What's the tensor product of the $2\times2$ matrix $\rho = \begin{bmatrix} 2/3 & 0.3 \\ 0.3 & 1/3 \\ \end{bmatrix}$ and $|\Psi\rangle = $ cos$(\theta)|0\rangle +$ sin$(\theta)|1\rangle$?

Initially I thought it should just be $\begin{bmatrix} 2/3|\Psi\rangle & 0.3|\Psi\rangle \\ 0.3|\Psi\rangle & 1/3|\Psi\rangle \\ \end{bmatrix}$ a $4\times2$ matrix.

But now I wonder because considering that the frist element of $\rho$ represents $2/3|0 \rangle \langle 0|$ does that mean that the first element of the tensor product matrix is $2/3$ cos$(\theta)|0 \rangle \langle 0|\otimes|0\rangle$ which is strange to interpret.

Should the correct method be writing down the density matrix of $|\Psi\rangle$ and computing the tensor product then? There seems to be no literature online on this issue.


I assume this is a state for two two-level substates $A$ and $B$, where \begin{equation} \rho_A = \begin{bmatrix} \frac{2}{3} & \frac{3}{10} \\ \frac{3}{10} & \frac{1}{3} \end{bmatrix} \, , \end{equation} and \begin{equation} \rho_B = |\psi \rangle \langle \psi|= \begin{bmatrix} \sin^2 \theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & \cos^2 \theta \end{bmatrix} \, , \end{equation} where I have assumed the order $|1\rangle,|0\rangle$ for the basis vectors. Now you can compute \begin{equation} \rho = \rho_A \otimes \rho_B \, . \end{equation}

  • $\begingroup$ Thanks! So what I suspected is correct, one doesn't directly tensor product the density matrix with the wave vector? $\endgroup$ – Abhijeet Melkani Feb 10 '18 at 6:11
  • $\begingroup$ No, I don't think there is much use for expressions like $\frac{2}{3} \sin \theta |11\rangle \langle 1| + ... $. $\endgroup$ – secavara Feb 10 '18 at 10:42

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