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How can I prove that $\mathrm{tr}(\rho^2) $ = 1 if and only if the state is pure?

My idea: I know how to show that $\mathrm{tr}(\rho^2) \leq 1$ and from there I am trying to show by contradiction that $\mathrm{tr}(\rho^2) = 1$ can only but true for pure state, but I am kind of stuck.

I just need a hint on how to prove this.

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2 Answers 2

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Since an arbitrary $\rho$ is self-adjoint, it has the spectral decomposition $\rho = \sum_n \rho_n |\psi_n><\psi_n|$, in terms of an orthonormal basis $\{ |\psi_n>\}$, which here we pick discrete for simplicity.

Hermiticity implies $\rho_n = \rho^*_n$. $\mathrm{Tr} \rho = 1$ implies $\sum_n \rho_n =1$. Semi-positivity implies $0 \leq \rho_n$. Together they imply $0 \leq \rho_n \leq 1$, which implies $\rho^2_n \leq \rho_n$. Hence, $\mathrm{Tr} \rho^2 = \sum_n \rho^2_n \leq \sum_n \rho_n = 1$ and so $\mathrm{Tr} \rho^2 \leq 1$ for a generic state, as you mention.

Now let's start assuming that $\mathrm{Tr} \rho^2 =1$. Following the inequalities we just wrote, this implies that $\rho^2_n = \rho_n$ for all $n$ i.e. $\rho^2 = \rho$. In particular this implies that $\rho_n = 1$ or $\rho_n = 0$. More precisely, due to the trace condition $\mathrm{Tr} \rho = 1$, only one $\rho_n$ is equal to one while the others vanish. This is a pure state.

The reverse implication is direct.

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  • $\begingroup$ Excellent answer, but I suggest beginning with "Now let's start assuming...." as this is the only info absolutely needed to answer the question. You could refer to the other results in the "Following the inequalities" sentence, as these are more background and could be replaced by "It can be shown". PS: the only reason I am being so pedantic is that this is a great answer to be a general reference for the demonstration of an important fundamental fact that people might want to cite. $\endgroup$ Jan 11, 2018 at 2:47
  • $\begingroup$ Thanks @WetSavannaAnimalakaRodVance for your suggestions and kindness. I'll take them into account. $\endgroup$
    – secavara
    Jan 11, 2018 at 2:53
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    $\begingroup$ Thanks a lot for your answer. It is very clear and straightforward :) $\endgroup$ Jan 11, 2018 at 10:01
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If you diagonalize the density matrix into eigenvectors $|\psi_i\rangle$ and eigenvalues (probabilities) $\lambda_i\in\mathbb{R}_+$, then the conditions $\text{tr}(\rho)=\text{tr}(\rho^2)=1$ give

$$\sum_{i}\lambda_i=1$$

and

$$\sum_{i}\lambda_i^2=1,$$

respectively. Showing that these equations are inconsistent unless only one of the $\lambda_i$ is nonzero is equivalent to showing that $\text{tr}(\rho^2)=1$ implies that the system is in a pure state.

Hint: Square the first condition and subtract away the second to reach a contradiction.

I hope this helps!

PS: Another way to rephrase this is to show that given any positive-definite operator $A$, then the inequality $\langle A\rangle^2-\langle A^2\rangle\leq 0$ is saturated if and only if $A=|\psi\rangle\langle\psi|$ for some vector $\psi\in\mathcal{H}$.

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  • $\begingroup$ Thanks a lot for your answer, when we rewrite the density matrix with the spectral analysis, are my new eigenvectors the same states I had before? $\endgroup$ Jan 11, 2018 at 10:00

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