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I am wanting to find the two qubit reduced density matrix on a 5 qubit system. I have the 5 qubit state in the form of a tensor product and I want to find the reduced density matrix of qubits 1 and 3. I have the following formula for the reduced density matrix for qubit 3:
\begin{align} \mathrm{Tr}_1\left|h\right>_{13}\left<h\right| = \sum_{i=0}^{1}\left<i|h\right>_{13}\left<h|i\right> \end{align}
I am unsure of how to extract $\left|h\right>_{13}$ from the 5 qubit state. Please help!
Doing a 5 to 2-qubit reduction is a little tedious, but we can illustrate how it works with a simpler example.
Let's take a 3-qubit state $\lvert\psi_{ABC}\rangle$ and boil it down to $\lvert\psi_{BC}\rangle$, say
$$\lvert\psi_{ABC}\rangle = \alpha\lvert 001\rangle + \beta\lvert 101\rangle + \gamma\lvert 011\rangle.$$
First of all, the density operator for the state $\lvert\psi_{ABC}\rangle$ is given by
$$\begin{align} \rho_{ABC} &= \lvert\psi_{ABC}\rangle\langle\psi_{ABC}\rvert \\ &= \left(\alpha\lvert 001\rangle + \beta\lvert 101\rangle + \gamma\lvert 011\rangle\right)\otimes\left(\alpha^*\langle 001\rvert + \beta^*\langle 101\rvert + \gamma^*\langle 011\rvert\right) \end{align}$$
Expanding this out, we get $$ \begin{align} \rho_{ABC} &= \alpha^*\alpha \lvert 001\rangle\langle 001\rvert + \beta^*\alpha \lvert 001\rangle\langle 101\rvert + \gamma^*\alpha \lvert 001\rangle\langle 011\rvert \\ &+\alpha^*\beta \lvert 101\rangle\langle 001\rvert + \beta^*\beta \lvert 101\rangle\langle 101\rvert + \gamma^*\beta \lvert 101\rangle\langle 011\rvert \\ &+\alpha^*\gamma \lvert 011\rangle\langle 001\rvert + \beta^*\gamma \lvert 011\rangle\langle 101\rvert + \gamma^*\gamma \lvert 011\rangle\langle 011\rvert \end{align} $$
The idea of the reduced density operator is to trace over the particles that you don't care about. For example, to find $\rho_{BC}$, we would trace over particle A:
$$\rho_{BC} = \text{Tr}_{A}\left(\rho_{ABC}\right)$$
Before we write out the whole thing, in symbols the trace over $A$ is:
$$\text{Tr}_{A}\left(\rho_{ABC}\right) = \langle 0_A\rvert\rho_{ABC}\lvert 0_A\rangle + \langle 1_A\rvert\rho_{ABC}\lvert 1_A\rangle.$$
Taking the trace will eliminate all terms where particle $A$ is not in the same state in the bra and ket. For example, the term $\beta\alpha^*\lvert101\rangle\langle001\rvert$ will disappear.
Carrying out the process, the terms that survive in the reduced density operator are
$$\begin{align} \rho_{BC} &= \alpha^*\alpha \lvert 01\rangle\langle 01\rvert + \gamma^*\alpha \lvert 01\rangle\langle 11\rvert \\ &+ \beta^*\beta \lvert 01\rangle\langle 01\rvert + \alpha^*\gamma \lvert 11\rangle\langle 01\rvert \\ &+ \gamma^*\gamma \lvert 11\rangle\langle 11\rvert \end{align}$$
If you want to further reduce to the $\rho_B$ density operator, then just trace over $C$, i.e.
$$\begin{align} \rho_B &= \text{Tr}_C \left(\rho_{BC}\right) \\ &= \langle 0_C\lvert\rho_{BC}\rvert 0_C\rangle + \langle 1_C\lvert\rho_{BC}\rvert 1_C\rangle \end{align}$$
In this case, all of the terms survive and the resulting density operator is
$$ \begin{align} \rho_B &= \left(\alpha^*\alpha + \beta^*\beta\right)\lvert0\rangle\langle0\rvert + \gamma^*\gamma\lvert 1\rangle\langle 1\rvert \\ & \gamma^*\alpha\lvert 0\rangle\langle 1\rvert+ \alpha^*\gamma\lvert 1\rangle\langle 0\rvert \end{align} $$
(If we were being strict with notation, then we should really write $\lvert 1_A\rangle$ as $\lvert 1\rangle\otimes I^{\otimes 2}$ where $I^{\otimes 2}$ is the identity operator on the $BC$ subspace. But we'll understand that when we write something like $\langle 1_A\rvert001\rangle$ it really means $\left(\langle 1_A\rvert 0_A\rangle\right)\otimes\lvert 01\rangle = \lvert 01\rangle$. Similarly, $\langle 001\rvert 1_A\rangle = \left(\langle 0_A\rvert 1_A\rangle\right)\langle01\rvert = 0$.)
