# How do I trace out the second qubit to find the reduced density operator? [duplicate]

I'm doing an exercise to trace out the second qubit to find the reduced density operator for the first qubit:

$$tr_2|11\rangle\langle00| = |1\rangle\langle0|\langle0|1\rangle$$

I'm just wondering if I do trace for the first qubit, should I have:

$$tr_1|11\rangle\langle00| = |1\rangle\langle0|\langle0|1\rangle$$ or $$tr_1|11\rangle\langle00| = \langle0|1\rangle|1\rangle\langle0|$$ ?

In the Nielsen-and-Chuang textbook, we have $$tr(|b_1\rangle\langle b_2|)=\langle b_2|b_1\rangle$$. Can I say the left and right hand side are just two ways to locate an element in a matrix? Thanks!!

Suppose you have the state $$|\psi\rangle = \dfrac{|00\rangle + |11\rangle}{\sqrt{2}} = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 0 \\ 0 \\ 1 \end{pmatrix}$$ then its density matrix representation is

$$\rho = |\psi \rangle \langle \psi | = \dfrac{1}{2} \begin{pmatrix} 1 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 \end{pmatrix}$$

Now, if we want to trace out the subsystem $$B$$ to find the density operator of the system $$A$$ denoted as $$\rho_A$$ then we can do the following:

$$\rho_A = Tr_B(\rho) = \dfrac{1}{2} \begin{pmatrix} Tr\begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix} & Tr\begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix}\\ Tr\begin{pmatrix} 0 & 0\\ 1 & 0 \end{pmatrix} & Tr\begin{pmatrix} 0 & 0\\ 0 & 1 \end{pmatrix} \end{pmatrix} = \dfrac{1}{2} \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}$$

It turns out that $$\rho_B = Tr_A(\rho)$$ is the same as $$\rho_A$$ here and from looking at the state, you might expect why that is the case.

More generally, giving a density operator

$$\rho = \begin{pmatrix} \rho_{11} & \rho_{12} & \rho_{13} & \rho_{14}\\ \rho_{21} & \rho_{22} & \rho_{23} & \rho_{24}\\ \rho_{31} & \rho_{32} & \rho_{33} & \rho_{34} \\ \rho_{41} & \rho_{42} & \rho_{43} & \rho_{44} \end{pmatrix}$$

then

$$\rho_A = Tr_B(\rho) = \begin{pmatrix} Tr\begin{pmatrix} \rho_{11} & \rho_{12}\\\rho_{21} & \rho_{22} \end{pmatrix} & Tr\begin{pmatrix} \rho_{13} & \rho_{14} \\ \rho_{23} & \rho_{24} \end{pmatrix}\\ Tr\begin{pmatrix} \rho_{31} & \rho_{32} \\ \rho_{41} & \rho_{42} \end{pmatrix} & Tr\begin{pmatrix}\rho_{33} & \rho_{34} \\ \rho_{43} & \rho_{44} \end{pmatrix} \end{pmatrix} = \begin{pmatrix} \rho_{11} + \rho_{22} & \rho_{13} + \rho_{24} \\ \rho_{31} + \rho_{42} & \rho_{33} + \rho_{44} \end{pmatrix}$$

and

$$\rho_B = Tr_A(\rho) = \begin{pmatrix} Tr\begin{pmatrix} \rho_{11} & \rho_{13}\\\rho_{31} & \rho_{33} \end{pmatrix} & Tr\begin{pmatrix} \rho_{12} & \rho_{14} \\ \rho_{32} & \rho_{34} \end{pmatrix}\\ Tr\begin{pmatrix} \rho_{21} & \rho_{23} \\ \rho_{41} & \rho_{43} \end{pmatrix} & Tr\begin{pmatrix}\rho_{22} & \rho_{24} \\ \rho_{42} & \rho_{44} \end{pmatrix} \end{pmatrix} = \begin{pmatrix} \rho_{11} + \rho_{33} & \rho_{12} + \rho_{34} \\ \rho_{21} + \rho_{43} & \rho_{22} + \rho_{44} \end{pmatrix}$$

• Thanks for the answer! How can we represent the subsystem A and B in the general case?
– ZR-
Jan 20, 2021 at 3:09
• @Zhengrong hmm... do you mean when the subsystem is different than 2-by-2 matrix? Jan 20, 2021 at 7:25

If you split your state into a bipartite system $$\rho_{AB} \in \mathcal{H}_A \otimes \mathcal{H}_B$$ then one general formula for a partial trace is given by:

$$\text{Tr}_B (\rho) = \sum_{j} (I_A \otimes \langle j |_B) \rho (I_A \otimes | j \rangle_B)$$

where $$\{ |j\rangle \}$$ is a basis for system $$B$$. In your case, for the first statement you can use this formula to find

\begin{align} \text{Tr}_B (|11\rangle\langle00|) &= \sum_{j} (I_A \otimes \langle j |_B) |11\rangle\langle00| (I_A \otimes | j \rangle_B) \\ &= (I_A \otimes \langle 0 |_B) |1\rangle_A |1\rangle_B \langle0|_A \langle0|_B (I_A \otimes | 0 \rangle_B) \\ &\qquad+ (I_A \otimes \langle 1 |_B) |1\rangle_A |1\rangle_B \langle0|_A \langle0|_B (I_A \otimes | 1 \rangle_B)\\ &= |1\rangle\langle0|_A (\langle 0|1\rangle\langle 0|0\rangle) + |1\rangle\langle 0|_A(\langle1|1\rangle\langle0|1\rangle) \\ &= |1\rangle\langle0|_A \langle 0| 1\rangle (\langle 0|0\rangle + \langle 1|1\rangle) \\ &= 0 \end{align} and you can do a similar calculation to derive the second statement.