The question asked is:

What is the Shannon channel capacity $C$ that is naturally associated to the two-spin quantum Hamiltonian $H = \boldsymbol{L\cdot S}$?

This question arises with a view toward providing a well-posed and concrete instantiation of Chris Ferrie's recent question titled Decoherence and measurement in NMR. It is influenced too by the guiding intuition of Anil Shaji and Carlton Caves' Qubit metrology and decoherence (arXiv:0705.1002) that "To make the analysis [of quantum limits] meaningful we introduce resources."

And finally, it is reasonable to hope that so simple and natural a question might have a rigorous answer that is simple and natural too---but to the best of my (imperfect) knowledge, no such answer is given in the literature.


Let Alice measure-and-control by arbitrary local operations a spin-$j_\text{S}$ particle on a local Hilbert space $\mathcal{S}$ having $\dim \mathcal{S} = 2j_\text{S}+1$, upon which spin operators $\{S_1,S_2,S_3\}$ are defined satisfying $[S_1,S_2] = i S_3$ as usual.

Similarly let Bob measure-and-control by arbitrary local operations a spin-$j_\text{L}$ particle on local Hilbert space $\mathcal{L}$ having $\dim \mathcal{L} = 2j_\text{L}+1$ upon which spin operators $\{L_1,L_2,L_3\}$ are defined satisfying $[L_1,L_2] = i L_3$ as usual.

Let the sole dynamical interaction between the spins — and thus the primary resource constraint acting upon the communication channel — be the Hamiltonian $H = \boldsymbol{L\cdot S}$ defined on the product space $\mathcal{S}\otimes \mathcal{L}$. Further allow Bob to communicate information to Alice by a classical communication channel of unbounded capacity, but let Alice have no channel of communication to Bob, other than the channel that is naturally induced by $H$.

Then the question asked amounts to this: what is the maximal Shannon information rate $C(j_\text{S},j_\text{L})$ (in bits-per-second) at which Alice can communicate (classical) information to Bob over the quantum channel induced by $H$?


In practical effect, this question asks for rigorous and preferably tight bounds on the channel capacity associated to single-spin microscopy. The sample-spin $S$ can be regarded as a sample spin that can be modulated in any desired fashion, and the receiver-spin $L$ can be regarded variously as a tuned circuit, a micromechanical resonator, or ferromagnetic resonator, as shown below:

Alice and Bob do spin microscopy

The analysis of the PNAS survey Spin Microscopy's Heritage, Achievements, and Prospects (2009) can be readily extended to yield the following conjectured asymptotic form:

$$\lim_{j_\text{S}\ll j_\text{L}} C(j_\text{S},j_\text{L})=\frac{j_\text{S}\,(j_\text{L})^{1/2}}{(2\pi)^{1/2}\log 2}$$

Note in particular that the dimensionality of Bob's receiver-spin Hilbert space $\mathcal{L}$ is $\mathcal{O}(\,j_\text{L})$; thus a Hilbert-space having exponentially large dimension is not associated to Bob's receiver. However it is perfectly admissible for Alice and Bob to (for example) collaborate in squeezing their respective spin states; in particular the question is phrased such that Alice may receive real-time instruction of unbounded complexity from Bob in doing so.

Preferred Form of the Answer

A closed-form answer giving a tight bound $C(j_\text{S},j_\text{L})$ is preferred, however a demonstration that (e.g.) $\mathcal{O}(C)$ is given by some closed asymptotic expression (as above) is acceptable.

It would also be very interesting, both from a fundamental physics point-of-view and from a medical research point-of-view, to have a better appreciation of whether the above conjectured capacity bound on spin imaging and spectroscopy can be substantially improved by any means whatsoever.

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    $\begingroup$ I downvoted this question as it seems like a nice question but it is not written in the clearest way. You talk a lot around the question and I am not sure why you included that image. I think if it is stated in a simple motivation-definition-question format it'd be a nice question. I am still not completely clear on why one should care about such an Hamiltonian for greater than spin-1/2 systems. I'll be happy to remove my downvote if you think I am being unreasonable. $\endgroup$
    – Matty Hoban
    Commented Nov 7, 2011 at 0:28
  • $\begingroup$ Hoban, downvotes trouble me far less than indifference! :) Yes, physics hovers uneasily between beautiful mathematics and beautiful experiments. Spins larger than 1/2 are motivated mainly experimentally, as mentioned in my comment appended to Aram Harrow's answer (below): it is that nature provides us with large-spin particles (in the form of ferromagnets) having a spatial spin density billions of times greater than can be achieved by (e.g.) trapped-ion condensates. The price to be paid for this spin density is that the effective Hilbert space dimension is far lower than said condensates. $\endgroup$
    – John Sidles
    Commented Nov 7, 2011 at 0:57

2 Answers 2


This is an open question.

The capacity of some related Hamiltonians was computed in quant-ph/0207052, and an upper bound was derived in 0704.0964, but the Hamiltonian you describe is an example of one for which we don't know the exact answer.

However, quant-ph/0207052 also contains a conjecture (eq 35) about the capacity you're interested in. Their conjecture is supported by numerical experiments, but they can't rule out that some block coding strategy could do better.

Edited in light of our conversation this afternoon: From our discussion, it sounds like you've proven a lower bound of $\Omega(j_S j_L)$ using some cat-like states, based on principles similar to those in quant-ph/0605013. I think I can prove an asymptotically matching upper bound of $O(j_S j_L)$. So the exact constant is still open (and I think a hard problem), but at least we know the scaling.

For this bound, first argue that the interaction can be modelled as $2j_S$ qubits for Alice and $2j_L$ qubits for Bob, each in a symmetric state. To prove an upper bound, we can relax the restriction that the qubits be in a symmetric state. Then the Hamiltonian you describe can be expressed as a sum of interactions between each qubit of Alice's and each qubit of Bob's, each of which has constant strength. These two-qubit Hamiltonians have capacities upper-bounded by a constant. One reference for this last claim is quant-ph/0205057, but it was probably known earlier.

  • $\begingroup$ Aram, these are wonderful references for which I am very grateful! With specific regard to quant-ph/0207052 (eq. 35) this result (AFAICT) applies solely to coupled qubits (that is, spin-1/2 particles), whereas when we look into the coming decades of quantum spin microscopy, we foresee spins coupled to ferromagnetic spheres having spin-j ~ 10^5. These spheres have (effectively) a state-space of dimensionality much larger than one qubit, yet much small than 10^5 qubits; in brief their state-space is large enough to squeeze, but too small for (e.g.) quantum computation. More please! :) $\endgroup$
    – John Sidles
    Commented Nov 7, 2011 at 0:05
  • $\begingroup$ Aram, as a followup, I am beginning to appreciate more of the reasons why this simple-to-state question is difficult to answer. For example, the case $j_S = 1/2$ and $j_L \gg 1$ large can be regarded as effectively describing a cavity QED experiment, with $\mathcal{S}$ the state-space of the transmitting two-level atom and $\mathcal{L}$ the state-space of the receiving high-quantum-efficiency detector; these systems are known to be tough to analyze theoretically and tough to build experimentally. Conversely, this means that a closed-form tight capacity bound would be quite valuable to know. $\endgroup$
    – John Sidles
    Commented Nov 7, 2011 at 10:48
  • $\begingroup$ @AramHarrow: This update sounds an awful lot like metrology. It seems like the Heisenberg limit may apply. $\endgroup$ Commented Nov 10, 2011 at 9:50
  • $\begingroup$ @JoeFitzsimons, I agree. I thought this sort of thing wasn't rigorously proven, but <a href=arxiv.org/abs/1004.3944>1004.3944</a> may be useful in this regard. $\endgroup$
    – Aram Harrow
    Commented Nov 21, 2011 at 8:42
  • $\begingroup$ @AramHarrow: Thank you for these wonderful comments! I've been busy running an experiment for the past couple of weeks (hence the slowness in responding), but I concur with the general thrust of your remarks, and have thereby gained a better appreciation of (what is likely to be) the order of the upper and lower bounds to the channel capacity that are (respectively) achieved by cat-state sensors and coherent-state sensors, and thus have rated it as an "answer". $\endgroup$
    – John Sidles
    Commented Dec 14, 2011 at 19:14

Aram's answer seems perfect, but since you are also asking about the case for higher dimensional systems, let me add that there is a simply way to get somewhat non-trivial upper and lower bounds on $C(j_S,j_L)$. As a lower bound, you can simply synthesize an arbitrary gate which implements communication between the quantum systems (for an explicit algorithm on constructing arbitrary gates see Nielsen et al, Phys. Rev. A 66, quant-ph/0109064).

A non-trivial upper bound is given by the Margolus-Levitin theorem. Their paper (quant-ph/9710043) gives the maximum number of orthogonal states a quantum system can pass through in a given period of time, or conversely, a minimum time (as a function of energy) required to from the initial state to an orthogonal state (which is necessarily a lower bound on the time required to perfectly transmit one bit).

  • $\begingroup$ Thank you Joe ... it will take a couple of days to process these references. A challenge with the Nielsen construction is "gate to rate", that is, the construction does not limit the energy associated to Hamiltonian resources; thus is not evident how to extract a channel capacity bound of any kind. A challenge with the Margolus-Levitan theorem is that the operations associated to the bound are (apparently) not local; thus the upper bound (that I get from applying it) is unphysically optimistic. Perhaps there are ways around these obstructions; this is what will take awhile to think about. $\endgroup$
    – John Sidles
    Commented Nov 7, 2011 at 22:03
  • $\begingroup$ Progress report: preliminary efforts to construct concrete two-spin communication channels that approach the Margolus-Levitin bound have (so far) foundered on the local-operation restriction. This contributes to an appreciation (on my part) that Aram Harrow's answer is correct in asserting that "This [two-qudit channel capacity] is an open question", largely because the locality-of-operation restriction has deep-and-subtle consequences. One emerging benefit is an increased appreciation that fundamental QIT constraints are associated to single-photon sources and efficient local detectors. $\endgroup$
    – John Sidles
    Commented Nov 8, 2011 at 13:18
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    $\begingroup$ The Bravyi paper I mentioned (0704.0964) can be thought of as applying the Margolus-Levitin principle to upper bound the entangling rate. The bound is constant * norm of Hamiltonian * log dimension. (Note that this is an upper bound for the communication rate. Also dimensional factors are inevitable for both communication rate and entanglement rate, because transmitting an n-bit message can mean a single pi/2 rotation.) Unfortunately his argument is incomplete because we don't know how to control how much ancillas can increase the capacity. $\endgroup$
    – Aram Harrow
    Commented Nov 9, 2011 at 7:55
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    $\begingroup$ @AramHarrow: Good point. However, it seems like you can bound the effective dimensionality via the number of splittings in the eigenvalues of local Hamiltonians. $\endgroup$ Commented Nov 9, 2011 at 8:17
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    $\begingroup$ @AramHarrow: To explain what I mean: Imagine you have two spin-1/2 systems. Then your coupling induces at most 4 distinct eigenspaces which pick up relative phase, independent of the number of ancillae, or how they are prepared prior to coupling. Given that the information transfer in periods of free evolution is determined entirely by the relative phases between these 4 spaces, this is identical to what can be achieved with two spins. The only subtlety then comes from how you can combine multiple uses of such a channel. $\endgroup$ Commented Nov 9, 2011 at 8:30

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