# Is partial trace the inverse operation of Kronecker product?

Computer science student here, who is interested in quantum information theory.

Suppose I have these pure states: $$\begin{bmatrix}1&0\\0&0\end{bmatrix}$$ and $$\begin{bmatrix}0&0\\0&1\end{bmatrix}$$

The Kronecker product of these is: $$\begin{bmatrix}0&0&0&0\\0&1&0&0\\0&0&0&0\\0&0&0&0\end{bmatrix}$$

By using partial trace, I can extract information of the original states: And if I were to tensor multiply them again, I'd get back the previous 4x4 matrix: Below is a short python code, which works for the above example. However it doesn't work for mixed states, as you can see on the right, in the console. If we were to tensor the two 2x2 density matrices extracted from the 4x4 matrix, we don't end up with the original 4x4 matrix. My question is that partial trace can only restore pure states without information loss?

• Quick comment : When the state of system with two sub-systems ($i=1,2$) is an entangled state, mutual information measure is finite i.e., $I[\rho_{1+2}^{}]=-\sum_{i \neq j =1,2}^{}Tr_{i}[Tr_{j}[\rho_{1+2}^{}]\ln Tr_{j}^{}[\rho_{1+2}^{}]]+Tr_{1+2}^{}[\rho_{1+2}^{}\ln \rho_{1+2}^{}] \neq 0$. But the direct product state you are constructing out of partially traced density matrices has clearly no mutual information. Oct 21, 2018 at 22:23
• I suggest you looking into Section 2.4.3 of Nielsen&Chuang, or any other standard text on quantum information. In short - partial trace over a certain subspace in the tensor product has the meaning of averaging over that subspace. (In what follows, by 'state' I mean density matrix). If your state is not entangled, i.e. is a tensor product state, then taking partial trace over a certain subspace is equivalent to throwing away the corresponding factor of the product, and leaving all the other terms unchanged. For entangled states, the situation is more involved. Oct 22, 2018 at 0:31

It is in the sense that given any pair of states $$\rho$$ and $$\sigma$$, you have \begin{align} \operatorname{Tr}_2(\rho\otimes\sigma)&=\rho,\\ \operatorname{Tr}_1(\rho\otimes\sigma)&=\sigma. \end{align} This happens because $$\rho\otimes\sigma$$ represents a product states, in which the two parts of the system are independent from each other.
For example, if $$E_+$$ and $$E_-$$ denote the two two-dimensional matrices you defined, then you can easily verify that taking the partial trace of $$E_+\otimes E_++E_-\otimes E_-$$ gives you something completely different than both $$E_+$$ and $$E_-$$.