2
$\begingroup$

This question is about the effect of available information on random quantum channels.

Suppose there are two black box devices.

Device 1. We have a black box device with a single qubit in it. Once we turn it on, this device does the following noisy quantum channel on the qubit $\rho \to (1-p) I \rho I^\dagger + p X \rho X^\dagger$, with certainty. Here $p \in (0,1)$. No measurement takes place after this device.

Device 2. This black box device has an identically prepared qubit in it. Furthermore, within this device there is a physical switch hidden that can be in state 'Heads' or 'Tails'. If the switch is in state Heads, nothing is guaranteed to happen; and if it is in state Tails, then the quantum bit-flip channel is guaranteed to happen. The switch is operated internally according to the outcome of a classical probability distribution, for example by a biased coin that flips Tails say with probability $q \in (0,1)$. Once we turn this device on, the device flips the coin internally, and therefore thus executes either $\rho \to \rho$ with probability $q$ or $\rho \to X \rho X^\dagger$ with probability $1-q$. No measurement takes place after this device.

We haven't yet but are about to turn on both devices.

Question 1. Suppose we will be informed of the outcome of the coin flip. If $p = q$, in this moment in time (so prior to turning on either device), are the two processes that are about to happen different or identical from a modeling perspective? Critically, assume that we do have access to all of the information above.

Question 2. Suppose we will not be informed of the outcome of the coin flip. What would be the answer now?

Improve this question
$\endgroup$
1
$\begingroup$

This question gets right to the heart of what information does a density matrix contain about the state of a qubit. Critically, it is a subjective state of knowledge. So, if I don't know the outcome of the coin flip, my best description of the system is the same in both cases.

At the moment that I learn the outcome of the coin flip, I have to update my description of the system based on the information that I have, in which case, I'll just record the pure state that I have. If I've done any operations on it, I have to propagate that updated state through the sequence of operations to have the best description of the state that I hold. If somebody else still doesn't know what the coin flip outcome was, they still hold a description based on their possible knowledge, i.e. the mixed state outcome.

Improve this answer
$\endgroup$
3
  • $\begingroup$ I am curious if there is more to it, and am wondering about your view on the following: I believe I understand your statement "if we don't know the outcome, then our best description of the system is the same in both cases". Here however, besides possibly knowing the outcome of the coin flip, we have in fact other meta information also available: that is, we know the second device operates according to a classical coin flip. Nothing is known about the internals of the first device. Generally, should we and/or can we express availability of meta information in our quantum mechanical models? $\endgroup$
    – Edin
    Jun 12 '20 at 13:24
  • $\begingroup$ Turn it the other way around. Does knowing that meta information alter, in any way, how you predict the outcomes of experiments that you perform on your state? In this case, it doesn't, so it shouldn't impact the description of the state that we hold. That might even be how you define the information to be "meta" $\endgroup$
    – DaftWullie
    Jun 12 '20 at 13:27
  • $\begingroup$ Maybe the following case helps to sharpen understanding... Consider two boxes. One contains 500 qubits in $|0\rangle$ state and 500 in $|1\rangle$. A second box contains 500 qubits in $|+\rangle$ and 500 in $|-\rangle$. You remove one photon and can perform any measurements you want on it. Is there any way that you can distinguish the two boxes using that single photon? (If I can remove more than one photon, does that change?) $\endgroup$
    – DaftWullie
    Jun 12 '20 at 13:31

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.