Prove that if P(a, b) factorizes, then the correlation between a and b is zero.
By definition:
\langle L \rangle = \sum_i \lambda_i P(\lambda_i)
Therefore:
\begin{aligned} & \langle \sigma_a \sigma_b \rangle = \sum_{i,j} a_i b_j P( a_i,b_j) \\ & \langle \sigma_a \rangle \langle \sigma_b \rangle = \sum_{i} a_iP(a_i) \sum_j b_j P(b_j) = \sum_{i,j} a_ib_jP(a_i)P(b_j) \end{aligned}
And the correlation is:
\langle \sigma_A \sigma_B \rangle - \langle \sigma_A \rangle \langle \sigma_B \rangle = \sum_{i,j} a_i b_j P( a_i,b_j) - \sum_{i,j} a_ib_jP(a_i)P(b_j)) = \sum_{i,j} a_ib_j\left(P( a_i,b_j) - P(a_i)P(b_j) \right) = 0
Show that if the two normalization conditions of Eqs. 6.4 are satisfied, then the state-vector of Eq. 6.5 is automatically normalized as well. In other words, show that for this product state, normalizing the overall state-vector does not put any additional constraints on the \alpha’s and the \beta’s.
If the systems are prepared (and assuming they are normalized) as:
\begin{aligned} & | \psi_A \rangle = \psi_{Au} | u \rangle + \psi_{Ad} | d \rangle \\ & | \psi_B \rangle = \psi_{Bu} | u \rangle + \psi_{Bd} | d \rangle \\ & \bar\psi_{Au}\psi_{Au} + \bar\psi_{Ad}\psi_{Ad} = 1 \\ & \bar\psi_{Bu}\psi_{Bu} + \bar\psi_{Bd}\psi_{Bd} = 1 \end{aligned}
then:
\begin{aligned} |\Psi_{AB} \rangle & = | \psi_A \rangle \otimes | \psi_B \rangle = \left( \psi_{Au} | u \rangle + \psi_{Ad} | d \rangle \right) \otimes \left( \psi_{Bu} | u \rangle + \psi_{Bd} | d \rangle \right) \\ & = \psi_{Au}\psi_{Bu} | u u \rangle + \psi_{Au}\psi_{Bd} | u d \rangle + \psi_{Ad}\psi_{Bu} | d u \rangle + \psi_{Ad}\psi_{Bd} | d d \rangle \end{aligned}
As the components of each state are normalized, the components of the combined system are normalized as well and therefore there are no additional constraints on the \psi_{ij}:
\begin{aligned} \langle \Psi_{AB} |\Psi_{AB} \rangle = & \left(\bar\psi_{Au}\bar\psi_{Bu} \langle u u | + \bar\psi_{Au}\bar\psi_{Bd} \langle u d | + \bar\psi_{Ad}\bar\psi_{Bu} \langle d u | + \bar\psi_{Ad}\bar\psi_{Bd} \langle d d | \right) \\ & \left(\psi_{Au}\psi_{Bu} | u u \rangle + \psi_{Au}\psi_{Bd} | u d \rangle + \psi_{Ad}\psi_{Bu} | d u \rangle + \psi_{Ad}\psi_{Bd} | d d \rangle\right) \\ = & \bar\psi_{Au}\bar\psi_{Bu}\psi_{Au}\psi_{Bu} \langle u u | u u \rangle + \bar\psi_{Au}\bar\psi_{Bd}\psi_{Au}\psi_{Bd} \langle u d | u d \rangle \\ & + \bar\psi_{Ad}\bar\psi_{Bu}\psi_{Ad}\psi_{Bu} \langle d u | d u \rangle + \bar\psi_{Ad}\bar\psi_{Bd}\psi_{Ad}\psi_{Bd} \langle d d | d d \rangle \\ & + \bar\psi_{Au}\bar\psi_{Bu}\psi_{Ad}\psi_{Bd} \langle u u | d d \rangle + \bar\psi_{Au}\bar\psi_{Bd}\psi_{Au}\psi_{Bu} \langle u d | u u \rangle \\ & + \bar\psi_{Ad}\bar\psi_{Bu}\psi_{Au}\psi_{Bd} \langle d u | u d \rangle + \bar\psi_{Ad}\bar\psi_{Bd}\psi_{Ad}\psi_{Bu} \langle d d | d u \rangle \\ & + \bar\psi_{Au}\bar\psi_{Bu}\psi_{Ad}\psi_{Bu} \langle u u | d u \rangle + \bar\psi_{Au}\bar\psi_{Bd}\psi_{Ad}\psi_{Bd} \langle u d | d d \rangle \\ & + \bar\psi_{Ad}\bar\psi_{Bu}\psi_{Au}\psi_{Bu} \langle d u | u u \rangle + \bar\psi_{Ad}\bar\psi_{Bd}\psi_{Au}\psi_{Bd} \langle d d | u d \rangle \\ & + \bar\psi_{Au}\bar\psi_{Bu}\psi_{Au}\psi_{Bd} \langle u u | u d \rangle + \bar\psi_{Au}\bar\psi_{Bd}\psi_{Ad}\psi_{Bu} \langle u d | d u \rangle \\ & + \bar\psi_{Ad}\bar\psi_{Bu}\psi_{Ad}\psi_{Bd} \langle d u | d d \rangle + \bar\psi_{Ad}\bar\psi_{Bd}\psi_{Au}\psi_{Bu} \langle d d | u u \rangle \\ = & \bar\psi_{Au}\bar\psi_{Bu}\psi_{Au}\psi_{Bu} 1 + \bar\psi_{Au}\bar\psi_{Bd}\psi_{Au}\psi_{Bd} 1 + \bar\psi_{Ad}\bar\psi_{Bu}\psi_{Ad}\psi_{Bu} 1 + \bar\psi_{Ad}\bar\psi_{Bd}\psi_{Ad}\psi_{Bd} 1 \\ & + \bar\psi_{Au}\bar\psi_{Bu}\psi_{Ad}\psi_{Bd} 0 + \bar\psi_{Au}\bar\psi_{Bd}\psi_{Au}\psi_{Bu} 0 + \bar\psi_{Ad}\bar\psi_{Bu}\psi_{Au}\psi_{Bd} 0 + \bar\psi_{Ad}\bar\psi_{Bd}\psi_{Ad}\psi_{Bu} 0 \\ & + \bar\psi_{Au}\bar\psi_{Bu}\psi_{Ad}\psi_{Bu} 0 + \bar\psi_{Au}\bar\psi_{Bd}\psi_{Ad}\psi_{Bd} 0 + \bar\psi_{Ad}\bar\psi_{Bu}\psi_{Au}\psi_{Bu} 0 + \bar\psi_{Ad}\bar\psi_{Bd}\psi_{Au}\psi_{Bd} 0 \\ & + \bar\psi_{Au}\bar\psi_{Bu}\psi_{Au}\psi_{Bd} 0 + \bar\psi_{Au}\bar\psi_{Bd}\psi_{Ad}\psi_{Bu} 0 + \bar\psi_{Ad}\bar\psi_{Bu}\psi_{Ad}\psi_{Bd} 0 + \bar\psi_{Ad}\bar\psi_{Bd}\psi_{Au}\psi_{Bu} 0 \\ = & \bar\psi_{Au}\bar\psi_{Bu}\psi_{Au}\psi_{Bu} + \bar\psi_{Au}\bar\psi_{Bd}\psi_{Au}\psi_{Bd} + \bar\psi_{Ad}\bar\psi_{Bu}\psi_{Ad}\psi_{Bu} + \bar\psi_{Ad}\bar\psi_{Bd}\psi_{Ad}\psi_{Bd} \\ = & \bar\psi_{Au}\psi_{Au}\big(\underbrace{\bar\psi_{Bu}\psi_{Bu} + \bar\psi_{Bd}\psi_{Bd}}_{1} \big) + \bar\psi_{Ad}\psi_{Ad}\big(\underbrace{\bar\psi_{Bu}\psi_{Bu} + \bar\psi_{Bd}\psi_{Bd}}_{1} \big) \\ = & \bar\psi_{Au}\psi_{Au} + \bar\psi_{Ad}\psi_{Ad} = 1 \end{aligned}
Prove that the state | sing \rangle cannot be written as a product state.
The singlet state can be written as:
| S \rangle = \frac{1}{\sqrt 2}\left( | ud \rangle - | du \rangle \right)
This cannot be written a product state; a generic state for a product state can be written as:
|\Psi_{AB} \rangle = \psi_{Au}\psi_{Bu} | u u \rangle + \psi_{Au}\psi_{Bd} | u d \rangle + \psi_{Ad}\psi_{Bu} | d u \rangle + \psi_{Ad}\psi_{Bd} | d d \rangle
For the singlet it would then give the equations:
\begin{aligned} & \psi_{Au}\psi_{Bu} | u u \rangle = 0 | u u \rangle \\ & \psi_{Au}\psi_{Bd} | u d \rangle = \frac{1}{\sqrt 2} | u d \rangle \\ & \psi_{Ad}\psi_{Bu} | d u \rangle = - \frac{1}{\sqrt 2} | d u \rangle \\ & \psi_{Ad}\psi_{Bd} | d d \rangle = 0 | d d \rangle \end{aligned}
From the above, if \psi_{Au}\psi_{Bu} = 0, at least either of \psi_{Au} or \psi_{Bu} must be 0, but then at least one between:
\begin{aligned} & \psi_{Au}\psi_{Bd} = \frac{1}{\sqrt 2} \\ & \psi_{Ad}\psi_{Bu} = - \frac{1}{\sqrt 2} \end{aligned}
cannot be true as one of the factors is 0.
There is another way to demonstrate it: for a single spin, it is not possible to have that the expectation along \sigma_x, \sigma_y and \sigma_z are all zero at the same time, as it exists a state \vec \sigma \cdot \vec n in which the expectation is +1 and therefore:
\langle \sigma_x \rangle^2 + \langle \sigma_y \rangle^2 + \langle \sigma_z \rangle^2 = 1
And these cannot be 0 at the same time. For example, computing the expectation of the signet for with \sigma_{Az}:
\begin{aligned} \langle \sigma_{Az} \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}
Computing the expectation for \sigma_{Ax}:
\begin{aligned} \langle \sigma_{Ax} \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}
Finally, computing the expectation for \sigma_{Ay}:
\begin{aligned} \langle \sigma_{Ay} \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
And therefore this cannot be separated in ta single system; a similar results would have been reached with \langle \sigma_{Bi} \rangle; this is precisely the meaning of entanglement, which is a system that cannot be separated into its single components.
Use the matrix forms of \sigma_z, \sigma_x, and \sigma_y and the column vectors for | u\} and |d\} to verify Eqs. 6.6. Then, use Eqs. 6.6 and 6.7 to write the equations that were left out of Eqs. 6.8. Use the appendix to check your answers.
Using the matrix form:
\begin{aligned} \sigma_z | u \} &= \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 1\\ 0 \end{pmatrix} = \begin{pmatrix} 1\\ 0 \end{pmatrix} = | u \} \\ \sigma_z | d \} &= \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 0\\ 1 \end{pmatrix} = \begin{pmatrix} 0\\ -1 \end{pmatrix} = -| d \} \\ \sigma_x | u \} &= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1\\ 0 \end{pmatrix} = \begin{pmatrix} 0\\ 1 \end{pmatrix} = | d \} \\ \sigma_x | d \} &= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0\\ 1 \end{pmatrix} = \begin{pmatrix} 1\\ 0 \end{pmatrix} = | u \} \\ \sigma_y | u \} &= \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \begin{pmatrix} 1\\ 0 \end{pmatrix} = \begin{pmatrix} 0\\ i \end{pmatrix} = i| d \} \\ \sigma_y | d \} &= \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \begin{pmatrix} 0\\ 1 \end{pmatrix} = \begin{pmatrix}-i\\ 0 \end{pmatrix} = -i| u \}. \end{aligned}
The changes for Bob are the same as above when replacing \sigma with \tau.
Following these relations for the operations on the product states, when applying Alice’s \sigma operators, the first letter in the ket vector is transformed according to equations above leaving Bob state unchanged:
\begin{array}{ll} \sigma_z | uu \rangle = | uu \rangle & \sigma_z | ud \rangle = | ud \rangle \\ \sigma_z | du \rangle = -| du \rangle & \sigma_z | dd \rangle = -| dd \rangle \\ \sigma_x | uu \rangle = | du \rangle & \sigma_x | ud \rangle = | dd \rangle \\ \sigma_x | du \rangle = | uu \rangle & \sigma_x | dd \rangle = | ud \rangle \\ \sigma_y | uu \rangle = i| du \rangle & \sigma_y | ud \rangle = i| dd \rangle \\ \sigma_y | du \rangle = -i| uu \rangle & \sigma_y | dd \rangle = -i| ud \rangle \end{array}
For Bob’s operators, leaving Alice’s state unchanged:
\begin{array}{ll} \tau_z | uu \rangle = | uu \rangle & \tau_z | du \rangle = | du \rangle \\ \tau_z | ud \rangle = -| ud \rangle & \tau_z | dd \rangle = -| dd \rangle \\ \tau_x | uu \rangle = | ud \rangle & \tau_x | du \rangle = | dd \rangle \\ \tau_x | ud \rangle = | uu \rangle & \tau_x | dd \rangle = | du \rangle \\ \tau_y | uu \rangle = i| ud \rangle & \tau_y | du \rangle = i| dd \rangle \\ \tau_y | ud \rangle = -i| uu \rangle & \tau_y | dd \rangle = -i| du \rangle \\ \end{array}
Prove the following theorem:
When any of Alice’s or Bob’s spin operators acts on a product state, the result is still a product state.
Show that in a product state, the expectation value of any component of \vec{\sigma} or \vec{\tau} is exactly the same as it would be in the individual single-spin states.
Considering the state of Alice as:
| \psi_A \rangle = \alpha_u | u \} + \alpha_d | d \}
And the state of Bob as:
|\psi_B \rangle = \beta_u | u \rangle + \beta_d | d \rangle
The product state is:
\begin{aligned} |\Psi_{AB} \rangle & = | \psi_A \rangle \otimes | \psi_B \rangle = \left( \psi_{Au} | u \} + \psi_{Ad} | d \} \right) \otimes \left( \psi_{Bu} | u \rangle + \psi_{Bd} | d \rangle \right) \\ & = \alpha_u \beta_u | u u \rangle + \alpha_u \beta_d | u d \rangle + \alpha_d \beta_u | d u \rangle + \alpha_d \beta_d | d d \rangle \end{aligned}
In general, if \sigma_i is one of Alice’s spin operators and | \Psi_{AB} \rangle = | a \} \otimes | b \rangle is a product state, then:
\sigma_i | \Psi_{AB} \rangle = (\sigma_i \otimes I) (| a \} \otimes | b \rangle) = | c_i \} \otimes | b \rangle = | \Psi_{C_iB} \rangle
Explicit the calculations, starting with \sigma_x:
\begin{aligned} (\sigma_x \otimes I) | \Psi_{AB} \rangle & = (\sigma_x \otimes I) (\alpha_u| u \} + \alpha_d | d \}) \otimes (\beta_u| u \rangle + \beta_d | d \rangle) = \sigma_x (\alpha_u| u \} + \alpha_d | d \}) \otimes I (\beta_u| u \rangle + \beta_d | d \rangle) \\ & = (\alpha_u \sigma_x | u \} + \alpha_d \sigma_x | d \}) \otimes (\beta_u I| u \rangle + \beta_d I | d \rangle) = (\alpha_u | d \} + \alpha_d | u \}) \otimes (\beta_u | u \rangle + \beta_d | d \rangle) \end{aligned}
Bob’s state is unchanged and this is still a product state as it is a tensor product (with a different state for Alice |\psi_c \rangle = \alpha_u | d \} + \alpha_d | u \}).
Then \sigma_y:
\begin{aligned} (\sigma_y \otimes I) | \Psi_{AB} \rangle & = (\sigma_y \otimes I) (\alpha_u| u \} + \alpha_d | d \}) \otimes (\beta_u| u \rangle + \beta_d | d \rangle) = \sigma_y (\alpha_u| u \} + \alpha_d | d \}) \otimes I (\beta_u| u \rangle + \beta_d | d \rangle) \\ & = (\alpha_u \sigma_y | u \} + \alpha_d \sigma_y | d \}) \otimes (\beta_u I| u \rangle + \beta_d I | d \rangle) = (i\alpha_u | d \} - i\alpha_d| u \}) \otimes (\beta_u | u \rangle + \beta_d | d \rangle) \end{aligned}
Bob’s state is unchanged and this is still a product state as it is a tensor product (with a different state for Alice |\psi_c \rangle = i\alpha_u | d \} - i\alpha_d| u \}).
Finally \sigma_z:
\begin{aligned} (\sigma_z \otimes I) | \Psi_{AB} \rangle & = (\sigma_z \otimes I) (\alpha_u| u \} + \alpha_d | d \}) \otimes (\beta_u| u \rangle + \beta_d | d \rangle) = \sigma_z (\alpha_u| u \} + \alpha_d | d \}) \otimes I (\beta_u| u \rangle + \beta_d | d \rangle)\\ & = (\alpha_u \sigma_z | u \} + \alpha_d \sigma_z | d \}) \otimes (\beta_u I| u \rangle + \beta_d I | d \rangle) = (\alpha_u | u \} - \alpha_d | d \}) \otimes (\beta_u | u \rangle + \beta_d | d \rangle) \end{aligned}
Bob’s state is unchanged and this is still a product state as it is a tensor product (with a different state for Alice |\psi_c \rangle = \alpha_u | u \} - \alpha_d | d \}).
Similarly for Bob’s spin operators:
\tau_i | \Psi_{AB} \rangle = (I \otimes \tau_i) (| a \} \otimes | b \rangle) = | a \} \otimes | d_i \rangle = | \Psi_{AD_i} \rangle
The same calculation as above can be easily repeated with simple substitutions.
For the expectation values:
For \sigma_x:
\begin{aligned} \langle \sigma_x \rangle & = \langle \Psi_{AB} |\sigma_x | \Psi_{AB} \rangle = \langle ab | \sigma_x \left[ \alpha_u \beta_u | uu \rangle + \alpha_u \beta_d | ud \rangle + \alpha_d \beta_u | du \rangle + \alpha_d \beta_d | dd \rangle \right] \\ & = \langle ab | \left[ \alpha_u \beta_u | du \rangle + \alpha_u \beta_d | dd \rangle + \alpha_d \beta_u | uu \rangle + \alpha_d \beta_d | ud \rangle \right] \\ & = \left[ \bar \alpha_u \bar \beta_u\langle uu | + \bar \alpha_u \bar \beta_d\langle ud | + \bar \alpha_d \bar \beta_u\langle du | + \bar \alpha_d \bar \beta_d\langle dd | \right] \\ & = \bar \alpha_u \bar \beta_u \alpha_d \beta_u + \bar \alpha_u \bar \beta_d \alpha_d \beta_d + \bar \alpha_d \bar \beta_u \alpha_u \beta_u + \bar \alpha_d \bar \beta_d \alpha_u \beta_d \\ & = (\bar \beta_u\beta_u)(\bar \alpha_u\alpha_d + \bar \alpha_d\alpha_u ) + (\bar \beta_d\beta_d)(\bar \alpha_u\alpha_d + \bar \alpha_d\alpha_u ) \\ & = (\bar \beta_u\beta_u + \bar \beta_d\beta_d)(\bar \alpha_u\alpha_d + \bar \alpha_d\alpha_u ) = \bar \alpha_u\alpha_d + \bar \alpha_d\alpha_u \end{aligned}
For \sigma_y:
\begin{aligned} \langle \sigma_y \rangle & = \langle \Psi_{AB} |\sigma_y | \Psi_{AB} \rangle = \langle ab | \sigma_y \left[ \alpha_u \beta_u | uu \rangle + \alpha_u \beta_d | ud \rangle + \alpha_d \beta_u | du \rangle + \alpha_d \beta_d | dd \rangle \right] \\ & = \langle ab | \left[ \alpha_u \beta_u i| du \rangle + \alpha_u \beta_d i| dd \rangle - \alpha_d \beta_u i| uu \rangle - \alpha_d \beta_d i| ud \rangle \right] \\ & = \left[ \bar \alpha_u \bar \beta_u\langle uu | + \bar \alpha_u \bar \beta_d\langle ud | + \bar \alpha_d \bar \beta_u\langle du | + \bar \alpha_d \bar \beta_d\langle dd | \right] \\ & = -i\bar \alpha_u \bar \beta_u \alpha_d \beta_u - i\bar \alpha_u \bar \beta_d \alpha_d \beta_d + i\bar \alpha_d \bar \beta_u \alpha_u \beta_u + i\bar \alpha_d \bar \beta_d \alpha_u \beta_d \\ & = i(\bar \beta_u\beta_u)(-\bar \alpha_u\alpha_d + \bar \alpha_d\alpha_u ) + i(\bar \beta_d\beta_d)(-\bar \alpha_u\alpha_d + \bar \alpha_d\alpha_u ) \\ & = (\bar \beta_u\beta_u + \bar \beta_d\beta_d)(-\bar \alpha_u\alpha_d + \bar \alpha_d\alpha_u ) = -i\bar \alpha_u\alpha_d + i\bar \alpha_d\alpha_u \end{aligned}
For \sigma_z:
\begin{aligned} \langle \sigma_z \rangle & = \langle \Psi_{AB} |\sigma_z | \Psi_{AB} \rangle = \langle ab | \sigma_z \left[ \alpha_u \beta_u | uu \rangle + \alpha_u \beta_d | ud \rangle + \alpha_d \beta_u | du \rangle + \alpha_d \beta_d | dd \rangle \right] \\ & = \langle ab | \left[ \alpha_u \beta_u | uu \rangle + \alpha_u \beta_d | ud \rangle - \alpha_d \beta_u | du \rangle - \alpha_d \beta_d | dd \rangle \right] \\ & = \left[ \bar \alpha_u \bar \beta_u\langle uu | + \bar \alpha_u \bar \beta_d\langle ud | + alpha_d^* \bar \beta_u\langle du | + \bar \alpha_d \bar \beta_d\langle dd | \right] \\ & = \bar \alpha_u \bar \beta_u \alpha_u \beta_u + \bar \alpha_u \bar \beta_d \alpha_u \beta_d - \bar \alpha_d \bar \beta_u \alpha_d \beta_u - \bar \alpha_d \bar \beta_d \alpha_d \beta_d \\ & = (\bar \beta_u\beta_u)(\bar \alpha_u\alpha_u - \bar \alpha_d\alpha_d ) + (\bar \beta_d\beta_d)(\bar \alpha_u\alpha_u - \bar \alpha_d\alpha_d ) \\ &= (\bar \beta_u\beta_u + \bar \beta_d\beta_d)(\bar \alpha_u\alpha_u - \bar \alpha_d\alpha_d ) = \bar \alpha_u\alpha_u - \bar \alpha_d\alpha_d \end{aligned}
Considering Alice system on its own:
\begin{aligned} \{\psi_A |\sigma_x|\psi_A \} & = (\bar \alpha_u \{u| + \bar \alpha_d \{d| ) \sigma_x (\alpha_u |u\} + \alpha_d |d\} ) \\ & = (\bar \alpha_u \{u| + \bar \alpha_d \{d| ) (\alpha_u |d\} + \alpha_d |u\} ) \\ & = \bar \alpha_u\alpha_d + \bar \alpha_d\alpha_u \\ \{\psi_A|\sigma_y|\psi_A \} & = (\bar \alpha_u \{u| + \bar \alpha_d \{d| ) \sigma_y (\alpha_u |u\} + \alpha_d |d\} ) \\ & = (\bar \alpha_u \{u| + \bar \alpha_d \{d| ) (\alpha_u i|d\} - \alpha_d i|u\} ) \\ & = -i \bar \alpha_u\alpha_d + i\bar \alpha_d\alpha_u \\ \{\psi_A|\sigma_z|\psi_A\} & = (\bar \alpha_u \{u| + \bar \alpha_d \{d| ) \sigma_z (\alpha_u |u\} + \alpha_d |d\} )\\ & = (\bar \alpha_u \{u| + \bar \alpha_d \{d| ) (\alpha_u |u\} - \alpha_d |d\} )\\ & = \bar \alpha_u\alpha_u - \bar \alpha_d\alpha_d \end{aligned}
And these values coincide.
Assume Charlie has prepared the two spins in the singlet state. This time, Bob measures \tau_y and Alice measure \sigma_x. What is the expectation value of \sigma_x\tau_y?
What does this say about the correlation between the two measurements?
Considering the singlet state:
| \text{sing} \rangle = \frac{1}{\sqrt 2}\left( | ud \rangle - | du \rangle \right)
Computing the composite state \sigma_{x}\tau_{y}:
\begin{aligned} \sigma_x\tau_y | \text{sing} \rangle & = \sigma_x\tau_y \frac{1}{\sqrt 2}\left( | ud \rangle - | du \rangle \right) = \sigma_x\left[\tau_y \frac{1}{\sqrt 2}\left( | ud \rangle - | du \rangle \right)\right] \\ & = \sigma_{x}\left[\frac{-i}{\sqrt 2}\left( | uu \rangle + | dd \rangle \right)\right] = \frac{-i}{\sqrt 2}\left( |ud \rangle + | du \rangle \right) = -i| T_1 \rangle \end{aligned}
The expectation value is:
\begin{aligned} \langle \sigma_x\tau_y \rangle & = \langle \bar S | \sigma_x\tau_y | \text{sing} \rangle \\ & =\frac{1}{\sqrt 2}\left( \langle ud | - \langle du | \right) \sigma_x\tau_y\frac{1}{\sqrt 2}\left( | ud \rangle - | du \rangle \right) \\ & = \frac{1}{2}\left( \langle dd | - \langle uu | \right)\left( - i|uu \rangle - i | dd \rangle \right) \\ & = \frac{-i}{2}\left( \langle dd | uu \rangle + \langle dd | dd \rangle - \langle uu | uu \rangle - \langle uu | dd \rangle \right) \\ & = \frac{-i}{2}\left( 0 + 1 - 1 - 0 \right) = 0 \end{aligned}
Since the correlation between two observable is:
\langle \sigma_x \tau_y \rangle - \langle \sigma_x \rangle \langle \tau_y \rangle
The expectation for \sigma_x is:
\begin{aligned} \langle \sigma_x \rangle = & \frac{1}{\sqrt 2}\left( \langle ud | - \langle du | \right) \sigma_x \frac{1}{\sqrt 2}\left( | ud \rangle - | du \rangle \right) = \frac{1}{2}\left( \langle ud | - \langle du | \right) \left[\sigma_x \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}
The expectation for \tau_y:
\begin{aligned} \langle \tau_y \rangle = & \frac{1}{\sqrt 2}\left( \langle ud | - \langle du | \right) \tau_y \frac{1}{\sqrt 2}\left( | ud \rangle - | du \rangle \right) = \frac{1}{2}\left( \langle ud | - \langle du | \right) \left[\tau_y \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{-i}{2}\left(\langle ud | uu \rangle + \langle ud | dd \rangle - \langle du | uu \rangle - \langle du | dd \rangle \right) \\ = & \frac{-i}{2}\left( 0 + 0 + 0 + 0 \right) = 0 \end{aligned}
Therefore:
\langle \sigma_x \tau_y \rangle - \langle \sigma_x \rangle \langle \tau_y \rangle = 0
So the two measurements are not correlated.
Next, Charlie prepares the spins in a different state, called | T_1 \rangle, where
| T_1 \rangle = \frac1{\sqrt2}\left(| ud \rangle + du \rangle \right)
In these examples, T stands for triplet. These triplet states are completely different from the states in the coin and die examples. What are the expectation values of the operators \sigma_z\tau_z, \sigma_x\tau_x, and \sigma_y\tau_y?
What a difference a sign can make!
Considering the state:
|T_1 \rangle = \frac{1}{\sqrt 2}\left( | ud \rangle + | du \rangle \right)
Computing the composite state \sigma_z\tau_z:
\begin{aligned} \sigma_z\tau_z | T_1 \rangle & = \sigma_z\tau_z\frac{1}{\sqrt 2}\left( | ud \rangle + | du \rangle \right) = \sigma_z\left[\tau_z \frac{1}{\sqrt 2}\left( | ud \rangle + | du \rangle \right)\right] \\ & = \sigma_z\left[\frac{1}{\sqrt 2}\left( -| ud \rangle + | du \rangle \right)\right] = \frac{1}{\sqrt 2}\left( -|ud \rangle - | du \rangle \right) = -| T_1 \rangle \end{aligned}
Therefore:
\sigma_z\tau_z | T_1 \rangle = - | T_1 \rangle
So |T_1 \rangle is eigenvector of the observable \sigma_z\tau_z with an eigenvalue -1; so every time that the product is measured it gives -1, nevertheless the expectation of the individual \langle \sigma_z\rangle and \langle \tau_z\rangle are both 0.
The expectation value is:
\begin{aligned} \langle \sigma_z\tau_z \rangle & = \langle \bar T_1 | \sigma_z\tau_z | T_1 \rangle \\ & =\frac{1}{\sqrt 2}\left( \langle ud | + \langle du | \right) \sigma_z\tau_z\frac{1}{\sqrt 2}\left( | ud \rangle + | du \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}
Considering the case \sigma_x\tau_x:
\begin{aligned} \sigma_x\tau_x | T_1 \rangle & = \sigma_x\tau_x\frac{1}{\sqrt 2}\left( | ud \rangle + | du \rangle \right) = \sigma_x\left[\tau_x \frac{1}{\sqrt 2}\left( | ud \rangle + | du \rangle \right)\right] \\ & = \sigma_x\left[\frac{1}{\sqrt 2}\left( | uu \rangle + | dd \rangle \right)\right] = \frac{1}{\sqrt 2}\left( |du \rangle + | ud \rangle \right) = +| T_1 \rangle \end{aligned}
So |T_1 \rangle is also an eigenvector of the observable \sigma_x\tau_x, this time with an eigenvalue +1; the expectation value is:
\begin{aligned} \langle \sigma_x\tau_x \rangle & = \langle \bar T_1 | \sigma_x\tau_x | T_1 \rangle \\ & =\frac{1}{\sqrt 2}\left( \langle ud | + \langle du | \right) \sigma_x\tau_x\frac{1}{\sqrt 2}\left( | ud \rangle + | du \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}
Considering the case \sigma_y\tau_y:
\begin{aligned} \sigma_y\tau_y | T_1 \rangle & = \sigma_y\tau_y\frac{1}{\sqrt 2}\left( | ud \rangle + | du \rangle \right) = \sigma_y\left[\tau_y \frac{1}{\sqrt 2}\left( | ud \rangle + | du \rangle \right)\right] \\ & = \sigma_y\left[\frac{1}{\sqrt 2}\left( -i| uu \rangle + i| dd \rangle \right)\right] = \frac{1}{\sqrt 2}\left( |du \rangle + | ud \rangle \right) = +| T_1 \rangle \end{aligned}
As could be expected, |T_1 \rangle is also eigenvector of the observable \sigma_y\tau_y with an eigenvalue +1 and an expectation:
\langle \sigma_y\tau_y \rangle = \langle \bar T_1 | \sigma_y\tau_y | T_1 \rangle = +1
Do the same for the other two entangled triplet states,
\begin{aligned} & | T_2 \rangle = \frac1{\sqrt2}\left( | uu \rangle + | dd \rangle \right) \\ & | T_3 \rangle = \frac1{\sqrt2}\left( | uu \rangle - | dd \rangle \right) \end{aligned}
Considering the state:
|T_2 \rangle = \frac{1}{\sqrt 2}\left( | uu \rangle + | dd \rangle \right)
Computing the composite state \sigma_z\tau_z:
\begin{aligned} \sigma_z\tau_z | T_2 \rangle & = \sigma_z\tau_z\frac{1}{\sqrt 2}\left( | uu \rangle + | dd \rangle \right) = \sigma_z\left[\tau_z \frac{1}{\sqrt 2}\left( | uu \rangle + | dd \rangle \right)\right] \\ & = \sigma_z\left[\frac{1}{\sqrt 2}\left( | uu \rangle - | dd \rangle \right)\right] = \frac{1}{\sqrt 2}\left( |uu \rangle + | dd \rangle \right) = + | T_2 \rangle \end{aligned}
So |T_2 \rangle is eigenvector of the observable \sigma_z\tau_z with an eigenvalue +1; so every time that the product is measured it gives +1, nevertheless the expectation of the individual \langle \sigma_z\rangle and \langle \tau_z\rangle are both 0.
The expectation value is:
\begin{aligned} \langle \sigma_z\tau_z \rangle & = \langle \bar T_2 | \sigma_z\tau_z | T_2 \rangle \\ & =\frac{1}{\sqrt 2}\left( \langle uu | + \langle dd | \right) \sigma_z\tau_z\frac{1}{\sqrt 2}\left( | uu \rangle + | dd \rangle \right) \\ & = \frac{1}{2}\left( \langle uu | + \langle uu | \right)\left( | uu \rangle + | dd \rangle \right) \\ & = \frac{1}{2}\left( \langle uu | uu \rangle + \langle dd | uu \rangle + \langle uu | dd \rangle + \langle dd | dd \rangle \right) \\ & = \frac{1}{2}\left( 1 + 0 + 0 + 1 \right) = +1 \end{aligned}
Considering the case \sigma_x\tau_x:
\begin{aligned} \sigma_x\tau_x | T_2 \rangle & = \sigma_x\tau_x\frac{1}{\sqrt 2}\left( | uu \rangle + | dd \rangle \right) = \sigma_x\left[\tau_x \frac{1}{\sqrt 2}\left( | uu \rangle + | dd \rangle \right)\right] \\ & = \sigma_x\left[\frac{1}{\sqrt 2}\left( | ud \rangle + | du \rangle \right)\right] = \frac{1}{\sqrt 2}\left( |uu \rangle - | dd \rangle \right) = + | T_2 \rangle \end{aligned}
So |T_2 \rangle is also an eigenvector of the observable \sigma_x\tau_x, with an eigenvalue +1 and an expectation:
\langle \sigma_x\tau_x \rangle = \langle \bar T_2 | \sigma_x\tau_x | T_2 \rangle = +1
Considering the case \sigma_y\tau_y:
\begin{aligned} \sigma_y\tau_y | T_2 \rangle & = \sigma_y\tau_y\frac{1}{\sqrt 2}\left( | uu \rangle + | dd \rangle \right) = \sigma_y\left[\tau_y \frac{1}{\sqrt 2}\left( | uu \rangle + | dd \rangle \right)\right] \\ & = \sigma_y\left[\frac{1}{\sqrt 2}\left( -i| ud \rangle + i| du \rangle \right)\right] = \frac{1}{\sqrt 2}\left( -|dd \rangle - | uu \rangle \right) = -| T_2 \rangle \end{aligned}
So |T_2 \rangle is also eigenvector of the observable \sigma_y\tau_y with an eigenvalue -1 and an expectation:
\begin{aligned} \langle \sigma_y\tau_y \rangle & = \langle \bar T_2 | \sigma_y\tau_y | T_2 \rangle \\ & =\frac{1}{\sqrt 2}\left( \langle uu | + \langle dd | \right) \sigma_y\tau_y\frac{1}{\sqrt 2}\left( | uu \rangle + | dd \rangle \right) \\ & = \frac{1}{2}\left( \langle uu | + \langle uu | \right)\left( -| uu \rangle - | dd \rangle \right) \\ & = \frac{1}{2}\left( -\langle uu | uu \rangle - \langle dd | uu \rangle - \langle uu | dd \rangle - \langle dd | dd \rangle \right) \\ & = \frac{1}{2}\left( -1 - 0 - 0 - 1 \right) = -1 \end{aligned}
Considering the state:
|T_3 \rangle = \frac{1}{\sqrt 2}\left( | uu \rangle - | dd \rangle \right)
Likewise done for the singlet |S\rangle it is possible to compute the expectation values; computing the composite state \sigma_z\tau_z:
\begin{aligned} \sigma_z\tau_z | T_3 \rangle & = \sigma_z\tau_z\frac{1}{\sqrt 2}\left( | uu \rangle - | dd \rangle \right) = \sigma_z\left[\tau_z \frac{1}{\sqrt 2}\left( | uu \rangle - | dd \rangle \right)\right] \\ & = \sigma_z\left[\frac{1}{\sqrt 2}\left( | uu \rangle + | dd \rangle \right)\right] = \frac{1}{\sqrt 2}\left( |uu \rangle - | dd \rangle \right) = + | T_3 \rangle \end{aligned}
So |T_3 \rangle is eigenvector of the observable \sigma_z\tau_z with an eigenvalue +1; so every time that the product is measured it gives +1, nevertheless the expectation of the individual \langle \sigma_z\rangle and \langle \tau_z\rangle are both 0.
The expectation value is:
\begin{aligned} \langle \sigma_z\tau_z \rangle & = \langle \bar T_3 | \sigma_z\tau_z | T_3 \rangle \\ & =\frac{1}{\sqrt 2}\left( \langle uu | - \langle dd | \right) \sigma_z\tau_z\frac{1}{\sqrt 2}\left( | uu \rangle - | dd \rangle \right) \\ & = \frac{1}{2}\left( \langle uu | - \langle dd | \right)\left( | uu \rangle - | dd \rangle \right) \\ & = \frac{1}{2}\left( \langle uu | uu \rangle - \langle dd | uu \rangle - \langle uu | dd \rangle + \langle dd | dd \rangle \right) \\ & = \frac{1}{2}\left( 1 - 0 - 0 + 1 \right) = +1 \end{aligned}
Considering the case \sigma_x\tau_x:
\begin{aligned} \sigma_x\tau_x | T_3 \rangle & = \sigma_x\tau_x\frac{1}{\sqrt 2}\left( | uu \rangle - | dd \rangle \right) = \sigma_x\left[\tau_x \frac{1}{\sqrt 2}\left( | uu \rangle - | dd \rangle \right)\right] \\ & = \sigma_x\left[\frac{1}{\sqrt 2}\left( | ud \rangle - | du \rangle \right)\right] = \frac{1}{\sqrt 2}\left( |uu \rangle - | dd \rangle \right) = - | T_3 \rangle \end{aligned}
So |T_3 \rangle is also an eigenvector of the observable \sigma_x\tau_x, with an eigenvalue -1 and an expectation:
\begin{aligned} \langle \sigma_x\tau_x \rangle & = \langle \bar T_3 | \sigma_x\tau_x | T_3 \rangle \\ & =\frac{1}{\sqrt 2}\left( \langle uu | - \langle dd | \right) \sigma_x\tau_x\frac{1}{\sqrt 2}\left( | uu \rangle - | dd \rangle \right) \\ & = \frac{1}{2}\left( \langle uu | - \langle dd | \right)\left( -| uu \rangle + | dd \rangle \right) \\ & = \frac{1}{2}\left( -\langle uu | uu \rangle + \langle dd | uu \rangle + \langle uu | dd \rangle - \langle dd | dd \rangle \right) \\ & = \frac{1}{2}\left( -1 + 0 + 0 - 1 \right) = -1 \end{aligned}
Considering the case \sigma_y\tau_y:
\begin{aligned} \sigma_y\tau_y | T_3 \rangle & = \sigma_y\tau_y\frac{1}{\sqrt 2}\left( | uu \rangle - | dd \rangle \right) = \sigma_y\left[\tau_y \frac{1}{\sqrt 2}\left( | uu \rangle - | dd \rangle \right)\right] \\ & = \sigma_y\left[\frac{1}{\sqrt 2}\left( -i| ud \rangle - i| du \rangle \right)\right] = \frac{1}{\sqrt 2}\left( -|dd \rangle + | uu \rangle \right) = +| T_3 \rangle \end{aligned}
So |T_3 \rangle is also eigenvector of the observable \sigma_y\tau_y with an eigenvalue +1 and an expectation:
\langle \sigma_y\tau_y \rangle = \langle \bar T_3 | \sigma_x\tau_x | T_3 \rangle = +1
Prove that the four vectors | sing \rangle, | T_1 \rangle, | T_2 \rangle, and | T_3 \rangle are eigenvectors of \vec{\sigma} \cdot \vec{\tau}. What are their eigenvalues?
From exercise 6.7:
(\vec{\sigma} \cdot \vec{\tau}) | T_1 \rangle = \sigma_x\tau_x |T_1 \rangle + \sigma_y\tau_y |T_1 \rangle + \sigma_z\tau_z |T_1 \rangle = + |T_1 \rangle + |T_1 \rangle - |T_1 \rangle = + |T_1 \rangle
From exercise 6.8:
\begin{aligned} & (\vec{\sigma} \cdot \vec{\tau}) | T_2 \rangle = \sigma_x\tau_x |T_2 \rangle + \sigma_y\tau_y |T_2 \rangle + \sigma_z\tau_z |T_2 \rangle = + |T_2 \rangle - |T_2 \rangle + |T_2 \rangle = + |T_2 \rangle \\ & (\vec{\sigma} \cdot \vec{\tau}) | T_3 \rangle = \sigma_x\tau_x |T_3 \rangle + \sigma_y\tau_y |T_3 \rangle + \sigma_z\tau_z |T_3 \rangle = - |T_3 \rangle + |T_3 \rangle + |T_3 \rangle = + |T_3 \rangle \end{aligned}
So the triplets have all a degenerated eigenvector +1 (from which the name “triplet”).
For the singlet:
\begin{aligned} \sigma_z\tau_z | \text{sing} \rangle & = \sigma_z\tau_z \frac{1}{\sqrt 2}\left( | ud \rangle - | du \rangle \right) = \sigma_z \left[\tau_z \frac{1}{\sqrt 2}\left( | ud \rangle - | du \rangle \right)\right] \\ & = \sigma_{Az}\left[\frac{1}{\sqrt 2}\left( -| ud \rangle - | du \rangle \right)\right] = \frac{1}{\sqrt 2}\left( -|ud \rangle + | du \rangle \right) = -| \text{sing} \rangle \\ \sigma_y\tau_y | \text{sing} \rangle & = \sigma_y\tau_y \frac{1}{\sqrt 2}\left( | ud \rangle - | du \rangle \right) = \sigma_y \left[\tau_y \frac{1}{\sqrt 2}\left( | ud \rangle - | du \rangle \right)\right] \\ & = \sigma_y \left[\frac{1}{\sqrt 2}\left( -i| uu \rangle - i| dd \rangle \right)\right] = \frac{1}{\sqrt 2}\left( | du \rangle - | ud \rangle \right) = -| \text{sing} \rangle \\ \sigma_z\tau_z | \text{sing} \rangle & = \sigma_z\tau_z \frac{1}{\sqrt 2}\left( | ud \rangle - | du \rangle \right) = \sigma_z \left[\tau_z \frac{1}{\sqrt 2}\left( | ud \rangle - | du \rangle \right)\right] \\ & = \sigma_z \left[\frac{1}{\sqrt 2}\left( -| ud \rangle - | du \rangle \right)\right] = \frac{1}{\sqrt 2}\left( -|ud \rangle + | du \rangle \right) = -| \text{sing} \rangle \end{aligned}
Computing the eigenequation:
(\vec{\sigma} \cdot \vec{\tau}) | \text{sing} \rangle = \sigma_x\tau_x |\text{sing} \rangle + \sigma_y\tau_y | \text{sing} \rangle + \sigma_z\tau_z | \text{sing} \rangle = - |\text{sing} \rangle - |\text{sing}\rangle - |\text{sing}\rangle = -3 |\text{sing}\rangle
So | \text{sing} \rangle is also an eigenvector with a single eigenvalue -3 (from which the name “singlet”).
A system of two spins has the Hamiltonian
\mathbf{H}^{(\ast)} = \frac{\hbar \omega}{2}\vec{\sigma} \cdot \vec{\tau}
What are the possible energies of the system, and what are the eigenvectors of the Hamiltonian?
Suppose the system starts in the state | uu \rangle. What is the state at any later time? Answer the same question for initial states | ud \rangle, | du \rangle, and | dd \rangle.
(\ast) in the book there is likely a \hbar missing as \omega generally does not have the dimension of energy
From exercise 6.9 it was found that \vec{\sigma} \cdot \vec{\tau} has four eigenvalues and eigenvectors and, since the space is four dimensional, these form a basis (exercise 3.1).
The eigenvalues (the energies) are -\frac{3}{2}\hbar \omega for the eigenvector |\text{sing} \rangle and \frac{1}{2}\hbar \omega for | T_1 \rangle, | T_2 \rangle and | T_3 \rangle.
Using Schrödinger Ket recipe starting from step 4. As the eigenvalues and eigenvectors are known, it is possible to start from the calculation of the initial coefficient \alpha_{j}(0):
for the initial state | uu \rangle:
\begin{aligned} & \langle \text{sing} | uu \rangle = \frac{1}{\sqrt 2}\left( \langle ud | - \langle du | \right) | uu \rangle = 0 \\ & \langle T_1 | uu \rangle = \frac{1}{\sqrt 2}\left( \langle ud | + \langle du | \right) | uu \rangle = 0 \\ & \langle T_2 | uu \rangle = \frac{1}{\sqrt 2}\left( \langle uu | + \langle dd | \right) | uu \rangle = \frac{1}{\sqrt 2} \\ & \langle T_3 | uu \rangle = \frac{1}{\sqrt 2}\left( \langle uu | - \langle dd | \right) | uu \rangle = \frac{1}{\sqrt 2} \end{aligned}
For t=0:
|\Psi(0)_{uu} \rangle = | uu \rangle = \sum_j \alpha_j(0) |e_j \rangle = \frac{1}{\sqrt 2} \left(| T_2 \rangle + | T_3 \rangle \right)
And for a generic time t:
|\Psi(t)_{uu}\rangle = \sum_j \alpha_j(0) e^{-i\frac{E_j}{\hbar}t}|e_j \rangle = \frac{1}{\sqrt 2} e^{-i\frac{\omega}{2}t} \left(| T_2 \rangle + | T_3 \rangle \right)
For the initial state | ud \rangle:
\begin{aligned} & \langle \text{sing} | ud \rangle = \frac{1}{\sqrt 2}\left( \langle ud | - \langle du | \right) | ud \rangle = \frac{1}{\sqrt 2} \\ & \langle T_1 | ud \rangle = \frac{1}{\sqrt 2}\left( \langle ud | + \langle du | \right) | ud \rangle = \frac{1}{\sqrt 2} \\ & \langle T_2 | ud \rangle = \frac{1}{\sqrt 2}\left( \langle uu | + \langle dd | \right) | ud \rangle = 0 \\ & \langle T_3 | ud \rangle = \frac{1}{\sqrt 2}\left( \langle uu | - \langle dd | \right) | ud \rangle = 0 \end{aligned}
For t=0:
|\Psi(0)_{ud} \rangle = | ud \rangle = \sum_j \alpha_j(0) |e_j \rangle = \frac{1}{\sqrt 2} \left(| \text{sing} \rangle + | T_1 \rangle \right)
And for a generic time t:
|\Psi(t)_{ud}\rangle = \sum_j \alpha_j(0) e^{-i\frac{E_j}{\hbar}t}|e_j \rangle = \frac{1}{\sqrt 2} \left( e^{i\frac{3\omega}{2}t} | \text{sing} \rangle + e^{-i\frac{\omega}{2}t} | T_1 \rangle \right)
For the initial state | du \rangle:
\begin{aligned} & \langle \text{sing} | du \rangle = \frac{1}{\sqrt 2}\left( \langle ud | - \langle du | \right) | du \rangle = -\frac{1}{\sqrt 2} \\ & \langle T_1 | du \rangle = \frac{1}{\sqrt 2}\left( \langle ud | + \langle du | \right) | du \rangle = \frac{1}{\sqrt 2} \\ & \langle T_2 | du \rangle = \frac{1}{\sqrt 2}\left( \langle uu | + \langle dd | \right) | du \rangle = 0 \\ & \langle T_3 | du \rangle = \frac{1}{\sqrt 2}\left( \langle uu | - \langle dd | \right) | du \rangle = 0 \end{aligned}
For t=0:
|\Psi(0)_{du} \rangle = | du \rangle = \sum_j \alpha_j(0) |e_j \rangle = \frac{1}{\sqrt 2} \left( | T_2 \rangle + | T_3 \rangle \right)
And for a generic time t:
|\Psi(t)_{du}\rangle = \sum_j \alpha_j(0) e^{-i\frac{E_j}{\hbar}t}|e_j \rangle = \frac{1}{\sqrt 2} \left( -e^{i\frac{3\omega}{2}t} | \text{sing} \rangle + e^{-i\frac{\omega}{2}t} | T_1 \rangle \right)
For the initial state | dd \rangle:
\begin{aligned} & \langle \text{sing} | dd \rangle = \frac{1}{\sqrt 2}\left( \langle ud | - \langle du | \right) | dd \rangle = 0 \\ & \langle T_1 | dd \rangle = \frac{1}{\sqrt 2}\left( \langle ud | + \langle du | \right) | dd \rangle = 0 \\ & \langle T_2 | dd \rangle = \frac{1}{\sqrt 2}\left( \langle uu | + \langle dd | \right) | dd \rangle = \frac{1}{\sqrt 2} \\ & \langle T_3 | dd \rangle = \frac{1}{\sqrt 2}\left( \langle uu | - \langle dd | \right) | dd \rangle = -\frac{1}{\sqrt 2} \end{aligned}
For t=0:
|\Psi(0)_{dd} \rangle = | dd \rangle = \sum_j \alpha_j(0) |e_j \rangle = \frac{1}{\sqrt 2} \left(| T_2 \rangle + | T_3 \rangle \right)
And for a generic time t:
|\Psi(t)_{dd}\rangle = \sum_j \alpha_j(0) e^{-i\frac{E_j}{\hbar}t}|e_j \rangle = \frac{1}{\sqrt 2} e^{-i\frac{\omega}{2}t} \left(| T_2 \rangle + | T_3 \rangle \right)