Quantum Cryptography Experiment
Aim:
Establish a shared key between Alice and Bob, and show how Alice and Bob can detect Eve, the eveasdropper.
Materials:
2 pencil cases
6 coins
Procedure:
1. Form a group of 4 people. Designate one person as Alice, one person as Bob, one person as Eve and one person as recorder. Designate one pencil case as H/V and the other pencil case as D/A.
Recorder:
2. Record the basis each person measures in and the result of measurement. Responsible for ensuring each member adheres to instructions.
Alice:
3. Toss a coin. The result tells you what basis to encode the qubit in (heads: H/V, tails: D/A).
4. Toss a coin. The result tells you what state to encode the qubit in (heads: 0, tails: 1).
5. Follow the following instruction for placing coins in the pencil cases:

Recorder:
6. Pass box to Eve/Bob.
Eve (Every 4th Trial):
7. Make a measurement. Toss a coin to decide what basis to measure in (heads: H/V, tails: D/A). The recorder allows Eve to look in the box that corresponds with measurement and record the value. The other box is then shaken to randomize bit.
Recorder:
8. Pass box to Bob.
Bob:
9. Make a measurement. Toss a coin to decide what basis to measure in (heads: H/V, tails: D/A). The recorder allows Bob to look in the box that corresponds with measurement and record the value.
10. Repeat step 2-8 twenty five times.
Refine the Secret Key:
11. Elimate any results where Bob and Alice measure in different basis.
12. To check for errors due to Eve, or someother sorce, such as interference, conduct a parity check.
Conducting a Parity Check: (The parity of a string of bits is simply the XOR of the bits in the string)
13. Divide the remaining bits of the key into subset string of length, l, so the chance of more than one error in a subset is less than one.
14. Alice and Bob both announce the result of the XOR on each subset. If the result of the parity check is the same for both Alice and Bob, they will both discard the last bit of the subset. The remaining bits will be added to the new key. If the result of the parity check is different for both Alice and Bob, they will both discard the entire subset.
15. The key that comes as a result of the parity check is the new secret key.
Calculations:
Raw Data
Alice's Raw Key: 0111000111111010011001011
Bob's Raw Key: 0111010111110010011010011
Key With Different Basis Removed
Alice's Key #1: 1110011101011001
Bob's Key #2: 1110011101011101
Parity Check
Length = 4
Figure 3: Parity Check
| Alice | Bob |
| Subset 1: 1110 Parity Check Result: 1 |
Subset 1: 1110 Parity Check Result: 1 |
| Subset 2: 0111 Parity Check Result: 1 |
Subset 2: 0111 Parity Check Result: 1 |
| Subset 3: 0101 Parity Check Result: 0 |
Subset 3: 0101 Parity Check Result: 0 |
| Subset 4: 1001 Parity Check Result: 0 |
Subset 4: 1101 Parity Check Result: 1 |
Figure 4: Parity Check Result
| Alice | Bob |
| Parity Check 1 Result: Same New Subset: 111 |
Parity Check 1 Result: Same New Subset: 111 |
| Parity Check 2 Result: Same New Subset: 011 |
Parity Check 2 Result: Same New Subset: 011 |
| Parity Check 3 Result: Same New Subset: 010 |
Parity Check 3 Result: Same New Subset: 010 |
| Parity Check 4 Result: Different New Subset: n/a |
Parity Check 4 Result: Different New Subset: n/a |
Final Key:
Alice's Final Key: 111011010
Bob's Final Key: 111011010
Observations:
Figure 1: Initial Observation Table
|
Alice |
Eve |
Bob |
|||
|
Basis |
Result |
Basis |
Result |
Basis |
Result |
|
D/A |
0 |
D/A |
0 |
H/V |
0 |
|
D/A |
1 |
|
|
D/A |
1 |
|
D/A |
1 |
|
|
D/A |
1 |
|
D/A |
1 |
|
|
D/A |
1 |
|
H/V |
0 |
H/V |
0 |
H/V |
0 |
|
H/V |
0 |
|
|
D/A |
1 |
|
D/A |
0 |
|
|
D/A |
0 |
|
H/V |
1 |
|
|
H/V |
1 |
|
H/V |
1 |
D/A |
1 |
D/A |
1 |
|
H/V |
1 |
|
|
H/V |
1 |
|
H/V |
1 |
|
|
D/A |
1 |
|
H/V |
1 |
|
|
H/V |
1 |
|
D/V |
1 |
H/A |
0 |
H/V |
0 |
|
H/V |
0 |
|
|
H/V |
0 |
|
H/V |
1 |
|
|
H/V |
1 |
|
D/A |
0 |
|
|
D/A |
0 |
|
H/V |
0 |
H/V |
0 |
D/A |
0 |
|
H/V |
1 |
|
|
H/V |
1 |
|
D/A |
1 |
|
|
D/A |
1 |
|
D/A |
0 |
|
|
H/V |
0 |
|
D/A |
0 |
H/V |
1 |
D/A |
1 |
|
D/A |
1 |
|
|
H/V |
0 |
|
D/A |
0 |
|
|
D/A |
0 |
|
D/A |
1 |
|
|
H/V |
1 |
|
D/A |
1 |
D/A |
1 |
D/A |
1 |
Figure 2: Observation Table with Different Basis Eliminated
|
Alice |
Eve |
Bob |
|||
|
Basis |
Result |
Basis |
Result |
Basis |
Result |
|
|
|
|
|
|
|
|
D/A |
1 |
|
|
D/A |
1 |
|
D/A |
1 |
|
|
D/A |
1 |
|
D/A |
1 |
|
|
D/A |
1 |
|
H/V |
0 |
H/V |
0 |
H/V |
0 |
|
|
|
|
|
|
|
|
D/A |
0 |
|
|
D/A |
0 |
|
H/V |
1 |
|
|
H/V |
1 |
|
|
|
|
|
|
|
|
H/V |
1 |
|
|
H/V |
1 |
|
|
|
|
|
|
|
|
H/V |
1 |
|
|
H/V |
1 |
|
|
|
|
|
|
|
|
H/V |
0 |
|
|
H/V |
0 |
|
H/V |
1 |
|
|
H/V |
1 |
|
D/A |
0 |
|
|
D/A |
0 |
|
|
|
|
|
|
|
|
H/V |
1 |
|
|
H/V |
1 |
|
D/A |
1 |
|
|
D/A |
1 |
|
|
|
|
|
|
|
|
D/A |
0 |
H/V |
1 |
D/A |
1 |
|
|
|
|
|
|
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|
D/A |
0 |
|
|
D/A |
0 |
|
|
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|
|
|
|
|
D/A |
1 |
D/A |
1 |
D/A |
1 |