Exercise 7.8

Alice’s density matrix is given by Eq. 7.23: \[ \rho_{a'a} = \sum_b \psi^*(a, b) \, \psi(a', b) \]

Bob’s density matrix is given by \[ \rho_{b'b} = \sum_a \psi^*(a, b) \, \psi(a, b') \]

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\[ \ket{\Psi_1} = \frac{1}{2}(\ket{uu} + \ket{ud} + \ket{du} + \ket{dd}) \]

The values of \(\psi(a, b)\) are: \[ \psi(u,u) = \psi^*(u,u) = \frac{1}{2}\] \[ \psi(u,d) = \psi^*(u,d) = \frac{1}{2}\] \[ \psi(d,u) = \psi^*(d,u) = \frac{1}{2}\] \[ \psi(d,d) = \psi^*(d,d) = \frac{1}{2}\]

The elements of the density matrix are: \[ \rho_{uu} = \psi^*(u,u) \, \psi(u,u) + \psi^*(u,d) \, \psi(u,d) = \frac{1}{2} \frac{1}{2} + \frac{1}{2} \frac{1}{2} = \frac{1}{2} \]

It is easy to see that all other matrix elements \(\rho_{ud}\), \(\rho_{du}\), \(\rho_{dd}\) also have a value of \(\frac{1}{2}\).

Hence, Alice’s density matrix is: \[ \rho_A = \begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix} \]

\[ \rho_A^2 = \begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix} \begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix} = \begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix} = \rho_A \]

Hence Alice’s density matrix corresponds to a pure state.

From the previous results it is easy to see that all matrix elements of Bob’s density matrix \(\rho_{b',b}\) have the value \(\frac{1}{2}\). Hence Bob’s density matrix looks exactly like Alice’s: \[ \rho_B = \begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix} = \rho_B^2 \]

And again this matrix corresponds to a pure state.

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Now for \(\Psi_2\), an entangled state (triplet): \[ \ket{\Psi_2} = \frac{1}{\sqrt{2}}(\ket{uu} + \ket{dd}) \]

The values of \(\psi(a, b)\) are: \[ \psi(u,u) = \psi^*(u,u) = \frac{1}{\sqrt{2}}\] \[ \psi(u,d) = \psi^*(u,d) = 0\] \[ \psi(d,u) = \psi^*(d,u) = 0\] \[ \psi(d,d) = \psi^*(d,d) = \frac{1}{\sqrt{2}}\]

The elements of Alice’s density matrix are: \[ \rho_{uu} = \psi^*(u,u) \, \psi(u,u) + \psi^*(u,d) \, \psi(u,d) = \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} = \frac{1}{2} \] \[ \rho_{ud} = \psi^*(u,u) \, \psi(d,u) + \psi^*(u,d) \, \psi(d,d) = 0 \] \[ \rho_{du} = \psi^*(d,u) \, \psi(u,u) + \psi^*(d,d) \, \psi(u,d) = 0 \] \[ \rho_{dd} = \psi^*(d,u) \, \psi(d,u) + \psi^*(d,d) \, \psi(d,d) = \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} = \frac{1}{2} \]

\[ \rho_A = \begin{pmatrix} \rho_{uu} & \rho_{ud} \\ \rho_{du} & \rho_{dd} \end{pmatrix} = \begin{pmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{pmatrix} \]

\[ \rho_A^2 = \begin{pmatrix} \frac{1}{4} & 0 \\ 0 & \frac{1}{4} \end{pmatrix} \ne \rho_A \]

hence the density matrix corresponds to a mixed or entangled state, as expected.

The elements of Bob’s density matrix are: \[ \rho_{uu} = \psi^*(u,u) \, \psi(u,u) + \psi^*(d, u) \, \psi(d, u) = \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} = \frac{1}{2} \] \[ \rho_{ud} = \psi^*(u,d) \, \psi(u,u) + \psi^*(d,d) \, \psi(d,u) = 0 \] \[ \rho_{du} = \psi^*(u,u) \, \psi(u, d) + \psi^*(d, u) \, \psi(d, d) = 0 \] \[ \rho_{dd} = \psi^*(u,d) \, \psi(u, d) + \psi^*(d, d) \, \psi(d, d) = \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} = \frac{1}{2} \]

\[ \rho_B = \begin{pmatrix} \rho_{uu} & \rho_{ud} \\ \rho_{du} & \rho_{dd} \end{pmatrix} = \begin{pmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{pmatrix} \]

\[ \rho_B^2 = \begin{pmatrix} \frac{1}{4} & 0 \\ 0 & \frac{1}{4} \end{pmatrix} \ne \rho_B \]

hence Bob’s density matrix also corresponds to an entangled state, as expected.

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\[ \ket{\Psi_3} = \frac{1}{5}(3\ket{uu}+4\ket{ud}) \]

The values of \(\psi(a, b)\) are: \[ \psi(u,u) = \psi^*(u,u) = \frac{3}{5}\] \[ \psi(u,d) = \psi^*(u,d) = \frac{4}{5}\] \[ \psi(d,u) = \psi^*(d,u) = 0\] \[ \psi(d,d) = \psi^*(d,d) = 0\]

The elements of Alice’s density matrix are: \[ \rho_{uu} = \psi^*(u,u) \, \psi(u,u) + \psi^*(u,d) \, \psi(u,d) = \frac{3}{5} \frac{3}{5} + \frac{4}{5} \frac{4}{5} = 1 \] \[ \rho_{ud} = \psi^*(u,u) \, \psi(d,u) + \psi^*(u,d) \, \psi(d,d) = 0 \] \[ \rho_{du} = \psi^*(d,u) \, \psi(u,u) + \psi^*(d,d) \, \psi(u,d) = 0 \] \[ \rho_{dd} = \psi^*(d,u) \, \psi(d,u) + \psi^*(d,d) \, \psi(d,d) = 0 \]

\[ \rho_A = \begin{pmatrix} \rho_{uu} & \rho_{ud} \\ \rho_{du} & \rho_{dd} \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \]

\[ \rho_A^2 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} = \rho_A \]

hence the density matrix corresponds to a pure state.

The elements of Bob’s density matrix are: \[ \rho_{uu} = \psi^*(u,u) \, \psi(u,u) + \psi^*(d, u) \, \psi(d, u) = \frac{3}{5} \frac{3}{5} = \frac{9}{25} \] \[ \rho_{ud} = \psi^*(u,d) \, \psi(u,u) + \psi^*(d,d) \, \psi(d,u) = \frac{4}{5} \frac{3}{5} = \frac{12}{25} \] \[ \rho_{du} = \psi^*(u,u) \, \psi(u, d) + \psi^*(d, u) \, \psi(d, d) = \frac{4}{5} \frac{3}{5} = \frac{12}{25} \] \[ \rho_{dd} = \psi^*(u,d) \, \psi(u, d) + \psi^*(d, d) \, \psi(d, d) = \frac{4}{5} \frac{4}{5} = \frac{16}{25} \]

\[ \rho_B = \begin{pmatrix} \rho_{uu} & \rho_{ud} \\ \rho_{du} & \rho_{dd} \end{pmatrix} = \begin{pmatrix} \frac{9}{25} & \frac{12}{25} \\ \frac{12}{25} & \frac{16}{25} \end{pmatrix} \]

\[ \rho_B^2 = \begin{pmatrix} \frac{9}{25} & \frac{12}{25} \\ \frac{12}{25} & \frac{16}{25} \end{pmatrix} \begin{pmatrix} \frac{9}{25} & \frac{12}{25} \\ \frac{12}{25} & \frac{16}{25} \end{pmatrix} = \begin{pmatrix} \frac{9}{25} & \frac{12}{25} \\ \frac{12}{25} & \frac{16}{25} \end{pmatrix} = \rho_B \]

hence Bob’s density matrix also corresponds to a pure state.