Write the tensor product I\otimes\tau_x as a matrix, and apply that matrix to each of the | uu \rangle, | ud \rangle, | du \rangle, and | dd \rangle column vectors. Show that Alice’s half of the state-vector is unchanged in each case. Recall that I is the 2\times2 unit matrix.
\mathbf I \otimes \tau_x:
\mathbf I \otimes \tau_x = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \otimes \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{bmatrix}
And applying it to |uu \rangle, |ud\rangle, |du \rangle and |dd\rangle:
\left(\mathbf I \otimes \tau_x\right) | uu \rangle = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} = | ud \rangle \left(\mathbf I \otimes \tau_x\right) | ud \rangle = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} = | uu \rangle \left(\mathbf I \otimes \tau_x\right) | du \rangle = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} = | dd \rangle \left(\mathbf I \otimes \tau_x\right) | dd \rangle = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix} = | du \rangle
Calculate the matrix elements of \sigma_z\otimes\tau_x by forming inner products as we did in Eq. 7.2.
Using the same steps:
\begin{aligned} \sigma_z \otimes \tau_x & = \begin{pmatrix} \langle uu | \sigma_z\tau_x | uu \rangle & \langle uu | \sigma_z\tau_x | ud \rangle & \langle uu | \sigma_z\tau_x | du \rangle & \langle uu | \sigma_z\tau_x | dd \rangle \\ \langle ud | \sigma_z\tau_x | uu \rangle & \langle ud | \sigma_z\tau_x | ud \rangle & \langle ud | \sigma_z\tau_x | du \rangle & \langle ud | \sigma_z\tau_x | dd \rangle \\ \langle du | \sigma_z\tau_x | uu \rangle & \langle du | \sigma_z\tau_x | ud \rangle & \langle du | \sigma_z\tau_x | du \rangle & \langle du | \sigma_z\tau_x | dd \rangle \\ \langle dd | \sigma_z\tau_x | uu \rangle & \langle dd | \sigma_z\tau_x | ud \rangle & \langle dd | \sigma_z\tau_x | du \rangle & \langle dd | \sigma_z\tau_x | dd \rangle \end{pmatrix} \\ & = \begin{pmatrix} \langle uu | \sigma_z | ud \rangle & \langle uu | \sigma_z | uu \rangle & \langle uu | \sigma_z | dd \rangle & \langle uu | \sigma_z | du \rangle \\ \langle ud | \sigma_z | ud \rangle & \langle ud | \sigma_z | uu \rangle & \langle ud | \sigma_z | dd \rangle & \langle ud | \sigma_z | du \rangle \\ \langle du | \sigma_z | ud \rangle & \langle du | \sigma_z | uu \rangle & \langle du | \sigma_z | dd \rangle & \langle du | \sigma_z | du \rangle \\ \langle dd | \sigma_z | ud \rangle & \langle dd | \sigma_z | uu \rangle & \langle dd | \sigma_z | dd \rangle & \langle dd | \sigma_z | du \rangle \end{pmatrix} \\ & = \begin{pmatrix} \langle uu|\sigma_z|ud\rangle & \langle uu|\sigma_z|uu\rangle & \langle uu|\sigma_z|-dd\rangle & \langle uu|\sigma_z|-du\rangle \\ \langle ud|\sigma_z|ud\rangle & \langle ud|\sigma_z|uu\rangle & \langle ud|\sigma_z|-dd\rangle & \langle ud|\sigma_z|-du\rangle \\ \langle du|\sigma_z|ud\rangle & \langle du|\sigma_z|uu\rangle & \langle du|\sigma_z|-dd\rangle & \langle du|\sigma_z|-du\rangle \\ \langle dd|\sigma_z|ud\rangle & \langle dd|\sigma_z|uu\rangle & \langle dd|\sigma_z|-dd\rangle & \langle dd|\sigma_z|-du\rangle \end{pmatrix} \\ & = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & -1 & 0 \end{pmatrix} \end{aligned}
Rewrite Eq. 7.10 in component form, replacing the symbols A, B, a, and b with the matrices and column vectors from Eqs. 7.7 and 7.8.
Perform the matrix multiplications Aa and Bb on the right-hand side. Verify that each result is a 4\times1 matrix.
Expand all three Kronecker products.
Verify the row and column sizes of each Kronecker product:
Perform the matrix multiplication on the left-hand side, resulting in a 4\times1 column vector. Each row should be the sum of four separate terms.
Finally, verify that the resulting column vectors on the left and right sides are identical.
The step will be all together because it is mechanical and the dimension are self-evident. Considering the expression:
\left(A \otimes B \right) \left(a \otimes b\right) = \left( A\,a \otimes B\,b\right)
Where A and B are 2 \times 2 matrices and a and b column vectors; this can be demonstrated using the Kronecker product; starting from the LHS:
\begin{aligned} \left(A \otimes B \right) \left(a \otimes b\right) & = \begin{bmatrix} A_{11} B_{11} & A_{11} B_{12} & A_{12} B_{11} & A_{12} B_{12} \\ A_{11} B_{21} & A_{11} B_{22} & A_{12} B_{21} & A_{12} B_{22} \\ A_{21} B_{11} & A_{21} B_{12} & A_{22} B_{11} & A_{22} B_{12} \\ A_{21} B_{21} & A_{21} B_{22} & A_{22} B_{21} & A_{22} B_{22} \end{bmatrix} \begin{bmatrix} a_{11}b_{11} \\ a_{11}b_{21} \\ a_{21}b_{11} \\ a_{21}b_{21} \end{bmatrix} \\ & = \begin{bmatrix} A_{11} B_{11} a_{11}b_{11} + A_{11} B_{12} a_{11}b_{21} +A_{12} B_{11} a_{21}b_{11} + A_{12} B_{12} a_{21}b_{21} \\ A_{11} B_{21} a_{11}b_{11} + A_{11} B_{22} a_{11}b_{21} +A_{12} B_{21} a_{21}b_{11} + A_{12} B_{22} a_{21}b_{21} \\ A_{21} B_{11} a_{11}b_{11} + A_{21} B_{12} a_{11}b_{21} +A_{22} B_{11} a_{21}b_{11} + A_{22} B_12{} a_{21}b_{21} \\ A_{21} B_{21} a_{11}b_{11} + A_{21} B_{22} a_{11}b_{21} +A_{22} B_{21} a_{21}b_{11} + A_{22} B_{22} a_{21}b_{21} \end{bmatrix} \end{aligned}
Then computing the RHS; the components are:
\begin{aligned} & A\,a = \begin{bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{bmatrix} \begin{bmatrix} a_{11} \\ a_{21} \end{bmatrix} = \begin{bmatrix} A_{11} a_{11} + A_{12} a_{21} \\ A_{21} a_{11} + A_{22} a_{21} \end{bmatrix} \\ & B\,b = \begin{bmatrix} B_{11} & B_{12} \\ B_{21} & B_{22} \end{bmatrix} \begin{bmatrix} b_{11} \\ b_{21} \end{bmatrix} = \begin{bmatrix} B_{11} b_{11} + B_{12} b_{21} \\ B_{21} b_{11} + B_{22} b_{21} \end{bmatrix} \end{aligned}
and finally computing the Kronecker product:
\begin{aligned} A\,a \otimes B\,b & = \begin{bmatrix} A_{11} a_{11} + A_{12} a_{21} \\ A_{21} a_{11} + A_{22} a_{21} \end{bmatrix} \begin{bmatrix} B_{11} b_{11} + B_{12} b_{21} \\ B_{21} b_{11} + B_{22} b_{21} \end{bmatrix} \\ & = \begin{bmatrix} A_{11} B_{11} a_{11}b_{11} + A_{11} B_{12} a_{11}b_{21} +A_{12} B_{11} a_{21}b_{11} + A_{12} B_{12} a_{21}b_{21} \\ A_{11} B_{21} a_{11}b_{11} + A_{11} B_{22} a_{11}b_{21} +A_{12} B_{21} a_{21}b_{11} + A_{12} B_{22} a_{21}b_{21} \\ A_{21} B_{11} a_{11}b_{11} + A_{21} B_{12} a_{11}b_{21} +A_{22} B_{11} a_{21}b_{11} + A_{22} B_12{} a_{21}b_{21} \\ A_{21} B_{21} a_{11}b_{11} + A_{21} B_{22} a_{11}b_{21} +A_{22} B_{21} a_{21}b_{11} + A_{22} B_{22} a_{21}b_{21} \end{bmatrix} \end{aligned}
The two terms are equal and therefore the equality is proven.
Calculate the density matrix for:
| \Psi \rangle = \alpha| u \rangle + \beta| d \rangle
Answer:
\begin{array}{ll} \psi(u) = \alpha; & \psi^*(u) = \alpha^* \\ \psi(d) \beta;& \psi^*(d) = \beta^* \end{array} \rho_{a'a} = \begin{pmatrix} \alpha^*\alpha & \alpha^*\beta \\ \beta^*\alpha& \beta^*\beta\\ \end{pmatrix}
Now try plugging in some numbers for \alpha and \beta. Make sure they are normalized to 1. For example, \alpha=\dfrac1{\sqrt2}, \beta=\dfrac1{\sqrt2}.
the density matrix for the wave function:
| \Psi \rangle = \alpha | u \rangle + \beta | d \rangle
Projecting along the | u \rangle and | d \rangle directions:
\begin{aligned} & \psi(u) = \langle u | \Psi \rangle = \alpha \\ & \bar \psi(u) = \bar \alpha \\ & \psi(d) = \langle d | \Psi \rangle = \beta \\ & \bar \psi(d) = \bar \beta \end{aligned}
and the density matrix is:
\rho = \rho_{a'a} = \begin{bmatrix} \rho_{uu} & \rho_{du} \\ \rho_{ud} & \rho_{dd} \end{bmatrix} = \begin{bmatrix} \bar\alpha\,\alpha & \bar\beta \, \alpha \\ \bar \alpha \, \beta & \bar\beta \, \beta \end{bmatrix}
It is possible to compute this matrix for some known states, with the requirement that the state is normalized to 1.
\begin{array}{ll} | u \rangle & \rho = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \\[12pt] | d \rangle & \rho = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \\[12pt] | r \rangle = \tfrac{1}{\sqrt 2}| u \rangle + \tfrac{1}{\sqrt 2}| d \rangle & \rho = \begin{bmatrix} \tfrac{1}{2} & \tfrac{1}{2} \\ \tfrac{1}{2} & \tfrac{1}{2} \end{bmatrix} \\[12pt] | l \rangle = \tfrac{1}{\sqrt 2}| u \rangle - \tfrac{1}{\sqrt 2}| d \rangle & \rho = \begin{bmatrix} \tfrac{1}{2} & -\tfrac{1}{2} \\ -\tfrac{1}{2} & \tfrac{1}{2} \end{bmatrix} \\[12pt] | i \rangle = \tfrac{1}{\sqrt 2}| u \rangle + \tfrac{i}{\sqrt 2}| d \rangle & \rho = \begin{bmatrix} \tfrac{i}{2} & -\tfrac{i}{2} \\ \tfrac{i}{2} & \tfrac{1}{2} \end{bmatrix} \\[12pt] | o \rangle = \tfrac{1}{\sqrt 2}| u \rangle - \tfrac{i}{\sqrt 2}| d \rangle & \rho = \begin{bmatrix} \tfrac{i}{2} & \tfrac{i}{2} \\ -\tfrac{i}{2} & \tfrac{1}{2} \end{bmatrix} \end{array}
\begin{pmatrix} a & 0 \\ 0 & b \\ \end{pmatrix}^2 = \begin{pmatrix} a^2 & 0 \\ 0 & b^2 \\ \end{pmatrix}
\rho = \begin{pmatrix} 1/3 & 0 \\ 0 & 2/3 \end{pmatrix}
Calculate
\begin{aligned} & \rho^2 \\ & \operatorname {Tr}(\rho) \\ & \operatorname {Tr}(\rho^2) \end{aligned}
Considering the matrix:
\begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix}
and computing the square:
\begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix}^2 = \begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix} \begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix} = \begin{bmatrix} a^2 & 0 \\ 0 & b^2 \end{bmatrix}
Considering the matrix:
\rho = \begin{bmatrix} \tfrac{1}{3} & 0 \\ 0 & \tfrac{2}{3} \end{bmatrix}
And computing:
\begin{aligned} &\rho^2 = \begin{bmatrix} \tfrac{1}{9} & 0 \\ 0 & \tfrac{4}{9} \end{bmatrix} \\ & \operatorname {Tr}\left(\rho \right) = 1 \\ & \operatorname {Tr}\left(\rho^2 \right) = \tfrac{5}{9} < 1 \end{aligned}
This density matrix represents a mixed state.
Use Eq. 7.22 to show that if \rho is a density matrix, then
\operatorname {Tr}(\rho) = 1.
Since:
P(a) = \rho_{aa}
then
\operatorname {Tr}\left(\rho \right) = \sum_{a} \rho_{aa} = \sum_{a} P(a) = 1.
as the sum of all the probabilities is equal to 1.
Use Eq. 7.24 to calculate \rho^2. How does this result confirm that \rho represents an entangled state? We’ll soon discover that there are other ways to check for entanglement.
The density matrix is:
\rho_A = \begin{bmatrix}\tfrac{1}{2} & 0 \\ 0 & \tfrac{1}{2} \end{bmatrix}
\operatorname {Tr}(\rho_A) = 1, \rho_A^2 is:
\rho_A^2 = \begin{bmatrix}\tfrac{1}{4} & 0 \\ 0 & \tfrac{1}{4} \end{bmatrix} \neq \rho_A
Since \operatorname {Tr}(\rho_A^2) < 1, the state is a mixed state.
\begin{aligned} & | \psi_1 \rangle = \frac12\left(| uu \rangle+| ud \rangle+| du \rangle+| dd \rangle\right) \\ & | \psi_2 \rangle = \frac1{\sqrt2}\left(| uu \rangle+| dd \rangle\right) \\ & | \psi_3 \rangle = \frac15\left(3| uu \rangle+4| ud \rangle\right) \\ \end{aligned}
Considering the state vector:
| \Psi \rangle = \frac{1}{2}(|uu \rangle + |ud \rangle + |du \rangle + |dd \rangle)
In this case all the four basis vectors have coefficient \frac{1}{2}; the state is also normalized because the length is 1; to calculate the density matrix the first step is to list the value of \psi(a,b):
\begin{aligned} & \psi(u,u) = \tfrac{1}{2} \\ & \psi(u,d) = \tfrac{1}{2} \\ & \psi(d,u) = \tfrac{1}{2} \\ & \psi(d,d) = \tfrac{1}{2} \end{aligned}
Then it is possible to compute:
\begin{aligned} & \rho_{Auu} = \bar \psi(u,u)\psi(u,u) + \bar \psi(u,d)\psi(u,d) = \frac{1}{2}\frac{1}{2}+\frac{1}{2}\frac{1}{2} = \frac{1}{2} \\ & \rho_{Aud} = \bar \psi(d,u)\psi(u,u) + \bar \psi(d,d)\psi(u,d) = \frac{1}{2}\frac{1}{2}+\frac{1}{2}\frac{1}{2} = \frac{1}{2} \\ & \rho_{Adu} = \bar \psi(u,u)\psi(d,u) + \bar \psi(u,d)\psi(d,d) = \frac{1}{2}\frac{1}{2}+\frac{1}{2}\frac{1}{2} = \frac{1}{2} \\ & \rho_{Add} = \bar \psi(d,u)\psi(d,u) + \bar \psi(d,d)\psi(d,d) = \frac{1}{2}\frac{1}{2}+\frac{1}{2}\frac{1}{2} = \frac{1}{2} \end{aligned}
The density matrix is:
\rho_A = \begin{bmatrix}\tfrac{1}{2} & \tfrac{1}{2} \\[6pt] \tfrac{1}{2} & \tfrac{1}{2} \end{bmatrix}
\operatorname {Tr}(\rho_A) = 1, \rho_A^2 is:
\rho_A^2 = \begin{bmatrix}\tfrac{1}{2} & \tfrac{1}{2} \\[6pt] \tfrac{1}{2} & \tfrac{1}{2} \end{bmatrix} \begin{bmatrix}\tfrac{1}{2} & \tfrac{1}{2} \\[6pt] \tfrac{1}{2} & \tfrac{1}{2} \end{bmatrix} = \begin{bmatrix}\tfrac{1}{2} & \tfrac{1}{2} \\[6pt] \tfrac{1}{2} & \tfrac{1}{2} \end{bmatrix} = \rho_A
Since \operatorname {Tr}(\rho_A^2) = \operatorname {Tr}(\rho) = 1, the state is a pure state and the eigenvalues are 1 and 0.
For the system B the computation will lead to the same exact matrix (swapping the indexes) and to the same conclusions:
\rho_B = \rho_A = \begin{bmatrix}\tfrac{1}{2} & \tfrac{1}{2} \\[6pt] \tfrac{1}{2} & \tfrac{1}{2} \end{bmatrix}
Considering the state vector:
| \Psi \rangle = \frac{1}{\sqrt 2}(|uu \rangle + |dd \rangle)
In this case two of the basis vector have coefficient \frac{1}{\sqrt 2}, while other two 0; the state is also normalized because the length is 1; to calculate the density matrix for the system A the first step is to list the value of \psi(a,b):
\begin{aligned} & \psi(u,u) = \tfrac{1}{\sqrt 2} \\ & \psi(u,d) = 0 \\ & \psi(d,u) = 0 \\ & \psi(d,d) = \tfrac{1}{\sqrt 2} \end{aligned}
Then it is possible to compute:
\begin{aligned} & \rho_{Auu} = \bar \psi(u,u)\psi(u,u) + \bar \psi(u,d)\psi(u,d) = \frac{1}{\sqrt 2}\frac{1}{\sqrt 2}+0\,0 = \frac{1}{2} \\ & \rho_{Aud} = \bar \psi(d,u)\psi(u,u) + \bar \psi(d,d)\psi(u,d) = \frac{1}{\sqrt 2}0+\frac{1}{\sqrt 2}0 = 0 \\ & \rho_{Adu} = \bar \psi(u,u)\psi(d,u) + \bar \psi(u,d)\psi(d,d) = 0\frac{1}{\sqrt 2}+\frac{1}{\sqrt 2}0 = 0 \\ & \rho_{Add} = \bar \psi(d,u)\psi(d,u) + \bar \psi(d,d)\psi(d,d) = 0\,0+\frac{1}{\sqrt 2}\frac{1}{\sqrt 2} = \frac{1}{2} \end{aligned}
The density matrix is the same as the one of the example 1:
\rho_A = \begin{bmatrix}\tfrac{1}{2} & 0 \\ 0 & \tfrac{1}{2} \end{bmatrix}
\operatorname {Tr}(\rho_A) = 1, \rho_A^2 is:
\rho_A^2 = \begin{bmatrix}\tfrac{1}{4} & 0 \\ 0 & \tfrac{1}{4} \end{bmatrix}
Since \operatorname {Tr}(\rho_A^2) < 1, the state is a mixed state, with a single degenerate eigenvalue \frac{1}{2}.
For the system B the computation will lead to the same exact matrix (swapping the indexes) and conclusions:
\rho_B = \rho_A = \begin{bmatrix}\tfrac{1}{2} & 0 \\ 0 & \tfrac{1}{2} \end{bmatrix}
Considering the state vector:
| \Psi \rangle = \frac{1}{5}(3|uu \rangle + 4|ud \rangle)
In this case two of the basis vector have coefficients non-zero but different, while other two 0; the state is also normalized because the length is 1; to calculate the density matrix for the system A the first step is to list the value of \psi(a,b):
\begin{aligned} & \psi(u,u) = \tfrac{3}{5} \\ & \psi(u,d) = \tfrac{4}{5} \\ & \psi(d,u) = 0 \\ & \psi(d,d) = 0 \end{aligned}
Then it is possible to compute:
\begin{aligned} & \rho_{Auu} = \bar \psi(u,u)\psi(u,u) + \bar \psi(u,d)\psi(u,d) = \tfrac{3}{5}\tfrac{3}{5} + \tfrac{4}{5}\tfrac{4}{5} = 1 \\ & \rho_{Aud} = \bar \psi(d,u)\psi(u,u) + \bar \psi(d,d)\psi(u,d) = \tfrac{3}{5} + 0\, \tfrac{4}{5} = 0\\ & \rho_{Adu} = \bar \psi(u,u)\psi(d,u) + \bar \psi(u,d)\psi(d,d) = \tfrac{3}{5}0 + 0\, \tfrac{4}{5} = 0 \\ & \rho_{Add} = \bar \psi(d,u)\psi(d,u) + \bar \psi(d,d)\psi(d,d) = 0\,0 + 0\,0 = 0 \end{aligned}
The density matrix is:
\rho_A = \begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix}
\operatorname {Tr}(\rho_A) = 1, \rho_A^2 is:
\rho_A^2 = \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix} = \rho_A
Since \operatorname {Tr}(\rho_A^2) =\operatorname {Tr}(\rho_A) = 1, the state is a pure state, with two eigenvalues 1 and 0.
For the system B this time the matrix is different as the indexes are not identical:
\begin{aligned} & \rho_{Buu} = \bar \psi(u,u)\psi(u,u) + \bar \psi(d,u)\psi(d,u) = \tfrac{3}{5}\tfrac{3}{5} + 0\,0 = \tfrac{9}{25} \\ & \rho_{Bud} = \bar \psi(u,d)\psi(u,u) + \bar \psi(d,d)\psi(d,u) = \tfrac{4}{5}\tfrac{3}{5} + 0\,0 = \tfrac{12}{25} \\ & \rho_{Bdu} = \bar \psi(u,u)\psi(u,d) + \bar \psi(d,u)\psi(d,d) = \tfrac{3}{5}\tfrac{4}{5} + 0\,0 = \tfrac{12}{25} \\ & \rho_{Bdd} = \bar \psi(u,d)\psi(u,d) + \bar \psi(d,d)\psi(d,d) = \tfrac{4}{5}\tfrac{4}{5} + 0\,0 = \tfrac{16}{25} \end{aligned}
The density matrix is:
\rho_B = \begin{bmatrix}\tfrac{9}{25} & \tfrac{12}{25} \\[6pt] \tfrac{12}{25} & \tfrac{16}{25} \end{bmatrix}
\operatorname {Tr}(\rho_A) = 1, \rho_A^2 is:
\rho_A^2 = \begin{bmatrix}\tfrac{9}{25} & \tfrac{12}{25} \\[6pt] \tfrac{12}{25} & \tfrac{16}{25} \end{bmatrix} \begin{bmatrix}\tfrac{9}{25} & \tfrac{12}{25} \\[6pt] \tfrac{12}{25} & \tfrac{16}{25} \end{bmatrix} = \begin{bmatrix}\tfrac{225}{625} & \tfrac{300}{625} \\[6pt] \tfrac{300}{625} & \tfrac{400}{625} \end{bmatrix} = \begin{bmatrix}\tfrac{9}{25} & \tfrac{12}{25} \\[6pt] \tfrac{12}{25} & \tfrac{16}{25} \end{bmatrix} = \rho_A
Since \operatorname {Tr}(\rho_B^2) =\operatorname {Tr}(\rho_B) = 1, the state is a pure state, with two eigenvalues 1 and 0.
Given any Alice observable \mathbf{A} and Bob observable \mathbf{B}, show that for a product state, the correlation C(A, B) is zero.
Lets assume that \mathbf A is an observable for the system A and \mathbf B is an observable for the system B; the correlation between them is defined in term of the average values (the expectation values) of the observable and their product:
C(A,B) = \langle \mathbf{AB} \rangle - \langle \mathbf{A} \rangle \langle \mathbf{B} \rangle
For a product state, this correlation is zero; it is possible to demonstrate it, writing:
\begin{aligned} & \langle \mathbf A \rangle = \langle \psi | \mathbf A | \psi \rangle \\ & \langle \mathbf B \rangle = \langle \phi | \mathbf B | \phi \rangle \\ & \langle \mathbf {AB} \rangle = \langle \Psi | \mathbf {AB} | \Psi \rangle = \langle \psi \otimes \phi | \mathbf {A \otimes B} | \psi \otimes \phi \rangle \end{aligned}
The decomposition of | \Psi \rangle as:
| \psi \otimes \phi \rangle
holds only because it is a product state; using the property:
(a \otimes b) (c \otimes d) = (ac \otimes bd)
it is possible to rewrite the latest expression as:
\langle \psi \otimes \phi | \mathbf {A \otimes B} | \psi \otimes \phi \rangle = (\langle \psi \otimes \phi ) | (\mathbf A | \psi \rangle \otimes \mathbf B | \phi \rangle)
Applying again the same property:
(\langle \psi \otimes \phi) | (\mathbf A | \psi \rangle \otimes \mathbf B | \phi \rangle) = \langle \psi | \mathbf A | \psi \rangle \otimes \langle \phi | \mathbf B | \phi \rangle = \langle \mathbf A \rangle \langle \mathbf B \rangle
So the correlation \langle \mathbf {AB} \rangle - \langle \mathbf A \rangle \langle \mathbf B \rangle is zero.
Therefore, if a system is in a state where it is possible to find any two observables \mathbf A and \mathbf B that are correlated (so C(\mathbf A, \mathbf b) \neq 0) then the state is entangled; as correlation are defined in the range [-1,+1], these extreme values represents the greatest possible entanglement.
Verify that the state-vector in 7.30 represents a completely untangled state.
The state:
\alpha_u |u,b \rangle + \alpha_d | d,l \rangle
Can we re-written as:
\alpha_u |u \rangle \otimes | b\} + \alpha_d | d \rangle \otimes | b\} = \left(\alpha_u |u \rangle + \alpha_d |d \rangle \right) \otimes | b\}
which is a product state, and as a consequence it is unentangled.
Calculate Alice’s density matrix for \sigma_z for the “near-singlet” state.
The density matrix is:
\begin{aligned} \rho_{Auu} = & \bar \psi_{uu}\psi_{uu} + \bar \psi_{ud}\psi_{ud} = 0\,0 + \sqrt{0.6}\sqrt{0.6} = 0.6 \\ \rho_{Aud} = & \bar \psi_{du}\psi_{uu} + \bar \psi_{dd}\psi_{ud} = \sqrt{0.6}\,0 + 0\,\sqrt{0.6} = 0 \\ \rho_{Adu} = & \bar \psi_{uu}\psi_{du} + \bar \psi_{ud}\psi_{dd} = 0\,\sqrt{0.4} + \sqrt{0.4}\,0 = 0 \\ \rho_{Add} = & \bar \psi_{du}\psi_{du} + \bar \psi_{dd}\psi_{dd} = \sqrt{0.4}\sqrt{0.4}+ 0\,0 = 0.4 \end{aligned}
In matrix form:
\rho_A = \begin{bmatrix} 0.6 & 0\\ 0 & 0.4 \end{bmatrix}
Verify the numerical values in each rap sheet.
Excessive locality, behavior of a classical system.
Each subsystem is fully characterized and there are no correlation between system A and B.
\alpha_u\beta_u | uu \rangle + \alpha_u\beta_d | ud \rangle + \alpha_d\beta_u | du \rangle + \alpha_d\beta_d | dd \rangle
\begin{aligned} & \bar \alpha_u \alpha_u + \bar \alpha_d \alpha_d = 1 \\ & \bar \beta_u \beta_u + \bar \beta_d \beta_d = 1 \end{aligned}
Each subsystem has exactly one nonzero eigenvalue equal to 1, which is equal to the wave function of the subsystem.
The wave function is factorized:
\psi(a, b) = \psi(a)\phi(b)
\begin{aligned} & \langle \sigma_{Ax}^2 \rangle + \langle \sigma_{Ay}^2 \rangle + \langle \sigma_{Az}^2 \rangle = 1 \\ & \langle \sigma_{Bx}^2 \rangle + \langle \sigma_{By}^2 \rangle + \langle \sigma_{Bz}^2 \rangle = 1 \end{aligned}
\langle \sigma_{Az}\sigma_{Bz} \rangle - \langle \sigma_{Az}\rangle \langle \sigma_{Bz} \rangle = 0
Non-locality, fully entanglement, particular behavior of a quantum system.
The composite system is fully characterized and there are no information about each single system A or system B.
\frac{1}{\sqrt 2}(|ud\rangle - |du\rangle)
\bar \psi_{uu} \psi_{uu} + \bar \psi_{ud} \psi_{ud} + \bar \psi_{du} \psi_{du} + \bar \psi_{dd} \psi_{dd} = 1
Fully composite system: \rho^2 = \rho and \operatorname {Tr}(\rho) =1.
Each subsystem: the density matrix is proportional to the unit matrix, having equal eigenvalues that add up to 1, hence each measurement outcome is equally likely.
\begin{array}{c} \rho^2 \neq \rho \\ \operatorname {Tr}(\rho) < 1 \end{array}
The wave function is not factorized:
\psi(a, b)
\begin{aligned} & \langle \sigma_{Az} \rangle = \langle \sigma_{Ax} \rangle = \langle \sigma_{Ay} \rangle = 0 \\ & \langle \sigma_{Bz} \rangle = \langle \sigma_{Bx} \rangle = \langle \sigma_{By} \rangle = 0 \\ & \langle \sigma_{Az} \sigma_{Bz} \rangle = \langle \sigma_{Ax} \sigma_{Bx} \rangle = \langle \sigma_{Ay} \sigma_{By} \rangle = -1 \end{aligned}
\langle \sigma_{Az}\sigma_{Bz} \rangle - \langle \sigma_{Az}\rangle \langle \sigma_{Bz} \rangle = -1
Indecision, trouble making distinction between | u \rangle and | d \rangle.
There are some information about the composite system and there are some information about each single system A or system B; the exact numbers depends from the particular state vector used.
\sqrt{0.6}|ud\rangle - \sqrt{0.4} |du\rangle
\bar \psi_{uu} \psi_{uu} + \bar \psi_{ud} \psi_{ud} + \bar \psi_{du} \psi_{du} + \bar \psi_{dd} \psi_{dd} = 1
Fully composite subsystem: \rho^2 = \rho and \operatorname {Tr}(\rho) =1.
Each subsystem:
\begin{array}{c} \rho^2 \neq \rho \\ \operatorname {Tr}(\rho) < 1 \end{array}
The density matrix of the system A and the one of system B might be different.
The wave function is not factorized:
\psi(a, b)
\begin{aligned} & \langle \sigma_{Az} \rangle = 0.2 \\ & \langle \sigma_{Ax} \rangle = \langle \sigma_{Ay} \rangle = 0 \\ & \langle \sigma_{Bz} \rangle = -0.2 \\ & \langle \sigma_{Bx} \rangle = \langle \sigma_{By} \rangle = 0 \\ & \langle \sigma_{Az} \sigma_{Bz} \rangle = -1 \\ & \langle \sigma_{Ax} \sigma_{Bx} \rangle = -2\sqrt{24}\\ & \langle \sigma_{Ay} \sigma_{By} \rangle = -2\sqrt{24} \end{aligned}
\langle \sigma_{Az}\sigma_{Bz} \rangle - \langle \sigma_{Az}\rangle \langle \sigma_{Bz} \rangle = -0.96
In general, the correlation is (-1,0) \cup (0,1), with -1, 0 and 1 excluded.
As it is not specified, for example it is possible to choose any two states which respect the normalization condition:
\begin{aligned} & | A \rangle = \frac{1}{\sqrt 2} | u \rangle + \frac{1}{\sqrt 2} | d \rangle \\ & | B \rangle = \frac{1}{2} | u \rangle + \frac{\sqrt 3}{2} | d \rangle \end{aligned}
These are respecting the normalization condition:
\begin{aligned} & \bar \alpha_u \alpha_u + \bar \alpha_d \alpha_d = \frac{1}{2} + \frac{1}{2} = 1 \\ & \bar \beta_u \beta_u + \bar \beta_d \beta_d = \frac{1}{4} + \frac{3}{4} = 1 \end{aligned}
Their product state is:
|\Psi \rangle = | AB \rangle = \frac{1}{2\sqrt 2} | uu \rangle + \frac{\sqrt 3}{2 \sqrt 2} | ud \rangle + \frac{1}{2 \sqrt 2} | du \rangle + \frac{\sqrt 3}{2 \sqrt 2} | dd \rangle
\begin{array}{ll} \psi_{uu} = \dfrac{1}{2\sqrt 2} & \psi_{ud} = \dfrac{\sqrt 3}{2\sqrt 2} \\[12pt] \psi_{du} = \dfrac{1}{2 \sqrt 2} & \psi_{dd} = \dfrac{\sqrt 3}{2 \sqrt 2} \end{array}
For the composite system, the density matrix is given by:
\begin{aligned} \rho = | \Psi \rangle \langle \Psi | & = \begin{bmatrix} \psi_{uu} \bar\psi_{uu} & \psi_{uu} \bar\psi_{ud} & \psi_{uu} \bar\psi_{du} & \psi_{uu} \bar\psi_{dd} \\ \psi_{ud} \bar\psi_{uu} & \psi_{ud} \bar\psi_{ud} & \psi_{ud} \bar\psi_{du} & \psi_{ud} \bar\psi_{dd} \\ \psi_{du} \bar\psi_{uu} & \psi_{du} \bar\psi_{ud} & \psi_{du} \bar\psi_{du} & \psi_{du} \bar\psi_{dd} \\ \psi_{dd} \bar\psi_{uu} & \psi_{dd} \bar\psi_{ud} & \psi_{dd} \bar\psi_{du} & \psi_{dd} \bar\psi_{dd} \end{bmatrix} \\ & = \begin{bmatrix} \dfrac{1}{2\sqrt 2} \dfrac{1}{2\sqrt 2} & \dfrac{1}{2\sqrt 2} \dfrac{\sqrt 3}{2 \sqrt 2} & \dfrac{1}{2\sqrt 2} \dfrac{1}{2\sqrt 2} & \dfrac{1}{2\sqrt 2} \dfrac{\sqrt 3}{2 \sqrt 2} \\[12pt] \dfrac{\sqrt 3}{2 \sqrt 2} \dfrac{1}{2\sqrt 2} & \dfrac{\sqrt 3}{2 \sqrt 2} \dfrac{\sqrt 3}{2 \sqrt 2} & \dfrac{\sqrt 3}{2 \sqrt 2} \dfrac{1}{2\sqrt 2} & \dfrac{\sqrt 3}{2 \sqrt 2} \dfrac{\sqrt 3}{2 \sqrt 2} \\[12pt] \dfrac{1}{2\sqrt 2} \dfrac{1}{2\sqrt 2} & \dfrac{1}{2\sqrt 2} \dfrac{\sqrt 3}{2 \sqrt 2} & \dfrac{1}{2\sqrt 2} \dfrac{1}{2\sqrt 2} & \dfrac{1}{2\sqrt 2} \dfrac{\sqrt 3}{2 \sqrt 2} \\[12pt] \dfrac{\sqrt 3}{2 \sqrt 2} \dfrac{1}{2\sqrt 2} & \dfrac{\sqrt 3}{2 \sqrt 2} \dfrac{\sqrt 3}{2 \sqrt 2} & \dfrac{\sqrt 3}{2 \sqrt 2} \dfrac{1}{2\sqrt 2} & \dfrac{\sqrt 3}{2 \sqrt 2} \dfrac{\sqrt 3}{2 \sqrt 2} \end{bmatrix} \\ & = \dfrac{1}{8} \begin{bmatrix} 1 & \sqrt 3 & 1 & \sqrt 3 \\ \sqrt 3 & 3 & \sqrt 3 & 3 \\ 1 & \sqrt 3 & 1 & \sqrt 3 \\ \sqrt 3 & 3 & \sqrt 3 & 3 \end{bmatrix} \end{aligned}
Computing \rho^2:
\begin{aligned} \rho^2 & = \dfrac{1}{64} \begin{bmatrix} 1 & \sqrt 3 & 1 & \sqrt 3 \\ \sqrt 3 & 3 & \sqrt 3 & 3 \\ 1 & \sqrt 3 & 1 & \sqrt 3 \\ \sqrt 3 & 3 & \sqrt 3 & 3 \end{bmatrix} \begin{bmatrix} 1 & \sqrt 3 & 1 & \sqrt 3 \\ \sqrt 3 & 3 & \sqrt 3 & 3 \\ 1 & \sqrt 3 & 1 & \sqrt 3 \\ \sqrt 3 & 3 & \sqrt 3 & 3 \end{bmatrix} \\ & = \dfrac{1}{64} \begin{bmatrix} 8 & 8 \sqrt 3 & 8 & 8 \sqrt 3 \\ 8 \sqrt 3 & 24 & 8 \sqrt 3 & 24 \\ 8 & \sqrt 3 & 8 & 8 \sqrt 3 \\ 8 \sqrt 3 & 24 & 8 \sqrt 3 & 24 \end{bmatrix} = \dfrac{1}{8} \begin{bmatrix} 1 & \sqrt 3 & 1 & \sqrt 3 \\ \sqrt 3 & 3 & \sqrt 3 & 3 \\ 1 & \sqrt 3 & 1 & \sqrt 3 \\ \sqrt 3 & 3 & \sqrt 3 & 3 \end{bmatrix} = \rho \end{aligned}
\operatorname {Tr}\left(\rho \right) = \operatorname {Tr}\left(\rho^2 \right) = 1 and therefore the system is in a pure state.
For the system A, the reduced density matrix is:
\begin{aligned} \rho_{Auu} = & \bar \psi_{uu}\psi_{uu} + \bar \psi_{ud}\psi_{ud} = \dfrac{1}{2\sqrt 2}\dfrac{1}{2\sqrt 2} + \dfrac{\sqrt 3}{2\sqrt 2}\dfrac{\sqrt 3}{2\sqrt 2} = \dfrac{1}{8} + \dfrac{3}{8} = \dfrac{1}{2} \\ \rho_{Aud} = & \bar \psi_{du}\psi_{uu} + \bar \psi_{dd}\psi_{ud} = \dfrac{1}{2\sqrt 2}\dfrac{1}{2\sqrt 2} + \dfrac{\sqrt 3}{2\sqrt 2}\dfrac{\sqrt 3}{2\sqrt 2} = \dfrac{1}{8} + \dfrac{3}{8} = \dfrac{1}{2} \\ \rho_{Adu} = & \bar \psi_{uu}\psi_{du} + \bar \psi_{ud}\psi_{dd} = \dfrac{1}{2\sqrt 2}\dfrac{1}{2\sqrt 2} + \dfrac{\sqrt 3}{2\sqrt 2}\dfrac{\sqrt 3}{2\sqrt 2} = \dfrac{1}{8} + \dfrac{3}{8} = \dfrac{1}{2} \\ \rho_{Add} = & \bar \psi_{du}\psi_{du} + \bar \psi_{dd}\psi_{dd} = \dfrac{1}{2\sqrt 2}\dfrac{1}{2\sqrt 2} + \dfrac{\sqrt 3}{2\sqrt 2}\dfrac{\sqrt 3}{2\sqrt 2} = \dfrac{1}{8} + \dfrac{3}{8} = \dfrac{1}{2} \end{aligned}
In matrix form:
\rho_A = \begin{bmatrix} \dfrac{1}{2} & \dfrac{1}{2} \\[12pt] \dfrac{1}{2} & \dfrac{1}{2} \end{bmatrix}
Computing \rho_A^2:
\rho_A^2 = \begin{bmatrix} \tfrac{1}{2} & \tfrac{1}{2} \\ \tfrac{1}{2} & \tfrac{1}{2} \end{bmatrix}\begin{bmatrix} \tfrac{1}{2} & \tfrac{1}{2} \\ \tfrac{1}{2} & \tfrac{1}{2} \end{bmatrix} = \begin{bmatrix} \tfrac{1}{2} & \tfrac{1}{2} \\ \tfrac{1}{2} & \tfrac{1}{2} \end{bmatrix} = \rho_A
\operatorname {Tr}\left(\rho_A \right) = \operatorname {Tr}\left(\rho_A^2 \right) = 1 and therefore the system is in a pure state.
Computing the eigenvalues:
\begin{aligned} | \rho_A - \lambda \, \mathbf I | & = \begin{vmatrix} \dfrac{1}{2} - \lambda & \dfrac{1}{2} \\[12pt] \dfrac{1}{2} & \dfrac{1}{2} - \lambda \end{vmatrix} = \left(\dfrac{1}{2} - \lambda\right)^2 - \dfrac{1}{4} = \lambda^2 - \lambda + \dfrac{1}{4} - \dfrac{1}{4} = \lambda (\lambda - 1) \end{aligned}
which gives the expected eigenvalues for a product state:
\lambda_1 = 1, \quad \lambda_2 = 0
Computing the eigenvector corresponding to \lambda_1 = 1:
\begin{aligned} & \left( \rho_A - \lambda_1 \mathbf I\right) \mathbf v_A = \mathbf 0 \\ & \begin{bmatrix} -\tfrac{1}{2} & \tfrac{1}{2} \\ \tfrac{1}{2} & -\tfrac{1}{2} \end{bmatrix} \begin{bmatrix} v_{A1} \\ v_{A2} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \end{aligned}
A vector which satisfies this equation is for example:
\mathbf v'_A = \begin{bmatrix} 1 \\ 1 \end{bmatrix}
This can be normalized as:
\mathbf v_A = \frac{\mathbf v'_A}{\left| \mathbf v'_A \right|} = \begin{bmatrix} \frac{1}{\sqrt 2} \\ \frac{1}{\sqrt 2} \end{bmatrix}
Therefore the eigenvector is:
\mathbf v_A = \frac{1}{\sqrt 2}|u \rangle + \frac{1}{\sqrt 2}|d \rangle = |A \rangle
As expected, the direction of the eigenvalue is the same as the original vector (taking \mathbf v''_A = - \mathbf v_A would be a valid normalized vector too).
For the system B the reduced density matrix is:
\begin{aligned} \rho_{Buu} = & \bar \psi_{uu}\psi_{uu} + \bar \psi_{du}\psi_{du} = \dfrac{1}{2\sqrt 2}\dfrac{1}{2\sqrt 2} + \dfrac{1}{2\sqrt 2}\dfrac{1}{2\sqrt 2} = \dfrac{1}{8} + \dfrac{1}{8} = \dfrac{1}{4} \\ \rho_{Bud} = & \bar \psi_{ud}\psi_{uu} + \bar \psi_{dd}\psi_{du} = \dfrac{\sqrt 3}{2\sqrt 2}\dfrac{1}{2\sqrt 2} + \dfrac{1}{2\sqrt 2}\dfrac{\sqrt 3}{2\sqrt 2} = \dfrac{\sqrt 3}{8} + \dfrac{\sqrt 3}{8} = \dfrac{\sqrt 3}{4} \\ \rho_{Bdu} = & \bar \psi_{uu}\psi_{ud} + \bar \psi_{du}\psi_{dd} = \dfrac{1}{2\sqrt 2}\dfrac{\sqrt 3}{2\sqrt 2} + \dfrac{1}{2\sqrt 2}\dfrac{\sqrt 3}{2\sqrt 2} = \dfrac{\sqrt 3}{8} + \dfrac{\sqrt 3}{8} = \dfrac{\sqrt 3}{4} \\ \rho_{Bdd} = & \bar \psi_{ud}\psi_{ud} + \bar \psi_{dd}\psi_{dd} = \dfrac{\sqrt 3}{2\sqrt 2}\dfrac{\sqrt 3}{2\sqrt 2} + \dfrac{\sqrt 3}{2\sqrt 2}\dfrac{\sqrt 3}{2\sqrt 2} = \dfrac{3}{8} + \dfrac{3}{8} = \dfrac{3}{4} \end{aligned}
In matrix form:
\rho_B = \begin{bmatrix} \dfrac{1}{4} & \dfrac{\sqrt 3}{4} \\[12pt] \dfrac{\sqrt 3}{4} & \dfrac{3}{4} \end{bmatrix}
Computing \rho_B^2:
\rho_B^2 = \begin{bmatrix} \tfrac{1}{4} & \tfrac{\sqrt 3}{4} \\ \tfrac{\sqrt 3}{4} & \tfrac{3}{4} \end{bmatrix} \begin{bmatrix} \tfrac{1}{4} & \tfrac{\sqrt 3}{4} \\ \tfrac{\sqrt 3}{4} & \tfrac{3}{4} \end{bmatrix} = \begin{bmatrix} \tfrac{1}{4} & \tfrac{\sqrt 3}{4} \\ \tfrac{\sqrt 3}{4} & \tfrac{3}{4} \end{bmatrix} = \rho_B
\operatorname {Tr}\left(\rho_B \right) = \operatorname {Tr}\left(\rho_B^2 \right) = 1 and therefore the system is in a pure state.
Computing the eigenvalues:
\begin{aligned} | \rho_B - \lambda \, \mathbf I | & = \begin{vmatrix} \dfrac{1}{4} - \lambda & \dfrac{\sqrt 3}{4} \\[12pt] \dfrac{\sqrt 3}{4} & \dfrac{3}{4} - \lambda \end{vmatrix} = \left(\dfrac{1}{4} - \lambda\right)\left(\dfrac{3}{4} - \lambda\right) - \dfrac{3}{16} = \lambda^2 - \lambda + \dfrac{3}{16} - \dfrac{3}{16} = \lambda (\lambda - 1) \end{aligned}
which gives the expected eigenvalues for a product state:
\lambda_1 = 1, \quad \lambda_2 = 0
The eigenvalues are not changing if computed by either system, the above was just a reconfirmation.
Computing the eigenvector corresponding to \lambda_1 = 1:
\begin{aligned} & \left( \rho_B - \lambda_1 \mathbf I\right) \mathbf v_B = \mathbf 0 \\ & \begin{bmatrix} -\tfrac{3}{4} & \tfrac{\sqrt 3}{4} \\ \tfrac{\sqrt 3}{4} & -\tfrac{1}{4} \end{bmatrix} \begin{bmatrix} v_{B1} \\ v_{B2} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \end{aligned}
A vector which satisfies this equation is for example:
\mathbf v'_B = \begin{bmatrix} 1 \\ \sqrt 3 \end{bmatrix}
This can be normalized as:
\mathbf v_B = \frac{\mathbf v'_B}{\left| \mathbf v'_B \right|} = \begin{bmatrix} \frac{1}{2} \\ \frac{\sqrt 3}{2} \end{bmatrix}
Therefore the eigenvector is:
\mathbf v_B = \frac{1}{2}|u \rangle + \frac{\sqrt 3}{2}|d \rangle = |B \rangle
Starting from the system A:
\begin{aligned} \langle \sigma_{Az} \rangle = & \langle AB | \sigma_{Az} | AB \rangle = \left( \frac{1}{2\sqrt 2} \langle uu | + \frac{\sqrt 3}{2 \sqrt 2} \langle ud | + \frac{1}{2 \sqrt 2} \langle du | + \frac{\sqrt 3}{2 \sqrt 2} \langle dd | \right)\\ & \sigma_{Az} \left( \dfrac{1}{2\sqrt 2} | uu \rangle + \frac{\sqrt 3}{2 \sqrt 2} | ud \rangle + \frac{1}{2 \sqrt 2} | du \rangle + \frac{\sqrt 3}{2 \sqrt 2} | dd \rangle \right) \\ = & \left( \frac{1}{2\sqrt 2} \langle uu | + \frac{\sqrt 3}{2 \sqrt 2} \langle ud | + \frac{1}{2 \sqrt 2} \langle du | + \frac{\sqrt 3}{2 \sqrt 2} \langle dd | \right) \\ & \left( \dfrac{1}{2\sqrt 2} | uu \rangle + \frac{\sqrt 3}{2 \sqrt 2} | ud \rangle - \frac{1}{2 \sqrt 2} | du \rangle - \frac{\sqrt 3}{2 \sqrt 2} | dd \rangle \right) = 0 \end{aligned} \begin{aligned} \langle \sigma_{Ax} \rangle = & \langle AB | \sigma_{Ax} | AB \rangle = \left( \frac{1}{2\sqrt 2} \langle uu | + \frac{\sqrt 3}{2 \sqrt 2} \langle ud | + \frac{1}{2 \sqrt 2} \langle du | + \frac{\sqrt 3}{2 \sqrt 2} \langle dd | \right)\\ & \sigma_{Ax} \left( \dfrac{1}{2\sqrt 2} | uu \rangle + \frac{\sqrt 3}{2 \sqrt 2} | ud \rangle + \frac{1}{2 \sqrt 2} | du \rangle + \frac{\sqrt 3}{2 \sqrt 2} | dd \rangle \right) \\ = & \left( \frac{1}{2\sqrt 2} \langle uu | + \frac{\sqrt 3}{2 \sqrt 2} \langle ud | + \frac{1}{2 \sqrt 2} \langle du | + \frac{\sqrt 3}{2 \sqrt 2} \langle dd | \right) \\ & \left( \dfrac{1}{2\sqrt 2} | du \rangle + \frac{\sqrt 3}{2 \sqrt 2} | dd \rangle + \frac{1}{2 \sqrt 2} | uu \rangle + \frac{\sqrt 3}{2 \sqrt 2} | ud \rangle \right) = \dfrac{1}{8} + \dfrac{3}{8} + \dfrac{3}{8} + \dfrac{1}{8} = 1 \end{aligned} \begin{aligned} \langle \sigma_{Ay} \rangle = & \langle AB | \sigma_{Ay} | AB \rangle = \left( \frac{1}{2\sqrt 2} \langle uu | + \frac{\sqrt 3}{2 \sqrt 2} \langle ud | + \frac{1}{2 \sqrt 2} \langle du | + \frac{\sqrt 3}{2 \sqrt 2} \langle dd | \right)\\ & \sigma_{Ay} \left( \dfrac{1}{2\sqrt 2} | uu \rangle + \frac{\sqrt 3}{2 \sqrt 2} | ud \rangle + \frac{1}{2 \sqrt 2} | du \rangle + \frac{\sqrt 3}{2 \sqrt 2} | dd \rangle \right) \\ = & \left( \frac{1}{2\sqrt 2} \langle uu | + \frac{\sqrt 3}{2 \sqrt 2} \langle ud | + \frac{1}{2 \sqrt 2} \langle du | + \frac{\sqrt 3}{2 \sqrt 2} \langle dd | \right) \\ & \left( \dfrac{i}{2\sqrt 2} | du \rangle + \frac{i \sqrt 3}{2 \sqrt 2} | dd \rangle - \frac{i}{2 \sqrt 2} | uu \rangle - \frac{i\sqrt 3}{2 \sqrt 2} | ud \rangle \right) = 0 \end{aligned}
That verifies:
\langle \sigma_{Ax}^2 \rangle + \langle \sigma_{Ay}^2 \rangle + \langle \sigma_{Az}^2 \rangle = 1
Computing for the system B:
\begin{aligned} \langle \sigma_{Bz} \rangle = & \langle AB | \sigma_{Bz} | AB \rangle = \left( \frac{1}{2\sqrt 2} \langle uu | + \frac{\sqrt 3}{2 \sqrt 2} \langle ud | + \frac{1}{2 \sqrt 2} \langle du | + \frac{\sqrt 3}{2 \sqrt 2} \langle dd | \right)\\ & \sigma_{Bz} \left( \dfrac{1}{2\sqrt 2} | uu \rangle + \frac{\sqrt 3}{2 \sqrt 2} | ud \rangle + \frac{1}{2 \sqrt 2} | du \rangle + \frac{\sqrt 3}{2 \sqrt 2} | dd \rangle \right) \\ = & \left( \frac{1}{2\sqrt 2} \langle uu | + \frac{\sqrt 3}{2 \sqrt 2} \langle ud | + \frac{1}{2 \sqrt 2} \langle du | + \frac{\sqrt 3}{2 \sqrt 2} \langle dd | \right) \\ & \left( \frac{1}{2\sqrt 2} | uu \rangle - \frac{\sqrt 3}{2 \sqrt 2} | ud \rangle + \frac{1}{2 \sqrt 2} | du \rangle - \frac{\sqrt 3}{2 \sqrt 2} | dd \rangle \right) = \dfrac{1}{8} - \dfrac{3}{8} - \dfrac{3}{8} + \dfrac{1}{8} = -\frac{1}{2} \end{aligned} \begin{aligned} \langle \sigma_{Bx} \rangle = & \langle AB | \sigma_{Bx} | AB \rangle = \left( \frac{1}{2\sqrt 2} \langle uu | + \frac{\sqrt 3}{2 \sqrt 2} \langle ud | + \frac{1}{2 \sqrt 2} \langle du | + \frac{\sqrt 3}{2 \sqrt 2} \langle dd | \right)\\ & \sigma_{Bx} \left( \dfrac{1}{2\sqrt 2} | uu \rangle + \frac{\sqrt 3}{2 \sqrt 2} | ud \rangle + \frac{1}{2 \sqrt 2} | du \rangle + \frac{\sqrt 3}{2 \sqrt 2} | dd \rangle \right) \\ = & \left( \frac{1}{2\sqrt 2} \langle uu | + \frac{\sqrt 3}{2 \sqrt 2} \langle ud | + \frac{1}{2 \sqrt 2} \langle du | + \frac{\sqrt 3}{2 \sqrt 2} \langle dd | \right) \\ & \left( \frac{1}{2\sqrt 2} | ud \rangle + \frac{\sqrt 3}{2 \sqrt 2} | uu \rangle + \frac{1}{2 \sqrt 2} | dd \rangle + \frac{\sqrt 3}{2 \sqrt 2} | du \rangle \right) = \dfrac{\sqrt 3}{8} + \dfrac{\sqrt 3}{8} + \dfrac{\sqrt 3}{8} + \dfrac{\sqrt 3}{8} = \dfrac{\sqrt 3}{2} \end{aligned} \begin{aligned} \langle \sigma_{By} \rangle = & \langle AB | \sigma_{By} | AB \rangle = \left( \frac{1}{2\sqrt 2} \langle uu | + \frac{\sqrt 3}{2 \sqrt 2} \langle ud | + \frac{1}{2 \sqrt 2} \langle du | + \frac{\sqrt 3}{2 \sqrt 2} \langle dd | \right)\\ & \sigma_{By} \left( \dfrac{1}{2\sqrt 2} | uu \rangle + \frac{\sqrt 3}{2 \sqrt 2} | ud \rangle + \frac{1}{2 \sqrt 2} | du \rangle + \frac{\sqrt 3}{2 \sqrt 2} | dd \rangle \right) \\ = & \left( \frac{1}{2\sqrt 2} \langle uu | + \frac{\sqrt 3}{2 \sqrt 2} \langle ud | + \frac{1}{2 \sqrt 2} \langle du | + \frac{\sqrt 3}{2 \sqrt 2} \langle dd | \right) \\ & \left( \frac{i}{2\sqrt 2} | ud \rangle - \frac{i\sqrt 3}{2 \sqrt 2} | uu \rangle + \frac{i}{2 \sqrt 2} | dd \rangle - \frac{i\sqrt 3}{2 \sqrt 2} | du \rangle \right) = -\dfrac{i\sqrt 3}{8} + \dfrac{i\sqrt 3}{8} - \dfrac{i\sqrt 3}{8} + \dfrac{i\sqrt 3}{8} = 0 \end{aligned}
That verifies:
\langle \sigma_{Bx}^2 \rangle + \langle \sigma_{By}^2 \rangle + \langle \sigma_{Bz}^2 \rangle = \dfrac{3}{4} + 0 + \dfrac{1}{4} = 1
\begin{aligned} \langle \sigma_{Az} \sigma_{Bz} \rangle = & \langle AB | \sigma_{Az} \sigma_{Bz} | AB \rangle = \left( \frac{1}{2\sqrt 2} \langle uu | + \frac{\sqrt 3}{2 \sqrt 2} \langle ud | + \frac{1}{2 \sqrt 2} \langle du | + \frac{\sqrt 3}{2 \sqrt 2} \langle dd | \right)\sigma_{Az}\\ & \sigma_{Bz} \left( \dfrac{1}{2\sqrt 2} | uu \rangle + \frac{\sqrt 3}{2 \sqrt 2} | ud \rangle + \frac{1}{2 \sqrt 2} | du \rangle + \frac{\sqrt 3}{2 \sqrt 2} | dd \rangle \right) \\ = & \left( \frac{1}{2\sqrt 2} \langle du | + \frac{\sqrt 3}{2 \sqrt 2} \langle dd | - \frac{1}{2 \sqrt 2} \langle uu | - \frac{\sqrt 3}{2 \sqrt 2} \langle ud | \right) \\ & \left( \dfrac{1}{2\sqrt 2} | ud \rangle - \frac{\sqrt 3}{2 \sqrt 2} | uu \rangle + \frac{1}{2 \sqrt 2} | dd \rangle - \frac{\sqrt 3}{2 \sqrt 2} | du \rangle \right) = 0 \end{aligned}
Previously calculated:
\begin{aligned} & \langle \sigma_{Az} \rangle = 0 \\ & \langle \sigma_{Bz} \rangle = -\frac{1}{2} \end{aligned}
That verifies:
\langle \sigma_{Az}\sigma_{Bz} \rangle - \langle \sigma_{Az}\rangle \langle \sigma_{Bz} \rangle = 0 - 0 \, \frac{1}{2} = 0
| \Psi \rangle = \frac{1}{\sqrt 2}(|ud\rangle - |du\rangle)
\begin{array}{ll} \psi_{uu} = 0 & \psi_{ud} = \dfrac{1}{\sqrt 2} \\[12pt] \psi_{du} = -\dfrac{1}{\sqrt 2} & \psi_{dd} = 0 \end{array}
For the composite system, the density matrix is given by:
\begin{aligned} \rho = | \Psi \rangle \langle \Psi | & = \begin{bmatrix} \psi_{uu} \bar\psi_{uu} & \psi_{uu} \bar\psi_{ud} & \psi_{uu} \bar\psi_{du} & \psi_{uu} \bar\psi_{dd} \\ \psi_{ud} \bar\psi_{uu} & \psi_{ud} \bar\psi_{ud} & \psi_{ud} \bar\psi_{du} & \psi_{ud} \bar\psi_{dd} \\ \psi_{du} \bar\psi_{uu} & \psi_{du} \bar\psi_{ud} & \psi_{du} \bar\psi_{du} & \psi_{du} \bar\psi_{dd} \\ \psi_{dd} \bar\psi_{uu} & \psi_{dd} \bar\psi_{ud} & \psi_{dd} \bar\psi_{du} & \psi_{dd} \bar\psi_{dd} \end{bmatrix} \\ & = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & \tfrac{1}{2} & -\tfrac{1}{2} & 0 \\ 0 & -\tfrac{1}{2} & \tfrac{1}{2} & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \end{aligned}
Computing \rho^2:
\begin{aligned} \rho^2 & = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & \tfrac{1}{2} & -\tfrac{1}{2} & 0 \\ 0 & -\tfrac{1}{2} & \tfrac{1}{2} & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & \tfrac{1}{2} & -\tfrac{1}{2} & 0 \\ 0 & -\tfrac{1}{2} & \tfrac{1}{2} & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \\ & = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & \tfrac{1}{2} & -\tfrac{1}{2} & 0 \\ 0 & -\tfrac{1}{2} & \tfrac{1}{2} & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} = \rho \end{aligned}
\operatorname {Tr}\left(\rho \right) = \operatorname {Tr}\left(\rho^2 \right) = 1 and therefore the system is in a pure state.
For the system A, the the reduced density matrix is:
\begin{aligned} \rho_{Auu} = & \bar \psi_{uu}\psi_{uu} + \bar \psi_{ud}\psi_{ud} = 0\,0 +\dfrac{1}{\sqrt 2}\dfrac{1}{\sqrt 2} = \dfrac{1}{2} \\ \rho_{Aud} = & \bar \psi_{du}\psi_{uu} + \bar \psi_{dd}\psi_{ud} = \dfrac{1}{\sqrt 2} \,0 + 0\,\dfrac{1}{\sqrt 2} = 0 \\ \rho_{Adu} = & \bar \psi_{uu}\psi_{du} + \bar \psi_{ud}\psi_{dd} = 0 \,\dfrac{1}{\sqrt 2} + \dfrac{1}{\sqrt 2}\,0 = 0 \\ \rho_{Add} = & \bar \psi_{du}\psi_{du} + \bar \psi_{dd}\psi_{dd} = \dfrac{1}{\sqrt 2}\dfrac{1}{\sqrt 2}+ 0\,0 = \dfrac{1}{2} \end{aligned}
In matrix form:
\rho_A = \begin{bmatrix} \dfrac{1}{2} & 0 \\[12pt] 0 & \dfrac{1}{2} \end{bmatrix}
Computing \rho_A^2:
\rho_A^2 = \begin{bmatrix} \tfrac{1}{2} & 0 \\ 0 & \tfrac{1}{2} \end{bmatrix} \begin{bmatrix} \tfrac{1}{2} & 0 \\ 0 & \tfrac{1}{2} \end{bmatrix} = \begin{bmatrix} \tfrac{1}{4} & 0 \\ 0 & \tfrac{1}{4} \end{bmatrix} \neq \rho_A
Since \operatorname {Tr}\left(\rho_A \right) = 1 and \operatorname {Tr}\left(\rho_B^2 \right) < 1 the system is in a entangled state.
The matrix is diagonal, the eigenvalues are on the diagonal:
\lambda_1 = \lambda_2 = \frac{1}{2}
For the system B, the the reduced density matrix is
\begin{aligned} \rho_{Buu} = & \bar \psi_{uu}\psi_{uu} + \bar \psi_{du}\psi_{du} = 0\,0 + \dfrac{1}{\sqrt 2}\dfrac{1}{\sqrt 2} = \dfrac{1}{2} \\ \rho_{Bud} = & \bar \psi_{ud}\psi_{uu} + \bar \psi_{dd}\psi_{du} = \dfrac{1}{\sqrt 2} \,0 + 0\, \dfrac{1}{\sqrt 2} = 0 \\ \rho_{Bdu} = & \bar \psi_{uu}\psi_{ud} + \bar \psi_{du}\psi_{dd} = 0\, \dfrac{1}{\sqrt 2} + 0\, \dfrac{1}{\sqrt 2} = 0 \\ \rho_{Bdd} = & \bar \psi_{ud}\psi_{ud} + \bar \psi_{dd}\psi_{dd} = \dfrac{1}{\sqrt 2}\dfrac{1}{\sqrt 2} + 0\,0 = \dfrac{1}{2} \end{aligned}
In matrix form:
\rho_B = \begin{bmatrix} \dfrac{1}{2} & 0 \\[12pt] 0 & \dfrac{1}{2} \end{bmatrix}
Computing \rho_B^2:
\rho_B^2 = \begin{bmatrix} \tfrac{1}{2} & 0 \\ 0 & \tfrac{1}{2} \end{bmatrix} \begin{bmatrix} \tfrac{1}{2} & 0 \\ 0 & \tfrac{1}{2} \end{bmatrix} = \begin{bmatrix} \tfrac{1}{4} & 0 \\ 0 & \tfrac{1}{4} \end{bmatrix} \neq \rho_B
Since \operatorname {Tr}\left(\rho_B \right) = 1 and \operatorname {Tr}\left(\rho_B^2 \right) < 1 the system is in a entangled state.
The matrix is diagonal, the eigenvalues are on the diagonal:
\lambda_1 = \lambda_2 = \frac{1}{2}
Starting from the system A:
\begin{aligned} \langle \sigma_{Az} \rangle = & \langle \bar \Psi | \sigma_{Az} | \Psi \rangle = \frac{1}{\sqrt 2}\left( \langle ud | - \langle du | \right) \sigma_{Az} \frac{1}{\sqrt 2}\left( | ud \rangle - | du \rangle \right) = \frac{1}{2}\left( \langle ud | - \langle du | \right) \left[\sigma_{Az} \left( | ud \rangle - | du \rangle \right)\right] \\ = & \frac{1}{2}\left( \langle ud | - \langle du | \right)\left( | ud \rangle + | du \rangle \right) = \frac{1}{2}\left(\langle ud | ud \rangle + \langle ud | du \rangle - \langle du | ud \rangle - \langle du | du \rangle \right) \\ = & \frac{1}{2}\left( 1 + 0 - 0 - 1 \right) = 0 \end{aligned} \begin{aligned} \langle \sigma_{Ax} \rangle = & \langle \bar \Psi | \sigma_{Ax} | \Psi \rangle = \frac{1}{\sqrt 2}\left( \langle ud | - \langle du | \right) \sigma_{Ax} \frac{1}{\sqrt 2}\left( | ud \rangle - | du \rangle \right) = \frac{1}{2}\left( \langle ud | - \langle du | \right) \left[\sigma_{Ax} \left( | ud \rangle - | du \rangle \right)\right] \\ = & \frac{1}{2}\left( \langle ud | - \langle du | \right)\left( | dd \rangle - | uu \rangle \right) = \frac{1}{2}\left(\langle ud | dd \rangle + \langle ud | uu \rangle - \langle du | dd \rangle - \langle du | uu \rangle \right) \\ = & \frac{1}{2}\left( 0 + 0 - 0 - 0 \right) = 0 \end{aligned} \begin{aligned} \langle \sigma_{Ay} \rangle = & \langle \bar \Psi | \sigma_{Ay} | \Psi \rangle = \frac{1}{\sqrt 2}\left( \langle ud | - \langle du | \right) \sigma_{Ay} \frac{1}{\sqrt 2}\left( | ud \rangle - | du \rangle \right) = \frac{1}{2}\left( \langle ud | - \langle du | \right) \left[\sigma_{Ay} \left( | ud \rangle - | du \rangle \right)\right] \\ = & \frac{1}{2}\left( \langle ud | - \langle du | \right)\left(i| dd \rangle +i| uu \rangle \right) = \frac{1}{2}\left(i\langle ud | dd \rangle + i\langle ud | uu \rangle + i\langle du | dd \rangle + i\langle du | uu \rangle \right) \\ = & \frac{1}{2}\left( 0 + 0 + 0i + 0i \right) = 0 \end{aligned}
Therefore:
\langle \sigma_{Az} \rangle = \langle \sigma_{Ax} \rangle = \langle \sigma_{Ay} \rangle = 0
Computing for the system B:
\begin{aligned} \langle \sigma_{Bz} \rangle = & \langle \bar \Psi | \sigma_{Bz} | \Psi \rangle = \frac{1}{\sqrt 2}\left( \langle ud | - \langle du | \right) \sigma_{Bz} \frac{1}{\sqrt 2}\left( | ud \rangle - | du \rangle \right) = \frac{1}{2}\left( \langle ud | - \langle du | \right) \left[\sigma_{Bz} \left( | ud \rangle - | du \rangle \right)\right] \\ = & \frac{1}{2}\left( \langle ud | - \langle du | \right)\left( -| ud \rangle - | du \rangle \right) = \frac{1}{2}\left(-\langle ud | ud \rangle - \langle ud | du \rangle + \langle du | ud \rangle + \langle du | du \rangle \right) \\ = & \frac{1}{2}\left( - 1 - 0 + 0 + 1 \right) = 0 \end{aligned} \begin{aligned} \langle \sigma_{Bx} \rangle = & \langle \bar \Psi | \sigma_{Bx} | \Psi \rangle = \frac{1}{\sqrt 2}\left( \langle ud | - \langle du | \right) \sigma_{Bx} \frac{1}{\sqrt 2}\left( | ud \rangle - | du \rangle \right) = \frac{1}{2}\left( \langle ud | - \langle du | \right) \left[\sigma_{Bx} \left( | ud \rangle - | du \rangle \right)\right] \\ = & \frac{1}{2}\left( \langle ud | - \langle du | \right)\left( | uu \rangle - | dd \rangle \right) = \frac{1}{2}\left(\langle ud | uu \rangle + \langle ud | dd \rangle - \langle du | uu \rangle - \langle du | dd \rangle \right) \\ = & \frac{1}{2}\left( 0 + 0 - 0 - 0 \right) = 0 \end{aligned} \begin{aligned} \langle \sigma_{By} \rangle = & \langle \bar \Psi | \sigma_{By} | \Psi \rangle = \frac{1}{\sqrt 2}\left( \langle ud | - \langle du | \right) \sigma_{By} \frac{1}{\sqrt 2}\left( | ud \rangle - | du \rangle \right) = \frac{1}{2}\left( \langle ud | - \langle du | \right) \left[\sigma_{By} \left( | ud \rangle - | du \rangle \right)\right] \\ = & \frac{1}{2}\left( \langle ud | - \langle du | \right)\left(-i| uu \rangle -i| dd \rangle \right) = \frac{1}{2}\left(-i\langle ud | uu \rangle - i\langle du | dd \rangle + i\langle du | uu \rangle + i\langle du | dd \rangle \right) \\ = & \frac{1}{2}\left( -0i - 0i + 0i + 0i \right) = 0 \end{aligned}
For the composite expectations:
\begin{aligned} \langle \sigma_{Az}\sigma_{Bz} \rangle & = \langle \bar \Psi | \sigma_{Az}\sigma_{Bz} | \Psi \rangle \\ & = \frac{1}{\sqrt 2}\left( \langle ud | - \langle du | \right) \sigma_{Az}\sigma_{Bz}\frac{1}{\sqrt 2}\left( | ud \rangle - | du \rangle \right) \\ & = \frac{1}{\sqrt 2}\left( \langle ud | - \langle du | \right) \sigma_{Az}\left[\sigma_{Bz}\frac{1}{\sqrt 2}\left( | ud \rangle - | du \rangle \right)\right] \\ & = \frac{1}{\sqrt 2}\left( \langle ud | - \langle du | \right) \left[ \sigma_{Az}\frac{1}{\sqrt 2}\left( - | ud \rangle - | du \rangle \right)\right] \\ & = \frac{1}{2}\left( \langle ud | - \langle du | \right)\left( - | ud \rangle + | du \rangle \right) \\ & = \frac{1}{2}\left( - \langle ud | ud \rangle + \langle du | ud \rangle + \langle ud | du \rangle - \langle du | du \rangle \right) \\ & = \frac{1}{2}\left( - 1 + 0 + 0 - 1 \right) = -1 \end{aligned} \begin{aligned} \langle \sigma_{Ax}\sigma_{Bx} \rangle & = \langle \bar \Psi | \sigma_{Ax}\sigma_{Bx} | \Psi \rangle \\ & = \frac{1}{\sqrt 2}\left( \langle ud | - \langle du | \right) \sigma_{Ax}\sigma_{Bx}\frac{1}{\sqrt 2}\left( | ud \rangle - | du \rangle \right) \\ & = \frac{1}{\sqrt 2}\left( \langle ud | - \langle du | \right) \sigma_{Ax}\left[\sigma_{Bx}\frac{1}{\sqrt 2}\left( | ud \rangle - | du \rangle \right)\right] \\ & = \frac{1}{\sqrt 2}\left( \langle ud | - \langle du | \right) \left[ \sigma_{Ax}\frac{1}{\sqrt 2}\left( | uu \rangle - | dd \rangle \right)\right] \\ & = \frac{1}{2}\left( \langle ud | - \langle du | \right)\left( - | ud \rangle + | du \rangle \right) \\ & = \frac{1}{2}\left( - \langle ud | ud \rangle + \langle du | ud \rangle + \langle ud | du \rangle - \langle du | du \rangle \right) \\ & = \frac{1}{2}\left( - 1 + 0 + 0 - 1 \right) = -1 \end{aligned} \begin{aligned} \langle \sigma_{Ay}\sigma_{By} \rangle & = \langle \bar \Psi | \sigma_{Ay}\sigma_{By} | \Psi \rangle \\ & = \frac{1}{\sqrt 2}\left( \langle ud | - \langle du | \right) \sigma_{Ay}\sigma_{By}\frac{1}{\sqrt 2}\left( | ud \rangle - | du \rangle \right) \\ & = \frac{1}{\sqrt 2}\left( \langle ud | - \langle du | \right) \sigma_{Ay}\left[\sigma_{By}\frac{1}{\sqrt 2}\left( | ud \rangle - | du \rangle \right)\right] \\ & = \frac{1}{\sqrt 2}\left( \langle ud | - \langle du | \right) \left[ \sigma_{Ay}\frac{1}{\sqrt 2}\left( -i| uu \rangle - i| dd \rangle \right)\right] \\ & = \frac{1}{2}\left( \langle ud | - \langle du | \right)\left( -i^2| du \rangle + i^2| ud \rangle \right) \\ & = \frac{1}{2}\left( \langle ud | - \langle du | \right)\left( - | ud \rangle + | du \rangle \right) \\ & = \frac{1}{2}\left( - \langle ud | ud \rangle + \langle du | ud \rangle + \langle ud | du \rangle - \langle du | du \rangle \right) \\ & = \frac{1}{2}\left( - 1 + 0 + 0 - 1 \right) = -1 \end{aligned}
Therefore:
\langle \sigma_{Bz} \rangle = \langle \sigma_{Bx} \rangle = \langle \sigma_{By} \rangle = 0
The composite expectation on the corresponding axis is -1.
Computing the cross expectations:
\begin{aligned} \langle \sigma_{Az}\sigma_{Bx} \rangle & = \langle \bar \Psi | \sigma_{Az}\sigma_{Bx} | \Psi \rangle \\ & = \frac{1}{\sqrt 2}\left( \langle ud | - \langle du | \right) \sigma_{Az}\sigma_{Bx}\frac{1}{\sqrt 2}\left( | ud \rangle - | du \rangle \right) \\ & = \frac{1}{\sqrt 2}\left( \langle ud | - \langle du | \right) \sigma_{Az}\left[\sigma_{Bx}\frac{1}{\sqrt 2}\left( | ud \rangle - | du \rangle \right)\right] \\ & = \frac{1}{\sqrt 2}\left( \langle ud | - \langle du | \right) \left[ \sigma_{Az}\frac{1}{\sqrt 2}\left( | uu \rangle - | dd \rangle \right)\right] \\ & = \frac{1}{2}\left( \langle ud | - \langle du | \right)\left( | uu \rangle + | dd \rangle \right) \\ & = \frac{1}{2}\left( \langle ud | uu \rangle - \langle du | uu \rangle + \langle ud | dd \rangle - \langle du | dd \rangle \right) \\ & = \frac{1}{2}\left( 0 - 0 + 0 - 0 \right) = 0 \end{aligned}
Similar results for the other combinations.
The correlations can be derived from the expectations computed above:
\langle \sigma_{Az}\sigma_{Bz} \rangle - \langle \sigma_{Az}\rangle \langle \sigma_{Bz} \rangle = -1 - 0\,0 = -1
| \Psi \rangle = \sqrt{0.6}|ud\rangle - \sqrt{0.4} |du\rangle
\begin{array}{ll} \psi_{uu} = 0 & \psi_{ud} = \sqrt{0.6} \\[12pt] \psi_{du} = - \sqrt{0.4} & \psi_{dd} = 0 \end{array}
For the composite system, the density matrix is given by:
\begin{aligned} \rho = | \Psi \rangle \langle \Psi | & = \begin{bmatrix} \psi_{uu} \bar\psi_{uu} & \psi_{uu} \bar\psi_{ud} & \psi_{uu} \bar\psi_{du} & \psi_{uu} \bar\psi_{dd} \\ \psi_{ud} \bar\psi_{uu} & \psi_{ud} \bar\psi_{ud} & \psi_{ud} \bar\psi_{du} & \psi_{ud} \bar\psi_{dd} \\ \psi_{du} \bar\psi_{uu} & \psi_{du} \bar\psi_{ud} & \psi_{du} \bar\psi_{du} & \psi_{du} \bar\psi_{dd} \\ \psi_{dd} \bar\psi_{uu} & \psi_{dd} \bar\psi_{ud} & \psi_{dd} \bar\psi_{du} & \psi_{dd} \bar\psi_{dd} \end{bmatrix} \\ & = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0.6 & -\sqrt {0.24} & 0 \\ 0 & -\sqrt {0.24} & 0.4 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \end{aligned}
Computing \rho^2:
\begin{aligned} \rho^2 & = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0.6 & -\sqrt {0.24} & 0 \\ 0 & -\sqrt {0.24} & 0.4 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0.6 & -\sqrt {0.24} & 0 \\ 0 & -\sqrt {0.24} & 0.4 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \\ & = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & (0.6)^2 + 0.24 & -(0.6+0.4)\sqrt {0.24} & 0 \\ 0 & -(0.6+0.4)\sqrt {0.24} & (0.4)^2 + 0.24 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0.6 & -\sqrt {0.24} & 0 \\ 0 & -\sqrt {0.24} & 0.4 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} = \rho \end{aligned}
\operatorname {Tr}\left(\rho \right) = \operatorname {Tr}\left(\rho^2 \right) = 1 and therefore the system is in a pure state.
For the system A, the reduced density matrix is:
\begin{aligned} \rho_{Auu} = & \bar \psi_{uu}\psi_{uu} + \bar \psi_{ud}\psi_{ud} = 0\,0 + \sqrt{0.6}\sqrt{0.6} = 0.6 \\ \rho_{Aud} = & \bar \psi_{du}\psi_{uu} + \bar \psi_{dd}\psi_{ud} = \sqrt{0.6}\,0 + 0\,\sqrt{0.6} = 0 \\ \rho_{Adu} = & \bar \psi_{uu}\psi_{du} + \bar \psi_{ud}\psi_{dd} = 0\,\sqrt{0.4} + \sqrt{0.4}\,0 = 0 \\ \rho_{Add} = & \bar \psi_{du}\psi_{du} + \bar \psi_{dd}\psi_{dd} = \sqrt{0.4}\sqrt{0.4}+ 0\,0 = 0.4 \end{aligned}
In matrix form:
\rho_A = \begin{bmatrix} 0.6 & 0\\ 0 & 0.4 \end{bmatrix}
Computing \rho_A^2:
\rho_A^2 = \begin{bmatrix} 0.6 & 0 \\ 0 & 0.4 \end{bmatrix} \begin{bmatrix} 0.6 & 0 \\ 0 & 0.4 \end{bmatrix} = \begin{bmatrix} 0.36 & 0 \\ 0 & 0.16 \end{bmatrix} \neq \rho_A
Since \operatorname {Tr}\left(\rho_A \right) = 1 and \operatorname {Tr}\left(\rho_A^2 \right) < 1 the system is in a entangled state.
The matrix is diagonal, the eigenvalues are on the diagonal:
\lambda_1 = 0.6, \quad \lambda_2 = 0.4
For the system B the reduced density matrix is:
\begin{aligned} \rho_{Buu} = & \bar \psi_{uu}\psi_{uu} + \bar \psi_{du}\psi_{du} = 0\,0 + \sqrt{0.4} \sqrt {0.4} = 0.4 \\ \rho_{Bud} = & \bar \psi_{ud}\psi_{uu} + \bar \psi_{dd}\psi_{du} = \sqrt {0.6}\,0 + 0\,\sqrt {0.4} = 0 \\ \rho_{Bdu} = & \bar \psi_{uu}\psi_{ud} + \bar \psi_{du}\psi_{dd} = 0\,\sqrt {0.6} + \sqrt {0.4}\,0 = 0 \\ \rho_{Bdd} = & \bar \psi_{ud}\psi_{ud} + \bar \psi_{dd}\psi_{dd} = \sqrt {0.6}\sqrt {0.6} + 0\,0 = 0.6 \end{aligned}
In matrix form:
\rho_B = \begin{bmatrix} 0.4 & 0\\ 0 & 0.6 \end{bmatrix}
Computing \rho_B^2:
\rho_B^2 = \begin{bmatrix} 0.4 & 0 \\ 0 & 0.6 \end{bmatrix} \begin{bmatrix} 0.4 & 0 \\ 0 & 0.6 \end{bmatrix} = \begin{bmatrix} 0.16 & 0 \\ 0 & 0.36 \end{bmatrix} \neq \rho_B
Since \operatorname {Tr}\left(\rho_B \right) = 1 and \operatorname {Tr}\left(\rho_B^2 \right) < 1 the system is in a entangled state.
The matrix is diagonal, the eigenvalues are on the diagonal:
\lambda_1 = 0.6, \quad \lambda_2 = 0.4
Starting from the system A:
\begin{aligned} \langle \sigma_{Az} \rangle = & \langle \bar \Psi | \sigma_{Az} | \Psi \rangle = \left( \sqrt{0.6}\langle ud | - \sqrt{0.4} \langle du | \right) \sigma_{Az} \left( \sqrt{0.6}|ud\rangle - \sqrt{0.4} |du\rangle \right) \\ = & \left( \sqrt{0.6}\langle ud | - \sqrt{0.4}\langle du | \right) \left[\sigma_{Az} \left( \sqrt{0.6}| ud \rangle - \sqrt{0.4}| du \rangle \right)\right] \\ = & \left(\sqrt{0.6}\langle ud | - \sqrt{0.4}\langle du | \right)\left( \sqrt{0.6}| ud \rangle + \sqrt{0.4} | du \rangle \right) \\ = & \left(0.6 \langle ud | ud \rangle + \sqrt{0.24}\langle ud | du \rangle - \sqrt{0.24} \langle du | ud \rangle - 0.4 \langle du | du \rangle \right) \\ = & \left( 0.6 + 0 - 0 - 0.4 \right) = 0.2 \end{aligned} \begin{aligned} \langle \sigma_{Ax} \rangle = & \langle \bar \Psi | \sigma_{Ax} | \Psi \rangle = \left( \sqrt{0.6}\langle ud | - \sqrt{0.4} \langle du | \right) \sigma_{Ax} \left( \sqrt{0.6}|ud\rangle - \sqrt{0.4} |du\rangle \right) \\ = & \left( \sqrt{0.6}\langle ud | - \sqrt{0.4}\langle du | \right) \left[\sigma_{Ax} \left( \sqrt{0.6}| ud \rangle - \sqrt{0.4}| du \rangle \right)\right] \\ = & \left(\sqrt{0.6}\langle ud | - \sqrt{0.4}\langle du | \right)\left( \sqrt{0.6}| dd \rangle - \sqrt{0.4} | uu \rangle \right) \\ = & \left(0.6 \langle ud | dd \rangle - \sqrt{0.24}\langle ud | uu \rangle - \sqrt{0.24} \langle du | uu \rangle + 0.4 \langle du | dd \rangle \right) \\ = & \left( 0 - 0 - 0 + 0 \right) = 0 \end{aligned} \begin{aligned} \langle \sigma_{Ay} \rangle = & \langle \bar \Psi | \sigma_{Ay} | \Psi \rangle = \left( \sqrt{0.6}\langle ud | - \sqrt{0.4} \langle du | \right) \sigma_{Ay} \left( \sqrt{0.6}|ud\rangle - \sqrt{0.4} |du\rangle \right) \\ = & \left( \sqrt{0.6}\langle ud | - \sqrt{0.4}\langle du | \right) \left[\sigma_{Ay} \left( \sqrt{0.6}| ud \rangle - \sqrt{0.4}| du \rangle \right)\right] \\ = & \left(\sqrt{0.6}\langle ud | - \sqrt{0.4}\langle du | \right)\left( \sqrt{0.6}i| dd \rangle + \sqrt{0.4} i| uu \rangle \right) \\ = & \left(0.6i \langle ud | dd \rangle + \sqrt{0.24}i\langle ud | uu \rangle - \sqrt{0.24}i \langle du | dd \rangle - 0.4i \langle du | uu \rangle \right) \\ = & \left( 0i + 0i - 0i - 0i \right) = 0 \end{aligned}
Computing for the system B:
\begin{aligned} \langle \sigma_{Bz} \rangle = & \langle \bar \Psi | \sigma_{Bz} | \Psi \rangle = \left( \sqrt{0.6}\langle ud | - \sqrt{0.4} \langle du | \right) \sigma_{Bz} \left( \sqrt{0.6}|ud\rangle - \sqrt{0.4} |du\rangle \right) \\ = & \left( \sqrt{0.6}\langle ud | - \sqrt{0.4}\langle du | \right) \left[\sigma_{Bz} \left( \sqrt{0.6}| ud \rangle - \sqrt{0.4}| du \rangle \right)\right] \\ = & \left(\sqrt{0.6}\langle ud | - \sqrt{0.4}\langle du | \right)\left( -\sqrt{0.6}| ud \rangle - \sqrt{0.4} | du \rangle \right) \\ = & \left(-0.6 \langle ud | ud \rangle - \sqrt{0.24}\langle ud | du \rangle + \sqrt{0.24} \langle du | ud \rangle + 0.4 \langle du | du \rangle \right) \\ = & \left( -0.6 - 0 + 0 + 0.4 \right) = -0.2 \end{aligned} \begin{aligned} \langle \sigma_{Bx} \rangle = & \langle \bar \Psi | \sigma_{Bx} | \Psi \rangle = \left( \sqrt{0.6}\langle ud | - \sqrt{0.4} \langle du | \right) \sigma_{Bx} \left( \sqrt{0.6}|ud\rangle - \sqrt{0.4} |du\rangle \right) \\ = & \left( \sqrt{0.6}\langle ud | - \sqrt{0.4}\langle du | \right) \left[\sigma_{Bx} \left( \sqrt{0.6}| ud \rangle - \sqrt{0.4}| du \rangle \right)\right] \\ = & \left(\sqrt{0.6}\langle ud | - \sqrt{0.4}\langle du | \right)\left( \sqrt{0.6}| uu \rangle - \sqrt{0.4} | dd \rangle \right) \\ = & \left(0.6 \langle ud | uu \rangle - \sqrt{0.24}\langle ud | dd \rangle - \sqrt{0.24} \langle du | dd \rangle - 0.4 \langle du | dd \rangle \right) \\ = & \left( 0 + 0 - 0 - 0 \right) = 0 \end{aligned} \begin{aligned} \langle \sigma_{By} \rangle = & \langle \bar \Psi | \sigma_{By} | \Psi \rangle = \left( \sqrt{0.6}\langle ud | - \sqrt{0.4} \langle du | \right) \sigma_{By} \left( \sqrt{0.6}|ud\rangle - \sqrt{0.4} |du\rangle \right) \\ = & \left( \sqrt{0.6}\langle ud | - \sqrt{0.4}\langle du | \right) \left[\sigma_{By} \left( \sqrt{0.6}| ud \rangle - \sqrt{0.4}| du \rangle \right)\right] \\ = & \left(\sqrt{0.6}\langle ud | - \sqrt{0.4}\langle du | \right)\left( -\sqrt{0.6}i| uu \rangle - \sqrt{0.4} i| dd \rangle \right) \\ = & \left(-0.6i \langle ud | uu \rangle - \sqrt{0.24}i\langle ud | dd \rangle + \sqrt{0.24}i \langle du | uu \rangle + 0.4i \langle du | dd \rangle \right) \\ = & \left( - 0i - 0i + 0i + 0i \right) = 0 \end{aligned}
For the composite expectations:
\begin{aligned} \langle \sigma_{Az}\sigma_{Bz} \rangle & = \langle \bar \Psi | \sigma_{Az}\sigma_{Bz} | \Psi \rangle \\ & = \left(\sqrt{0.6}\langle ud | - \sqrt{0.4}\langle du | \right) \sigma_{Az}\sigma_{Bz}\left( \sqrt{0.6}|ud\rangle - \sqrt{0.4} |du\rangle \right) \\ & = \left(\sqrt{0.6}\langle ud | - \sqrt{0.4}\langle du | \right) \sigma_{Az}\left[\sigma_{Bz}\left( \sqrt{0.6}|ud\rangle - \sqrt{0.4} |du\rangle \right) \right] \\ & = \left(\sqrt{0.6}\langle ud | - \sqrt{0.4}\langle du | \right) \left[ \sigma_{Az}\left( -\sqrt{0.6}|ud\rangle - \sqrt{0.4} |du\rangle \right)\right] \\ & = \left(\sqrt{0.6}\langle ud | - \sqrt{0.4}\langle du | \right) \left( -\sqrt{0.6}|ud\rangle + \sqrt{0.4} |du\rangle \right) \\ & = \left( - 0.6 \langle ud | ud \rangle + \sqrt{0.24} \langle du | ud \rangle + \sqrt{0.24} \langle ud | du \rangle - 0.4\langle du | du \rangle \right) \\ & = \left( - 0.6 + 0 + 0 - 0.4 \right) = -1 \end{aligned} \begin{aligned} \langle \sigma_{Ax}\sigma_{Bx} \rangle & = \langle \bar \Psi | \sigma_{Ax}\sigma_{Bx} | \Psi \rangle \\ & = \left(\sqrt{0.6}\langle ud | - \sqrt{0.4}\langle du | \right) \sigma_{Ax}\sigma_{Bx}\left( \sqrt{0.6}|ud\rangle - \sqrt{0.4} |du\rangle \right) \\ & = \left(\sqrt{0.6}\langle ud | - \sqrt{0.4}\langle du | \right) \sigma_{Ax}\left[\sigma_{Bx}\left( \sqrt{0.6}|ud\rangle - \sqrt{0.4} |du\rangle \right) \right] \\ & = \left(\sqrt{0.6}\langle ud | - \sqrt{0.4}\langle du | \right) \left[ \sigma_{Ax}\left( \sqrt{0.6}|uu\rangle - \sqrt{0.4} |dd\rangle \right)\right] \\ & = \left(\sqrt{0.6}\langle ud | - \sqrt{0.4}\langle du | \right) \left( \sqrt{0.6}|du\rangle - \sqrt{0.4} |ud\rangle \right) \\ & = \left( 0.6 \langle ud | du \rangle - \sqrt{0.24} \langle du | du \rangle - \sqrt{0.24} \langle ud | ud \rangle + 0.4\langle du | ud \rangle \right) \\ & = \left( 0 - \sqrt{0.24} - \sqrt{0.24} + 0 \right) = -2 \sqrt{0.24} \end{aligned} \begin{aligned} \langle \sigma_{Ay}\sigma_{By} \rangle & = \langle \bar \Psi | \sigma_{Ay}\sigma_{By} | \Psi \rangle \\ & = \left(\sqrt{0.6}\langle ud | - \sqrt{0.4}\langle du | \right) \sigma_{Ay}\sigma_{By}\left( \sqrt{0.6}|ud\rangle - \sqrt{0.4} |du\rangle \right) \\ & = \left(\sqrt{0.6}\langle ud | - \sqrt{0.4}\langle du | \right) \sigma_{Ay}\left[\sigma_{By}\left( \sqrt{0.6}|ud\rangle - \sqrt{0.4} |du\rangle \right) \right] \\ & = \left(\sqrt{0.6}\langle ud | - \sqrt{0.4}\langle du | \right) \left[ \sigma_{Ay}\left( -i\sqrt{0.6}|uu\rangle - i\sqrt{0.4} |dd\rangle \right)\right] \\ & = \left(\sqrt{0.6}\langle ud | - \sqrt{0.4}\langle du | \right)\left( -i^2 \sqrt{0.6}|du\rangle + i^2 \sqrt{0.4} |ud\rangle \right) \\ & = \left(\sqrt{0.6}\langle ud | - \sqrt{0.4}\langle du | \right)\left(- \sqrt{0.4} |ud\rangle + \sqrt{0.6}|du\rangle \right) \\ & = - \sqrt{0.24} \langle ud | ud \rangle - 0.4 \langle ud | ud \rangle + 0.6 \langle ud | du \rangle - \sqrt{0.24} \langle du | du \rangle \\ & = \left( 0 - \sqrt{0.24} - \sqrt{0.24} + 0 \right) = -2 \sqrt{0.24} \end{aligned}
Computing the cross expectations:
\begin{aligned} \langle \sigma_{Az}\sigma_{Bx} \rangle & = \langle \bar \Psi | \sigma_{Az}\sigma_{Bx} | \Psi \rangle \\ & = \left(\sqrt{0.6}\langle ud | - \sqrt{0.4}\langle du | \right) \sigma_{Az}\sigma_{Bx}\left( \sqrt{0.6}|ud\rangle - \sqrt{0.4} |du\rangle \right) \\ & = \left(\sqrt{0.6}\langle ud | - \sqrt{0.4}\langle du | \right) \sigma_{Az}\left[\sigma_{Bx}\left( \sqrt{0.6}|ud\rangle - \sqrt{0.4} |du\rangle \right) \right] \\ & = \left(\sqrt{0.6}\langle ud | - \sqrt{0.4}\langle du | \right) \left[ \sigma_{Az}\left( \sqrt{0.6}|uu\rangle - \sqrt{0.4} |dd\rangle \right)\right] \\ & = \left(\sqrt{0.6}\langle ud | - \sqrt{0.4}\langle du | \right) \left( \sqrt{0.6}|uu\rangle + \sqrt{0.4} |dd\rangle \right) \\ & = \left( 0.6 \langle ud | uu \rangle - \sqrt{0.24} \langle du | dd \rangle + \sqrt{0.24} \langle ud | uu \rangle - 0.4\langle du | dd \rangle \right) \\ & = \left( 0 - 0 + 0 - 0 \right) = 0 \end{aligned}
The correlations can be derived from the expectations computed above:
\begin{aligned} & \langle \sigma_{Az}\sigma_{Bz} \rangle - \langle \sigma_{Az}\rangle \langle \sigma_{Bz} \rangle = -1 -(0.2\cdot-0.2)=- 0.96 \\ & \langle \sigma_{Ax}\sigma_{Bx} \rangle - \langle \sigma_{Ax}\rangle \langle \sigma_{Bx} \rangle = -2 \sqrt{0.24}\approx - 0.98 \\ & \langle \sigma_{Ay}\sigma_{By} \rangle - \langle \sigma_{Ay}\rangle \langle \sigma_{By} \rangle = -2 \sqrt{0.24} \approx - 0.98 \end{aligned}