7
$\begingroup$

I understand the mathematical construction of the Choi-Jamiolkowski isomorphism aka channel-state duality. It all makes sense formally, yet I still struggle to grasp its physical (or quantum-informational) meaning. Does the isomorphism between quantum states and quantum channels in any sense establish some kind of connection relating the constitution (state) of systems to evolution (channel) of other systems?


Cross-posted on physics.SE

$\endgroup$
  • 3
    $\begingroup$ If you give me the state dual of a Clifford channel, I can teleport a new state through it in order to apply the channel to the new state. $\endgroup$ – Craig Gidney Jun 19 at 23:12
5
$\begingroup$

Mathematically it is a relationship between a bipartite linear operator vector space $L(X\otimes Y)$ and a superoperator vector space $C(X): L(X)\to L(Y)$ (maps of linear operators to linear operators). Bipartite density matrices are contained in the former, and quantum channels in the latter. The real "physical" meaning of the isomorphism for quantum information theory is that a superoperator is a valid quantum channel if and only if its Choi-state is a valid density matrix. You can read more about properties of the Choi-representation (and other representations of quantum channels) in my review article [1].

This also has several quantum computing applications outside of quantum information theory. The most common two are probably

  1. Quantum process tomography
  2. Gate teleportation (especially in measurement based quantum computing)

Matthew Leiffer has a good blog post [2] which describes point 2 (and talks more generally about the Choi-Jamiolkowski isomorphism in the context of some of his research on conditional states).

For point 1. If if one wants to experimentally reconstruct a description of a quantum channel they basically need to reconstruct the Choi-matrix representation of the channel. This is typically done by preparing a tomographically complete set of input states (states that span the input state space), and perform a tomographyically complete set of measurements (measurements with outcomes projectors that span the output state space) and use the resulting probabilities to reconstruct the description of the channel.

From the point of view of measurement probabilities, the "channel" is indistinguishable from a bipartite "state" that we would have performed state tomography on, and so we can reconstruct the description as if we were simply reconstructing a bipartite density matrix using state tomography (there are some subtleties concerning the trace preserving property which I won't get into). You can see an example of process tomography in this Qiskit Ignis tutorial notebook [3].

The is also another way to do process tomography called Ancilla assisted process tomography [4] which is a physical implementation of the Choi-Jamiolkowski isomorphism. You could prepare a maximally entangled bell state, send half of it through the channel to be investigated (and do nothing to the other half), and then perform state tomography of the full output to directly reconstruct the Choi-state for the unknown channel. This is rarely used in practice as it is typically less accurate than the standard method due to errors in preparing the maximally entangled input state.

[1] Quant. Inf. Comp. 15, 0579-0811 (2015) arXiv:1111.6950 [quant-ph]

[2] http://mattleifer.info/2011/08/01/the-choi-jamiolkowski-isomorphism-youre-doing-it-wrong/

[3] qiskit-iqx-tutorials/qiskit/advanced/ignis/6b_process_tomography.ipynb

[4] Physical Review Letters 90, 193601 (2003), arXiv:quant-ph/0303038

$\endgroup$
  • $\begingroup$ great answer, thanks! $\endgroup$ – glS Oct 10 at 16:13
  • 1
    $\begingroup$ Yes, thank you very much Christopher. Btw, I had actually read your publication [1] when I first studied superoperators and their representations - it was very helpful. I understand the mathematical procedure; guess I am still hoping for some further insight as to why this construction works or what it implies beyond the mere encoding of the evolution law in a state. Sometimes there is some additional/different operational perspective on the same math - that is what I was angling for... $\endgroup$ – quantumorsch Oct 10 at 18:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.