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.

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

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 $\frac{1}{8}({|1 + i|}^{2}+ {|1 + i|}^2) = \frac{1}{2}^{2}) = \frac{1}{2}$

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

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 $\frac{1}{2}$ .

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

Exercise 1

Exercise 2

Exercise 3

Exercise 4

Exercise 5

Exercise 6

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Exercise 1

Exercise 2

Exercise 3

Exercise 4

Exercise 5

Exercise 6

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