Work Check: Simple entanglement problem

[WWCY]

Homework Statement

Could someone assist me in skimming through my work for this problem? Many thanks!

I attached an image of the problem below. Also, I only need help for the first part (part a), cheers.

Homework Equations

General entangled state vector of a two-particle system:
$$|\psi \rangle = \sum_{i,j} \psi_{i,j} |\phi_i \Omega |\nu_j \rangle$$
where $| \phi_i \rangle$ and $| \Omega_j \rangle$ are basis states of A and B Hilbert spaces respectively

Density operator that acts on the tensor-product space:
$$\hat{\rho} = |\psi \rangle \langle \psi |$$

Partial trace over A states gives reduced density matrix that acts on B Hilbert space:
$$\hat{\rho} _B = \text{Tr}_A \big( \ \hat{\rho} \ \big) = \sum_{i'} \big( \langle \phi_{i'} | \otimes \hat{I} \big) \ \hat{\rho} \ \big( | \phi_{i'} \rangle \otimes \hat{I} \big) = \sum_{j, j'} | \Omega_j \rangle \big( \sum_{i} \psi_{i,j} \psi ^* _{i,j'} \big) \langle \Omega _{j'} | $$

The Attempt at a Solution

The reduced density matrix equation seems to imply that there are 9 matrix elements to compute. When written in matrix form, with $|p\rangle = (1,0,0)^{T}$, $|q\rangle = (0,1,0)^{T}$, $|r\rangle = (0,0,1)^{T}$. This gave me

$$\hat{\rho}_B = \begin{pmatrix}
1/4 & 1/4 & 1/2\sqrt{3} \\
1/4 & 5/12 & 1/2\sqrt{3} \\
1/2\sqrt{3} & 1/2\sqrt{3} & 1/3
\end{pmatrix}$$

and the probabilities of measurements on particle B are given by its diagonals. Is this right?

 

[mark726]

Yes, your approach and solution for the reduced density matrix are correct. However, I would suggest using a normalized state vector for your calculations, such as $$|\psi \rangle = \frac{1}{\sqrt{6}} \big( |p\rangle \otimes |q\rangle + |q\rangle \otimes |r\rangle + |r\rangle \otimes |p\rangle \big)$$
This will result in a simpler matrix for the reduced density matrix, which you can then easily normalize to obtain the probabilities of measurements on particle B. Additionally, for future reference, you can also use the density operator for a pure state, which is simply $$\hat{\rho} = |\psi \rangle \langle \psi |$$ without the summation. Keep up the good work!