What follows are my notes on the matter, all of which are either well-known or wrong. Skip to the end for the very simple formulation with no mention of quantum states. Even better advice is to just Read Nick Harvey’s survey.

** How I understand quantum measurement **

A (mixed) quantum state on the space is an Hermatian matrix satisfying ( is positive semi-definite) and . Suppose we also have a resolution of the identity: A collection of PSD matrices with . Then when we perform the measurement , the probability of obtaining outcome is for .

A *pure state* is given by a density matrix of the form for some . Equivalently, a state is pure if and only if it cannot be written as a non-trivial convex combination of other states, i.e. if for , then . The pure states are the extreme points of the convex set of all states.

** Quantum logic **

There is another way to think about quantum measurement. Let be a (separable) Hilbert space and consider a projection onto some closed subspace. A *measure on projections* assigns to every such projection a value in . We say that is *-additive* if whenever and the projections are mutually orthogonal, we have . If this only holds for finite sums of projections, we say that is *finitely additive.* We should also have .

Note that we have allowed here the possibility that is infinite-dimensional. In the case where the measure is -additive, it is completely specified by its value at one-dimensional projections, i.e. the values where denotes projection onto . Now we can state Gleason’s theorem: Our two notions of quantum measurement coincide.

Theorem 1Let be a Hilbert space of dimension at least 3 and let be a -additive measure on projections. Then there exists a unique linear operator such that for any projection , we have .

In other words, every -additive measure on projections arises from a mixed state.

** Restricted purity **

Note that a measure on projections corresponds to a pure state if and only if it cannot be written as a non-trivial convex combination of measures on projections. In this way, we can also define purity with respect to a restricted family of projections.

Consider, for example, the Hilbert space of complex sequences, and let denote the standard basis. Suppose we consider only the projections onto each of the basis elements. Say that a *-measure* is a measure on all projections of the form for . Such a measure only gives us probabilities for measurements performed in the standard basis. (The algebra generated by is a canonical example of a “maximal abelian subalgebra,” i.e. MASA.) We can say that a -measure is *pure* if it cannot be written as a non-trivial convex combination of other -measures. What are the pure -measures? Well, the only -additive pure -measures are the point measures for some . If for , then by -additivity, we can split into a convex combination of two measures (one supported only on and the other supported off ).

But when -additivity is dropped, some tricky things can happen (see the next section). It turns out that, in this case, there are more pure states than just (in fact, the cardinality of the finitely-additive pure states is gigantic, ). Now we are in position to state the Kadison-Singer problem.

**Kadison-Singer Problem:** Is it the case that every pure, finitely-additive -measure has a unique extension to a finitely-additive measure on all the projections of ?

The answer is clearly positive for -additive measures (in which case the unique extension corresponds to ). The answer is also positive for finitely-additive -measures, as recently proved by Marcus, Spielman, and Srivistava.

In their original work, Kadison and Singer showed that the answer is false if one constructs the MASA using the continuous Hilbert space instead of . One can take the subalgebra of multiplication operators for . Here the corresponding family of projections are the maps for . Unlike in the case above, this subalgebra is not “discrete.” It turns out that every maximal abelian subalgebra of bounded operators on a separable Hilbert space is either finite-dimensional, the discrete one, the continuous one, or a direct sum of algebras isomorphic to these.

Note that states corresponding to finitely additive (but not -additive measures) do not have density matrices and are often considered to be “unphysical” because they would take an infinite amount of energy to prepare.

** A -world teeming with pure states **

In the above setting, we said there are finitely-additive pure -measures other than point measures. To see what one of these must look like, consider the following construction. Let denote the projection onto and define similarly. Define the measures and . Then we can write . This is a non-trivial convex combination unless either or .

Indeed, for every partition , a pure -measure can only put non-zero weight on projections involving coordinates of exactly one of or . Furthermore, if then . In other words, for every set of projections, the measure takes only values 0 or 1. Such finitely-additive measures are exactly given by ultrafilters on . The finitely-additive pure -measures are in one-to-one correspondence with such ultrafilters, the set of which can be identified with , the Stone-Cech compactification of the naturals. See Terry Tao’s notes on ultafilters and Stone-Cech compactification.

Take the Hilbert space with standard basis , and let denote the set of all its closed subspaces. Consider a map . We say that is a *finitely-additive measure* if and whenever is a finite collection of mutually orthogonal subspaces, we have . Say that such a measure is *pure* if it cannot be written as a non-trivial convex combination of two finitely-additive measures. A canonical example of a pure measure involves fixing a unit vector and setting where denotes the orthogonal projection onto .

Now let be the set of all the subspaces of of the form for some subset , i.e. all the coordinate subspaces. We can define “finitely-additive” and “pure” for measures on in the analogous way.

**Kadison-Singer problem:** Can every finitely-additive pure measure on be extended uniquely to a finitely-additive measure on ?

Note that the real question is about uniqueness of the extension. There always exists a pure extension; see the comments.

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**Open Question:** Do there exist constants such that for any , the following holds? Let be a collection of orthonormal bases and define . Then there are subsets with and .

[Some additional notes: One piece of intuition for why the question should have a positive resolution is that these orthonormal bases which together comprise at most vectors cannot possibly “fill” -dimensional space in a way that achieves -dimensional isoperimetry. One would seem to need points for this.

One can state an equivalent question in terms of vertex expansion. Say that a graph on vertices is a *vertex expander* if for all subsets with . Here, denotes all the nodes that are in or are adjacent to . Then one can ask whether there exists a 1-1 mapping from to for some orthonormal bases such that the endpoints of every edge are mapped at most apart (as ).]

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- Derandomiziation of the maximum of a Gaussian process was solved by Raghu Meka in this FOCS’12 paper.
- Conjecture 1 from that post was solved by Jian Ding in the special case of trees and bounded-degree graphs. The conjecture is still open in general.
- Siu On Chan solved the PSD Lifting question in the comments of the post. Since then, the invention of the short code gave a much more sophisticated way of derandomizing Unique Games instances.
- A PTAS for TSP in doubling spaces was solved by Bartal, Gottlieb, and Krauthgamer in this STOC 2012 paper.
- The Dimension reduction in L
_{1}question is still largely open, though Andoni, Naor, and Neiman have reportedly made some progress in an unpublished manucsript (see Remark 5.2 here). - Capacity scaling in PSD flows was solved, in the non-uniform case, in a STOC 2010 paper with my student Mohammad Moharrami. The case of uniform (all-pairs) flows is still open.
- No progress has been made on improving the quantitative dependence on h in the low-diameter graph partitioning problem.

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Here is an interesting and elementary (to state) open question whose resolution would give an optimal higher-order Cheeger esimate. The goal is to replace the factor in Theorem 1 of the previous post with a factor, or even an optimal bound of .

Fix and let denote a probability measure on the -dimensional sphere . For the purpose of this question, one can assume that is supported on a finite number of points. Suppose also that is *isotropic* in the sense that, for every ,

Now our goal is to find, for some , a collection of (measurable) subsets such that for each , and such that for , we have

In Lemma 5 of the previous post, we proved that this is possible for and . In this paper, we improve the estimate on somewhat, though the best-known is still for some .

Question:

Is it possible to achieve ? One could even hope for and .

Two extreme cases are when (1) is uniform on , or (2) is uniform on the standard basis . In the first case, one can easily achieve (in fact, one could even get sets satisfying these bounds). In the second case, taking each set to contain a single basis vector achieves and .

If one only wants to find, say, sets instead of sets, then it is indeed possible to achieve and . *[Update: This is actually impossible for the reasons stated above. To get the corresponding bounds, we do a random projection into dimensions. This only preserves the original Euclidean distances on average.]* Furthermore, for , this bound on is asymptotically the best possible.

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*[Update: The application of Miclo can also be done using the higher-order Cheeger inequalities of Louis, Raghavendra, Tetali, and Vempala.]*

The main simplification comes from doing the random partitioning non-optimally with axis-parallel cubes. For ease of notation, we will deal only with regular graphs, but there will be no quantitative dependence on the degree and this assumption can be removed (see the full paper).

Suppose that is a connected, -vertex, -regular graph. Define the Laplacian by , where is the adjacency matrix of . We will think of as acting on , the space of functions equipped with the norm. is positive semi-definite with spectrum

We we define the Rayleigh quotient of a function by

where the numerator is summed over edges of . By the variational principle for eigenvalues, we have

For a subset , we define the *expansion of * by , where is the indicator function of .

Finally, for , we define the *-way expansion constant of * by

where the minimum is over collections of disjoint, non-empty subsets of .

The classical discrete Cheeger inequality asserts that

We will now prove the following generalization. See the full paper for a discussion of the surrounding issues and better quantitative bounds.

First, let’s prove the (easy) LHS of (2). Suppose we have which are disjoint and non-empty and which satisfy . Then certainly is a -dimensional subspace of . On the other hand, every satisfies because if , then

where we have used the fact that if and , then

But now using (1), the subspace witnesses the fact that .

To prove the more difficult RHS of (2), we will use the following discrete Cheeger inequality with boundary conditions.

Lemma 2For any , there is a subset such that

*Proof:* Let . Observe that for each , one has . For , let denote the edges of with exactly one endpoint in . Then we have

On the other hand, thus

implying there exists a such that .

In light of the preceding lemma, to prove the RHS of (2), it suffices to find disjointly supported functions such that for each . Then Lemma 2 guarantees the existence of disjoint subsets of vertices satisfying our desired conclusion.

Toward this end, let denote orthonormal functions satisfying for each , i.e. consider the first eigenfunctions of the Laplacian. Define the map via

We also put . (Since , the function takes value everywhere, hence is never zero and this is well-defined.)

The next lemma shows that, in order to find disjointly supported functions, it suffices to find “separated sets.”

Lemma 3Suppose that for some and , there exist subsets satisfying the conditions:

- For every , we have , and
- For every , if and , then

Then there are disjointly supported functions such that .

*Proof:* For each , we define maps by

Observe that, by the triangle inequality, for , we have

Observe that since is identically on , we have

Next, we argue that the functions are disjointly supported. This immediately implies that the are disjointly supported. If for some , then there are points and such that and . But then by the triangle inequality, , violating condition (ii).

Finally, we bound . For any and , we have

Now using (3),

where we have used the fact that for any non-zero vectors , we have

Using (6) and the fact that in (5) yields

Finally, observe that

Combining this with (7) and (4) shows that

We will first make the task of finding separated sets slightly easier.

Lemma 4Suppose that for some and , there are subsets which satisfy:

- .
- For every , if and , then
- For every ,

Then there are sets satisfying the assumption of Lemma 3 with .

*Proof:* We can form the desired sets by taking disjoint unions of the sets . Begin with the family and repeatedly replace two sets and by their union if they each satisfy .

When we finish, we are left with a collection of sets each member of which satisfies , and possibly one set for which . By (i), we must end with at least sets which each satisfy .

Our final lemma, which finishes the proof of Theorem 1, simply asserts that such sets exist. We will no longer need the fact that comes from eigenfunctions.

Lemma 5Suppose that are orthornormal in . Letand . Then there is an and subsets such that

- .
- For every , if and , then
- For every ,

*Proof:* Consider the matrix which has columns . For any , we have

since the columns of are orthornormal

For a subset of the unit sphere in , we put . Fix some and let denote the diameter of , then by (8), we have

Now, let be a partition of into axis-parallel squares of side length . For any such square , we let denote the set of points which are at Euclidean distance at least from every side of . Observe that

Consider now the collection of sets . Since , (9) implies that . Furthermore, by construction, for any and with , we have

Thus the collection of sets satisfy both conditions (ii) and (iii) of the lemma. We are left to verify (i).

Note that . If we choose a uniformly random axis-parallel translation of the partition , then (10) implies

In particular, there exists some fixed partition that achieves this bound. For this partition , the sets satisfy all three conditions of the lemma, completing the proof.

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**Bernoulli process.**

Consider a finite subset and define the value

where are i.i.d. random . This looks somewhat similar to the corresponding value

where are i.i.d. standard normal random variables. But while can be characterized (up to universal constant factors) by the Fernique-Talagrand majorizing measures theory, no similar control was known for . One stark difference between the two cases is that depends only on the distance geometry of , i.e. whenever is an affine isometry. On the other hand, can depend heavily on the coordinate structure of .

There are two basic ways to prove an upper bound on . One is via the trivial bound

The other uses the fact that the tails of Gaussians are “fatter” than those of Bernoullis.

This can be proved easily using Jensen’s inequality.

Talagrand’s Bernoulli conjecture is that can be characterized by these two upper bounds.

Bernoulli conjecture:There exists a constant such that for every , there are two subsets such thatand

Note that this is a “characterization” because given such sets and , equations (1) and (2) imply

The set controls the “Gaussian” part of the Bernoulli process, while the set controls the part that is heavily dependent on the coordinate structure.

This beautiful problem finally appears to have met a solution. While I don’t know of any applications yet in TCS, it does feel like something powerful and relevant.

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Anyone interested in expander graphs should take a look at Amir Yehudayoï¬€’s recent article in the SIGACT newsletter: Proving expansion in three steps. Amir uses a ping-pong lemma argument (which is essentially the same as the “combinatorial lemma” in the preceding post) to exemplify the “opening” of the three part strategy of Bourgain and Gamburd to proving expansion in Cayley graphs.

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Given any two invertible, integral matrices , one can consider the family of graphs , where contains edges from every to each of

The Gabber-Galil graphs correspond to the choice and .

Consider also the countably infinite graph with vertex set and edges

Using some elementary Fourier analysis and a discrete Cheeger inequality, one can prove the following relationship (we use to represent the transpose of a matrix ).

Theorem 1For any , if has positive Cheeger constant, then is a family of expander graphs.

We recall that an infinite graph with uniformly bounded degrees has positive Cheeger constant if there is a number such that every finite subset has at least edges with exactly one endpoint in .

While Theorem 1 may not seem particularly powerful, it turns out that in many interesting cases, proving a non-trivial lower bound on the Cheeger constant of is elementary. For the Gabber-Galil graphs, the argument is especially simple. The following argument is, in fact, significantly simpler than the analysis of the previous post.

In the next lemma, let where and and let be the corresponding edge set. We will prove that has positive Cheeger constant.

*Proof:* Define . This is the first quadrant, without the -axis and the origin. Define similarly by rotating by , , and degrees, respectively, and note that we have a partition .

Let . We will show that . Since our graph is invariant under rotations of the plane by , this will imply our goal:

It is immediate that . Furthermore, we have because maps points in above (or onto) the line and maps points of below the line . Furthermore, and are bijections, thus In particular, this yields , as desired.

One can generalize the Gabber-Galil graphs in a few different ways. As a prototypical example, consider the family for any . An elementary analysis yields the following characterization.

Theorem 3For any , it holds that is an expander family if and only if .

For instance, the preceding theorem implies that if has order dividing then is not a family of expander graphs, but if has order or infinite order (the other possibilities) and then the graphs are expanders.

Here is a different sort of generalization. Let be the reflection across the line . The Gabber-Galil graphs can also be seen as where and . We can characterize expansion of these graphs as well.

Theorem 4For any , it holds that is an expander family if and only if .

It is an interesting open problem to find a complete characterization of the pairs such that is a family of expanders.

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If you’re thirsting for more open problems, I suggest this list on analysis of Boolean functions, compiled by Ryan O’Donnell in conjunction with the 2012 Simons Symposium.

Finally, in the realm of re-proving theorems, let me mention this absolutely gorgeous proof by Lovett and Meka of Spencer’s six standard deviations suffice theorem in discrepancy theory. The proof is constructive (i.e., there exists a polynomial-time algorithm to construct the 2-coloring) as in recent groundbreaking work of Bansal, but unlike Bansal’s proof is independent of Spencer’s bound. It also happens to be elementary and beautiful; my thanks to Aravind Srinivasan for pointing it out.

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The idea is to first start with an initial “expanding object,” and then try to construct a family of graphs out of it. First, consider the infinite graph with vertex set . The edges are given by two maps and , where and . So the edges are and . Clearly is not adjacent to anything. Except for the origin, this graph is an expander in the following sense.

Lemma 1For any subset , we have

where denotes the edges leaving .

The following simple proof is inspired by a paper of Linial and London.

*Proof:* First consider a subset that does not intersect the coordinate axes. Let represent the four quadrants of , and let . Consider that and . The latter fact follows because if then . Similarly, and and , while all the respective pairs and and are disjoint.

Combining this with the fact that and are bijections on immediately shows that is at least

Dealing with the coordinate axes is not too hard. Suppose now that is arbitrary. Let . The preceding argument shows . We will show that . Averaging these two inequalities with weights and completes the proof.

Let , and . Then we have the bounds:

The first equation follows since each element of is connected to exactly two elements of , and each element of is connected to exactly two elements of . For instance, is connected to and , while is connected to and .

The second follows because every point of is connected to two unique points of , e.g. is connected to and . The final inequality follows from the fact that from our first argument (since does not touch the axes), and because has no edges to . Summing these three inequalities yields .

Of course, is not a finite graph, so for a parameter , we define the graph with vertex set , where . There are four types of edges in : A vertex is connected to the vertices

This yields a graph of degree at most 8.

Theorem 2There is a constant such that for every ,

where is the second-smallest eigenvalue of the Laplacian on .

Of course, the graphs now have torsion, and thus our expansion result for is not, a priori, very useful. The non-trivial idea of the proof is that the torsion doesn’t matter, making Lemma 1 applicable. This is not difficult to show using some Fourier analysis. It turns out to be better though, if we first move to the continuous torus.

Recall that,

Let be the 2-dimensional torus equipped with the Lebesgue measure, and consider the complex Hilbert space

equipped with the inner product .

We might also define a related value,

Note that this is just defined as a number; the eigenvalue notation is merely suggestive, but we have not introduced an operator on the torus.

Claim 1There is some such that for any , we have \

*Proof:* We simply sketch a proof, which is rather intuitive. Suppose we are given some map such that . Define its continuous extension as follows: Under the natural embedding of into , every point sits inside a grid square with four corners . Writing a local convex combination , we define

Now it is elementary to verify that . It is also easy to verify that for some . (Even if , we still get a contribution from to on this square because we are taking a weighted average.)

Finally, for any , there is a path of length in connecting each of the corners of ‘s square to the corners of ‘s square. A similar fact holds for . In fact, this is the only place where we need to use the fact that edges of the form and are present in . Thus any contribution to can be charged against a term in , and similarly for .

Now our goal is to show that . We will use the Fourier transform to “remove the torsion.” The point is that and , being shift operators, will act rather nicely on the Fourier basis.

We recall that if and we define by , then forms an orthonormal Hilbert basis for . In particular, every can be written uniquely as

where . The Fourier transform is defined as a map , where . In particular, is a linear isometry.

Define and . Then for any , we have

In particular, for any , we have and . The final thing to note is that . So now if we simply apply the Fourier transform (a linear isometry) to the expression in (1), we get a reformulation that is precisely

Here we have simply replaced by in (1), and then written .

But now recall our initial infinite graph on . If we denote by the Laplacian on , then we can rewrite this as,

In other words, it is precisely the first Dirichlet eigenvalue on , subject to the boundary condition .

But now the discrete Cheeger inequality tells us that

where is the minimal expansion of any set not containing the origin. Thus we have indeed unwrapped the torsion and returned to our initial question. Lemma 1 shows that , yielding the desired lower bound on .

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