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Week 5: MULTI-QUBIT QUANTUM SYSTEMS | ECE 802-730
ECE 802-730 Quantum
Week 5: MULTI-QUBIT QUANTUM SYSTEMS | ECE 802-730
ECE 802-730 Quantum

Week 5: MULTI-QUBIT QUANTUM SYSTEMS | ECE 802-730

Coursework focused on Chapter 3 of Quantum Computing: A Gentle Introduction.

Exercise 1

Show that the state:

|W_n\rangle = \frac{1}{\sqrt{2}}((|0…001\rangle + |0…010\rangle + |0…100\rangle + ··· + |1…000\rangle)

is entangled, with respect to the decomposition into the n qubits, for every n > 1.

|W_n\rangle = a_{1}|0\rangle + b_{1}|1\rangle \otimes a_{2}|0\rangle + b_{2}|1\rangle \otimes … \otimes a_{n}|0\rangle + b_{n}|1\rangle,
a_{1} … a_{n-1}b_{1}|1\rangle = a_{1} … b_{n-1}a_{n} = … = b_{1} … a_{n-1}a_{n} = \frac{1}{sqrt{n}}

∴ b_i \neq 0 for every element of i in the set [1, n]

AND a_j \neq 0 for every element of j in the set [1, n]

Since the coefficients of all states that don’t appear in the tensor products of |W_{n}\rangle must be zero,
then there is a contradiction of values in the state – hence entanglement.

Exercise 2

Is the state \frac{1}{\sqrt{2}}(|0|+\rangle + |1|-\rangle) entangled?

\frac{1}{\sqrt{2}}(|0|+\rangle + |1|-\rangle) = \frac{1}{\sqrt{2}} (|00\rangle + |01\rangle + (|10\rangle - |01\rangle)

IF

\frac{1}{\sqrt{2}}(|00\rangle + |01\rangle + (|10\rangle - |01\rangle) = (a_{1}|0\rangle + b_{1}|1\rangle \otimes (a_{1}|0\rangle + b_{1}|1\rangle

THEN

a_{1}a_{2} = \frac{1}{2}

a_{1}b_{2} = \frac{1}{2}

a_{2}b_{2} = \frac{1}{2}

b_{1}b_{2} = -\frac{1}{2}

Since this cannot be due to the necessity of opposite signs for sets a_{1}a_{2} and b_{1}b_{2}. Since this is a contradiction the state of \frac{1}{\sqrt{2}}(|0|+\rangle + |1|-\rangle) is entangled.

Exercise 3

Write the following states in terms of the Bell basis.
The Bell basis for a 2-qubit system is: ({|\Phi^{+}\rangle, |\Phi^{−}\rangle, |\Psi^{+}\rangle, |\Psi^{−}\rangle)

a. |00\rangle

= \frac{1}{\sqrt{2}}(|\Phi^{+}\rangle + |\Phi^{-}\rangle)

b. |+\rangle|-\rangle

= \frac{1}{2}(|00\rangle - |01\rangle + |10\rangle - |11\rangle) = \frac{1}{\sqrt{2}}(|\Phi^{-}\rangle - |\Psi^{-}\rangle)

c. \frac{1}{\sqrt{3}}(|00\rangle + |01\rangle + |10\rangle)

= \frac{1}{\sqrt{6}}|\Phi^{+}\rangle + \frac{1}{\sqrt{6}}|\Phi^{-}\rangle \sqrt{\frac{2}{3}}|\Psi^{+}\rangle

Exercise 4

Give an example of a two-qubit state that is a superposition with respect to the standard basis but that is not entangled.

IF

|\Psi\rangle = \frac{1}{2}(|00\rangle + |01\rangle + |10\rangle + |11\rangle)

THEN

|\Psi\rangle = (\frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle) \otimes (\frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle)

Exercise 5

a. Show that the four-qubit state |\Psi\rangle = \frac{1}{2}(|00\rangle + |11\rangle + |22\rangle + |33\rangle) of example 3.2.3 is entangled with respect to the decomposition into two two-qubit subsystems consisting of the first and second qubits and the third and fourth qubits.

|\Psi\rangle = (a_{0}|0\rangle + a_{1}|1\rangle + a_{2}|2\rangle + a_{3|}3\rangle) \otimes (b_{0}|0\rangle + b_{1}|1\rangle + b_{2}|2\rangle + b_{3}|3\rangle)

IF values of a_{0} and b_{0} are both respectively non-zero (must equate \frac{1}{2})

AND a_{0} & b_{0} = 0

THEN this is a logical impossibility and therefore entangled.

b. For the four decompositions into two subsystems consisting of one and three qubits, say whether |\Psi\rangle is entangled or unentangled with respect to each of these decompositions.

IF written in terms of two subsystems, such that:

|Psi\rangle = (a_{0}|0\rangle + a_{1}|1\rangle) \otimes (b_{0}|000\rangle + b_{1}|001\rangle + b_{2}|010\rangle + b_{3}|011\rangle + b_{4}|100\rangle + b_{5}|101\rangle + b_{6}|110\rangle + b_{7}|111\rangle

AND

a_{0}b_{0} = \frac{1}{2},

AND

a_{1}b_{2} is not zero,

AND

a_{0}b_{2} = 0,

AND

b_{2} is not zero

THEN

This is a logical impossibility with the composition of these subsystems within this configuration or any of the other three as well


∴ |\Psi\rangle must be entangled, regardless of subsystem combination

Exercise 6

a. For the standard basis, the Hadamard basis, and the basis B = \{\frac{1}{\sqrt{2}}|0\rangle + i|1\rangle, |0\rangle, - |1\rangle\}, determine the probability of each outcome when the second qubit of a two qubit system in the state of |00\rangle is measured in each of these bases.

Hadamard basis state of |00\rangle = \frac{1}{2}(|+\rangle|+\rangle + |+\rangle|-\rangle + |-\rangle|+\rangle + |-\rangle|-\rangle

∴ When measured w/ Hadamard basis the second qubit is equivalent to |+\rangle and |-\rangle with equal probability of \frac{1}{2}.

For the basis B, |u\rangle = \frac{1}{\sqrt{2}}(|0\rangle + i|1\rangle) AND |v\rangle = = \frac{1}{\sqrt{2}}(|0\rangle - i|1\rangle)

∴

|0\rangle = \frac{1}{\sqrt{2}}(|u\rangle + |v\rangle)

Such that,

|00\rangle = \frac{1}{2}(|u\rangle(|u\rangle + |u\rangle(|v\rangle + |v\rangle(|u\rangle + |v\rangle(|v\rangle)

Where there is a \frac{1}{2} probability of the second bit equaling |u\rangle or |v\rangle

b. Determine the probability of each outcome when the second qubit of the state |00\rangle is first measured in the Hadamard basis and then in the basis B of part a).

IF measured with Hadamard basis we have:

|00\rangle = \frac{1}{2}(|+\rangle|+\rangle + |+\rangle|-\rangle + |-\rangle|+\rangle + |-\rangle|-\rangle

The second bit will be measured as:

\frac{1}{\sqrt{2}}(|+\rangle|+\rangle + |-\rangle|+\rangle) represented as 'X' OR \frac{1}{\sqrt{2}}(|+\rangle|-\rangle + |-\rangle|-\rangle) represented as 'Y' with equal probability.

IF

|+\rangle|+\rangle= \frac{1}{2} (−i|u\rangle|u\rangle + |u\rangle|v\rangle + |v\rangle|u\rangle + i|v\rangle|v\rangle),

AND 

|-\rangle|+\rangle= \frac{1}{2} (|u\rangle|u\rangle + i|u\rangle|v\rangle - i|v\rangle|u\rangle + |v\rangle|v\rangle)

THEN if case 1 is the first measurement,

X= \frac{1}{2\sqrt{2}}((1 - i)|u\rangle|u\rangle + (1 + i)|u\rangle|v\rangle + (1 - i)|v\rangle|u\rangle + (1 + i)|v\rangle|v\rangle)]

∴ the second measurement in this case would yield |u\rangle with a probability of \frac{1}{8}({|1 - i|}^{2}+ {|1 - i|}^2) = \frac{1}{2},

AND |v\rangle with a probability of \frac{1}{8}({|1 + i|}^{2}+ {|1 + i|}^2) = \frac{1}{2}^{2}) = \frac{1}{2}

OR

IF

|+\rangle|-\rangle= \frac{1}{2}(u\rangle|u\rangle-i|u\rangle|v\rangle + i|v\rangle|u\rangle+|v\rangle|v\rangle),

AND 

|-\rangle|-\rangle = \frac{1}{2}(i|u\rangle|u\rangle + |u\rangle|v\rangle + |v\rangle|u\rangle - i|v\rangle|v\rangle)

THEN if case 1 is the first measurement,

Y= \frac{1}{2\sqrt{2}}((1 + i)|u\rangle|u\rangle + (1 - i)|u\rangle|v\rangle + (1 + i)|v\rangle|u\rangle + (1-i)|v\rangle|v\rangle)]

∴ the second measurement in this case would yield |u\rangle with a probability of \frac{1}{8}(|1 + i|^{2} + |1 + i|^{2}) = \frac{1}{2}  

AND

|v\rangle with a probability of \frac{1}{8}(|1 - i|^{2} + |1 - i|^{2}) = \frac{1}{2}

b. Determine the probability of each outcome when the second qubit of the state |00\rangle is first. Measured in the Hadamard basis and then in the standard basis.

Exactly the same as the last scenario and results in equal probability of \frac{1}{2}.

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