# I Debunk these news stories (QM Radar)

1. Sep 23, 2016

### .Scott

Normally when I cite something, it's to an article I consider credible, but this one is not:
The specific assertion in question is this paragraph:
The article also asserts that Lockheed has patented this method. Who knows, perhaps they have.

Here's a paragraph from another news source http://www.extremetech.com/extreme/...could-render-entire-us-stealth-fleet-obsolete:
My suggestion is that if you're a physics graduate, this is too simple to debunk. Let's see analysis from QM rookies.

Also, if there really is such a functional thing as quantum radar, perhaps someone could provide a real description of how it works.

2. Sep 23, 2016

### Staff: Mentor

3. Sep 23, 2016

### .Scott

Thanks for the link. The patent makes sense. Certainly what is described in the news articles is very different than what is described in the patent.

4. Sep 23, 2016

### Strilanc

Seems plausible enough to me. Not the way the news story butchers it of course, but just the concept "radar gets better with entanglement".

A simplification of the radar 'problem' is "I send out a qubit A. Some qubit comes back. It might be A or it might be some noise qubit C. Which is it?".

If you use a qubit A in a pure state, like $|0\rangle$ or $|0\rangle+|1\rangle$, then you can distinguish it from a noise qubit C with about 75% fidelity. That's because about half of noise qubits will give a wrong measurement result that gives them away, and then you get another half by just randomly guessing, leaving only 25% wrong. We can compute this easily by using the trace distance, which is 0.5 for A [a pure state] vs C [the maximally mixed state].

But now lets use a signal qubit that's entangled with a kept-at-home qubit B.

The trace distance between an EPR pair AB, and the "some noise qubit got into my EPR" pair CB is 0.75. Then random guessing gets us up to succeeding 87.5% of the time.

So entanglement does provide an advantage. If you think of distinguishing A from C (or AB from CB) as a noisy channel, like someone trying to send you a message by having a plane present or not, then adding B to the system increased the bandwidth from ~0.19 bits per attempt up to ~0.46 bits per attempt.

Of course it only starts getting really crazy when you start throwing in counterfactual-bomb-detector stuff, augmented with Grover's algorithm, to make negligible the chance of the enemy plane detecting it's being pinged.

Python code:

Code (Text):

import numpy as np
import math

def trace_dist(A, B):
return sum(np.abs(np.linalg.eig(A-B)[0]))*0.5

def prob_distinguish(A, B):
return trace_dist(A, B)*0.5 + 0.5

def H(p):
q = 1-p
return -p*math.log(p, 2) - q*math.log(q, 2)

I2 = np.eye(2)/2
I4 = np.eye(4)/4
EPR = np.mat([[1,0,0,1],[0,0,0,0],[0,0,0,0],[1,0,0,1]])/2.0
Z=np.mat([[1,0],[0,0]])
X=np.mat([[1,1],[1,1]])/2.0

print(prob_distinguish(I2, Z))
# 0.75
print(prob_distinguish(I4, EPR))
# 0.875
print(1 - H(0.75))
# 0.18872187554086717
print(1 - H(0.875))
# 0.4564355568004036

The wikipedia page Quantum Radar gives some good references. For example, one is by Seth Lloyd: Enhanced Sensitivity of Photodetection via Quantum Illumination, Science 321, 1463-1465 (2008) ([4])]).

Last edited: Sep 23, 2016
5. Sep 23, 2016

### .Scott

The descriptions provided in the news articles do not mention anything about receiving radar returns. At least by my reading, they imply that the information is being deduced solely from the "B" photons. That's the part that is wrong.

6. Sep 23, 2016

### Strilanc

Well, yeah. But news articles get that kind of fundamental thing wrong every time they mention anything quantum. They're basically never worth reading; you have to go the source.

The underlying concept "make radar better with entanglement" is meaningful. Doing it without bouncing the photons back is obviously junk. It would violate the no-communication theorem.

7. Sep 24, 2016

### nsaspook

Last edited: Sep 24, 2016