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Week 4: SINGLE QUBIT QUANTUM SYSTEMS | ECE 802-730
Week 4: SINGLE QUBIT QUANTUM SYSTEMS | ECE 802-730

Week 4: SINGLE QUBIT QUANTUM SYSTEMS | ECE 802-730

Coursework focused on Chapter 2 of Quantum Computing: A Gentle Introduction.

Exercise 1

Let the direction |v\rangle of polaroid B’s preferred axis be given as a function of \theta, |v\rangle = \cos{\theta} |\rightarrow\rangle + \sin{\theta} |\uparrow\rangle, and suppose that the polaroids A and C remain horizontally and vertically polarized as in the experiment of Section 2.1.1 of Quantum Computing: A Gentle Introduction. What fraction of photons reach the
screen? Assume that each photon generated by the laser pointer has random polarization.

50% of photons emitted from the laser are blocked and 50% pass through polaroid A with a state of |\rightarrow\rangle. As polaroid B's preferred axis is already provided as a function of \theta, we have:

|v\rangle = \cos{\theta}|\rightarrow\rangle + \sin{\theta}|\uparrow\rangle

which can be used to represent the fraction of photons that are permitted to pass through the second filter as:

|\langle\rightarrow|v\rangle|^2 = |\langle\rightarrow|[\cos{\theta}|\rightarrow\rangle + \sin{\theta}|\uparrow\rangle]|^2

Which simplifies down to:

|\langle\rightarrow|v\rangle|^2 = |\cos{\theta}\leftarrow|\rightarrow\rangle + \sin{\theta}\langle\rightarrow|\uparrow\rangle|^2

|\langle\rightarrow|v\rangle|^2 = \cos^2{\theta} in the |v\rangle state

Photons passing through polaroid C will be allowed to pass through when in a |\uparrow\rangle state, which we represent as:

|\langle v |\uparrow|^2 = |\cos{\theta}\langle\rightarrow | \uparrow\rangle + \sin{\theta}\langle\uparrow | \uparrow\rangle|^2

|\langle v | \uparrow\rangle|^2 = \sin^2{\theta} in the |\uparrow\rangle state

Combined the resultant probability of photons passing through polaroid A through C is:

P = .5\cos^2{\theta}\sin^2{\theta}

Exercise 2

Which pairs of expressions for quantum states represent the same state? For those pairs that represent different states, describe a measurement for which the probabilities of the two outcomes differ for the two states and give these probabilities.

a. |0\rangle and -|0\rangle are the same state.

b. |1\rangle and i|1\rangle are the same state.

c. \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) and \frac{1}{\sqrt{2}}(-|0\rangle + i|1\rangle)

|\psi\rangle = ({\frac{1}{\sqrt{2}}} , {\frac{1}{\sqrt{2}}}) and |v\rangle = (-{\frac{1}{\sqrt{2}}} , {\frac{i}{\sqrt{2}}})

P_\psi = \langle v | \psi\rangle^2

P_\psi = |\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} \cdot \frac{i}{\sqrt{2}}|^2

P_\psi = |\frac{1}{2} - \frac{i}{2}|^2

P_\psi = \frac{1}{2}

d. \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) and \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)

|\psi\rangle =  ({\frac{1}{\sqrt{2}}} , {\frac{1}{\sqrt{2}}}) and |v\rangle = ({\frac{1}{\sqrt{2}}} , {-\frac{1}{\sqrt{2}}})

P_\psi = \langle v | \psi\rangle^2

P_\psi = |\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}|^2

P_\psi = |\frac{1}{2} + \frac{i}{2}|^2

P_\psi = 0 same state

e. \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle) and \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)

|\psi\rangle =  ({\frac{1}{\sqrt{2}}} , {-\frac{1}{\sqrt{2}}}) and |v\rangle = ({-\frac{1}{\sqrt{2}}} , {-\frac{i}{\sqrt{2}}})

P_\psi = \langle v | \psi\rangle^2

P_\psi = |\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}|^2

P_\psi = |\frac{1}{2} + \frac{i}{2}|^2

P_\psi = 1 same state

f. \frac{1}{\sqrt{2}}(|0\rangle + i|1\rangle) and \frac{1}{\sqrt{2}}(i|1\rangle - |0\rangle)

|\psi\rangle = ({\frac{1}{\sqrt{2}}} , {\frac{i}{\sqrt{2}}}) and |v\rangle = ({-\frac{1}{\sqrt{2}}} , {-\frac{i}{\sqrt{2}}})

P_\psi = \langle v | \psi\rangle^2

P_\psi = |\frac{1}{\sqrt{2}} \cdot \frac{i}{\sqrt{2}} - \frac{i}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}|^2

P_\psi = |\frac{i}{2} - \frac{i}{2}|^2

P_\psi = 0

g. \frac{1}{\sqrt{2}}(|+\rangle + |-\rangle) and |0\rangle

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

= \frac{1}{2}(|0\rangle + |1\rangle) + \frac{1}{2}(|0\rangle - |1\rangle))

= |0\rangle same state

h. \frac{1}{\sqrt{2}}(|i\rangle - |-i\rangle) and |1\rangle

= \frac{1}{\sqrt{2}}(\frac{1}{\sqrt{2}}(|0\rangle + i|1\rangle) - \frac{1}{\sqrt{2}}(|0\rangle - i|1\rangle))

= \frac{1}{2}(|0\rangle + i|1\rangle) - \frac{1}{2}(|0\rangle - i|1\rangle))

= i|1\rangle same state

i. \frac{1}{\sqrt{2}}(|i\rangle + |-i\rangle) and \frac{1}{\sqrt{2}}(|-\rangle + |+\rangle)

= \frac{1}{\sqrt{2}}(\frac{1}{\sqrt{2}}(|0\rangle + i|1\rangle) + \frac{1}{\sqrt{2}}(|0\rangle - i|1\rangle))

= \frac{1}{2}(|0\rangle + i|1\rangle) + \frac{1}{2}(|0\rangle - i|1\rangle))

= |0\rangle

and

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

= \frac{1}{2}(|0\rangle + |1\rangle) + \frac{1}{2}(|0\rangle - |1\rangle))

= |0\rangle

\frac{1}{\sqrt{2}}(|i\rangle + |-i\rangle) and \frac{1}{\sqrt{2}}(|-\rangle + |+\rangle) same state

j. \frac{1}{\sqrt{2}}(|0\rangle + e^{\frac{i\pi}{4}}|1\rangle) and \frac{1}{\sqrt{2}}(e^{-\frac{i\pi}{4}}|0\rangle + |1\rangle)

|\psi\rangle = (\frac{1}{\sqrt{2}}, \frac{e^{\frac{i\pi}{4}}}{\sqrt{2}})

and

|v\rangle = (\frac{e^{\frac{-i\pi}{4}}}{\sqrt{2}},\frac{1}{\sqrt{2}}) \rightarrow (\frac{e^{\frac{i\pi}{4}}}{\sqrt{2}},\frac{1}{\sqrt{2}})

P_\psi = {\langle v | \psi\rangle}^2

P_\psi = |\frac{e^{\frac{i\pi}{4}}}{2} + \frac{e^{\frac{i\pi}{4}}}{2}|^2

P_\psi = |e^{\frac{i\pi}{4}}|^2

e^{ix} = \cos{x} + i{\sin{x}}

P_\psi = |{\frac{\sqrt{2}}{2} + i{\frac{\sqrt{2}}{2}}|^2 = 1

Exercise 3

Which states are superpositions with respect to the standard basis, and which are not? For each state that is a superposition, give a basis with respect to which it is not a superposition. Which of these are in states of superposition with regard to the Hadamard basis?

a. |+\rangle

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

This state is a superposition with regard to the standard basis (|0\rangle, |1\rangle), but not the Hadamard basis of |+\rangle, |-\rangle) because it’s a basis vector of the Hadamard basis.

b. \frac{1}{\sqrt{2}}(|+\rangle + |-\rangle)

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

= \frac{1}{2}(|0\rangle + |1\rangle) + \frac{1}{2}(|0\rangle - |1\rangle)

= |0\rangle

This state is not a superposition with regard to the standard basis, but is a superposition with regard to the Hadamard basis.

c. \frac{1}{\sqrt{2}}(|+\rangle - |-\rangle)

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

= \frac{1}{2}(|0\rangle + |1\rangle) - \frac{1}{2}(|0\rangle - |1\rangle)

= |0\rangle

This state is not a superposition with regard to the standard basis, but is a superposition with regard to the Hadamard basis.

d. \frac{\sqrt{3}}{2}(|+\rangle - \frac{1}{2}|-\rangle)

= \frac{\sqrt{3}}{2}(\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) - \frac{1}{2}(\frac{1}{\sqrt{2}}|0\rangle - |1\rangle))

= \frac{\sqrt{3}-1}{2}|0\rangle + \frac{\sqrt{3}+1}{2}|1\rangle)

This is a superposition state, but is a superposition with regard to the Hadamard basis.

e. \frac{1}{\sqrt{2}}(|i\rangle - |i\rangle)

= \frac{1}{\sqrt{2}}(\frac{1}{\sqrt{2}}(|0\rangle + i|1\rangle) - \frac{1}{\sqrt{2}}(\frac{1}{\sqrt{2}}|0\rangle - i|1\rangle))

= \frac{1}{2}(|0\rangle + i|1\rangle) - \frac{1}{2}(|0\rangle - i|1\rangle)

= i|1\rangle

This state is not a superposition with regard to the standard basis, but is a superposition with regard to the Hadamard basis.

f. \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)

This state is a superposition with regard to the standard basis (|0\rangle , |1\rangle), but not the Hadamard basis of (|+\rangle, |-\rangle) because it’s a basis vector of the Hadamard basis.

Exercise 4

For each pair consisting of a state and a measurement basis, describe the possible measurement outcomes and give the probability for each outcome.

a. \frac{\sqrt{3}}{2}|0\rangle - \frac{1}{2}|1\rangle, \{|0\rangle, |1\rangle\}

P_{|0\rangle} = |\frac{\sqrt{3}}{2}|0\rangle|^2 = \frac{3}{4}

P_{|1\rangle} = |\frac{-1}{2}|1\rangle|^2 = \frac{1}{4}

b. \frac{\sqrt{3}}{2}|1\rangle - \frac{1}{2}|0\rangle, \{|0\rangle, |1\rangle\}

P_{|0\rangle} = |\frac{-1}{2}|1\rangle|^2 = \frac{1}{4}

P_{|1\rangle} = |\frac{\sqrt{3}}{2}|1\rangle|^2 = \frac{3}{4}

c. |i\rangle , \{|0\rangle, |1\rangle\}

= \frac{1}{\sqrt{2}}(|0\rangle - i|1\rangle), \{|0\rangle, |1\rangle\}

P_{|0\rangle} = |\frac{1}{\sqrt{2}}|0\rangle|^2 = \frac{1}{2}

P_{|0\rangle} = |-\frac{1}{\sqrt{2}}i|1\rangle|^2 = \frac{1}{2}

d. |0\rangle , \{|+\rangle, |-\rangle\}

= |0\rangle , \{(\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle), \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle))\}

= |P\rangle_{|+\rangle} = \frac{1}{2}

= |P\rangle_{|-\rangle} = \frac{1}{2}

e. \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle), \{|i\rangle, -|i\rangle\}

\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle), \{\frac{1}{\sqrt{2}}(|0\rangle + i|1\rangle)\}, \frac{1}{\sqrt{2}}(|0\rangle - i|1\rangle)\}

\frac{1+i}{2}|i\rangle + \frac{1-i}{2}|-i\rangle

P_{-i} = \frac{1}{2}

P_{i} = \frac{1}{2}

f. |1\rangle, \{|i\rangle, -|i\rangle\}

-\frac{i}{\sqrt{2}}|i\rangle, \frac{i}{2}|-i\rangle

P_{-i} = \frac{1}{2}

P_{i} = \frac{1}{2}

g. |+\rangle, \{\frac{1}{2}|0\rangle + \frac{\sqrt{3}}{2}|1\rangle, \frac{\sqrt{3}}{2}|0\rangle\ - \frac{1}{2}|1\rangle\}

|b_{1}\rangle = \frac{1}{2}(|0\rangle + \frac{\sqrt{3}}{2}|1\rangle)

|b_{2}\rangle = \frac{\sqrt{3}}{2}(|0\rangle - \frac{1}{2}|1\rangle)

|+\rangle = \frac{\sqrt{3} + 1}{2 \sqrt{2}}|b_{1}\rangle + \frac{\sqrt{3} - 1}{2 \sqrt{2}}|b_{2}\rangle

P_{|b_{1} \rangle = \frac{4 + 2\sqrt{3}}{8}

P_{|b_{2} \rangle = \frac{4 - 2\sqrt{3}}{8}

Exercise 5

Diagram, engineering drawing

Description automatically generated

a. Show that the surface of the Bloch sphere can be parametrized in terms of two real-valued parameters, the angles \Theta and \Psi illustrated in the figure above. The parametrization is in one-to-one correspondence with points on the sphere, and therefore single-qubit quantum states, in the range \Theta \in [0, \pi] and \Phi \in [0, 2\pi] except for the points corresponding to |0\rangle and |1\rangle.

|\psi\rangle = \alpha |0\rangle + \beta |1\rangle

= r_{\alpha}e^{i\psi\alpha}|0\rangle + r_{\beta}e^{i\psi\beta}|1\rangle

= e^{i\psi\alpha}(r_{a}|0\rangle + e^{i(\phi\beta - \phi\alpha)}r_{\beta}|1\rangle

Global Phase:

\gamma = \phi_\alpha and \phi = \phi_\beta - \phi_\alpha

Normalization Condition:

\langle{\psi}|{\psi}\rangle = 1

r^{2}_{\alpha} + r^{2}_{\beta} = 1

r_{\alpha} = \cos{\theta} and r_{\beta} = \sin{\theta}

|\psi\rangle = e^{i\gamma}(\cos{\frac{\theta}{2}}|0\rangle + e^{i\phi}\sin{\frac{\theta}{2}}|1\rangle)

b. What are \Theta and \Psi for each of the states |+\rangle , |−\rangle , |i\rangle , and |−i\rangle?

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

|-\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |-\rangle)

|-\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |-\rangle)

|i\rangle = \frac{1}{\sqrt{2}}(|0\rangle + i|-\rangle)

|-i\rangle = \frac{1}{\sqrt{2}}(|0\rangle - i|-\rangle)

|+\rangle \rightarrow \Theta = \frac{\pi}{2} , \Psi = 0

|-\rangle \rightarrow \Theta = \frac{\pi}{2} , \Psi = \pi

|i\rangle \rightarrow \Theta = \frac{\pi}{2} , \Psi = \frac{\pi}{2}

|-i\rangle \rightarrow \Theta = \frac{\pi}{2} , \Psi = \frac{3\pi}{2}

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