An insight about the Partial Trace

The Partial Trace is an important tool for analyzing the state of one or more qubits in a larger system. In this article, I'd like to share an insight that I've had involving the Partial Trace of Density Matrices.

Let's start with a simple pure state, where Alice holds qubit \(\left| u \right\rangle\) and Bob holds qubit \(\left| v \right\rangle\). The state of the total system can then be expressed as:

\[ \begin{aligned} \left| \psi \right\rangle &= \sum_{i,j \ \in\ \{0,1\}} \alpha_{ij} \left| ij \right\rangle \\[5pt] &= \alpha_{00}\left| 00 \right\rangle \ +\ \alpha_{01}\left| 01 \right\rangle \ +\ \alpha_{10}\left| 10 \right\rangle \ +\ \alpha_{11}\left| 11 \right\rangle \end{aligned} \]

where the various \(\alpha_{ij}\) represent the amplitudes of the basis states. The values of these amplitudes can be obtained by expressing the state in a slightly more elaborate fashion:

\[ \begin{aligned} \left| \psi \right\rangle &= \left| u \right\rangle \otimes \left| v \right\rangle = \begin{bmatrix} u_0\\ u_1 \end{bmatrix} \otimes \begin{bmatrix} v_0\\ v_1 \end{bmatrix} = \begin{bmatrix} u_0 \begin{bmatrix} v_0\\ v_1 \end{bmatrix}\\ u_1 \begin{bmatrix} v_0\\ v_1 \end{bmatrix}\\ \end{bmatrix} = \begin{bmatrix} u_0 v_0\\ u_0 v_1\\ u_1 v_0\\ u_1 v_1 \end{bmatrix} \\[5pt] &= u_0 v_0\left| 00 \right\rangle \ +\ u_0 v_1\left| 01 \right\rangle \ +\ u_1 v_0\left| 10 \right\rangle \ +\ u_1 v_1\left| 11 \right\rangle \end{aligned} \]

The density matrix, which describes the total system, can then be computed by taking the outer product of \( \left| \psi \right\rangle \) with its conjugate transpose:

\[ \begin{aligned} \rho &= \left| \psi \right\rangle\!\left\langle \psi \right| = \begin{bmatrix} u_0 v_0\\ u_0 v_1\\ u_1 v_0\\ u_1 v_1 \end{bmatrix} \begin{bmatrix} \overline{u_0 v_0} & \overline{u_0 v_1} & \overline{u_1 v_0} & \overline{u_1 v_1} \end{bmatrix} \\[5pt] &= \begin{bmatrix} u_0 v_0 \overline{u_0 v_0} & u_0 v_0 \overline{u_0 v_1} & u_0 v_0 \overline{u_1 v_0} & u_0 v_0 \overline{u_1 v_1}\\ u_0 v_1 \overline{u_0 v_0} & u_0 v_1 \overline{u_0 v_1} & u_0 v_1 \overline{u_1 v_0} & u_0 v_1 \overline{u_1 v_1}\\ u_1 v_0 \overline{u_0 v_0} & u_1 v_0 \overline{u_0 v_1} & u_1 v_0 \overline{u_1 v_0} & u_1 v_0 \overline{u_1 v_1}\\ u_1 v_1 \overline{u_0 v_0} & u_1 v_1 \overline{u_0 v_1} & u_1 v_1 \overline{u_1 v_0} & u_1 v_1 \overline{u_1 v_1} \end{bmatrix} \hspace{20pt}\text{(Matrix 1)} \\[5pt] &= \begin{bmatrix} \alpha_{00} \overline{\alpha_{00}} & \alpha_{00} \overline{\alpha_{01}} & \alpha_{00} \overline{\alpha_{10}} & \alpha_{00} \overline{\alpha_{11}}\\ \alpha_{01} \overline{\alpha_{00}} & \alpha_{01} \overline{\alpha_{01}} & \alpha_{01} \overline{\alpha_{10}} & \alpha_{01} \overline{\alpha_{11}}\\ \alpha_{10} \overline{\alpha_{00}} & \alpha_{10} \overline{\alpha_{01}} & \alpha_{10} \overline{\alpha_{10}} & \alpha_{10} \overline{\alpha_{11}}\\ \alpha_{11} \overline{\alpha_{00}} & \alpha_{11} \overline{\alpha_{01}} & \alpha_{11} \overline{\alpha_{10}} & \alpha_{11} \overline{\alpha_{11}} \end{bmatrix} \hspace{20pt}\text{(Matrix 2)} \end{aligned} \]

The Partial Trace
Taking the partial trace over a density matrix gives us a reduced density matrix, which expresses the state of the system from the perspective of an observer who can only see a part of it. For example, if we want to express the state of the system from Bob's perspective, we take the partial trace over Alice's qubit. The canonical way of computing the partial trace is:

\[ \begin{aligned} \rho^B = Tr_A \left( \rho \right) =& \sum_{i,j \ \in\ \{0,1\}} \left( \sum_{k \ \in\ \{0,1\}} \alpha_{ki} \overline{\alpha_{kj}} \right) \left| i \right\rangle\!\left\langle j \right| \\[5pt] =&\ \left( \alpha_{00}\overline{\alpha_{00}} + \alpha_{10}\overline{\alpha_{10}} \right) \left| 0 \right\rangle\!\left\langle 0 \right|\ + \\[5pt] &\ \left( \alpha_{00}\overline{\alpha_{01}} + \alpha_{10}\overline{\alpha_{11}} \right) \left| 0 \right\rangle\!\left\langle 1 \right|\ + \\[5pt] &\ \left( \alpha_{01}\overline{\alpha_{00}} + \alpha_{11}\overline{\alpha_{10}} \right) \left| 1 \right\rangle\!\left\langle 0 \right|\ + \\[5pt] &\ \left( \alpha_{01}\overline{\alpha_{01}} + \alpha_{11}\overline{\alpha_{11}} \right) \left| 1 \right\rangle\!\left\langle 1 \right| \\[5pt] =&\begin{bmatrix} \alpha_{00}\overline{\alpha_{00}} + \alpha_{10}\overline{\alpha_{10}} & \alpha_{00}\overline{\alpha_{01}} + \alpha_{10}\overline{\alpha_{11}}\\ \alpha_{01}\overline{\alpha_{00}} + \alpha_{11}\overline{\alpha_{10}} & \alpha_{01}\overline{\alpha_{01}} + \alpha_{11}\overline{\alpha_{11}} \end{bmatrix} \hspace{20pt}\text{(Matrix 3)} \end{aligned} \]

This method is often described as "summing over all possible states of Alice's qubit". Indeed, we add up all amplitudes where Alice's component is either \(u_0\) (the ones that are \(u_0 v_* \overline{u_0 v_*} \) or \(\alpha_{0*} \overline{\alpha_{0*}} \)) or \(u_1\) (the ones that are \(u_1 v_* \overline{u_1 v_*} \) or \(\alpha_{1*} \overline{\alpha_{1*}} \)) ("\(*\)" denotes a wildcard here: \(0\) or \(1\)).

Conversely, computing the system from Alice's perspective yields:

\[ \begin{aligned} \rho^A = Tr_B \left( \rho \right) =& \sum_{i,j \ \in\ \{0,1\}} \left( \sum_{k \ \in\ \{0,1\}} \alpha_{ik} \overline{\alpha_{jk}} \right) \left| i \right\rangle\!\left\langle j \right| \\[5pt] =&\ \left( \alpha_{00}\overline{\alpha_{00}} + \alpha_{01}\overline{\alpha_{01}} \right) \left| 0 \right\rangle\!\left\langle 0 \right|\ + \\[5pt] &\ \left( \alpha_{00}\overline{\alpha_{10}} + \alpha_{01}\overline{\alpha_{11}} \right) \left| 0 \right\rangle\!\left\langle 1 \right|\ + \\[5pt] &\ \left( \alpha_{10}\overline{\alpha_{00}} + \alpha_{11}\overline{\alpha_{01}} \right) \left| 1 \right\rangle\!\left\langle 0 \right|\ + \\[5pt] &\ \left( \alpha_{10}\overline{\alpha_{10}} + \alpha_{11}\overline{\alpha_{11}} \right) \left| 1 \right\rangle\!\left\langle 1 \right| \\[5pt] =&\begin{bmatrix} \alpha_{00}\overline{\alpha_{00}} + \alpha_{01}\overline{\alpha_{01}} & \alpha_{00}\overline{\alpha_{10}} + \alpha_{01}\overline{\alpha_{11}}\\ \alpha_{10}\overline{\alpha_{00}} + \alpha_{11}\overline{\alpha_{01}} & \alpha_{10}\overline{\alpha_{10}} + \alpha_{11}\overline{\alpha_{11}} \end{bmatrix} \hspace{20pt}\text{(Matrix 4)} \end{aligned} \]

The Geometric Interpretation
The partial trace of a density matrix can also be interpreted in a geometric fashion. If we take the density matrix as given by Matrix 2 and add some colour to it, we get:

\[ \begin{aligned} \begin{bmatrix} \bbox[#cc99ff, 3px]{\alpha_{00} \overline{\alpha_{00}}} & \bbox[#99ff99, 3px]{\alpha_{00} \overline{\alpha_{01}}} & \alpha_{00} \overline{\alpha_{10}} & \alpha_{00} \overline{\alpha_{11}}\\ \bbox[#99ccff, 3px]{\alpha_{01} \overline{\alpha_{00}}} & \bbox[#ff9999, 3px]{\alpha_{01} \overline{\alpha_{01}}} & \alpha_{01} \overline{\alpha_{10}} & \alpha_{01} \overline{\alpha_{11}}\\ \alpha_{10} \overline{\alpha_{00}} & \alpha_{10} \overline{\alpha_{01}} & \bbox[#cc99ff, 3px]{\alpha_{10} \overline{\alpha_{10}}} & \bbox[#99ff99, 3px]{\alpha_{10} \overline{\alpha_{11}}}\\ \alpha_{11} \overline{\alpha_{00}} & \alpha_{11} \overline{\alpha_{01}} & \bbox[#99ccff, 3px]{\alpha_{11} \overline{\alpha_{10}}} & \bbox[#ff9999, 3px]{\alpha_{11} \overline{\alpha_{11}}} \end{bmatrix} \end{aligned} \]

If we now "take out" the two coloured submatrices and add them together, we get:

\[ \begin{aligned} \begin{bmatrix} \bbox[#cc99ff, 3px]{\alpha_{00} \overline{\alpha_{00}}} & \bbox[#99ff99, 3px]{\alpha_{00} \overline{\alpha_{01}}}\\ \bbox[#99ccff, 3px]{\alpha_{01} \overline{\alpha_{00}}} & \bbox[#ff9999, 3px]{\alpha_{01} \overline{\alpha_{01}}} \end{bmatrix} + \begin{bmatrix} \bbox[#cc99ff, 3px]{\alpha_{10} \overline{\alpha_{10}}} & \bbox[#99ff99, 3px]{\alpha_{10} \overline{\alpha_{11}}}\\ \bbox[#99ccff, 3px]{\alpha_{11} \overline{\alpha_{10}}} & \bbox[#ff9999, 3px]{\alpha_{11} \overline{\alpha_{11}}} \end{bmatrix} = \begin{bmatrix} \bbox[#cc99ff, 3px]{\alpha_{00} \overline{\alpha_{00}} + \alpha_{10} \overline{\alpha_{10}}} & \bbox[#99ff99, 3px]{\alpha_{00} \overline{\alpha_{01}} + \alpha_{10} \overline{\alpha_{11}}}\\ \bbox[#99ccff, 3px]{\alpha_{01} \overline{\alpha_{00}} + \alpha_{11} \overline{\alpha_{10}}} & \bbox[#ff9999, 3px]{\alpha_{01} \overline{\alpha_{01}} + \alpha_{11} \overline{\alpha_{11}}} \end{bmatrix} \end{aligned} \]

The result is exactly Bob's reduced density matrix as given by Matrix 3! For Alice's reduced density matrix, we have to draw the colours a bit differently:

\[ \begin{aligned} \begin{bmatrix} \bbox[#cc99ff, 3px]{\alpha_{00} \overline{\alpha_{00}}} & \alpha_{00} \overline{\alpha_{01}} & \bbox[#99ff99, 3px]{\alpha_{00} \overline{\alpha_{10}}} & \alpha_{00} \overline{\alpha_{11}}\\ \alpha_{01} \overline{\alpha_{00}} & \bbox[#cc99ff, 3px]{\alpha_{01} \overline{\alpha_{01}}} & \alpha_{01} \overline{\alpha_{10}} & \bbox[#99ff99, 3px]{\alpha_{01} \overline{\alpha_{11}}}\\ \bbox[#99ccff, 3px]{\alpha_{10} \overline{\alpha_{00}}} & \alpha_{10} \overline{\alpha_{01}} & \bbox[#ff9999, 3px]{\alpha_{10} \overline{\alpha_{10}}} & \alpha_{10} \overline{\alpha_{11}}\\ \alpha_{11} \overline{\alpha_{00}} & \bbox[#99ccff, 3px]{\alpha_{11} \overline{\alpha_{01}}} & \alpha_{11} \overline{\alpha_{10}} & \bbox[#ff9999, 3px]{\alpha_{11} \overline{\alpha_{11}}} \end{bmatrix} \end{aligned} \]

Then, to obtain Alice's reduced density matrix as given by Matrix 4:

\[ \begin{aligned} \begin{bmatrix} \bbox[#cc99ff, 3px]{\alpha_{00} \overline{\alpha_{00}}} & \bbox[#99ff99, 3px]{\alpha_{00} \overline{\alpha_{10}}}\\ \bbox[#99ccff, 3px]{\alpha_{10} \overline{\alpha_{00}}} & \bbox[#ff9999, 3px]{\alpha_{10} \overline{\alpha_{10}}} \end{bmatrix} + \begin{bmatrix} \bbox[#cc99ff, 3px]{\alpha_{01} \overline{\alpha_{01}}} & \bbox[#99ff99, 3px]{\alpha_{01} \overline{\alpha_{11}}}\\ \bbox[#99ccff, 3px]{\alpha_{11} \overline{\alpha_{01}}} & \bbox[#ff9999, 3px]{\alpha_{11} \overline{\alpha_{11}}} \end{bmatrix} = \begin{bmatrix} \bbox[#cc99ff, 3px]{\alpha_{00} \overline{\alpha_{00}} + \alpha_{01} \overline{\alpha_{01}}} & \bbox[#99ff99, 3px]{\alpha_{00} \overline{\alpha_{10}} + \alpha_{01} \overline{\alpha_{11}}}\\ \bbox[#99ccff, 3px]{\alpha_{10} \overline{\alpha_{00}} + \alpha_{11} \overline{\alpha_{01}}} & \bbox[#ff9999, 3px]{\alpha_{10} \overline{\alpha_{10}} + \alpha_{11} \overline{\alpha_{11}}} \end{bmatrix} \end{aligned} \]

So far, pretty basic textbook stuff. But why does this work? Why do these summations and their geometric interpretation lead to how Alice and Bob observe the total system from their own perspective?

Reversing the tensor product
I'm sure the insight that I got when thinking about this question isn't new, but since I couldn't find it in any of the textbooks or articles that I studied, I decided to write it up. Here goes:

To get the density matrix of the total system, as given by Matrix 1, we took the tensor product of \( \left| v \right\rangle \) and \( \left| w \right\rangle \), which yielded \( \left| \psi \right\rangle \), and then we took the outer product of \( \left| \psi \right\rangle \) with its conjugate transpose.

We could, however, also do it the other way around. Instead of computing the outer product of the tensor product, we could also compute the tensor product of the outer products. These are equivalent. So, instead of:

\[ \begin{aligned} \rho = \left| \psi \right\rangle\!\left\langle \psi \right| = \left( \left| u \right\rangle \otimes \left| v \right\rangle \right) \left( \left\langle u \right| \otimes \left\langle v \right| \right) \end{aligned} \]

we get:

\[ \begin{aligned} \rho = \left| u \right\rangle\!\left\langle u \right| \otimes \left| v \right\rangle\!\left\langle v \right| \end{aligned} \]

Of course, this does imply that the qubits are unentangled, because if they are entangled, they won't have their own separate state vectors and you can't create the outer products of those either. Despite this limitation, let's explore this equation in detail and see what insight it can bring us:

\[ \begin{aligned} \rho &= \left| u \right\rangle\!\left\langle u \right| \otimes \left| v \right\rangle\!\left\langle v \right| \\[5pt] &= \begin{bmatrix} u_0\\ u_1 \end{bmatrix} \begin{bmatrix} \overline{u_0} & \overline{u_1} \end{bmatrix} \otimes \begin{bmatrix} v_0\\ v_1 \end{bmatrix} \begin{bmatrix} \overline{v_0} & \overline{v_1} \end{bmatrix} \\[5pt] &= \begin{bmatrix} u_0 \overline{u_0} & u_0 \overline{u_1} \\ u_1 \overline{u_0} & u_1 \overline{u_1} \\ \end{bmatrix} \otimes \begin{bmatrix} v_0 \overline{v_0} & v_0 \overline{v_1} \\ v_1 \overline{v_0} & v_1 \overline{v_1} \\ \end{bmatrix} \\[5pt] &= \begin{bmatrix} u_0 \overline{u_0} \begin{bmatrix} v_0 \overline{v_0} & v_0 \overline{v_1} \\ v_1 \overline{v_0} & v_1 \overline{v_1} \\ \end{bmatrix} & u_0 \overline{u_1} \begin{bmatrix} v_0 \overline{v_0} & v_0 \overline{v_1} \\ v_1 \overline{v_0} & v_1 \overline{v_1} \\ \end{bmatrix} \\ u_1 \overline{u_0} \begin{bmatrix} v_0 \overline{v_0} & v_0 \overline{v_1} \\ v_1 \overline{v_0} & v_1 \overline{v_1} \\ \end{bmatrix} & u_1 \overline{u_1} \begin{bmatrix} v_0 \overline{v_0} & v_0 \overline{v_1} \\ v_1 \overline{v_0} & v_1 \overline{v_1} \\ \end{bmatrix} \\ \end{bmatrix} \hspace{20pt}\text{(Matrix 5)} \\[5pt] &= \begin{bmatrix} u_0 \overline{u_0} v_0 \overline{v_0} & u_0 \overline{u_0} v_0 \overline{v_1} & u_0 \overline{u_1} v_0 \overline{v_0} & u_0 \overline{u_1} v_0 \overline{v_1}\\ u_0 \overline{u_0} v_1 \overline{v_0} & u_0 \overline{u_0} v_1 \overline{v_1} & u_0 \overline{u_1} v_1 \overline{v_0} & u_0 \overline{u_1} v_1 \overline{v_1}\\ u_1 \overline{u_0} v_0 \overline{v_0} & u_1 \overline{u_0} v_0 \overline{v_1} & u_1 \overline{u_1} v_0 \overline{v_0} & u_1 \overline{u_1} v_0 \overline{v_1}\\ u_1 \overline{u_0} v_1 \overline{v_0} & u_1 \overline{u_0} v_1 \overline{v_1} & u_1 \overline{u_1} v_1 \overline{v_0} & u_1 \overline{u_1} v_1 \overline{v_1} \end{bmatrix} \hspace{20pt}\text{(Matrix 6)} \end{aligned} \]

Matrix 6 is obviously equivalent to Matrix 1: since complex multiplication is commutative, for each entry of Matrix 1, just move the u's to the left and the v's to the right and you end up with Matrix 6.

But let's focus on Matrix 5. According to the geometric interpretation for computing Bob's reduced density matrix, we should add up the following submatrices:

\[ \begin{aligned} \begin{bmatrix} \bbox[#ffcc99, 3px]{ u_0 \overline{u_0} \begin{bmatrix} v_0 \overline{v_0} & v_0 \overline{v_1} \\ v_1 \overline{v_0} & v_1 \overline{v_1} \\ \end{bmatrix}} & u_0 \overline{u_1} \begin{bmatrix} v_0 \overline{v_0} & v_0 \overline{v_1} \\ v_1 \overline{v_0} & v_1 \overline{v_1} \\ \end{bmatrix} \\ u_1 \overline{u_0} \begin{bmatrix} v_0 \overline{v_0} & v_0 \overline{v_1} \\ v_1 \overline{v_0} & v_1 \overline{v_1} \\ \end{bmatrix} & \bbox[#ffcc99, 3px]{ u_1 \overline{u_1} \begin{bmatrix} v_0 \overline{v_0} & v_0 \overline{v_1} \\ v_1 \overline{v_0} & v_1 \overline{v_1} \\ \end{bmatrix}} \\ \end{bmatrix} \end{aligned} \]

which gives us:

\[ \begin{aligned} \rho^B = u_0 \overline{u_0} \begin{bmatrix} v_0 \overline{v_0} & v_0 \overline{v_1} \\ v_1 \overline{v_0} & v_1 \overline{v_1} \\ \end{bmatrix} \ + \ u_1 \overline{u_1} \begin{bmatrix} v_0 \overline{v_0} & v_0 \overline{v_1} \\ v_1 \overline{v_0} & v_1 \overline{v_1} \\ \end{bmatrix} \end{aligned} \]

The inner product of a qubit with itself must always be equal to 1, since the total probability of all outcomes must be 1. Since \( u_0 \) and \( u_1 \) are the amplitudes of Alice's qubit, we know that \( u_0 \overline{u_0} + u_1 \overline{u_1} = 1 \). Therefore, the reduced density matrix of Bob is equal to:

\[ \begin{aligned} \rho^B = \begin{bmatrix} v_0 \overline{v_0} & v_0 \overline{v_1} \\ v_1 \overline{v_0} & v_1 \overline{v_1} \\ \end{bmatrix} \end{aligned} \]

This is exactly the same as the outer product of Bob's qubit with its conjugate transpose or, in other words, the density matrix of Bob's qubit in isolation. Similarly, if we use Matrix 5 to determine Alice's reduced density matrix, we get:

\[ \begin{aligned} \begin{bmatrix} \bbox[#ffcc99, 3px]{ u_0 \overline{u_0} } \begin{bmatrix} \bbox[#ffcc99, 3px]{ v_0 \overline{v_0} } & v_0 \overline{v_1} \\ v_1 \overline{v_0} & \bbox[#ffcc99, 3px]{ v_1 \overline{v_1} } \\ \end{bmatrix} & \bbox[#ffcc99, 3px]{ u_0 \overline{u_1} } \begin{bmatrix} \bbox[#ffcc99, 3px]{ v_0 \overline{v_0} } & v_0 \overline{v_1} \\ v_1 \overline{v_0} & \bbox[#ffcc99, 3px]{ v_1 \overline{v_1} } \\ \end{bmatrix} \\ \bbox[#ffcc99, 3px]{ u_1 \overline{u_0} } \begin{bmatrix} \bbox[#ffcc99, 3px]{ v_0 \overline{v_0} } & v_0 \overline{v_1} \\ v_1 \overline{v_0} & \bbox[#ffcc99, 3px]{ v_1 \overline{v_1} } \\ \end{bmatrix} & \bbox[#ffcc99, 3px]{ u_1 \overline{u_1} } \begin{bmatrix} \bbox[#ffcc99, 3px]{ v_0 \overline{v_0} } & v_0 \overline{v_1} \\ v_1 \overline{v_0} & \bbox[#ffcc99, 3px]{ v_1 \overline{v_1} } \\ \end{bmatrix} \\ \end{bmatrix} \end{aligned} \]

which gives us:

\[ \begin{aligned} \rho^A = \begin{bmatrix} u_0 \overline{u_0} \left( v_0 \overline{v_0} + v_1 \overline{v_1} \right) & u_0 \overline{u_1} \left( v_0 \overline{v_0} + v_1 \overline{v_1} \right) \\ u_1 \overline{u_0} \left( v_0 \overline{v_0} + v_1 \overline{v_1} \right) & u_1 \overline{u_1} \left( v_0 \overline{v_0} + v_1 \overline{v_1} \right) \end{bmatrix} \end{aligned} \]

Analogous to Alice's amplitudes, we know that \( v_0 \overline{v_0} + v_1 \overline{v_1} = 1 \). Therefore, the reduced density matrix of Alice is equal to:

\[ \begin{aligned} \rho^A = \begin{bmatrix} u_0 \overline{u_0} & u_0 \overline{u_1} \\ u_1 \overline{u_0} & u_1 \overline{u_1} \\ \end{bmatrix} \end{aligned} \]

Like Bob's, Alice's reduced density matrix is equal to the density matrix of Alice's qubit in isolation.

So, if we take Matrix 5 (which is a blueprint for calculating the density matrix by taking the tensor product of the separate components) as a base, and apply the geometric interpretation of the partial trace (which is equivalent to the summation notation), we can see that the partial trace effectively reverses the calculation of the density matrix! It yields the separate components that formed the density matrix of the total system in the first place! Of course, this only literally applies when the two qubits are unentangled, but it still provides a nice insight into why the partial trace works the way it does.

If we look at simple multiplication, then 2*10 = 20, but if you have 20, you don't know whether it came from 2*10 or 4*5. That information is lost when you do the multiplication: the value 20 is only a single number and therefore contains less information than the original components combined. With density matrices that are created from qubits, this does not apply. Since the density matrix of the total system has 16 complex values, it makes sense that those values contain enough information to reconstruct the 2 qubits with 2 complex values each that created it in the first place.

I hope this small insight of mine is useful to you. Thanks for reading!