m values. This follows from the fact that our inner product is a bilinear form. Turns out that, for any function on real or complex vector spaces of dimension 3 or more, any finitely additive continuous measure can be expanded in terms of this set of spherical harmonics. In 3 dimensions all finitely additive measures must be rotationally invariant, so the spherical harmonics give the transformations that encode this invariance. The spherical harmonics are in some sense "built in" to these measures.

In 2 dimensional vector spaces we have finitely additive measures which are not required to be rotationally invariant in 3 dimensions, only 2 dimensions, which means cylindrically symmetric functions will be allowed. Cylindrically symmetric functions include $\cos(n\theta)$ for example and thus have no expression in terms of quadratic terms. Without the constraint imposed by transformation under the spherical harmonics, there's no way to relate a bilinear form to the measure.

My physical argument for disallowing these measures as states would be the following: Any 2-dimensional Hilbert space we study in all likelihood is a subspace of a larger dimensional Hilbert space. In those spaces these 2-dimensional measures would not be allowed since they lack 3-dimensional rotational invariance. If these states cannot be embedded into a larger space, then they can't be physical.

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  • $\begingroup$ Thanks for your time on this one. I'm still unsure whether it will answer my question, as I need more time to think about it, but your contribution provides me with some fresh insights. $\endgroup$ – Physics Footnotes Jun 25 '16 at 3:30
  • $\begingroup$ I'm not sure it's a satisfying answer either but it might be the nature of the question. In C* algebras and their representations the states are defined a priori, so a Gleason measure with no state isn't a state. The theorem is often used to justify probabilistic interpretations of QM as it puts the set of finitely additive measures in correspondence with the set of states. The important implication is that every state has a representation as a probability measure. Not every probability measure needs to correspond to a state, that's my take on it. $\endgroup$ – Daniel Kerr Jun 25 '16 at 3:48
  • $\begingroup$ That is a plausible argument, which I hadn't looked into. What I would really like to see though is a physical reason that the exceptional Gleason measures should not be considered possible quantum states. $\endgroup$ – Physics Footnotes Jun 26 '16 at 3:28
  • $\begingroup$ Physically the argument is that the measure does not correspond to a linear operator. If I have time I will try to flesh out the counterexample so it's more apparent. The issue is that some of these measures are not implemented by a linear operator on a vector space, that means they violate the postulates of quantum mechanics. It cannot be a state for the same reason that you exclude all nonlinear density "operators" one could write down. $\endgroup$ – Daniel Kerr Jun 26 '16 at 4:55
  • $\begingroup$ I understand what you're saying, but it's long been my gut feeling that measures on a lattice provide a more probabilistically native foundation for QM than linear operators. $\endgroup$ – Physics Footnotes Jun 26 '16 at 5:04

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