On teleportation: A thought experiment

I would be glad to know your Lordship's opinion whether when my brain has lost its original structure, and when some hundred years after the same materials are fabricated so curiously as to become an intelligent being, whether, I say that being will be me; or, if, two or three such beings should be formed out of my brain; whether they will all be me, and consequently one and the same intelligent being.

— Thomas Reid letter to Lord Kames, 1775

If only nature could provide some way to distinguish between identical and the same then one could answer Thomas Reid's question.  If a reconstructed Thomas Reid were the same being then that being would be him; if it were an identical being then it would merely be a person with the same memories, but without any continuity of experience linking it to the original Thomas Reid.  Incredibly, it turns out that nature does provide a way!

A well known feature of modern physics is that if you perform the same experiment twice you do not get the same results.  This isn't particularly obvious with experiments on a human scale like dropping lead weights from the leaning tower of Pisa, but on the atomic scale it is significant.  Einstein railed against this idea with his famous God does not play dice remark, believing instead that there must be some hidden variables.  This is another way of saying that there must be small but significant differences in how the experiments start, rather than a fundamental randomness involved in choosing the outcome.  In the end Einstein was hoisted by his own petard when Bell demonstrated that the accepted mathematics of quantum theory permitted either hidden variables, or Einstein's own postulate of no interaction faster than light, but not both!

If we cannot predict the outcome of experiments, then what can we do?  It turns out the answer is: calculate the odds.  And Feynman came up with an elegant way of illustrating how we do this.

The Feynman diagram above illustrates an experiment where you start with an electron and a positron (left) and some time later you end up with an electron and a positron (right) but with locations swapped.  (Note that by convention arrows are drawn backwards for positrons so that the positron at the top left actually moves towards the centre where it collides with the electron.)  The diagram shows the two particles annihilating, resulting in a photon, which eventually decays back into an electron and a positron.  This diagram is associated with a complex number called the probability amplitude for the diagram.  You calculate this probability amplitude by multiplying together probability amplitudes for all the individual parts.  There are seven of these, each of which can be calculated using a simple formula involving some physical constants:

1. electron moves from the bottom to the middle in the time shown
2. positron moves from the top to the middle in the time shown
3. electron and positron annihilate emitting a photon
4. photon goes from place shown to place shown
5. photon decays into positron and electron
6. electron goes from middle to top in time shown
7. positron goes from middle to bottom in time shown

 So how do you go from the probability amplitude thus calculated to the probability for the outcome.  The answer is almost just square the amplitude (so that a probability amplitude of $\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i$ would give a probability of $1$).  Almost, but not quite.  The subtlety is that there are multiple ways in which the starting condition shown above can evolve into the end condition and we can draw a Feynman diagram for each of these.  The diagrams below demonstrate two more:

Each Feynman diagram consistent with the starting conditions and the end conditions generates its own probability amplitude and these are added together to produce the overall probability amplitude.  It is this overall probability amplitude whose amplitude must be squared to obtain the probability - a number between zero and one which tells us how likely it is to obtain the outcome given the initial state.  It is important to note that since probability amplitudes have phase as well as magnitude it is possible to end up with a probability of zero despite the fact that all the components are non-zero - this is known as destructive interference.

So the rule is: sum then square.  But what if we are interested in the probability of multiple different outcomes.  So for example, if you throw two dice and you want to know the probability of getting a total of 3 then there are really two categories of outcome.  Either you get a 1 on the first die and 2 on the second or you get a 2 on the first and a 1 on the second.  You cannot really describe these as the same outcome as the dice are not identical.  In all likelihood there will be a nick or some mark which - if you cared enough - you could use to determine which was the first die and which was the second.  So in this case we are not talking about one outcome but many.  The rule now is: calculate the overall probability for each of the many outcomes and add them together.  Note that we are now adding probabilities - squares of magnitudes of probability amplitudes - and not the probability amplitudes themselves, so there is no way for the individual outcomes to interfere destructively.  In short, if there are multiple different outcomes the rule is: square then sum.

Now, here's an interesting question: if you permute a set of particles do you get a different (but identical) state, or do you get the same state?  Permutation means simply that you move the fundamental particles around so that you end up with particles in different locations to where they began, but with the same set of positions (spins, polarisations, &c.) for each type of particle.  Well, how on Earth could one possibly answer that question? It isn't even clear whether or not the question is meaningful!  But it is, and you can...!

Suppose that permuting particles led to a different state.  Then the two diagrams below would describe different outcomes of the same experiment.  The first outcome is 'electron "Bob" at A and electron "Alice" at B', and the second outcome is 'electron "Bob" at B and electron "Alice" at A'.  As we are discussing different outcomes we would have to use the square then sum rule to work out the probability of getting one electron at A and one electron at B, given the starting condition.

But what if permuting particles led to the same state?  In that case
we would have to use the sum then square rule to work out the probability of getting the (sole) outcome.  This is great because the two different ways of calculating the probability give different results.  This means it is possible to perform experiments, make measurements, and determine once and for all: does permuting the particles lead to a different (but identical) state or does it lead to the same state?  And this has been done... and the answer is: Permuting the particles results in the same state!

Now here is a thought experiment, based on Derek Parfit's "teletransporter".  Suppose we built a pair of doors in different locations, and as you walk through the first door you are scanned, and then - simultaneously - the particles making up your body are taken apart and dumped on a slag heap, and at the other location a heap of particles is used to reconstruct it.  At your original location others will see you disappear into a puff of smoke and at the other location you will be seen to appear.  But is the being that walks out of the teletransporter you?  It will certainly look like you, and have the same memories as you, but will there be a continuity of experience linking you and it?

To answer this question it is helpful to imagine what happens if the machine goes wrong so that although the first door still scans, and the second door still reconstructs, the first door fails to disassemble!  There will then be two of you and it is clear that these will represent two experiential lineages.  And what if the first door does disassemble you, but after a delay of 1 minute?  It is quite clear then, that you will die, and that thing the other side of the world, that continues as if you, is in fact something else.

To see what this has to do with the Feynman diagrams we need to take Parfit's thought experiment one step further.  Suppose you physically merge the two doors so that you are destroyed and recreated (from new particles) at the same time and in the same place.  Philosophically this isn't very different, and our reasoning above would lead us to conclude that you do indeed die, and only appear to others to continue living - it is in fact someone else who walks out the other side.  But now let's make the door randomly choose to either a) do nothing, or b) destroy and recreate you as you walk through.  And critically, let's make the door throw away the evidence so that after the fact it is impossible to determine which of the two options it chose.  Then, at random, you either die (and someone different but identical is created), or you live.  But these two outcomes are mere permutations of each other, in terms of the fundamental particles.  And as we have seen, permuting particles gives the same state, not merely an identical state.  So living from one moment to the next is no different to dying and being rebuilt, and this tells us that the continuity of experience we all feel is just an illusion existing in the moment.

POSTSCRIPT

I have just come across the following quote, which seems apposite

No man ever steps in the same stream twice, for it is not the same stream, and he is not the same man

- Heraclitus