Reduced density matrix and partial trace

Reduced density matrix

Suppose we have two quantum systems \(a, b\), with dimension \(N_a, N_b\) respectively. Then the Hilbert space of \(a+b\) is of dimension \(N=N_aN_b\). Suppose we have a density matrix

$$\hat\rho=\sum_{i,j}\rho_{ij}\lvert i\rangle \langle j\rvert=\sum_{i,j,k,l}\rho_{ijkl}\lvert i\rangle_a\lvert j\rangle_b \langle k\rvert_a\langle l\rvert_b$$

Then the reduced density matrix of \(a\) is defined as

$$\hat\rho_a=\mathrm{tr}_b\hat\rho=\sum_i \langle i\rvert_b\hat\rho\lvert i\rangle_b$$

i.e. reduced density matrix problem is equivalent to partial trace problem.

Tensor

In fact, if we take \(\hat\rho\) as a 4-tensor \(\rho_{ijkl}\), then the reduced density matrix is

$$\rho^{(a)}_{ij}=\delta^{\mu\nu}\rho_{i\mu k\nu}$$

For simple density matrix \(\rho=\lvert \psi\rangle \langle \psi\rvert\), the reduced matrix is

$$\rho^{(a)}_{ik}=\delta^{jl}\rho_{ijkl}=\delta^{jl}\psi_{ij}\psi^+_{lk}=\sum_i |\langle i_b\lvert \psi\rangle|^2=[\psi\psi^+]_{ik}$$

Here we are taking \(\psi\) as an \(N_a\times N_b\) matrix.

For general case, if we find decomposition

$$\rho=\sum_c \lambda_c\lvert \psi_c\rangle \langle \psi_c\rvert,\quad \sum_c \lambda_c=1$$

then we have

$$\rho^{(a)}_{ik}=\left[\sum_c\lambda_c\psi_c\psi^+_c\right]_{ik}$$
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