- #1

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Calculate the density matrix for ##|\Psi\rangle = \alpha|u\rangle + \beta|d\rangle##.

Answer:

$$ \psi(u) = \alpha, \quad \psi^*(u) = \alpha^* \\

\psi(d) = \beta, \quad \psi^*(d) = \beta^*\\

\rho_{a'a} = \begin{pmatrix}

\alpha^*\alpha & \alpha^*\beta \\

\beta^*\alpha & \beta^*\beta

\end{pmatrix}$$

From my understanding, if we use ##\rho_{a'a} = \psi(a')\psi^*(a)##, then I notice that the matrix is equivalent to:

$$\rho_{a'a} = \begin{pmatrix}

\psi^*(u)\psi(u) & \psi^*(u)\psi(d) \\

\psi^*(d)\psi(u) & \psi^*(d)\psi(d)

\end{pmatrix} = \begin{pmatrix}

\rho_{uu} & \rho_{du} \\

\rho_{ud} & \rho_{dd}

\end{pmatrix}

$$

I was wondering how the indices affect the position of each ##\rho_{a'a}## in the density matrix.

In other words, why is the matrix not:

$$

\rho_{a'a} = \begin{pmatrix}

\alpha^*\alpha & \beta^*\alpha \\

\alpha^*\beta & \beta^*\beta

\end{pmatrix}$$