# POVMs that do not require enlargement of the Hilbert space

+ 10 like - 0 dislike
481 views

The usual justification for regarding POVMs as fundamental measurements is via Neumark's theorem, i.e., by showing that they can always be realized by a projective measurement in a larger Hilbert space.

That justification is sometimes problematic because for some applications is important not to enlarge the Hilbert space, so as to guarantee that the result you proved via POVMs is really about a Hilbert space of that dimension, and not just a shadow of a larger Hilbert space.

So, my question is, how to implement POVMs without enlarging the Hilbert space?

The only strategy I know is doing PVMs stochastically and grouping outcomes; for instance, the POVM $$\bigg\{\frac{1}{2}|0\rangle\langle0|,\frac{1}{2}|1\rangle\langle1|,\frac{1}{2}|+\rangle\langle+|,\frac{1}{2}|-\rangle\langle-|\bigg\}$$ can be implemented by measuring either $\{|0\rangle\langle0|,|1\rangle\langle1|\}$ or $\{|+\rangle\langle+|,|-\rangle\langle-|\}$ with probability $1/2$; by grouping the outcomes one can then measure the POVM $$\bigg\{\frac{1}{2}|0\rangle\langle0|+\frac{1}{2}|+\rangle\langle+|,\frac{1}{2}|1\rangle\langle1|+\frac{1}{2}|-\rangle\langle-|\bigg\}$$ or $\{I/2,I/2\}$

But this class can't be the whole set of POVMs; the Mercedes-Benz POVM (which has three outcomes proportional to $|0\rangle$ and $\frac{1}{2}|0\rangle \pm \frac{\sqrt{3}}{2}|1\rangle$) clearly can't be implemented this way. Is there a neat characterization of this class? Is there published research on it? Even better, is there another (more powerful) way of implementing POVMs without enlarging the Hilbert space?

This post has been migrated from (A51.SE)
retagged Mar 7, 2014

+ 5 like - 0 dislike

It seems that perhaps you are missing a crucial piece of the puzzle. The stochastic approach you describe is equivalent to performing a unitary which is chosen stochastically (essentially applying a superoperator) and then measuring in some fixed basis. In principle, however, you can go further, by stochastically choosing whether or not to measure at all, thereby producing a class of non-projective measurements. Once non-projective measurements are a possibility, the range of possible measurements expands greatly, as you may well want to make a sequence of such measurements.

Since the only operations open to you in this closed system are unitaries and projective measurements on its subsystems, and you are capable of randomized classical computation, I believe the entire set of measurements which can be performed are exactly those which can be performed as follows:

• Take $R$ to be some private classical register initially in the all zero state, and an additional bit $H$, also initially set to zero.
• Then for each round $i$:
1. Select a unitary operator $U_i$ via a classical randomized calculation on $R$, and write the result to $R$.
2. Apply $U_i$ to the quantum system.
3. Decide which if any subsystem of the quantum system to measure via a classical randomized calculation on $R$, and write the result to $R$.
4. If so decided perform the measurement on the required subsystem and store the result in $R$.
5. Decide via classical randomized computation on $R$ whether or not to set $H$ to 1.
6. If $H=1$ halt, other wise proceed to round $i+1$.

The above should hold for any system, however if you are considering qubits, 3 can be simplified to simply deciding whether or not to measure a designated qubit. The reason for choosing which subsystem to measure is because they might have relatively prime dimensionality, in which case the dimensionality of the measured subsystem likely matters.

I believe the above class includes the Mercedes-Benz measurement you mention as being impossible via your method.

This post has been migrated from (A51.SE)
answered Jan 18, 2012 by (3,575 points)
I don't see how choosing whether to measure or not would produce a weak measurement; could you expand on provide a reference? It seems to me that not measuring with probability $p$ would just create another effect $p I$ in the POVM.

This post has been migrated from (A51.SE)
@MateusAraújo: I should have been more careful with my language. You do indeed simply add $pI$ to a particular round. However this is an important difference to your procedure, and allows you to construct a significantly wider range of POVMs: In each round you can choose to implement one POVM element + the scaled identity. Repeating as necessary you can obtain measurements consisting of many POVM elements.

This post has been migrated from (A51.SE)
If that's the case, I don't see how one can implement more measurements than via "my" method, since I can measure scaled identity simply by grouping outcomes, and output cardinality is not a problem. Specifically, how would you go about implement the Mercedes-Benz POVM?

This post has been migrated from (A51.SE)
@MateusAraújo: In your case the effective identity operator still collapses the state onto whatever basis you picked, so repeated measurements do nothing. However, my method does not collapse the state in the case of an identity term, which is why multiple rounds makes sense.

This post has been migrated from (A51.SE)
Ok, but I still don't see how multiple rounds of identity (in the case of qubits) will get you useful information.

This post has been migrated from (A51.SE)
+ 1 like - 0 dislike

One step-by-step approach is to (1) regard the POVM as the coarse-grained approximation to the Lindbladian operator on the Hilbert space $\mathcal{H}$ of interest, then (2) then write the Lindbladian in terms of (Stratonovich) stochastic differential forms, then (3) pullback the forms onto any Kählerian manifold that seems suitable (a matrix product state, for example), together with (4) the metric and symplextic forms of the parent space $\mathcal{H}$.

This mathematically natural and computationally efficient pullback framework allows you to integrate dynamical trajectories on pretty much any Kählerian state-space you like, having any dimension you need, (including classical Bloch spheres and classical harmonic oscillators).

A good tutorial on the required Lindbladian forms is Carlton Caves' on-line notes Completely positive maps, positive maps, and the Lindblad form.

This post has been migrated from (A51.SE)
answered Jan 24, 2012 by (485 points)
+ 0 like - 0 dislike

I've found a paper that seems relevant to the question: Classical randomness in quantum measurements, by Giacomo Mauro D’Ariano, Paoloplacido Lo Presti, and Paolo Perinotti. The authors are in describing which POVMs are "pure", i.e., can't be implemented by stochastically choosing between other POVMs.

To do that, they provide a simple algorithm for decomposing a given POVM into a mixture of pure POVMs. Since PVMs are a special case of pure POVMs, one can use their algorithm to decide whether a given POVM can or not be implemented by mixing PVMs.

They also prove an interesting bound on the number of outcomes a POVM can have and still be pure: $d^2$. For some applications, this shows that it's useless to consider POVMs with more than $d^2$ outcomes.

This post has been migrated from (A51.SE)
answered Jan 25, 2012 by (270 points)

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOverflo$\varnothing$Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.