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.

Figure 2.3 from Quantum Computing: A Gentle Introduction

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

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 $|1\rangle$ .