2019-02-08 01:55:46 8 Comments

Suppose I have two photons $A$ and $B$, which are entangled.

As I have had it explained to me, entanglement works more or less as follows:

I observe or measure $A$, and find $A=S_a$, where $S_a$ is some state.

Next, I measure $B$ in the exact same way, and find $B = S_a$ as well.

However, if I return to measure $A$ in the instant after measuring $B$, I will find that the state of $A$ has become stochastic once more.

As I understand it, quantum based encryption works as follows: I entangle a set of photons with each other in a one-to-one and onto correspondence such that $\mathbf{A}$ and $\mathbf{B}$ consist of photons on either side of the correspondence, and $f(\mathbf{A}): \mathbf{A} \rightarrow \mathbf{B}$ together with $f^{-1}$ defines the entanglement.

I observe all photons in $\mathbf{A}$, and send all photons in $\mathbf{A}$ and $\mathbf{B}$ as follows: $\mathbf{A}$ precedes $\mathbf{B}$.

If someone receives the message, they will first read off the states of the $A$-side photons, and then observe the $B$ side photons. Since the entangled global state still exists at this point, they will observe a correlation (I don’t think it will be perfect after transmission, but rather statistically significant) between the $\mathbf{A}$ side and the $\mathbf{B}$ side photons.

However, noting the state of the $\mathbf{B}$ side photons will collapse the global entangled state, and the correlation between $\mathbf{A}$ side and $\mathbf{B}$ side photons will vanish for the next observer.

Thus, if one intercepts the message, the intended receiver will know whether it was intercepted—this apparently counts even for copies of the information.

The question is as follows: suppose I split the signal and pass it through a double slit (as a stand in for whatever is actually done in an optical Fourier transform) or some other mechanism to convert the signal from a coherent state to a diffracted wave state (squeezed).

Next, I record the signal in a hologram placed immediately behind the device that performs the optical Fourier transform (as explained in the answer here).

Since an optical Fourier transform is possible and reversible, the state saved in the hologram must(?) be in a wave state, as versus a particle state.

Then, I wait a week or two, and replay the recorded message with the exact wave state of the original entangled headers preserved back through the transform, and attempt to decrypt it.

Has the encryption header that detects interception been bypassed?

Or does recording the data in the hologram collapse the entanglement?

As a follow up (and willing to post in a new question), if the entanglement collapses, wouldn’t there be a physical change in the hologram which would take energy to accomplish?

On the other hand, if recording in the hologram counts as an observation, and there need never be a collapsing state, then is the invertibility of the optical Fourier transform challenged?

Or is the hologram capable of making a copy of the entangled state without destroying it?

(In short: which horse wins?)

**Added Later** (Will reorganize and incorporate into the question as a whole in a bit; on a cell phone)

A good reference is the no cloning theorem, which explains how no randomized quantum state can be copied. However, the notion of “theorem” in the context of a model used to describe physical reality is not good.

Theorems only rest, ultimately, on axioms. So the claim implies there are axioms (beliefs) in the scientific model “quantum theory,” and that seems odd.

And so my question becomes: is a hologram a medium which may be an edge case to the no cloning “theorem”?

And the reasoning is follows:

A double slit diffracts incoming light unless at least one half of it is measured at an individual photon level immediately behind the double slit, whereupon no diffraction occurs. — **For this reason, I hypothesize the double slit does not, on its own, disturb an entangled state**

Next, a hologram may perfectly record, at an individual photon level, the output of a double slit without destroying the diffraction (answer referenced above); this is a known property used in analog optical processing for useful tasks.

Further, the hologram may be used to “time reverse” the incident signal through a double slit such that an I disturbed diffraction produces a coherent output — **For this reason, I hypothesize the hologram records everything that was incident on it without disturbing its wave state**

Arbitrarily, I dismiss the idea of a “theorem” in the domain of a physics "theory"—it has only the weight of the math, and rests on the “belief” that the underlying model is correct: i.e., the authority of math theorems rest absolutely on assumed axioms, while the authority of a physics theory rests on induction in terms of experimental results, and so a model can be a theory, but it can only produce postulates; it is not declarative with respect to reality (which a math theorem is). In other words: the food does not come from the grocery store.

Finally, the information about exactly how the quantum keys are distributed in a quantum cryptography setup is very obfuscated by pop science simplifications, but I understand that detection of whether a key has been observed more or less follows from state transference and the no-cloning theorem.

Above I described a simple example of how it could work.

Without getting too formal with references and research: with respect to state transference, let us take it on faith that it happens with respect to QuBits at least, and photons as well (due to the no cloning property: the mixed state is the only state, there is not more than one result when one looks at $A$, then $B$), as the source on this is a real life expert in the field.

**In response to S. Mcgrew's answer**

Suppose we relax the "magic" of "entanglement" to the simple case of a markov chain coupling: two stochastic processes, when constrained to transition from state to state the exact same way for a sufficient period of time (measured in state transitions) to cause both stochastic processes to have the same global state.

Sufficient period of timeThe number of state transitions necessary to completely couple a given system

Once photon A is coupled with photon B, photon A and B will share the same stochastic state through time if quantum mechanical state transitions are deterministic, but approximable with probability.

This causes an observed "entangelement." The coupling, however, is very sensitive, and any interaction with the underlying state of the photon will modify the transition probabilities at the time of inspection such that A will no longer be coupled with B henceforth.

Then, at some time removed from the measurement of A, we measure B and observe that B had the same state as A (it was coupled and there have been no state transitions in B since A was measured).

However, we did not measure B and A in the *exact* same way, so that A and B are no longer coupled after measurement of both.

Any "observation" necessarily breaks entanglement by causing a state transition in A that is statistically independent of B.

Coping with the no-copy theoremI do not doubt there is much experimental evidence for this property, which I have not read, so I leave it at: experimental evidence at that size scale might not be able to prove a negative, and I don't see it as a reason not to try something simple, like the markov chain, to explain the observed behavior.

With this model of entanglement as a coupling (rather than any teleportation or spooky action at a distance), it seems very possible to man in the middle a quantum key distribution through the use of holograms recording diffracted signal (i.e., recording the optical fourier transform).

### Related Questions

#### Sponsored Content

#### 3 Answered Questions

### [SOLVED] Entanglement and the double slit experiment

**2012-02-14 19:05:13****Paul Merrifield****2162**View**3**Score**3**Answer- Tags: quantum-mechanics photons quantum-entanglement double-slit-experiment wavefunction-collapse

#### 1 Answered Questions

### [SOLVED] Quantum Key Distribution via Modified Double Slit Experiment

**2019-07-13 23:43:42****Thor****45**View**0**Score**1**Answer- Tags: quantum-information quantum-entanglement double-slit-experiment

#### 1 Answered Questions

### [SOLVED] can entangled photons produce an interference pattern?

**2019-01-19 13:56:25****kishdude****139**View**3**Score**1**Answer- Tags: quantum-mechanics quantum-entanglement double-slit-experiment interference coherence

#### 1 Answered Questions

### [SOLVED] Entangled photon absorption redshift/blueshift scattering

**2018-04-30 18:55:57****Árpád Szendrei****100**View**1**Score**1**Answer- Tags: quantum-entanglement

#### 1 Answered Questions

### [SOLVED] Does the Delayed Choice Quantum Eraser require the signal and idler photons to be entangled?

**2016-02-16 00:48:17****brian kirkby****108**View**0**Score**1**Answer- Tags: quantum-mechanics quantum-entanglement double-slit-experiment quantum-eraser

#### 3 Answered Questions

### [SOLVED] Reconciling Understanding of Quantum Entanglement and Superposition

**2017-06-23 14:04:40****W. Randy King****171**View**3**Score**3**Answer- Tags: quantum-mechanics quantum-entanglement double-slit-experiment superposition

#### 1 Answered Questions

### [SOLVED] Can entanglement swapping be performed on already-entangled photons, and if so, can it preserve this entanglement over the swap?

**2012-08-24 11:07:49****Andrew Palfreyman****646**View**4**Score**1**Answer- Tags: quantum-mechanics quantum-entanglement

#### 4 Answered Questions

### [SOLVED] Interference and which-path information

**2013-04-02 09:55:56****user1247****1982**View**8**Score**4**Answer- Tags: quantum-mechanics quantum-entanglement double-slit-experiment

## 1 comments

## @S. McGrew 2019-02-08 05:48:50

I can't comment on the protocols for quantum key distribution; only on the entanglement aspects of your question.

If A and B are entangled and you measure the state of A, then you know the state of B. If Mary measures A and George measures B, and both Mary and George know that A and B are entangled, then George knows that his B measurement agrees with Mary's A measurement.

It doesn't really matter

howA and B are measured, as long as the measurements extract the same type of information about the state of A or B. Once a measurement result is obtained, it will not change due to a later measurement. So, if Mary records a hologram which contains the result of a state measurement on A, and if the hologram is kept in a "Schrodinger's cat" box (during and after the recording) so Mary can't see it and it cannot interact with anything in Mary's world, and then if George measures B, George can predict with certainty that Mary's hologram of A agrees with George's measurement of B. Nothing changes at Mary's end due to George's measurement of B. A measurement "collapses" a mixed-state wavefunction by selecting at random one of the states that comprise the mixed state.I can't think of a way that "recording an entangled state" is physically meaningful. The reason entanglement is useful for generating and distributing quantum keys is in the fact that measurements of a mixed-state A a and mixed-state B with which A is entangled will come out in agreement -- regardless of when or where the measurements are done -- as long as the entanglement is not disturbed. Disturbing the entanglement amounts to letting either A (or B) interact with anything that can randomly alter the state of A (or B); like another particle just like A (or B) that is itself in a mixed state.

An optical Fourier transform does not change light from particles to waves or vise versa. It just rearranges the waves.

## @Christopher 2019-02-08 12:27:17

it sounds like what you are saying is that recording a signal in a hologram is

notenough to disturb the entangled state, so long as the recording is performed in a schrodinger’s box, otherwise the hologram recording will disturb the entangled state?## @Christopher 2019-02-08 12:28:39

And regarding the state transfer from A to B when A is read, then B is read: this is something I got from a primary source. But maybe it was related to QuBits and not photons?

## @Christopher 2019-02-08 14:24:03

Supposing you are right, I am adding an edit to the question

## @S. McGrew 2019-02-08 14:44:40

There is a huge amount of stuff written about Schroedinger's cat and related gedankenexperiments. A measurement changes the state of the measuring instrument to correspond to the state of the measured object. If the measuring instrument itself is considered to be a quantum object, then when it measures a mixed-state object, the instrument acquires a mixed state. If kept totally isolated from the rest of the universe, the instrument remains in the mixed state until it is read by an observer.

## @S. McGrew 2019-02-08 14:44:48

According to the Many Worlds view, the observer then acquires the mixed state, and in effect splits into multiple timelines, one for each of the components of the mixed state. It's a very messy process.