Ex4 14

$\def\abs#1{|#1|}\def\i{\mathbf {i}}\def\ket#1{|{#1}\rangle}\def\bra#1{\langle{#1}|}\def\braket#1#2{\langle{#1}|{#2}\rangle}\def\tr{\mathord{\mbox{tr}}}\mathbf{Exercise\ 4.14}$

a) Give the outcomes and their probabilities for measurement of each of the standard basis elements with respect to the Bell decomposition of Example 4.2.6.

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

vb314 31 Jul 2017 07:18

It appears to be easier to do part b) first and then use it to produce answers for part a).

Part b)

Projectors for the four Bell states are easily derived from the states definitions as

(1)

\begin{align} P_{\Phi^+} = \frac{1}{2} \big( \ket{00}\bra{00} + \ket {11}\bra{00} + \ket{00}\bra{11} + \ket{11}\bra{11} \big) \end{align}

(2)

\begin{align} P_{\Phi^-} = \frac{1}{2} \big( \ket{00}\bra{00} - \ket {11}\bra{00} - \ket{00}\bra{11} + \ket{11}\bra{11} \big) \end{align}

(3)

\begin{align} P_{\Psi^+} = \frac{1}{2} \big( \ket{01}\bra{01} + \ket {10}\bra{01} + \ket{01}\bra{10} + \ket{10}\bra{10} \big) \end{align}

(4)

\begin{align} P_{\Psi^-} = \frac{1}{2} \big( \ket{01}\bra{01} - \ket {10}\bra{01} - \ket{01}\bra{10} + \ket{10}\bra{10} \big) \end{align}

Choosing eigenvalues 1, -1, 2 and 2, we could write the measurement operator as a neat

(5)

\begin{align} M = \ket{11}\bra{00} + \ket{00}\bra{11} + 2 \ket{10}\bra{01} + 2 \ket{01}\bra{10} \end{align}

however it's not directly useful for our needs here.

Considering the four projectors applied to a general state

$\ket{\psi} = a\ket{00} + b\ket{01} + c\ket{10} + d\ket{11}$

we get

(6)

\begin{align} P_{\Phi^+} \ket{\psi} = \frac{a + d}{2} \big( \ket{00} + \ket{11} \big) \end{align}

(7)

\begin{align} P_{\Phi^-} \ket{\psi} = \frac{a - d}{2} \big( \ket{00} - \ket{11} \big) \end{align}

(8)

\begin{align} P_{\Psi^+} \ket{\psi} = \frac{b + c}{2} \big( \ket{01} + \ket{10} \big) \end{align}

(9)

\begin{align} P_{\Psi^-} \ket{\psi} = \frac{b - c}{2} \big( \ket{01} - \ket{10} \big) \end{align}

(note that these are not normalized) and their probabilities

(10)

\begin{align} \bra{\psi} P_{\Phi^+} \ket{\psi} = \frac{(a + d)^2}{2} \end{align}

(11)

\begin{align} \bra{\psi} P_{\Phi^-} \ket{\psi} = \frac{(a - d)^2}{2} \end{align}

(12)

\begin{align} \bra{\psi} P_{\Psi^+} \ket{\psi} = \frac{(b + c)^2}{2} \end{align}

(13)

\begin{align} \bra{\psi} P_{\Psi^-} \ket{\psi} = \frac{(b - c)^2}{2} \end{align}

Part a)

For $\ket{\psi} = \ket{00}$, substituting $a=1$ and all other parameters 0 in the above (and not forgetting to normalize), we get the outcomes

$\ket{s_1} = \frac{1}{\sqrt{2}} \big( \ket{00} + \ket{11} \big) = \ket{\Phi^+}$

and

$\ket{s_2} = \frac{1}{\sqrt{2}} \big( \ket{00} - \ket{11} \big) = \ket{\Phi^-}$,

each with probability $\frac{1}{2}$.

Analogously, measurement outcomes for $\ket{01}$ are $\ket{\Psi^+}$ and $\ket{\Psi^-}$, the outcomes for $\ket{10}$ are $\ket{\Psi^+}$ and $-\ket{\Psi^-}$, and the outcomes for $\ket{11}$ are $\ket{\Phi^+}$ and $-\ket{\Phi^-}$, all with probability $\frac{1}{2}$.

Last edited on 01 Aug 2017 05:48 by vb314 Show more

HornetMaX 09 Apr 2020 22:10

In your expressions (10), (11), (12), (13) you should keep in mind $a, b, c, d$ are complex:

(1)

\begin{align} \bra{\psi}P_{\Phi^+}\ket{\psi} & = \frac{|a+d|^2}{2} \end{align}

(2)

\begin{align} \bra{\psi}P_{\Phi^-}\ket{\psi} & = \frac{|a-d|^2}{2} \end{align}

(3)

\begin{align} \bra{\psi}P_{\Psi^+}\ket{\psi} & = \frac{|b+c|^2}{2} \end{align}

(4)

\begin{align} \bra{\psi}P_{\Psi^-}\ket{\psi} & = \frac{|b-c|^2}{2} \end{align}