Exercise 2.6 Discussion

$\def\i{\mathbf {i}}\def\ket#1{|{#1}\rangle}\def\bra#1{\langle{#1}|}\def\braket#1#2{\langle{#1}|{#2}\rangle}\mathbf{Exercise\ 2.6}$

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

a) $\frac{\sqrt 3}{2}\ket 0 - \frac{1}{2}\ket 1$, $\{\ket 0, \ket 1\}$

b) $\frac{\sqrt 3}{2}\ket 1 -\frac{1}{2}\ket 0$, $\{\ket 0, \ket 1\}$

c) $\ket{-\i}$, $\{\ket 0, \ket 1\}$

d) $\ket{0}$, $\{\ket{+}, \ket{-}\}$

e) $\frac{1}{\sqrt 2}\left(\ket{0} - \ket 1 \right)$, $\{\ket{\i}, \ket{-\i}\}$

f) $\ket{1}$, $\{\ket{\i}, \ket{-\i}\}$

g) $\ket{+}$, $\{\frac{1}{2}\ket{0} + \frac{\sqrt 3}{2}\ket 1, \frac{\sqrt 3}{2}\ket{0} - \frac{1}{2}\ket 1 \}$

page revision: 1, last edited: 12 Nov 2011 05:43

The measurement basis determines the outcomes, so in all cases the possible outcomes are the same as the measurement basis states. The probability that a measurement of qubit state $\ket{\psi}$ will yield a measurement basis state $\ket{b}$ as its outcome is the square of the magnitude of the component of $\ket{\psi}$ along $\ket{b}$. The probability $p_{\ket{b}}$ of obtaining outcome $\ket{b}$ is given below for each part of the exercise.

a.$p_{\ket{0}} = \tfrac{3}{4}$ and $p_{\ket{1}} = \tfrac{1}{4}$b.$p_{\ket{0}} = \tfrac{1}{4}$ and $p_{\ket{1}} = \tfrac{3}{4}$c.$p_{\ket{0}} = \tfrac{1}{2}$ and $p_{\ket{1}} = \tfrac{1}{2}$d.$\ket{0} = \tfrac{1}{\sqrt{2}}\big(\ket{+}+\ket{-}\big)$, thus $p_{\ket{+}} = \tfrac{1}{2}$ and $p_{\ket{-}} = \tfrac{1}{2}$e.$\tfrac{1}{\sqrt{2}}\big(\ket{0} - \ket{1}\big) = \tfrac{1+\mathbf{i}}{2}\ket{\mathbf{i}} + \tfrac{1-\mathbf{i}}{2}\ket{-\mathbf{i}}$, thus $p_{\ket{\mathbf{i}}} = \tfrac{1}{2}$ and $p_{\ket{-\mathbf{i}}} = \tfrac{1}{2}$f.$\ket{1} = \tfrac{-\mathbf{i}}{\sqrt{2}}\ket{\mathbf{i}} + \tfrac{\mathbf{i}}{\sqrt{2}}\ket{-\mathbf{i}}$, thus $p_{\ket{\mathbf{i}}} = \tfrac{1}{2}$ and $p_{\ket{-\mathbf{i}}} = \tfrac{1}{2}$g.Let $\ket{b_{1}} = \big(\tfrac{1}{2}\ket{0}+\tfrac{\sqrt{3}}{2}\ket{1}\big)$ and $\ket{b_{2}} = \big(\tfrac{\sqrt{3}}{2}\ket{0} - \tfrac{1}{2}\ket{1}\big)$.Then $\ket{+} = \tfrac{\sqrt{3}+1}{2\sqrt{2}}\ket{b_{1}} + \tfrac{\sqrt{3}-1}{2\sqrt{2}}\ket{b_{2}}$. Thus, $p_{\ket{b_{1}}} = \tfrac{4+2\sqrt{3}}{8}$ and $p_{\ket{b_{2}}} = \tfrac{4-2\sqrt{3}}{8}$.

ReplyOptionsa.We measure the state $\ket{\psi} = \frac{\sqrt3}{2} \ket{0} - \frac{1}{2} \ket{1}$ in the standard computational basis $\{\ket{0}, \ket{1}\}$. There are two possible results:

1. Result “$\ket{0}$” is obtained with probability $\left|\braket{\psi}{0}\right|^2 = \frac{3}{4}$. The state after measurement is $\ket{0}$.

2. Result “$\ket{1}$” is obtained with probability $\left|\braket{\psi}{1}\right|^2 = \frac{1}{4}$. The state after measurement is $\ket{1}$.

b.We measure the state $\ket{\psi} = \frac{\sqrt3}{2} \ket{1} - \frac{1}{2} \ket{0}$ in the standard computational basis $\{\ket{0}, \ket{1}\}$. There are two possible results:

1. Result “$\ket{0}$” is obtained with probability $\left|\braket{\psi}{0}\right|^2 = \frac{1}{4}$. The state after measurement is $\ket{0}$.

2. Result “$\ket{1}$” is obtained with probability $\left|\braket{\psi}{1}\right|^2 = \frac{3}{4}$. The state after measurement is $\ket{1}$.

c.We measure the state $\ket{\psi} = \ket{\mathbf{-i}}$ in the standard computational basis $\{\ket{0}, \ket{1}\}$. There are two possible results:

1. Result “$\ket{0}$” is obtained with probability $\left|\braket{\psi}{0}\right|^2 = \frac{1}{2}$. The state after measurement is $\ket{0}$.

2. Result “$\ket{1}$” is obtained with probability $\left|\braket{\psi}{1}\right|^2 = \frac{1}{2}$. The state after measurement is $\ket{1}$.

d.We measure the state $\ket{\psi} = \ket{0}$ in the Hadamard basis $\{\ket{+}, \ket{-}\}$. Note that we can rewrite the state in the desired basis as $\ket{\psi} = \frac{1}{\sqrt2}(\ket{+} + \ket{-})$. There are two possible results:

1. Result “$\ket{+}$” is obtained with probability $\left|\braket{\psi}{+}\right|^2 = \frac{1}{2}$. The state after measurement is $\ket{+}$.

2. Result “$\ket{-}$” is obtained with probability $\left|\braket{\psi}{-}\right|^2 = \frac{1}{2}$. The state after measurement is $\ket{-}$.

e.We measure the state $\ket{\psi} = \frac{1}{\sqrt2}(\ket{0} - \ket{1})$ in the basis $\{\ket{\mathbf{i}}, \ket{\mathbf{-i}}\}$. Note that we can rewrite the state in the desired basis as $\ket{\psi} = \frac{1+i}{2} \ket{\mathbf{i}} + \frac{1-i}{2} \ket{\mathbf{-i}}$. There are two possible results:

1. Result “$\ket{\mathbf{i}}$” is obtained with probability $\left|\braket{\psi}{\mathbf{i}}\right|^2 = \frac{1}{2}$. The state after measurement is $\ket{\mathbf{i}}$.

2. Result “$\ket{\mathbf{-i}}$” is obtained with probability $\left|\braket{\psi}{\mathbf{-i}}\right|^2 = \frac{1}{2}$. The state after measurement is $\ket{\mathbf{-i}}$.

f.We measure the state $\ket{\psi} = \ket{1}$ in the basis $\{\ket{\mathbf{i}}, \ket{\mathbf{-i}}\}$. Note that we can rewrite the state in the desired basis as $\ket{\psi} = \frac{-i}{\sqrt2} \ket{\mathbf{i}} + \frac{i}{\sqrt2} \ket{\mathbf{-i}}$. There are two possible results:

1. Result “$\ket{\mathbf{i}}$'' is obtained with probability $\left|\braket{\psi}{\mathbf{i}}\right|^2 = \frac{1}{2}$. The state after measurement is $\ket{\mathbf{i}}$.

2. Result ``$\ket{\mathbf{-i}}$” is obtained with probability $\left|\braket{\psi}{\mathbf{-i}}\right|^2 = \frac{1}{2}$. The state after measurement is $\ket{\mathbf{-i}}$.

g.We measure the state $\ket{\psi} = \ket{+}$ in the basis $\{\ket{a}, \ket{a^\perp}\}$, where $\ket{a} = \frac{1}{2} \ket{0} + \frac{\sqrt3}{2} \ket{1}$ and $\ket{a^\perp} = \frac{\sqrt3}{2} \ket{0} - \frac{1}{2} \ket{1}$. Note that we can rewrite the state in the desired basis as $\ket{\psi} = \frac{\sqrt3 + 1}{2 \sqrt2} \ket{a} + \frac{\sqrt3 - 1}{2 \sqrt2} \ket{a^\perp}$. There are two possible results:

1. Result “$\ket{a}$” is obtained with probability $\left|\braket{\psi}{a}\right|^2 = \frac{2+\sqrt3}{4}$. The state after measurement is $\ket{a}$.

2. Result “$\ket{a^\perp}$” is obtained with probability $\left|\braket{\psi}{a^\perp}\right|^2 = \frac{2-\sqrt3}{4}$. The state after measurement is $\ket{a^\perp}$.

ReplyOptions