- #1
doggydan42
- 170
- 18
I am reading Leonard Susskind's Theoretical Minimum book on Quantum Mechanics. Excercise 7.4 is as follows:
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:
$$
\rho_{a'a} = \begin{pmatrix}
\alpha^*\alpha & \beta^*\alpha \\
\alpha^*\beta & \beta^*\beta
\end{pmatrix}$$
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:
$$
\rho_{a'a} = \begin{pmatrix}
\alpha^*\alpha & \beta^*\alpha \\
\alpha^*\beta & \beta^*\beta
\end{pmatrix}$$