The kernel of this linear map is the set of solutions to the equation A x = 0, where 0 is understood as the zero vector.

But what's the physical meaning of the kernel of density matrix?


1 Answer 1


Suppose the system is in some mixed state described by the density operator: $$\hat { \rho}=\sum_{k=1}^N \rho_k |\psi_k\rangle\langle\psi_k|$$ where $N$ is the dimensionality of the Hilbert space. Now for every vector $|x\rangle \in \mathrm{Ker}(\hat \rho)$, we have by definition: $$\hat { \rho}|x\rangle=\sum_{k=1}^N \rho_k |\psi_k \rangle\langle\psi_k|x\rangle=0$$ Because of the linear independence of the states, this means that $\langle\psi_k|x\rangle=0 $ for all $k \in \{1,...,N\}$, i.e. the kernel of the density operator is the subspace corresponding to all vectors orthogonal to the ensemble of pure states which constitute the overall mixed state. Physically, this means that the kernel is the subspace spanned by states which the system has zero probability of being in. In other words, every state inside the kernel has zero probability of occurrence, and vise versa.

If you're unsure about the vise versa part, consider a state $|x\rangle $ which has zero probability of occurrence; we can easily show that this state belongs to the kernel of the density operator. To do this, recall that the probability of being in a state is just the trace of the density operator multiplied by the projector onto that state; i.e. $$P=\mathrm{Tr}(\hat \rho |x\rangle \langle x|) = \sum_{k=1}^N \langle\psi_k| \ (\hat \rho |x\rangle \langle x|) \ |\psi_k \rangle=\sum_{k=1}^N \langle\psi_k| \hat \rho |x\rangle \langle x |\psi_k\rangle$$ $$P=\sum_{k=1}^N \rho_k\langle \psi_k |x\rangle \langle x |\psi_k\rangle =\sum_{k=1}^N \rho_k \ |\langle \psi_k |x\rangle |^2$$ Where I have used the fact that the density operator is hermitian. Now setting $P=0$ , because all the terms in the above sum are non-negative (remember $\rho_k$ is a probability), they all have to be zero for the sum to be zero; which means $\langle \psi_k|x\rangle =0 $ . Thus: $$\hat { \rho}|x\rangle =\sum_{k=1}^N \rho_k |\psi_k \rangle \langle \psi_k|x\rangle =0$$ Which means that $|x\rangle \in \mathrm{Ker}(\hat \rho)$.

  • $\begingroup$ Then does it means that the support is the subspace spanned by states which the system has non-zero probability of being in $\endgroup$
    – Sirui Lu
    Dec 8, 2017 at 3:36
  • $\begingroup$ Yes. As was shown a state has zero probability of occurrence if and only if it is in the kernel, this means that any other state must not be in the kernel, i.e. it must be in the support of the operator. $\endgroup$ Dec 8, 2017 at 4:09

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