Quantum State Transformations

[Genady]

(From: Rieffel, Eleanor G.; Polak, Wolfgang H.. Quantum Computing: A Gentle Introduction.)

Take v in S₁ and w in S₂: 〈v|w〉 = 0.

〈Uv|Uw〉 = 〈v|U^†U|w〉 = 〈v|w〉 = 0.

Too easy.

[Genady]

a. U =

1	0	0	0
0	1	0	0
0	0	0	1
0	0	1	0

 

b. Same as a but in the {|+〉, |-〉} basis.

c. If such U existed, it would transform U(|00〉) = |00〉 and U(|10〉) = |11〉.

Then U(|+〉|0〉) = U(1/√2 (|0〉 + |1〉)|0〉) = 1/√2 (|00〉 + |11〉) ≠ |++〉, i.e. |+〉 would not be cloned.

 

 

Edited October 16, 2023 by Genady

[Genady]

Continuing exercise 5.2 above.

d. Construct the operator as follows:

U = 

/* required cloning */

|0+ 0+〉 〈0+ 0+| + |0- 0-〉 〈0- 0+| + |1+ 1+〉 〈1+ 0+| + |1- 1-〉 〈1- 0+| +

/* transpose */

                              |0- 0+〉 〈0- 0-| + |1+ 0+〉 〈1+ 1+| + |1- 0+〉 〈1- 1-| +

/* diagonal */

                             |0+ 0-〉 〈0+ 0-| + |0- 0+〉 〈0- 0+| + |0- 0-〉 〈0- 0-| + 

|0+ 1+〉 〈0+ 1+| + |0+ 1-〉 〈0+ 1-| + |0- 1+〉 〈0- 1+| + |0- 1-〉 〈0- 1-| +

|1+ 0+〉 〈1+ 0+| + |1+ 0-〉 〈1+ 0-| + |1- 0+〉 〈1- 0+| + |1- 0-〉 〈1- 0-| + 

|1+ 1+〉 〈1+ 1+| + |1+ 1-〉 〈1+ 1-| + |1- 1+〉 〈1- 1+| + |1- 1-〉 〈1- 1-|

 

e. This set includes c, thus impossible to clone.

[Genady]

I've made errors in the 'diagonal' part of the construction for the question 'd' above. One needs to be careful not to include diagonal terms in rows or columns which already have terms in them. Thus, the correction:

/* diagonal */

                             |0+ 0-〉 〈0+ 0-| +                            

|0+ 1+〉 〈0+ 1+| + |0+ 1-〉 〈0+ 1-| + |0- 1+〉 〈0- 1+| + |0- 1-〉 〈0- 1-| +

                             |1+ 0-〉 〈1+ 0-| +                             |1- 0-〉 〈1- 0-| + 

                             |1+ 1-〉 〈1+ 1-| + |1- 1+〉 〈1- 1+|

[Genady]

(For the reference, the BB84 is described here: https://www.scienceforums.net/topic/132653-quantum-key-distribution/)

I assume that Eve always measures her qubit in the standard basis. Let's see what happens:

1. Alice's bit is 0.

1.1. Alice and Bob pick the standard basis. She sends |0〉. Eve applies C_not: |0〉|0〉 ↦ |0〉|0〉.

         Bob gets 0. Correct. Eve gets 0. Correct.

1.2. Alice and Bob pick the Hadamard basis. Alise sends 1/√2(|0〉+|1〉). Eve applies C_not: 1/√2(|0〉+|1〉)|0〉 ↦ 1/√2(|0〉|0〉+|1〉|1〉). 

1.2.1. Bob gets 0. Correct. Eve gets 0. Correct.

1.2.2. Bob gets 0. Correct. Eve gets 1. Error.

1.2.3. Bob gets 1. Error. Eve gets 0. Correct.

1.2.4. Bob gets 1. Error. Eve gets 1. Error.

2. Alice's bit is 1.

2.1. Alice and Bob pick the standard basis. She sends |1〉. Eve applies C_not: |1〉|0〉 ↦ |1〉|1〉.

         Bob gets 1. Correct. Eve gets 1. Correct.

2.2. Alice and Bob pick the Hadamard basis. Alise sends 1/√2(|0〉-|1〉). Eve applies C_not: 1/√2(|0〉-|1〉)|0〉 ↦ 1/√2(|0〉|0〉-|1〉|1〉). 

2.2.1. Bob gets 0. Error. Eve gets 0. Error.

1.2.2. Bob gets 0. Error. Eve gets 1. Correct.

1.2.3. Bob gets 1. Correct. Eve gets 0. Error.

1.2.4. Bob gets 1. Correct. Eve gets 1. Correct.

 

Looks like Eve gets 75% of Alice's bits correctly and introduces 25% errors in the bits measured by Bob, the same as in the original scheme.

Am I missing something?

 

[Genady]

For the reference:

K(0) = T(0) = R(0) = I. Thus, K(0)T(0)R(0)T(0) = I.

T(π/2) =  i *

1	0
0	-1

R(π/2) = 

0	1
-1	0

T(π/2)R(π/2)T(0) = iX. X = -iT(π/2)R(π/2)T(0).

R(π/4) = 1/√2*

1	1
-1	1

T(π/2)R(π/4)T(0) = iH. H = -iT(π/2)R(π/4)T(0).

 

 

[Genady]

 

a. 

b. 

c. 

d. 

e. 

f.  

[Genady]

An n-qubit cat state is the state 1/√2 (|00 . . . 0〉 + |11 . . . 1〉). Design a circuit that, upon input of |00 . . . 0〉, constructs a cat state.

========

My idea is first to transform the first qubit to the superposition using the Hadamard gate and then to propagate the result to the rest of the qubits with the Cnot gates:

 

 

 

 

 

[TheVat]

I think you accidentally produced a wormhole that moved the site forward in time one week.