# Quantum Intro (Was: Re: Universal non-point of light)

Eliezer Yudkowsky (sentience@pobox.com)
Wed, 05 Feb 1997 21:42:36 -0600

> If you don't need to hit it with a photon, then you aren't restricted b=
y
> the speed of light are you?

Sure you do. The way it works is that you "could have" hit it with a
photon, but you "didn't really" hit it with a photon. For those readers
who do not savvy quantum counterfactuals, the five-minute explanation of
the basic counterfactual split-photon experiment:

1 =

| Arrows indicate direction. =

A/--------/---2 A single photon leaves START,
| D| and is reflected from mirrors. =

|C | What is the chance that the photon
START---/--------/B will reach target 1, or target 2? =

There are four mirrors: {A, B, C, D}. {A, B} are "fully silvered"
mirrors: They always reflect a photon. {C, D} are "half-silvered"
mirrors: They reflect a photon 50% of the time.

Under "classical" physics, the sort we're used to seeing, the photon has
a 50% chance of being reflected from C. If reflected from C, it will
bounce off A and thus reach D. Now we roll the dice again: It has a
50% chance of being reflected again and reaching 1, and a 50% chance of
reaching 2. Thus, since the photon only has a 50% chance of going down
path A (the path intercepted by mirror A), the total contribution of
path a is a 25% chance of reaching 1 and a 25% chance of reaching 2.

Path B is likewise. The photon has a 50% chance of going down path B. =

Then it reaches mirror D and has a 50% chance of being reflected and
reaching 2, and a 50% chance of continuing to 1. So path B contributes
a 25% chance of reaching 1 and a 25% chance of reaching 2.

Summing up the contributions, we find that the chance of reaching 1 is
(25% + 25% =3D 50%), and that of reaching 2 is (25% + 25% =3D 50%). Thus=
,
under classical physics, a photon has an equal chance of reaching A or
B.

Quantum physics is different, and weirder. Probabilities in quantum
physics are measured by complex numbers, of the form (a + bi). Now we
perform the calculation again, but this time, mirrors not only change
the *path* of the photon - assuming they reflect it - but they also
change the *phase* of the photon by 90=B0 or a quarter-turn. In practice=
,
they multiply the probability of the photon being in a particular place
by i, the square root of negative one. What this all means will soon
become clear.

Now recalculate the probabilities. A photon is emitted from START, with
a probability of 1, or 1 + 0i, of being there. It hits mirror C. Now
the photon has (1 + 0i)*(i)=3D(0 + i) or i "probability" of going down
path A, and a 1 "probability" of going down path B. In the branch of
reality in which the photon goes down path A, the photon hits the mirror
A and reflects again, so the photon now has an (i)*(i)=3D(-1) quantum
"probability" of getting to D. The photon hits D and "splits" again,
part of it getting to target 2 (without being reflected) with a
"probability" of -1, and part hitting target 1 with a "probability" of
-i. So far, therefore, the photon has a -1 "probability" of being at
target 2 and a -i "probability" of being at target 1.

Interlude: The reason "probability" has quotes around it is because
there is no such thing of a -50% chance of something happening. In
quantum physics these are called probability amplitudes. We will soon
see how probability amplitudes are converted to genuine, normal,
positive, real, probabilities. But for now, don't be concerned by
probabilities like -1.

Now let's continue with the branch of reality in which the photon goes
down path B, with a probability of 1. It gets reflected by B, and
continues with a probability of i. It gets to D. Along one branch, it
is not reflected and gets to target 1 with a probability of i. Along
another branch, it is reflected and gets to target 2 with a probability
of (i)*(i)=3D(-1). So for branch B, the contributions are an i
probability of getting to target 1 and a -1 probability of getting to
target 2.

Now we sum the branches of reality, just like with normal
probabilities. Branch A gives a -i probability of getting to target 1,
and branch B gives an i probability. i + -i equals 0. They cancel
out. There is probability amplitude zero of getting to target 1. =

Branch A and branch B both give -1 probability amplitude of the photon
getting to target 2. -1 + -1 =3D -2, for a -2 probability amplitude of
the photon getting to 2.

Now convert probability amplitudes to probabilities. This we do by
taking the "squared modulus", which for a complex number (a + bi), is
equal to a squared plus b squared, or (a * a) + (b * b). The chance of
getting to target 1 is 0, or zero squared plus zero squared. The chance
of getting to target 2 is 4, or -2 squared plus zero squared. 4 chances
of getting to target 2, 0 chances of getting to target 1, so there's a
100% chance of getting to target 2.

That is, quantum physics says that instead of a 50% chance of getting to
either target, every single photon "splits" into two separate branches
of reality, goes down *both* paths, *interferes* with *itself*, and
winds up at target 2.

But suppose we remove mirror C, so that all photons go down path B? Or
suppose we place a rock in the middle of path A? Then the two paths no
longer cancel out. There's an i probability of getting to target 1, for
a (1*1) or 1 chance of getting there. There's a -1 probability of
getting to target 2, or (-1*-1)=3D1 chance of getting there. A 50% chanc=
e
of getting to either target.

What's worse, just *measuring* the photon will influence the probability
that it gets to its target - even if it doesn't deflect the photon by so
much as an inch. Because when you measure a system, all probability
amplitudes are *immediately* converted to probabilities and the system
"collapses" or is "reduced" to a single branch of reality. The photon
is suddenly in only one place in only one branch of reality. This is
known as "the collapse of the wavefunction" or "state-vector reduction".

So if we replace mirror A with a "measuring device" that reflects the
photon as usual, but also lights up a little device that says the photon
has passed, the photon has a 50% chance of reaching target 1 or 2, just
like in classical physics. Because when mirror A measures the system -
when hit by a photon in the branch of reality where the photon goes down
path A - the system is "reduced". The photon has an i probability of
being on path A, and a 1 probability of being on path B. Squared
amplitudes - 1 chance A, 1 chance B. So the photon will wind up on
either A or B with a 50% chance - but not on both A and B with i and 1
probability. There are no longer two branches of reality. The photon
does split again when it reaches target 1 or target 2, but it has a -1
probability for 1 and an i probability for 2, which works out to 50/50
again.

As you'll recall from way back in the beginning, you can use this trick
to observe an object without touching it. Take the Elitzur-Vaidman
bomb-testing problem. We have been supplied with a large number of
bombs, with detonators so sensitive that a single photon reflecting from
one will set off the bomb. The problem is that most of our bombs are
duds. The detonating mechanism doesn't work. And it's impossible to
test the detonating mechanism without reflecting a photon from it. =

Question: How do you produce a bomb that is guaranteed to work...
without blowing yourself up?

Answer: This problem, as presented, is *impossible* in classical
physics. But quantum physics can provide a solution. =

Replace mirror A... with a bomb. Since the bomb constitutes a
"measuring instrument" - that is, it blows up if a photon strikes it,
but doesn't blow up otherwise - the photon now has a chance of
reflecting off B, and winding up at target 1. (And an equal chance of
reaching target 2... and of course it's only 50% that it reaches
*either* target instead of blowing up the bomb.)

Thus we can determine whether the bomb is a measuring instrument
*without* *even* *touching* *it*! However, although you can measure
something without a photon having touched it in that branch of reality
which turns out to be real, it still has to be possible that the photon
*could* have touched it, or rather it has to be the case that the photon
*did* touch it in an alternate branch of reality. Thus this trick does
not, Mike Lorrey, constitute FTL.

On the other hand, if you have a photon that is 'locked' to the one
inside the system, but in another galaxy, measuring the Andromeda photon
will cause the photon in the experimental apparatus to collapse as
well. Once collapsed, it has a chance of reaching target 1, which it
did not have before. Thus if you can achieve this sort of 'lock',
which, as I understand, you can - using the recent discovery as reported
in SciAm, you can synchronize the quantum states of large numbers of
photons - you have instantaneous communication, somebody wins a Nobel
Prize, and all hell breaks loose in the field of physics.

Even if you can't 'lock' photons, the experimentally confirmed Bell
inequalities demonstrate that performing a *particlar* measurement can
influence the outcome of another measurement performed on the other side
of the galaxy. There are various reasons why you can't use Bell to
transmit information, but any way you look it at, quantum mechanics and
relativity are having screaming catfights.

I hope that cleared everything up.
-- =

sentience@pobox.com Eliezer S. Yudkowsky
http://tezcat.com/~eliezer/singularity.html
http://tezcat.com/~eliezer/algernon.html
Disclaimer: Unless otherwise specified, I'm not telling you
everything I think I know.