Exercise 7.12

This rather long exercise is a wrap up and review of the chapters about entanglement. We’ll go through the “Rap Sheets”, proving the propositions and verifying the numerical values.

Product State

\[ \ket{\Psi} = \alpha_u \beta_u \ket{uu} + \alpha_u \beta_d \ket{ud} + \alpha_d \beta_u \ket{du} + \alpha_d \beta_d \ket{dd} \]

Alice’s density matrix has exactly one nonzero eigenvalue, which equals 1.

Proof:

Since the wave function of the composite system is factorized \[ \psi(a, b) = \psi_A(a) \psi_B(b) \]

Alice’s density matrix is given by Eq. (7.25) in the texbook: \[ \rho_{a'a} = \psi_A^*(a) \, \psi_A(a') \]

Alice’s state vector is given by \[ \ket{\Psi_A} = \alpha_u \ket{u} + \alpha_d \ket{d} \] Hence the values of the wave function are \[\psi_A(u) = \alpha_u\] \[\psi_A^*(u) = \alpha^*_u\] \[\psi_A(d) = \alpha_d\] \[\psi_A^*(d) = \alpha^*_d\]

Alice’s density matrix is given by \[ \rho_A = \begin{pmatrix} \alpha^*_u \alpha_u & \alpha^*_d \alpha_u \\ \alpha^*_u \alpha_d & \alpha^*_d \alpha_d \end{pmatrix} \]

Now we solve for the eigenvalue \(\lambda\): \[ \rho_A \ket{\lambda} = \lambda \ket{\lambda} \]

Using the standard procedure \[ \det (\rho_A - \lambda \mathrm{I}) = 0 \]

\[ \begin{vmatrix} \alpha_u^* \alpha_u - \lambda & \alpha_d^* \alpha_u \\ \alpha_u^* \alpha_d & \alpha_d^* \alpha_d - \lambda \end{vmatrix} = 0 \]

we obtain the characteristic polynomial: \[ \lambda^2 - \lambda (\alpha_u^* \alpha_u + \alpha_d^* \alpha_d) = \lambda^2 - \lambda = 0 \]

The two eigenvalues are \(\lambda_1 = 0\) and \(\lambda_2 = +1\).

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The eigenvector with this nonzero eigenvalue is the wave function of Alice’s subsystem.

To be precise, there is an infinite number of eigenvectors, and the wave function \(\psi_A\) is not a solution to the eigenvector equation, but the state vector \(\ket{\Psi_A}\) is.

Proof: \[ \rho_A \ket{\Psi_A} = \ket{\Psi_A} \] We show this simply by plugging in: \[ \begin{pmatrix} \alpha^*_u \alpha_u & \alpha^*_d \alpha_u \\ \alpha^*_u \alpha_d & \alpha^*_d \alpha_d \end{pmatrix} \begin{pmatrix} \alpha_u \\ \alpha_d \end{pmatrix} = \begin{pmatrix} \alpha_u \\ \alpha_d \end{pmatrix} \]

\[\alpha^*_u \alpha_u \alpha_u + \alpha_d^* \alpha_u \alpha_d = \alpha_u \Rightarrow \alpha^*_u \alpha_u + \alpha^*_d \alpha_d = 1 \] \[\alpha^*_u \alpha_d \alpha_u + \alpha_d^* \alpha_d \alpha_d = \alpha_d \Rightarrow \alpha^*_u \alpha_u + \alpha^*_d \alpha_d = 1\]

We get the normalization condition, which proves that \(\ket{\Psi_A}\) is an eigenvector.

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\[ \braket{\sigma_x}^2 + \braket{\sigma_y}^2 + \braket{\sigma_z}^2 = 1 \]

Proof:

We’ll use the density matrix to compute the expectation values, see Eq. (7.13) in the textbook: \[ \braket{\mathrm{L}} = \mathrm{Tr} \, \rho \mathrm{L} \]

In a product state, Alice’s operator only acts on her part of the system and doesn’t affect Bob’s. Hence it is sufficient to only work with Alice’s wave function.

\[ \begin{flalign*} \braket{\sigma_x} &= \mathrm{Tr} \, \rho_A \sigma_x &\\ &= \mathrm{Tr} \begin{pmatrix} \alpha^*_u \alpha_u & \alpha^*_d \alpha_u \\ \alpha^*_u \alpha_d & \alpha^*_d \alpha_d \end{pmatrix} \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}&\\ &= \mathrm{Tr} \begin{pmatrix} \alpha^*_d \alpha_u & \alpha^*_u \alpha_u \\ \alpha^*_d \alpha_d & \alpha^*_u \alpha_d \end{pmatrix}\\ &= \alpha^*_d \alpha_u + \alpha^*_u \alpha_d \end{flalign*} \] \[ \braket{\sigma_x}^2 = (\alpha^*_d)^2 \alpha_u^2 + 2 \alpha^*_d \alpha_u \alpha^*_u \alpha_d + (\alpha^*_u)^2 \alpha_d^2 \]

\[ \begin{flalign*} \braket{\sigma_y} &= \mathrm{Tr} \, \rho_A \sigma_y &\\ &= \mathrm{Tr} \begin{pmatrix} \alpha^*_u \alpha_u & \alpha^*_d \alpha_u \\ \alpha^*_u \alpha_d & \alpha^*_d \alpha_d \end{pmatrix} \begin{pmatrix} 0 & -\mathrm{i}\\ \mathrm{i} & 0 \end{pmatrix}&\\ &= \mathrm{Tr} \begin{pmatrix} \mathrm{i} \alpha^*_d \alpha_u & -\mathrm{i} \alpha^*_u \alpha_u \\ \mathrm{i} \alpha^*_d \alpha_d & -\mathrm{i} \alpha^*_u \alpha_d \end{pmatrix}\\ &= \mathrm{i} \alpha^*_d \alpha_u - \mathrm{i} \alpha^*_u \alpha_d \end{flalign*} \] \[ \braket{\sigma_y}^2 = - (\alpha^*_d)^2 \alpha_u^2 + 2 \alpha^*_d \alpha_u \alpha^*_u \alpha_d - (\alpha^*_u)^2 \alpha_d^2 \]

\[ \begin{flalign*} \braket{\sigma_z} &= \mathrm{Tr} \, \rho_A \sigma_z &\\ &= \mathrm{Tr} \begin{pmatrix} \alpha^*_u \alpha_u & \alpha^*_d \alpha_u \\ \alpha^*_u \alpha_d & \alpha^*_d \alpha_d \end{pmatrix} \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}&\\ &= \mathrm{Tr} \begin{pmatrix} \alpha^*_u \alpha_u & -\alpha^*_d \alpha_u \\ \alpha^*_u \alpha_d & -\alpha^*_d \alpha_d \end{pmatrix}\\ &= \alpha^*_u \alpha_u - \alpha^*_d \alpha_d \end{flalign*} \] \[ \braket{\sigma_z}^2 = (\alpha^*_u)^2 \alpha_u^2 - 2 \alpha^*_d \alpha_u \alpha^*_u \alpha_d + (\alpha^*_d)^2 \alpha_d^2 \]

If we add up the squares of the expectation values, some terms cancel: \[ \begin{flalign*} \braket{\sigma_x}^2 + \braket{\sigma_y}^2 + \braket{\sigma_z}^2 &= (\alpha^*_u)^2 \alpha_u^2 + (\alpha^*_d)^2 \alpha_d^2 + 2 \alpha^*_d \alpha_u \alpha^*_u \alpha_d &\\ &= (\alpha^*_u \alpha_u + \alpha^*_d \alpha_d)^2\\ &= 1 \end{flalign*} \] The same applies for Bob’s \(\tau\) operator.

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Correlation: \[ \braket{\sigma_z \tau_z} - \braket{\sigma_z} \braket{\tau_z} = 0 \]

Proof: see Exercise 7.9

Singlet State

\[ \ket{\Psi} = \frac{1}{\sqrt{2}}(\ket{ud} - \ket{du}) \]

The density matrix for the composite system is as follows: \[ \rho_{ab, a'b'} = \psi^*(a',b')\, \psi(a,b) \]

The non-zero values of the wave function are: \[ \psi(u, d) = \psi^*(u, d) = \frac{1}{\sqrt{2}} \] \[ \psi(d, u) = \psi^*(d, u) = -\frac{1}{\sqrt{2}} \]

\[ \rho = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & \frac{1}{2} & -\frac{1}{2} & 0 \\ 0 & -\frac{1}{2} & \frac{1}{2} & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \]

It is easy to show that \[\rho^2 = \rho\] and \[ \mathrm{Tr}(\rho) = 1 \] Hence the full composite system is a pure state.

For Alice’s subsystem the 2x2 density matrix is given by: \[ \rho_A = \rho_{a',a} = \sum_b \psi^*(a, b)\, \psi(a', b) = \begin{pmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{pmatrix} \] \[ \rho_A^2 \ne \rho_A \] Hence Alice’s subsystem is a mixed state.

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Expectation Values: \[ \begin{flalign*} \braket{\sigma_z} &= \mathrm{Tr} \, \rho_A \sigma_z &\\ &=\mathrm{Tr} \begin{pmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\\ &=\mathrm{Tr} \begin{pmatrix} \frac{1}{2} & 0 \\ 0 & -\frac{1}{2} \end{pmatrix}\\ &= 0 \end{flalign*} \]

\[ \begin{flalign*} \braket{\sigma_x} &= \mathrm{Tr} \, \rho_A \sigma_x &\\ &=\mathrm{Tr} \begin{pmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\\ &=\mathrm{Tr} \begin{pmatrix} 0 & \frac{1}{2} \\ \frac{1}{2} & 0 \end{pmatrix}\\ &= 0 \end{flalign*} \]

\[ \begin{flalign*} \braket{\sigma_y} &= \mathrm{Tr} \, \rho_A \sigma_y &\\ &=\mathrm{Tr} \begin{pmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{pmatrix} \begin{pmatrix} 0 & -\mathrm{i} \\ \mathrm{i} & 0 \end{pmatrix}\\ &=\mathrm{Tr} \begin{pmatrix} 0 & -\frac{\mathrm{i}}{2} \\ \frac{\mathrm{i}}{2} & 0 \end{pmatrix}\\ &= 0 \end{flalign*} \]

The same holds for the components of Bob’s \(\tau\) operator.

\[ \begin{flalign*} \braket{\tau_z \sigma_z} &= \mathrm{Tr} \, \rho (\tau_z \otimes \sigma_z) &\\ &= \mathrm{Tr} \, \rho \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\\ &= \mathrm{Tr} \, \rho \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}\\ &= \mathrm{Tr} \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & \frac{1}{2} & -\frac{1}{2} & 0 \\ 0 & -\frac{1}{2} & \frac{1}{2} & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}\\ &= \mathrm{Tr} \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & -\frac{1}{2} & \frac{1}{2} & 0 \\ 0 & \frac{1}{2} & -\frac{1}{2} & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}\\ &= -1 \end{flalign*} \]

\[ \begin{flalign*} \braket{\tau_x \sigma_x} &= \mathrm{Tr} \, \rho (\tau_x \otimes \sigma_x) &\\ &= \mathrm{Tr} \, \rho \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \otimes \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\\ &= \mathrm{Tr} \, \rho \begin{pmatrix} 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0 \end{pmatrix}\\ &= \mathrm{Tr} \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & \frac{1}{2} & -\frac{1}{2} & 0 \\ 0 & -\frac{1}{2} & \frac{1}{2} & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0 \end{pmatrix}\\ &= \mathrm{Tr} \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & -\frac{1}{2} & \frac{1}{2} & 0 \\ 0 & \frac{1}{2} & -\frac{1}{2} & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}\\ &= -1 \end{flalign*} \]

\[ \begin{flalign*} \braket{\tau_y \sigma_y} &= \mathrm{Tr} \, \rho (\tau_y \otimes \sigma_y) &\\ &= \mathrm{Tr} \, \rho \begin{pmatrix} 0 & -\mathrm{i} \\ \mathrm{i} & 0 \end{pmatrix} \otimes \begin{pmatrix} 0 & -\mathrm{i} \\ \mathrm{i} & 0 \end{pmatrix}\\ &= \mathrm{Tr} \, \rho \begin{pmatrix} 0 & 0 & 0 & -1\\ 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 0\\ -1 & 0 & 0 & 0 \end{pmatrix}\\ &= \mathrm{Tr} \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & \frac{1}{2} & -\frac{1}{2} & 0 \\ 0 & -\frac{1}{2} & \frac{1}{2} & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & 0 & -1\\ 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 0\\ -1 & 0 & 0 & 0 \end{pmatrix}\\ &= \mathrm{Tr} \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & -\frac{1}{2} & \frac{1}{2} & 0 \\ 0 & \frac{1}{2} & -\frac{1}{2} & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}\\ &= -1 \end{flalign*} \]

Hence the correlation is \[ \braket{\sigma_z \tau_z} - \braket{\sigma_z} \braket{\tau_z} = - 1 \]

Near-Singlet State

\[ \Psi = \sqrt{0.6}\ket{ud} - \sqrt{0.4}\ket{du} \]

The density matrix for the composite system is as follows: \[ \rho_{ab, a'b'} = \psi^*(a',b')\, \psi(a,b) \]

The non-zero values of the wave function are: \[ \psi(u, d) = \psi^*(u, d) = \sqrt{0.6} \] \[ \psi(d, u) = \psi^*(d, u) = -\sqrt{0.4} \]

\[ \rho = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0.6 & -\sqrt{0.24} & 0 \\ 0 & -\sqrt{0.24} & 0.4 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \]

\[ \rho^2 = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0.0864 & -\sqrt{0.24} & 0 \\ 0 & -\sqrt{0.24} & 0.4 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \ne \rho \]

\[ \mathrm{Tr}(\rho^2) < 1 \] Hence the full composite system is a mixed state.

The density matrix for Alice’s subsystem is already known from Exercise 7.11: \[ \rho_A = \begin{pmatrix} 0.6 & 0 \\ 0 & 0.4 \end{pmatrix} \]

\[ \begin{flalign*} \braket{\sigma_z} &= \mathrm{Tr} \, \rho_A \sigma_z &\\ &= \mathrm{Tr} \begin{pmatrix} 0.6 & 0 \\ 0 & 0.4 \end{pmatrix} \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}\\ &= \mathrm{Tr} \begin{pmatrix} 0.6 & 0\\ 0 & -0.4 \end{pmatrix}\\ &= 0.6 - 0.4 \\ &= 0.2 \end{flalign*} \]

\[ \begin{flalign*} \braket{\sigma_x} &= \mathrm{Tr} \, \rho_A \sigma_x &\\ &= \mathrm{Tr} \begin{pmatrix} 0.6 & 0 \\ 0 & 0.4 \end{pmatrix} \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\\ &= \mathrm{Tr} \begin{pmatrix} 0 & 0.6\\ 0.4 & 0 \end{pmatrix}\\ &= 0 \end{flalign*} \]

\[ \begin{flalign*} \braket{\sigma_y} &= \mathrm{Tr} \, \rho_A \sigma_y &\\ &= \mathrm{Tr} \begin{pmatrix} 0.6 & 0 \\ 0 & 0.4 \end{pmatrix} \begin{pmatrix} 0 & -\mathrm{i}\\ \mathrm{i} & 0 \end{pmatrix}\\ &= \mathrm{Tr} \begin{pmatrix} 0 & -0.6\mathrm{i}\\ 0.4\mathrm{i} & 0 \end{pmatrix}\\ &= 0 \end{flalign*} \]

In order to compute the expectation values for the components of Bob’s \(\tau\) operator, we need to know the density matrix \(\rho_B\) of Bob’s subsystem:

\[ \rho_B = \rho_{b'b} = \sum_a \psi^*(a, b) \, \psi(a, b') \]

\[ \rho_{uu} = \psi^*(u, u)\, \psi(u, u) + \psi^*(d, u) \, \psi(d, u) = 0.4 \]

\[ \rho_{ud} = \psi^*(u, d)\, \psi(u, u) + \psi^*(d, d) \, \psi(d, u) = 0 \]

\[ \rho_{du} = \psi^*(u, u)\, \psi(u, d) + \psi^*(d, u) \, \psi(d, d) = 0 \]

\[ \rho_{dd} = \psi^*(u, d)\, \psi(u, d) + \psi^*(d, d) \, \psi(d, d) = 0.6 \]

Hence \[ \rho_B = \begin{pmatrix} 0.4 & 0 \\ 0 & 0.6 \end{pmatrix} \]

\[ \begin{flalign*} \braket{\tau_z} &= \mathrm{Tr} \, \rho_B \tau_z &\\ &= \mathrm{Tr} \begin{pmatrix} 0.4 & 0 \\ 0 & 0.6 \end{pmatrix} \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}\\ &= \mathrm{Tr} \begin{pmatrix} 0.4 & 0\\ 0 & -0.6 \end{pmatrix}\\ &= 0.4 - 0.6 \\ &= -0.2 \end{flalign*} \]

\[ \begin{flalign*} \braket{\tau_x} &= \mathrm{Tr} \, \rho_B \tau_x &\\ &= \mathrm{Tr} \begin{pmatrix} 0.4 & 0 \\ 0 & 0.6 \end{pmatrix} \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\\ &= \mathrm{Tr} \begin{pmatrix} 0 & 0.4\\ 0.6 & 0 \end{pmatrix}\\ &= 0 \end{flalign*} \]

\[ \begin{flalign*} \braket{\tau_y} &= \mathrm{Tr} \, \rho_B \tau_y &\\ &= \mathrm{Tr} \begin{pmatrix} 0.4 & 0 \\ 0 & 0.6 \end{pmatrix} \begin{pmatrix} 0 & -\mathrm{i}\\ \mathrm{i} & 0 \end{pmatrix}\\ &= \mathrm{Tr} \begin{pmatrix} 0 & -0.4\mathrm{i}\\ 0.6\mathrm{i} & 0 \end{pmatrix}\\ &= 0 \end{flalign*} \]

\[ \begin{flalign*} \braket{\tau_z \sigma_z} &= \mathrm{Tr} \, \rho (\tau_z \otimes \sigma_z) &\\ &= \mathrm{Tr} \, \rho \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\\ &= \mathrm{Tr} \, \rho \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}\\ &= \mathrm{Tr} \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0.6 & -\sqrt{0.24} & 0 \\ 0 & -\sqrt{0.24} & 0.4 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}\\ &= \mathrm{Tr} \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & -0.6 & \sqrt{0.24} & 0 \\ 0 & \sqrt{0.24} & -0.4 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}\\ &= -1 \end{flalign*} \]

\[ \begin{flalign*} \braket{\tau_x \sigma_x} &= \mathrm{Tr} \, \rho (\tau_x \otimes \sigma_x) &\\ &= \mathrm{Tr} \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0.6 & -\sqrt{0.24} & 0 \\ 0 & -\sqrt{0.24} & 0.4 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0 \end{pmatrix}\\ &= \mathrm{Tr} \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & -\sqrt{0.24} & 0.6 & 0 \\ 0 & 0.4 & -\sqrt{0.24} & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}\\ &=-2 \sqrt{0.24} \end{flalign*} \]

Correlation: \[ \braket{\sigma_z \tau_z} - \braket{\sigma_z} \braket{\tau_z} = -1 - (0,2 \, (-0.2)) = -1 +0.04 = -0.96 \]