Ron Garret’s “The Quantum Conspiracy: What Popularizers of QM Don’t Want You to Know” Google Tech Talk

Way back in January 2011, Ron Garret gave a Google Tech Talk titled “The Quantum Conspiracy: What Popularizers of QM Don’t Want You to Know”.

I’m just watching it again now, for the third time, hoping that if I write down my thoughts as I go along I will actually be able to make sense of it this time!

The best introduction to quantum mechanics I have found so far is a series of recordings of the great Richard Feynman talking about “The Quantum Mechanical View of Reality” (which you can find here:    ).

Ron splits his talk up into four steps:

1. Review the usual QM story
3. Do some math and show how that resolves the contradiction
4. Tell a new story based on the math

I’ll split this post up to follow those steps so you can watch along with me.

Step 1: Review the usual QM story

Quantum mystery #1: The two-slit experiment

Ron starts by explaining the standard “two-slit” experiment, which I’m going to assume you are familiar with. If not, then this video explains it well.

The root of the apparent weirdness lies in the measurement problem, which gives rise to the so-called wave-particle duality.

Any modification that we make to this experiment that allows us to determine even in principal which of these slits this particle went through destroys the interference.Ron Garret

It is worth noting that this holds for any particle, any “measurement” and any equivalent “split/combine” experiment. This is not some obscure effect, it is possible that similar effects are happening around us all the time in ways we do not usually notice (or perhaps take for granted?) at the macro scale.

Ron goes on to attack the Copenhagen interpretation, in which the wave function is assumed to “collapse” at some point and “become” a particle.

By asking the question “how and when does this collapse happen?”, he starts with some intuition that any form of “collapse” is by nature irreversible and discontinuous. This contradicts the mathematics of quantum mechanics, in which quantum effects are continuous and reversible in time.

Quantum mystery #2: The “Quantum Eraser”

He then moves on to introduce an example of a “quantum eraser” experiment with one split and one combine: In his eraser, he places one detector at a place that would represent a “dark fringe” and another at a place that would represent a “bright fringe” if an interference pattern was produced.

Taking a measurement after the split but before the combine destroys the interference pattern and each detector detects the same amount of particles. The interesting bit is that this measurement can be “erased” by destroying the information that the measurement determined before the particle hits the combining step. This restores the interference pattern that the detectors witness. The mind boggles…

There’s a really interesting article that appeared in Scientific American in 2007 that explains how to make a DIY quantum eraser. The corresponding web article can still be found here. This is what Ron is demonstrating in his talk with the polarised light filters.

The claim Ron makes is that the implication of this erasure is that it does not make sense to say that the wave function “collapses at the time of measurement”, since erasing that measurement in some sense restores the wave function so that wave information must have been preserved somewhere in the system.

Quantum mystery #3: Entanglement

He then moves on to an example of quantum entanglement: Here, a UV laser and “down converter” (some kind of crystal structure) is used to produce photons of varying wavelengths, sending them off in opposite directions, in a 1:1 ratio.

By splitting at each end, using, for example, one of the polarisation measurements from earlier, we find that the LU/RD and LD/RU detectors are perfectly correlated, because of conservation laws.

This is what Einstein famously called “Spooky action at a distance”Ron Garret

What this seems to show is that, independent of the distance of separation between two entangled particles, a measurement of one instantaneously changes a perfectly correlated aspect of the particle’s entangled twin.

At first glance, this would seem to imply that faster than light communication is possible, however, it is widely believed that this is not the case since we have no control over what the result of a measurement will actually be.

Now, Ron says that he is going to show us how the story presented so far leads to a contradiction.

So far we have that:

• A split/combine experiment produces interference
• Any which-way measurement destroys interference
• Some which-way measurements can be erased, restoring interference
• Measurements on entangled particles are perfectly correlated

What they don’t want you to know:

All of these things cannot possibly be true!Ron Garret

He presents a thought experiment paradox which he dubs the Einstein-Podolsky-Rosen-Garret Paradox: In this experiment, two “two-slit” experiments are fed by quantum entangled photons produced by the entanglement process he described earlier.

The question is, what happens if we take a measurement on the left side. Will this destroy interference on the right?

• If the answer is “yes”, then we have faster than light communications, by using measurement on the left to produce a signal on the right, which is assumed to be impossible
• But if the answer is “no”, then we know the position of the particle but we still have interference, which violates a fundamental principle of quantum mechanics

There is one more possibility, however. Perhaps there was no interference, to begin with! Maybe entanglement counts as a measurement that destroys interference?

Unfortunately, Ron claims that even in this case, we are not out of the woods yet. We can put in an eraser on the right and destroy the entanglement, leading us back to producing interference but still knowing the position of the particle and so still leading to a contradiction as before.

Step 3: Do some math and show how that resolves the contradiction

After some preliminaries about the wave function $\Psi$ and the “hack” of extracting the probability of the wave function by “squaring” the magnitude $|\Psi|^2$, we move on to some two-slit math, making use of Dirac notation.

Note that all the non scalar quantities in an amplitude expression represent complex numbers. Also, “squaring” the magnitude of a complex number $\Psi$ is like taking the inner product of that number with itself, where the inner product is defined as the complex conjugate operation and so $\langle \Psi|\Psi \rangle = |\Psi|^2 .$

Also, remember the properties of the inner product, in particular: $\langle A + B|A + B \rangle = |A|^2 + |B|^2 + \langle A|B \rangle + \langle B|A\rangle.$

Without detectors

The amplitude of the particle without measurement is $(\Psi_U + \Psi_L)/\sqrt{2},$ where $\Psi_U$ is the amplitude of the upper detector and $\Psi_L$ is the amplitude of the lower detector. $\sqrt{2}$ is a normalisation constant used to ensure probabilities sum to 1.

The resulting probability is $(|\Psi_U|^2 + |\Psi_L|^2 + \Psi_U{}^*\Psi_L + \Psi_L{}^*\Psi_U)/2 .$

The term $\Psi_U{}^*\Psi_L + \Psi_L{}^*\Psi_U$ is an interference term made up of the sum of two complex products involving the complex conjugates ${}^*\Psi_L$ and ${}^*\Psi_U$ of the original detector amplitudes. This is the only part of the probability expression that can contribute negatively.

With detectors

The amplitude of the particle with measurement is $(\Psi_U|D_U\rangle + \Psi_L|D_L\rangle)/\sqrt{2},$ where the new term $|D_U\rangle$ is the amplitude of the detector indicating a particle at the upper slit and $|D_L\rangle$ is the amplitude of the detector indicating a particle at the lower slit.

The resulting probability is $(|\Psi_U|^2 + |\Psi_L|^2 + \Psi_U{}^*\Psi_L \langle D_U|D_L\rangle + \Psi_L{}^*\Psi_U \langle D_L|D_U\rangle)/2 .$

Again, we have an interference term $\Psi_U{}^*\Psi_L \langle D_U|D_L\rangle + \Psi_L{}^*\Psi_U \langle D_L|D_U\rangle$ that involves two new quantities. $\langle D_U|D_L\rangle$ is the amplitude of the detector switching spontaneously from the $U$ state to the $L$ state. Likewise, $\langle D_L|D_U\rangle$ is the amplitude of the detector switching spontaneously from the $L$ state to the $U$ state.

If the detector is working properly, then both the $\langle D_U|D_L\rangle$ and $\langle D_L|D_U\rangle$ terms will be $0 .$ This means that the probability would be $(|\Psi_U|^2 + |\Psi_L|^2)/2$, with no interference term at all.

In some sense, this demonstrates that “measurement” is a continuum! With imperfect knowledge of the actual state (because of imperfect detectors), the interference term is slowly introduced.

Entangled particles

Ron states that the amplitude of the entangled particle experiment in the EPRG paradox is given by $(\lvert\uparrow\downarrow\rangle + \lvert\downarrow\uparrow\rangle)/\sqrt{2},$ which is the amplitude for the left particle to be in the up state and the right particle to be in the down state $\lvert\uparrow\downarrow\rangle$ superimposed with the amplitude for the left particle to be in the down state and the right particle to be in the up state $\lvert\downarrow\uparrow\rangle .$

Changing the notation, this is equivalent to $(\Psi_{LU}|RD\rangle + |\Psi_{LD}|RU\rangle)/\sqrt{2} .$ This is similar to our earlier two-slit with detectors amplitude.

Recall that the LU/RD and LD/RU detectors are perfectly correlated. This means that an observation of $\Psi_{LD}$, for example, gives us all the information about $\Psi_{RU} .$ Likewise, $\Psi_{LU}$ gives us all the information about $\Psi_{RD} .$

Entanglement and measurement are the same phenomenon!Ron Garret

It is as if entanglement and measurement of LU and LD is the same as measuring RU and RD directly without entanglement: $(\Psi_{RD}|RD\rangle + |\Psi_{RU}|RU\rangle)/\sqrt{2} .$

Quantum eraser

After measurement, but before erasure, the amplitude is $(|U\rangle|H\rangle + |L\rangle|V\rangle)/\sqrt{2})$, where $|U\rangle|H\rangle$ is an upper photon that is horizontally polarised and $|L\rangle|V\rangle$ is a lower photon that is vertically polarised.

The corresponding probability is: $\displaystyle (\langle U|U\rangle \langle H|H\rangle + \langle L|L\rangle \langle V|V\rangle + \langle U|L\rangle \langle H|V\rangle + \langle L|U\rangle \langle V|H\rangle)/2 .$

However, this reduces to $( \langle U|U\rangle + \langle L|L\rangle )/2,$ with no inteference, since the inteference terms are zero because the polarisation is assumed to be stable and so $\langle H|V\rangle = 0, \langle V|H\rangle = 0, \langle H|H\rangle = 1, \langle V|V\rangle = 1 .$

After erasure, by filtering in at $45^\circ ,$ the amplitude is given by $(|U\rangle + |L\rangle)(|H\rangle + |V\rangle)/2\sqrt{2},$ which is a photon that is either in the upper $|U\rangle$ or lower $|L\rangle$ slit and is either horizontally $|H\rangle$ or vertically $|V\rangle$ polarised. $|H\rangle + |V\rangle$ means polarised at $45^\circ .$

Note that the probability normalisation constant is $2\sqrt{2}$ not $\sqrt{2}$ as before. It turns out that the total probability of the amplitude is not one, but a half!

It turns out this is because we have not accounted for half of the photons, those that were filtered by the eraser.

The filtered out photons have a different amplitude $(|U\rangle + |L\rangle)(|H\rangle - |V\rangle)/2\sqrt{2} .$ The $|H\rangle - |V\rangle$ term means “filter out” polarisation at $45^\circ$, but along a different axis than the $|H\rangle + |V\rangle$ “filter in” polarisation. Both sets of photons (both the filtered in and the filtered out) interfere with themselves. The filtered in photons display interference fringes. The filtered out photons display interference “anti-fringes”. These fringes sum together to produce “non-interference”.

You can see this with a little bit of algebra. The overall amplitude after erasure is $\displaystyle \begin{array}{rcl} && ((|U\rangle + |L\rangle)(|H\rangle + |V\rangle) + (|U\rangle + |L\rangle)(|H\rangle - |V\rangle))/2\sqrt{2} \\ &=& (|U|\rangle |H\rangle + |U\rangle |V\rangle + |L\rangle |H\rangle + |L\rangle |V\rangle + |U\rangle |H\rangle - |L\rangle |V\rangle + |L\rangle |H\rangle - |U\rangle |V\rangle )/2\sqrt{2} \\ &=& (|U\rangle |H\rangle + |L\rangle |H\rangle)/\sqrt{2} \\ \end{array}$

The corresponding probability is $(\langle U|U\rangle + \langle L|L\rangle + \langle U|L\rangle + \langle L|U\rangle)/2 .$

Which always has an interference term $\langle U|L\rangle + \langle L|U\rangle .$

So quantum erasers don’t “erase” anything, and they don’t produce interference either, they just “filter out” interference that was already there.Ron Garret

I think what Ron is trying to highlight in the quote above is that the math shows us that all the eraser does is filter out the $|U\rangle |V\rangle + |L\rangle |V\rangle - |L\rangle |V\rangle - |U\rangle |V\rangle$ interference, leaving only the self interference of $\langle U|L\rangle + \langle L|U\rangle.$ It does not “add in” any interference of it’s own, rather it just highlights interference that is already in the system through this “filtering out” process.

Next, he points out that you can observe this “cancelling out” phenomenon in the laboratory in an Einstein-Podolsky-Rosen experiment, by recording the $U$ and $D$ detector states, then sending this information classically (over a wire for example) to the right-hand side of the experiment. You can then look at the $U$ and $D$ photons separately and notice that there is one “interference pattern” and another “anti-interference pattern” that you usually do not see because, in the combined effect, these patterns cancel out. Coming back to the original point of this section, we have resolved the contradiction, since:

• Entanglement does “count” as measurement
• There is no interference in the EPRG experiment
• Interference is not “produced” by the use of an eraser

Step 4: Tell a new story based on the math

Here, Ron begins to introduce his “quantum information theory”, or “zero-worlds” interpretation of quantum mechanics. This is an extension of classical information theory that uses complex numbers.

He introduces the entropy measure of a system, $H(A) = -\Sigma_a \text{P}(a) \text{log} \text{P}(a),$ which is $0$ when a system is definitely in a single state and $\text{log}(N)$ when a system has an equal probability of being in $N$ states.

Some entropy measures of two systems are then introduced.

The joint entropy is $H(AB) = -\Sigma_{a,b} \text{P}(ab) \text{log} \text{P}(ab).$

The conditional entropy is $H(A|B) = -\Sigma_{a,b} \text{P}(a|b) \text{log} \text{P}(a|b).$

The information entropy (more commonly known as the mutual information) is $I(A:B) = I(B:A) = H(AB) - H(A|B) - H(B|A),$ which is the information about A contained in B and is always in the range $[0, 1].$

This is extended into the complex plane by using the Von Neuman entropy of a system, $S(A) = -\text{Tr}(\rho_A \text{log} \rho_A)$ where $\rho_A$ is the quantum density matrix, $\text{Tr}$ is the matrix trace operator and $\text{log}$ is the natural matrix logarithm.

Ron makes a key point here that the information entropy is no longer restricted to the range $[0, 1].$

He then illustrates what happens in terms of the mutual information when we consider a system with three mutually entangled particles, where $A$ and $B$ are the “measurement apparatus” and $C$ is the particle we are interested in measuring: By ignoring the particle $C$, the resulting system looks exactly like a “coin with a sensor”. The two particles $A$ and $B$ are perfectly correlated.

This can be extended to any macroscopic system with a very large number of entangled particles: This is exactly what Feynman was talking about in one of his workshops!

The only reason the macroscopic world seems so deterministic and classical is that many entanglements are being made all of the time in a messy web that is almost impossible to “undo” in practice, but in principle, there is nothing that says this is not possible.

Closing thoughts

If you made it this far then well done, I didn’t realise there was so much content packed into this talk.

My takeaways can be summarised as follows:

• Measurement is the same phenomenon as entanglement
• Quantum effects only present themselves when there is a high degree of uncertainty (e.g. not much entanglement)
• In our daily lives, we have been habituated into becoming very familiar with the consequences of a very high degree of entanglement
• On the flip side, we are very unfamiliar with the consequences of a very low degree of entanglement, but that should not stop us from accepting what really does happen

I’ll finish with another quote from Ron that made a lot of sense to me after learning all of this:

“Spooky action at a distance” is no more (and no less) mysterious than “spooky action across time.” Both are produced by the same physical mechanism.Ron Garret