Week 6: MEASUREMENT OF MULTI-QUBIT SYSTEMS | ECE 802-730

Coursework focused on Chapter 4 & 5 of Quantum Computing: A Gentle Introduction.

Exercise 4.1

Give the matrix, in the standard basis, for the following operators:

a. $|0\rangle \langle 0| \rightarrow \left(^1_0\right)\left(10\right) = \left(1000\right)$

b. $|+\rangle \langle 0| - i|-\rangle \langle 1|$

$= \frac{1}{\sqrt{2}}((|0\rangle + |1\rangle)\langle0|) - i(\frac{1}{\sqrt{2}}((|0\rangle - |1\rangle)\langle1|)$

$= \frac{1}{\sqrt{2}}(1 0 0 0)(0 0 1 0) - i(\frac{1}{\sqrt{2}}(0 1 0 0)(0001)$

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

$= \frac{1}{\sqrt{2}}(1 -i 1 i)$

c. $|00\rangle\langle00| + |01\rangle\langle01|$

$= (1 0 0 0)(1 0 0 0) + (0 1 0 0)(0 1 0 0)$

$=(1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0)$

d. $|00\rangle\langle00| + |01\rangle\langle01| + |11\rangle\langle01| + |10\rangle\langle11|$

$= (1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0) + (0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0)$

$+ (0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0) + (0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0)$

$=(1 0 0 0 0 1 0 0 0 0 0 1 0 1 0 0)$

Exercise 4.2

Write the following operators in bra/ket notation:

a. The Hadamard operator: $H = \left(\frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} -\frac{1}{\sqrt{2}}\right)$

$= \frac{1}{\sqrt{2}}\left(|0\rangle\langle0| + |0\rangle\langle1| + |1\rangle\langle0| - |1\rangle\langle1|\right)$

b. $X = \left(0 1 1 0\right)$

$= |0\rangle\langle1| + |1\rangle\langle0|$

c. $Y = (0 1 - 1 0)$

$= |0\rangle\langle1| - |1\rangle\langle0|$

d. $Z = \left(1 0 0 -1\right)$

$= |0\rangle\langle0| - |1\rangle\langle1|$

e. $\left(23 0 0 0 0 -5 0 0 0 0 0 0 0 0 0 9\right)$

Exercise 4.7

Using the projection operator formalism:

a. Compute the probability of each of the possible outcomes of measuring the first qubit of an arbitrary two-qubit state in the Hadamard basis $\left{|+\rangle, |-\rangle\right}$

$P_{+} = |+\rangle\langle+| = \frac{1}{2}\left(|0\rangle\langle0| + |1\rangle\langle0| + |1\rangle\langle1|\right)$

$P_{-} = \frac{1}{2}\left(|0\rangle\langle0| - |0\rangle\langle1| - |01\rangle\langle0| - |1\rangle\langle1|\right)$

$|\Psi\rangle = a_{00}|00\rangle + a_{01}|01\rangle + a_{10}|10\rangle +a_{11}|11\rangle$

$I = |0\rangle\langle0| + |1\rangle\langle1|$

$P_{+} \otimes I = \frac{1}{2}\left(|0\rangle\langle0| + |0\rangle\langle1| + |1\rangle\langle0| + |1\rangle\langle1|\right)$

$P_{+} \otimes I = \frac{1}{2}\left(|00\rangle\langle00| + |00\rangle\langle10| + |10\rangle\langle00| + |10\rangle\langle10| + |01\rangle\langle01| + |01\rangle\langle11| + |11\rangle\langle01| + |11\rangle\langle11|$

$|\left(P_{+} \otimes I\left)|\Psi\rangle^{2} = | \frac{1}{2}\left[a_{00}\left(|00\rangle + |10\rangle\right) + a_{01}\left(|01\rangle + |11\rangle\right) + a_{10}\left(|00\rangle + |10\rangle\right) + a_{11}\left(|01\rangle + |11\rangle\right)\right]|^{2}$

$= \frac{1}{4}|\left(a_{00} + a_{10}\right)|00\rangle + \left(a_{01} + a_{11}\right)|01\rangle + \left(a_{00} + a_{10}\right)|10\rangle + \left(a_{01} + a_{11}\right)|11\rangle|^{2}$

$= \frac{1}{2}\left(|a_{00} + a_{10}|^{2} + |a_{01} + a_{11}|^{2}\right)$

AND

$P_{-} \otimes I = \frac{1}{2}\left(|0\rangle\langle0| - |0\rangle\langle1| - |1\rangle\langle0| - |1\rangle\langle1|\right) \otimes |0\rangle\langle0| + |1\rangle\langle1|$

$P_{+} \otimes I = \frac{1}{2}\left(|00\rangle\langle00| - |00\rangle\langle10| - |10\rangle\langle00| + |10\rangle\langle10| + |01\rangle\langle01| - |01\rangle\langle11| - |11\rangle\langle01| + |11\rangle\langle11|$

$|\left(P_{+} \otimes I\left)|\Psi\rangle^{2} = | \frac{1}{2}\left[a_{00}\left(|00\rangle - |10\rangle\right) + a_{01}\left(|01\rangle - |11\rangle\right) + a_{10}\left(-|00\rangle + |10\rangle\right) + a_{11}\left(-|01\rangle + |11\rangle\right)\right]|^{2}$

$= \frac{1}{4}|\left(a_{00} - a_{10}\right)|00\rangle + \left(a_{01} - a_{11}\right)|01\rangle + \left(-a_{00} + a_{10}\right)|10\rangle + \left(-a_{01} + a_{11}\right)|11\rangle|^{2}$

$= \frac{1}{2}\left(|a_{00} - a_{10}|^{2} + |a_{01} - a_{11}|^{2}\right)$

Probability of measuring $|+\rangle = |\left(P_{+} \otimes I\right)|\Psi\rangle|^{2} = \frac{1}{2}\left(|a_{00} + a_{10}|^{2} + |a_{01} + a_{11}|^{2}\right)$

Probability of measuring $|-\rangle = |\left(P_{-} \otimes I\right)|\Psi\rangle|^{2} = \frac{1}{2}\left(|a_{00} - a_{10}|^{2} + |a_{01} - a_{11}|^{2}\right)$

b. Compute the probability of each outcome for such a measurement on the state $|\Psi^{+}\rangle = \frac{1}{\sqrt{2}}\left(|00\rangle + |11\rangle\right)$

$a_{00} = a_{11} = \frac{1}{\sqrt{2}}$ and $a_{01} = a_{10} = 0$

Probability of measuring $|+\rangle = |\left(P_{+} \otimes I\right)|\Psi\rangle|^{2} = \frac{1}{2}\left(|a_{00} + a_{10}|^{2} + |a_{01} + a_{11}|^{2}\right) = \frac{1}{2}$

Probability of measuring $|-\rangle = |\left(P_{-} \otimes I\right)|\Psi\rangle|^{2} = \frac{1}{2}\left(|a_{00} - a_{10}|^{2} + |a_{01} - a_{11}|^{2}\right) = \frac{1}{2}$

Exercise 4.14

Give the outcomes and their probabilities for measurement of a general two-qubit state $|\Psi\rangle = a_{00}|00\rangle + a_{01}|01\rangle + a_{10}|10\rangle + a_{11}|11\rangle$ with respect to the Bell decomposition.

$P_{\Phi}^{+} = \frac{1}{2}\left(|00\rangle\langle00| + |11\rangle\langle00| + |00\rangle\langle11| + |11\rangle\langle11|\right)$

$P_{\Phi}^{-} = \frac{1}{2}\left(|00\rangle\langle00| - |11\rangle\langle00| - |00\rangle\langle11| + |11\rangle\langle11|\right)$

$P_{\Psi}^{+} = \frac{1}{2}\left(|00\rangle\langle00| + |11\rangle\langle00| + |00\rangle\langle11| + |11\rangle\langle11|\right)$

$P_{\Psi}^{-} = \frac{1}{2}\left(|00\rangle\langle00| - |11\rangle\langle00| - |00\rangle\langle11| + |11\rangle\langle11|\right)$

$|\Psi\rangle = a|00\rangle + b|01\rangle + c|10\rangle + d|11\rangle$

$P_{\Phi}^{+|\Psi\rangle = \frac{a + d}{2}\left(|00\rangle + |11\rangle\right)$

$P_{\Phi}^{-|\Psi\rangle = \frac{a - d}{2}\left(|00\rangle - |11\rangle\right)$

$P_{\Psi}^{+|\Psi\rangle = \frac{b + c}{2}\left(|01\rangle + |10\rangle\right)$

$P_{\Psi}^{-|\Psi\rangle = \frac{b - c}{2}\left(|01\rangle + |10\rangle\right)$

$\langle\Psi|P_{\Phi}^{+}|\Psi\rangle = \frac{|a + d|^{2}}{2}$

$\langle\Psi|P_{\Phi}^{-}|\Psi\rangle = \frac{|a - d|^{2}}{2}$

$\langle\Psi|P_{\Psi}^{+}|\Psi\rangle = \frac{|b + c|^{2}}{2}$

$\langle\Psi|P_{\Psi}^{-}|\Psi\rangle = \frac{|b - c|^{2}}{2}$

Exercise 5.4

Prove that the following are decompositions for some of the standard gates.

$I = K(0)T(0)R(0)T(0) \rightarrow IIII = I$

$X = -iT(\frac{\pi}{2})R(\frac{\pi}{2})T(0)$

$(T(\frac{\pi}{2}) = (i 0 0 -i)$

$R(\frac{\pi}{2}) = (0 1 -1 0)$

$-iT(\frac{\pi}{2})R(\frac{\pi}{2})T(0) = -i(i 0 0 -i)(0 1 -1 0)I = -i(0 i i 0) = (0 1 1 0) = X$

$H = -iT\left(\frac{\pi}{2}\right)R\left(\frac{\pi}{4}\right)T(0)$

$-iT\left(\frac{\pi}{2}\right)R\left(\frac{\pi}{4}\right)T(0) = -i(0 i i 0)\left(\frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} -\frac{1}{\sqrt{2}}\right)I$

$= (0 i i 0)\left(\frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} -\frac{1}{\sqrt{2}}\right) = \left(\frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}}\right)$

$H = \left(\frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}}\right)$

Exercise 5.6

a. Show that $R(\alpha)$ is a rotation of $2(\alpha)$ about the y-axis of the Bloch sphere

$R(\alpha) = ($ $cos$ $cos \alpha$ $sin$ $sin \alpha -$ $sin$ $sin \alpha$ $cos$ $cos \alpha)$

$\sigma_{y} = (0 -i i 0)$

$\sigma^{\rightarrow} = (\hat{\sigma}x, \hat{\sigma}y, \hat{\sigma}z)$

$e^{i\theta(\hat{\pi}\cdot\overrightarrow{\sigma})} = \hat{I}cos{\theta} + i(\hat{n} \cdot \overrightarrow{\sigma})sin{\theta}$

IF $\hat{n} = \hat{y}$ AND $\theta = \alpha$

$e^{i\alpha\hat{\sigma}_{y}} = \hat{I}$ $cos$ $cos{\alpha}} + i\hat{\sigma}_{y}$ $sin$ $sin{\alpha} = (cos$ $cos{\alpha} 0 0 cos$ $cos{\alpha}) + (0 \alpha i sin$ $sin{\alpha} 0)$

Therefore, $R(\alpha) = (cos$ $cos{\alpha}$ $sin$ $sin{\alpha} - sin$ $sin{\alpha}$ $cos$ $cos{\alpha})$

b. Show that $T(\beta)$ is a rotation of $2(\beta)$ about the z-axis of the Bloch sphere

$T(\beta) = (e^{i\beta}$ $0$ $0$ $e^{-i\beta})$

$\sigma_{z} = (1$ $0$ $0$ $-1)$

$e^{i\theta(\hat{n}\cdot\overrightarrow{\sigma})} = \hat{I}cos{\theta} + i(\hat{n} \cdot \overrightarrow{\sigma})sin{\theta}$

IF $\hat{n} = \hat{z}$ AND $\theta = \beta$

$e^{i\Beta\hat{\sigma}_{z}} = \hat{I} cos$ $cos{\beta}) + i\hat{sigma}_{z} sin$ $sin{\beta}$

$= (cos$ $cos{\beta}$ $0$ $0$ $cos$ $cos{\beta}) + i(sin$ $sin{\beta}$ $0$ $0$ $-sin$ $sin{\beta})$

$= (cos$ $cos{\beta} + i sin$ $sin{\beta}$ $0$ $0$ $cos$ $cos{\beta} - i sin$ $sin{\beta})$

Therefore, $T\left(\beta) = e^{i\beta}$ $0$ $0$ $e^{-i\beta}\right)$