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Saturday, 30 November 2013

The density matrix is in the eye of the beholder

Consider the following scenario. Alice first flips a biased coin that has outcome $0$ with probability $\cos^2(\pi/8)$ and $1$ with probability $\sin^2(\pi/8)$. If the coin value is $0$ she creates the state $\ket{0}$ and if the coin value is $1$ she creates the state $\ket{1}$. Then Alice sends the state that she created to Bob, but does not send the coin value.

From Alice's perspective (who knows the coin value), the density matrix of the state she created will be either
\[
\rho_A^0=\ket{0}\bra{0}=\begin{pmatrix} 1 &0 \\ 0 & 0 \end{pmatrix} \ \ \ \text{or} \ \ \ \rho_A^1=\ket{1}\bra{1}=\begin{pmatrix} 0 &0 \\ 0 & 1 \end{pmatrix}.
\]
 What is the density matrix of the state from Bob's perspective (who does not know the coin value)?

 Assuming that Bob at least knows that the biased coin results in outcome $0$ with probability $\cos^2(\pi/8)$ and $1$ with probability $\sin^2(\pi/8)$, and also that Alice sends him either the state $\ket{0}\bra{0}$ or the state $\ket{1}\bra{1}$ depending on whether or not the coin value was $0$ or $1$, respectively, then Bob's density matrix $\rho_B$ will be given as the statistical mixture of the two cases:
\[\begin{align*}
 \rho_B=&\cos^2(\pi/8)\ket{0}\bra{0}+\sin^2(\pi/8)\ket{0}\bra{0} \\
 =&\cos^2(\pi/8)\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}+\sin^2(\pi/8)\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}  \\
 =&\begin{pmatrix} \cos^2(\pi/8) & 0 \\ 0 &\sin^2(\pi/8) \end{pmatrix}.
\end{align*}\]

Suppose that, upon receiving the state from Alice, Bob measures it in the computational basis. The measurement process yields a classical bit and an output state $\ket{0}$ or $\ket{1}$. Will Bob's density matrix for the state (with Bob knowing the classical measurement outcome) be the same as Alice's?

 When Bob measures his density matrix $\rho_B$ the resulting density matrix will either be
\[ \begin{align*}
 \rho_B^0=\ket{0}\bra{0} = \rho_A^0 \ \ \ &\text{if Alice flipped} \  0, \\
 \rho_B^1=\ket{1}\bra{1} = \rho_A^1 \ \ \ &\text{if Alice flipped}  \ 1.
 \end{align*}\]

 In either case, Bob's resulting density matrix, $\rho_B^0$ or $\rho_B^1$, after the measurement is the same as Alice's density matrix, $\rho_B^0$ or $\rho_B^1$, depending on the coin value Alice flipped in the original state's construction.

 Suppose that we modify the above scenario to one where Alice flips a fair coin (where outcomes $0$ and $1$ occur with probability $1/2$ each) and if the coin value is $0$ she creates the state $\ket{\psi_0}=\cos(\pi/8)\ket{0}+\sin(\pi/8)\ket{1}$ and if the coin value is $1$ she creates the state $\ket{\psi_1}=\cos(\pi/8)\ket{0}+\sin(\pi/8)\ket{1}$. Alice send the state (but not the coin value) to Bob.

From Alice's perspective (who knows the coin value), the density matrix of the state she created will be either $\rho_A^0= \ket{\psi_0}\bra{\psi_0}$ or  $\rho_A^1\ket{\psi_1}\bra{\psi_1}$ , where \[\begin{align*}
\rho_A^0=&(\cos(\pi/8)\ket{0}+\sin(\pi/8)\ket{1})(\cos(\pi/8)\bra{0}+\sin(\pi/8)\bra{1}) \\
 =&\cos^2(\pi/8)\ket{0}\bra{0}+\cos(\pi/8)\sin(\pi/8)\ket{0}\bra{1}+\cos(\pi/8)\sin(\pi/8)\ket{1}\bra{0}+\sin^2(\pi/8)\ket{1}\bra{1} \\
 =&\cos^2(\pi/8)\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}+\cos(\pi/8)\sin(\pi/8)\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}+ \cos(\pi/8)\sin(\pi/8)\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}+\sin^2(\pi/8)\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}  \\
 =&\begin{pmatrix} \cos^2(\pi/8) & \cos(\pi/8)\sin(\pi/8) \\ \cos(\pi/8)\sin(\pi/8) &\sin^2(\pi/8)\end{pmatrix},
\end{align*}\]
 or
 \[ \begin{align*}
\rho_A^1=&(\cos(\pi/8)\ket{0}-\sin(\pi/8)\ket{1})(\cos(\pi/8)\bra{0}-\sin(\pi/8)\bra{1}) \\
 =&\cos^2(\pi/8)\ket{0}\bra{0}-\cos(\pi/8)\sin(\pi/8)\ket{0}\bra{1}-\cos(\pi/8)\sin(\pi/8)\ket{1}\bra{0}+\sin^2(\pi/8)\ket{1}\bra{1} \\
 =&\cos^2(\pi/8)\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}-\cos(\pi/8)\sin(\pi/8)\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}- \cos(\pi/8)\sin(\pi/8)\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}+\sin^2(\pi/8)\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}  \\
 =&\begin{pmatrix} \cos^2(\pi/8) & -\cos(\pi/8)\sin(\pi/8) \\ -\cos(\pi/8)\sin(\pi/8) &\sin^2(\pi/8)\end{pmatrix}.
\end{align*}\]

 Then the density matrix of the state from Bob's perspective $\rho_B$ (who does not know the coin value) will be the statistical mixture
\[ \begin{align*}
 \rho_B=&\frac{1}{2}\rho_A^0+\frac{1}{2}\rho_A^1 \\
 =&\frac{1}{2}\begin{pmatrix} \cos^2(\pi/8) & \cos(\pi/8)\sin(\pi/8) \\ \cos(\pi/8)\sin(\pi/8) &\sin^2(\pi/8)\end{pmatrix}+\frac{1}{2}\begin{pmatrix} \cos^2(\pi/8) & -\cos(\pi/8)\sin(\pi/8) \\ -\cos(\pi/8)\sin(\pi/8) &\sin^2(\pi/8)\end{pmatrix} \\
 =&\begin{pmatrix} \cos^2(\pi/8) & 0 \\ 0 &\sin^2(\pi/8)\end{pmatrix}.
 \end{align*}\]

Suppose that, upon receiving the state from Alice, Bob measures it in the computational basis, yielding a classical bit and an output state $\ket{0}$ or $\ket{1}$. Bob knows the classical bit outcome from his measurement, but does not reveal this to Alice. Will Bob's density matrix for the output state be the same as Alice's.

 If Bob measures his state $\rho_B$, the resulting state will be either
\[ \begin{align*}
 \rho_B^0=\ket{0}\bra{0}, \ \ \ &\text{with probability} \ \cos^2(\pi/8), \\
& \text{or} \\
 \rho_B^1=\ket{1}\bra{1}, \ \ \ &\text{with probability} \ \sin^2(\pi/8).
 \end{align*}\]

 However, if Alice does not learn of the resulting state produced by Bob's  measurement, than her lack of knowledge yields the density matrix $\rho'_A$ consisting of the statistical mixture given by

 \[\begin{align*}
 \rho'_A=&\cos^2(\pi/8)\ket{0}\bra{0} +\sin^2(\pi/8)\ket{1}\bra{1} \\
 &=\begin{pmatrix} \cos^2(\pi/8) & 0 \\ 0 &\sin^2(\pi/8)\end{pmatrix}.
 \end{align*}\]
  Thus, Bob's density matrix is not the same as Alice's regardless of the measurement outcome: $\rho_B^0\neq \rho'_A$ and $\rho_B^1\neq \rho'_A$.

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