PHYS 360 Quantum Mechanics
Lecture 11
Quantum Key Distribution
READ Rieffel & Polak Ch. 2
READ Beck Chapter 17.2
Friday Feb 6, 2026
PHYS 360 – Lecture 11:
The quantum mechanics of polarized photons,
and quantum key distribution.
See Chapter 2 of Rieffel & Polak, “Quantum Computing – A Gentle Introduction”
Also, there are real products on the market for this:
Did you take PHYS 340 Modern Physics Lab?
Do you recall the polarization set-up you used before tackling the Quantum
Entanglement Demonstrator?
It turns out this was a trick to get you to do a truly quantum experiment, even
though it looks classical. This forms the basis for Quantum Key Distribution.
Why does the middle polarizer allow light to pass?
Here is the same thing from our PA analyzers:
Here is a result from a Stern-Gerlach experiment. Note the similarity:
Experiment 3:
• If z- and x-directions are independent, then starting with only +z should end up with only +z.
• Experiment shows that the x Spin Analyzer must have changed z-properties, because now we see
both +z and –z in the final beams.
|↑ۧ = the unit vector for vertical polarization
An arbitrary polarization can be expressed as
|→ۧ = the unit vector for horizontal polarization
|ۧ =  |↑ۧ + b |→ۧ
| ۧ = the unit vector for + diagonal polarization
For example,
| ۧ = the unit vector for – diagonal polarization
| ۧ=
1
2
|↑ۧ +
1
2
|→ۧ , a superposition state.
When a photon with polarization |ۧ =  |↑ۧ + b |→ۧ meets a polarizer with preferred axis |↑ۧ, the
photon will be transmitted with probability  2 and will be absorbed with probability  2 .
Furthermore, any photon that passes through the polarizer will now be polarized in the direction of
the polarizer.
• Note the probabilistic nature of the interaction, and
• the resulting change of state.
When Polarizers A and C are inserted, no light reached the screen because all photons transmitted by
A are in the |→ۧ state, which has zero transmission probability by Polarizer C.
See Chapter 2 of Rieffel & Polak, “Quantum Computing – A Gentle Introduction”
Also note, had Polarizer C been in any other orientation, some light would have gotten through.
Polarizer B has the preferred axis | ۧ . Let’s now rewrite the horizontal polarization as:
|→ۧ =
1
2
| ۧ-
1
2
| ۧ.
Hence any photon transmitted by Polarizer A has an amplitude
1
2
in the + diagonal direction and
will be transmitted with probability ½.
Photons transmitted by Polarizer B are in the state | ۧ =
1
2
|↑ۧ +
transmission by Polarizer C is ½.
Summary:
• Polarization is modelled by a unit vector.
• Interaction is probabilistic and
• depends on amplitude in the polarizer direction.
• Either photon is absorbed or is transmitted with polarization
aligned with the polarizer.
These are axioms of quantum mechanics, not derivable from
physical principles, but observed in experiments.
See Chapter 2 of Rieffel & Polak, “Quantum Computing – A Gentle Introduction”
1
2
|→ۧ , so the probability of
What is a Quantum Bit?
• The space of all possible polarization states is an example of a qubit.
• A qubit has a continuum of possible values.
• The set of all possible states of a physical system is called the space state.
• Any quantum mechanical system that can be modelled by a two-dimensional complex
vector space can be viewed as a qubit.
• This means all possible states can be represented as a superposition of just two states.
• These two states are the orthonormal basis vectors.
• BUT the same space can be described by more than one set of basis vectors:
ۧ = 1 1 ۧ + 2 |2 ۧ = 1 |1 ۧ + 2 |2 ۧ.
For example:
ۧ = 1 ↑ۧ + 2 |→ۧ = 1 | ۧ + 2 | ۧ.
In general, we write: ۧ = 1 0ۧ + 2 |ۧ.
See Chapter 2 of Rieffel & Polak, “Quantum Computing – A Gentle Introduction”
A single qubit has an infinite number of possible values, whereas a classical
bit only has two. Does that mean a single qubit can store lots of classical
information?
No.
Information is obtained by measurement, and this means only one of two
possible basis states is observed: a single measurement at most yields a
single classical bit of information.
Also, because measurement changes the states, one cannot make two
measurements of the original state of the qubit.
See Chapter 2 of Rieffel & Polak, “Quantum Computing – A Gentle Introduction”
A Quantum Key Distribution Protocol (BB84).
The goal is to share a random sequence (0010110100110...)
between Alice and Bob, the key to reading encoded messages.
Photon polarization is used. Alice sends photon, Bob receives
them, but it is possible for eaves-dropper Eve to intercept,
read, and re-broadcast them to Bob. Quantum mechanics
allows to determine if eaves-dropping has occurred.
Using the quantum channel, Alice sends single photons
with one of these polarization states:
0 = |↑ۧ
 = | → ۧ.
When Bob receives them, he uses a polarimeter with two
detector, D0 and D1, which tells him if the photon is in
polarization state 0 or 1. Under ideal conditions, Bob will
record the exact string (0010110100110...) sent by Alice.
See Chapter 2 of Rieffel & Polak, “Quantum Computing – A Gentle Introduction”
But this is highly vulnerable to eavesdropping,
especially if Eve correctly guesses Alice’s basis states.
Alice uses a polarizing device set to either Horizontal or Vertical to fix each photon’s polarization.
Now assume that she can choose to use the diagonal basis set:
1
1
0 = | ۧ= 2 |↑ۧ + 2 |→ۧ
1
 = | ۧ 2 |↑ۧ −
1
2
|→ۧ.
Now the photons are in a superposition of Horizontal & Vertical states but are pure diagonal
(Hadamard) states.
What happens if Bob measures these with his Horizontal/Vertical polarizers? Each of these
photons will randomly be measured as either 0 or 1.
What if Bob measures these after rotating his polarimeter detection system by 45 degrees? Then
he will record the exact sequence sent by Alice (0010110100110...).
OK, now let Alice transmit a string of photons with random
polarizations (0010110100110...), but NOW she also randomly
switches from the standard basis to the Hadamard basis.
Bob knows this is happening, but the choice of basis for each
photon is kept secret from him. To maximize security, he also
randomly switches between standard and Hadamard bases.
What happens when their bases agree? Then Bob always
detects the state (0 or 1) sent by Alice.
Standard basis:
What happens when their bases disagree? Then Bob is
randomly right about half the time, wrong the other half.
0 = |↑ۧ
 = |→ ۧ
Hadamard basis:
1
1
0 = | ۧ= |↑ۧ + |→ۧ
2
=| ۧ=
1
2
2
|↑ۧ −
1
2
| → ۧ.
Standard basis:
0 = |↑ۧ
 = |→ۧ
Hadamard basis:
1
1
0 = | ۧ= |↑ۧ + |→ۧ
2
=| ۧ=
1
2
2
|↑ۧ −
1
2
|→ۧ.
After Bob has recorded his string of numbers, Alice communicates with Bob using a
public, classical channel (telephone, snail mail, Facebook DM, or other insecure options).
She tells Bob her basis settings for each of her photons. (She does NOT give the values of
the photon polarizations!) Bob determines which bits they used the same basis and
throws away all the other bits. They now have a shared string of random bits that can be
used as a cryptography key.
But what about Eve, the eavesdropper?
Assume that Eve can intercept Alice’s string of bits (photon polarizations). She knows the basis
sets might be randomized, so she randomizes her polarimeter settings for each photon also.
She records the values of each bit, and the basis used.
She then sends to Bob an identical stream of bits, using the same basis set from her own
measurement. Bob measures these bits with his own randomized sequence of basis settings,
as usual.
What happens when Bob get’s Alice’s data on her basis set settings? He expects that half the
time they will agree, but now he gets a different result.
Since Eve will be wrong half the time in her choice of basis set, that means that half of the
photons sent on to Bob were measured incorrectly. This includes half of the bits where Alice
and Bob used the same basis.
Bob is expecting 50% of his bits to agree with Alice. He shares a subset of his ”good” bits with
Alice over the public channel. (These bits will be discarded, since they were sent over an
insecure channel.) These should agree 100% of the time. If they do not, they should suspect
an eaves-dropper, discard all the bits, and start over.
But wait, why doesn’t Eve just make a copy of Alice’s bit stream and send
one copy on to Bob while she analyzes her own copy?
This is prevented by the quantum “no-cloning” theorem, to be discussed
later. (Or see Beck Ch 17.3.)
Quantum Key Distribution using BB84
NOTE: This problem was readily done using Excel. Feel free to use any other programming medium.
A. Generate a string of one hundred random 0’s and 1’s. (In Excel, this would put one digit (0 or 1) in one hundred consecutive
cells in one row.) Label these “Alice’s bits”.
B. Generate a string of one hundred S’s and H’s, corresponding to a random selection of the Standard basis or the Hadamard
basis for each of the photons that Alice sent. Label these “Alice’s bases”. (This would be the second row of the Excel file.)
C. Now Bob measures the polarization of these photons. Generate a different string of one hundred S’s and H’s, corresponding
to a random selection of the Standard basis or the Hadamard basis for each one of Bob’s measurements.
D. Now compute the string of bits (0’s and 1’s) that Bob might measure. (Some of these measurements might be probabilistic,
so there is not one unique answer.)
E. After Alice tells Bob her string of S’s and H’s and Bob discards the bits where they did not agree with his choice, what is the
resulting random quantum key? How many bits long is it? Does this agree with what Alice sent?
F. Next, the eavesdropper Eve intercepts Alice’s bits, reads them with her own random set of bases, and retransmits them in
those same bases to an unsuspecting Bob. When he gets Alice’s bases and discards the bits where their bases did not match,
how long is his key? How good is his agreement with what Alice sent? Can he tell that there was an eavesdropper?
BB84 Quantum Key Distribution Homework Problem: