Nielsen and Chuang [10e, p. 74] introduce the Kronecker product $A\otimes_K B$ as a matrix representation of the tensor product $A\otimes B$ of the operators $A$ and $B$ (for clarity I use a subscript $K$ to distinguish the Kronecker product of matrices from the tensor product). My question is about whether there is such a representation for biproduct mixed states in multipartite quantum systems.

Suppose I have Hilbert spaces $H_A$, $H_B$, $H_C$ of a tripartite system e.g. three qubits. An example of a fully separable state is the product $\rho^A \otimes \rho^B \otimes \rho^C$ where the density operators $\rho^A$, $\rho^B$, $\rho^C$ lie in the respective Hilbert spaces. Examples of biseperable states might be $\rho^A \otimes \rho^{BC} $ and $\rho^B \otimes \rho^{AC}$ where the density operators $\rho^{BC}$ and $\rho^{AC}$ now lie in the composite Hilbert spaces $H_B\otimes H_C$ and $H_A\otimes H_C$ respectively.

I wouldn't be surprised if you told me the matrix representation of the product $\rho^A \otimes \rho^B \otimes \rho^C$ was $\rho^A \otimes_K \rho^B \otimes_K \rho^C$, or that the representation of $\rho^A \otimes \rho^{BC} $ was $\rho^A \otimes_K \rho^{BC} $ (where $\rho^{BC}$ is a square matrix of side length $\text{dim}(H_B)\times \text{dim}(H_C)$). However, I don't think the matrix representation of $\rho^B \otimes \rho^{AC}$ could possibly be of the form $\rho^B \otimes_K \rho^{AC}$, because the spaces are 'in the wrong order.'

Mathematically, I understand the Kronecker product as a matrix representation with respect to bases $U_A,U_B,U_C$ (say) of the subsystems in a given order. So if I take $\rho^A \otimes_K \rho^B \otimes_K \rho^C$ to be the matrix representation of $\rho^A\otimes \rho^B \otimes \rho^C$, I feel like I would need to apply some permutation to the $\text{dim}(H_A) \times \text{dim}(H_B) \times \text{dim}(H_C) $ matrix $\rho^B \otimes_K \rho^{AC}$ in order for it to represent $\rho^B \otimes \rho^{AC}$. Is there a consistent way of doing this? Perhaps I am misunderstanding the use of the tensor product in quantum information, in which case please let me know.


1 Answer 1


I think you pretty much got it right. Yes you use the "Kronecker product representation" to represent tensor products as matrices/vectors for generic operators/vectors, and thus in particular for density matrices or ket states. Yes, it also technically doesn't work too nicely if you have to put together things like some $\rho^{AC}$ and $\rho^B$. Yes, an easy way to represent the actual state still using Kronecker products is using permutation operations. E.g. you could write in this example $$\rho^{ABC} = W_{BC}(\rho^{AC}\otimes \rho^B),$$ where $W_{BC}$ denotes the operator swapping spaces $B$ and $C$. You can write this explicitly as $$W_{BC} \equiv \sum_{ijk} |i\rangle\!\langle i|\otimes |j\rangle\!\langle k|\otimes |k\rangle\!\langle j|.$$ Alternatively you could decompose $\rho^{AC}$ as a linear combination of simple tensors, some $\rho^{AC}=\sum_{ijk\ell} c_{ijk\ell} (|i\rangle\!\langle k|\otimes |j\rangle\!\langle \ell|)$, and then $$\rho^{ABC} = \sum c_{ijk\ell} (|i\rangle\!\langle k|\otimes \rho^B\otimes |j\rangle\!\langle \ell|).$$ Note that here $|i\rangle\!\langle k|$ is an operator in $H_A$ while $|j\rangle\!\langle \ell|$ is an operator in $H_C$.

In practice, you might often find that people are not too careful about this. Especially if you keep writing the space indices (your $A,B,C$ here) you can often just write $\rho^{ABC}=\rho^{AC}\otimes\rho^B$ with the implicit understanding that you swapped two spaces to do it. As long as you're consistent you'll still get the correct results.

Similar things are done to write the matrix representation of many-qubit gates between non-adjacent qubits. See e.g. How to derive the CNOT matrix for a 3-qubit system where the control & target qubits are not adjacent? and links therein.

  • $\begingroup$ Many thanks @gls for the detailed response (also for linking the question - Quantum Computing is not my field of study so I would have not have noticed it as relevant otherwise). May I check that there is nothing stopping me representing $W_{BC}$ as a matrix (using Kronecker products) and thus evaluating $\rho^{ABC}$ explicitly given matrices for $\rho^{AC}$ and $\rho^B$? $\endgroup$
    – pip
    Commented Nov 6, 2022 at 19:33
  • $\begingroup$ yes of course, $W_{BC}$ is (can be represented as) a matrix, which you can compute with the rules you already know. Whenever you see bra-ket notation, everything can always be represented as vectors and/or matrices $\endgroup$
    – glS
    Commented Nov 6, 2022 at 19:39
  • $\begingroup$ That's great, and an excellent answer - your comment about people not worrying too much about this was particularly apt: I came across the 'implicit understanding' in literature and was surprised by this, hence my question :). Thanks again. $\endgroup$
    – pip
    Commented Nov 6, 2022 at 19:53
  • $\begingroup$ I realise there is a small part wrong in your answer (or at least clarification needed): Applying the operator $W_{BC}$ from the left-hand side only swaps the kets; in order to fully swap spaces one needs to apply this operator from both the left and the right. $\endgroup$
    – pip
    Commented Dec 7, 2022 at 17:34

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