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Let $\rho$ and $\rho'$ be $n\times n$ density matrices, and suppose that for every observable $A$ and every $\lambda$ in the spectrum of $A$ we have $$ \text{tr}(\rho P_{\lambda})=\text{tr}(\rho' P_{\lambda}), $$ where $P_{\lambda}$ is the orthogonal projection onto the $\lambda$-eigenspace of $A$. Does it then necessarily follow that $\rho'=U\rho U^{\dagger}$ for some unitary $U$?

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3 Answers 3

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Let $H$ denote a finite-dimensional complex Hilbert space, $\rho$ and $\rho^\prime$ be two density matrices, i.e. positive semi-definite operators with unit trace and $A$ an arbitrary hermitian operator. Its spectral representation reads $$ A=\sum\limits_\lambda \lambda(A)\, P_\lambda (A) \quad .$$

If the equality in the question holds for all $A$, then $\rho^\prime = \rho$. To see this, multiply the equality with $\lambda(A)$ and sum over all $\lambda$. We then find

$$\mathrm{Tr}\rho A = \mathrm{Tr}\rho^\prime A \quad, $$

which we can write as $$\mathrm{Tr} (\rho-\rho^\prime)A = 0 \quad \forall A \quad .$$

Now there are several ways to see that this implies $\rho-\rho^\prime =0$. For example, pick $A:=\rho-\rho^\prime$. Then $\mathrm{Tr} A^2 = 0$. But $A^2 \geq 0$ and thus $A=0$.

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Actually $\rho=\rho'$. Indeed, specializing $A= P= |\psi\rangle \langle \psi|$ for $||\psi||=1$, the hypothesis implies $\langle\psi| (\rho-\rho')\psi\rangle =0$. Linearity permits to relax the requirment $||\psi||=1$. By polarization, in turn, it implies $\langle\psi| (\rho-\rho')\phi\rangle =0$ for every pair of vectors $\psi,\phi$. Fixing $\phi$ and choosing $\psi:= (\rho-\rho')\phi$, we conclude that, for every vector $\phi$ $$|| (\rho-\rho')\phi||^2 =0 $$ which means $\rho-\rho'=0$. The result is valid also for infinite dimensional Hilbert spaces.

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    $\begingroup$ Thanks for the answer! Do you know of some weaker form of the hypotheses which yields equality up to unitary equivalence rather than equality? $\endgroup$
    – dezign
    Jun 29 at 8:59
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    $\begingroup$ A very stupid condition is that the states are both pure. So your question is maningful only for mixed states. A condition is that $\rho$ abd $\rho'$ have the same spectrum and each eigenspace is non-degenerated. $\endgroup$ Jun 29 at 9:34
  • $\begingroup$ So superposing two different sequences of mutually orthogonal pure states with the same classical probability distribution (and different ptobabilities for different pure states in each sequence) leads to a pair of unitary equivalent mixed states. $\endgroup$ Jun 29 at 9:41
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The answers from Jason Funderberker and Valter Moretti are perfect at explaining mathematically how the equivalence of all possible measurements requires the two density matrices to be the same.

However, something I think is important to highlight is that this is not some kind of mathematical accident, it is an essential part of a good theroy. If two things are completely impossible to distinguish from one another in real life (with experimental measurements), then it is really bad if the theory says the two are in fact distinct. If the two could not be distinguished by the experimentalists, then why did the theorist trying to fit a model to there results insist that the two were in fact different? Its is a big no-no.

If density matrices did not have that property they would not be a good representation of the system (and our ignorance of the system), and they would need to be replaced with a different approach.

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  • $\begingroup$ But isn't it possible that nature may have properties which are not distinguishable by measurement (e.g. hidden variables)? Moreover, the question only addresses PVMs. A priori it could be the case that two states are equal if and only if they agree on all POVMs, but it turns out that being equal on PVMs is sufficient. $\endgroup$
    – dezign
    Jul 2 at 8:36
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    $\begingroup$ If nature has properties that are not distinguishable by any measurement then we (as human beings) would not know that those properites existed - so we could not put them in theory (except maybe by dumb luck). The entire point hidden variables is to explain why two states propared the same way are actually observed to be different in a real measurment. The POVMs is a very good point. A theory where to be indentical two states needed the stronger condition of fully matching on all POVMs would be very reasonable by my standards. I fully agree that the fact that PVMs are enough is a non-trivial. $\endgroup$
    – Dast
    Jul 4 at 10:02

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