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Imagine I have two potentially entangled qudits and measure the first one using some Hermitian operator $M$ that has only one eigenstate per eigenvalue and get some outcome $m$. Can I know for sure the state of my two qudits i.e their respective state vectors? Or is it a probability distribution so a mixed state ?

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In general, the post-measurement state may be mixed. However, if the joint state of the qudits is pure before measurement, then the post-measurement state is also pure. The converse is not true.

Equivalent condition

Let $\rho$ denote the joint state of the two qudits before measurement and write the eigendecomposition of the observable $M$ as $$ M=\sum_m\lambda_m|\psi_m\rangle\langle\psi_m|\tag1 $$ where the eigenvalues $\lambda_m$ are all distinct since $M$ is assumed to be non-degenerate. From the measurement postulate, the post-measurement state $\sigma_m$ is $$ \begin{align} \sigma_m&\sim(|\psi_m\rangle\langle\psi_m|\otimes I)\rho(|\psi_m\rangle\langle\psi_m|\otimes I)\\ &=|\psi_m\rangle\langle\psi_m|\otimes\langle\psi_m|\rho|\psi_m\rangle \end{align}\tag2 $$ where $\sim$ means that the operator on the left hand side is a scalar multiple of the operator on the right hand side. The post-measurement state is pure if and only if $\mathrm{rank}(\sigma_m)=1$. Substituting from $(2)$, we have

$$ \begin{align} \mathrm{rank}(\sigma_m)&=\mathrm{rank}(|\psi_m\rangle\langle\psi_m|\otimes\langle\psi_m|\rho|\psi_m\rangle)\\ &=\mathrm{rank}(|\psi_m\rangle\langle\psi_m|)\,\mathrm{rank}(\langle\psi_m|\rho|\psi_m\rangle)\\ &=\mathrm{rank}(\langle\psi_m|\rho|\psi_m\rangle) \end{align}\tag3 $$ so for non-degenerate $M$ the post-measurement state is pure if and only if $$ \mathrm{rank}(\langle\psi_m|\rho|\psi_m\rangle)=1.\tag4 $$

Example 1: Pure input

For any linear operators $A$ and $B$, we have $\mathrm{rank}(AB)\le \min(\mathrm{rank}(A),\mathrm{rank}(B))$. Therefore, if the input state $\rho$ is pure, then the post-measurement state is also pure.

Example 2: Maximally mixed input

If $\rho\sim I$, then $\mathrm{rank}(\langle\psi_m|\rho|\psi_m\rangle)=d>1$ where $d$ is the dimension of the second qudit's Hilbert space. Thus, in this case the post-measurement state is not pure.

Correlations

The next two examples demonstrate that in some circumstances if the qudits are correlated, either classically or quantumly, then we can infer the post-measurement state of one qudit from the measurement outcome on the other. Below, we assume that $|\psi_m\rangle=|0\rangle$.

Example 3: Classical correlation

If $\rho\sim|00\rangle\langle 00|+|11\rangle\langle 11|$, i.e. the input state exhibits perfect classical correlation, then $\mathrm{rank}(\langle 0|\rho|0\rangle)=\mathrm{rank}(|0\rangle\langle 0|)=1$ and the post-measurement state is pure. Note that the choice of the measurement basis is significant.

Example 4: Quantum correlation

If $\rho=|\phi\rangle\langle\phi|$ where $|\phi\rangle\sim|00\rangle+|11\rangle$, i.e. the input state is maximally entangled, then $\mathrm{rank}(\langle 0|\rho|0\rangle)=\mathrm{rank}(|0\rangle\langle 0|)=1$ and the post-measurement state is once again pure. In this case, the state is pure regardless of the choice of the measurement basis as long as the joint state is pure before measurement.

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