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I know that in general a "non-pure" state described by : $$ \rho = \sum_i p_i |\psi_i\rangle\langle \psi_i| $$ can't be written as $ \rho = | \phi\rangle\langle \phi|$.

But if we exclude the obvious when all the $ | \psi_i \rangle $ are identical, is it still possible?

In fact for me it is not obvious at first sight if I have $$ \rho = \sum_i p_i |\psi_i\rangle\langle\psi_i| $$ to be sure that it is or it is not a pure state without calculating $ \textrm{Tr}(\rho^2)$ for example. And I don't know if excepted the obvious case with identical $ | \psi_i \rangle $, such a state is necessarily non pure ?

So to summarise: if I have a density matrix state with different $ |\psi_i \rangle$, do you agree with me if I say that it still can be a pure state (and the only way to know it is to compute $\textrm{Tr}(\rho^2)$ )?

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    $\begingroup$ You can answer your question by computing $\mathrm{Tr}(\rho^2)$ for the generic sum and think about whether or not that can be 1 if more than one of the $p_i$ is non-zero. $\endgroup$ – ACuriousMind Dec 7 '16 at 14:14
  • $\begingroup$ Hint: For counterexamples, look at the Bloch sphere. $\endgroup$ – Qmechanic Dec 7 '16 at 14:23
  • $\begingroup$ I tried to do this but I don't see how to simplify the sum : $Tr(\rho^2)=\sum_{k,k',p} p_k p_k' <u_p|\psi_k><\psi_k|\psi_k'><\psi_k'|u_p>$ $\endgroup$ – StarBucK Dec 7 '16 at 14:23
  • $\begingroup$ If there is more than one $p_i>0$, and the corresponding $\psi_i$ are different, the resulting $\rho$ has rank $\ge2$. $\endgroup$ – Norbert Schuch Dec 7 '16 at 16:49
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Let me rephrase your question.

Suppose that $\rho$ is a pure state, i.e., it is written as $$\rho =|\psi\rangle \langle \psi|$$ for some unit vector $\psi$.

Your question is the following.

Q1. Is it possible to find a set of vectors $\phi_1, \ldots, \phi_n$ satisfying $$n>1\:,$$ $||\phi_i||=1$, possibly $\langle \phi_i|\phi_j \rangle \neq 0$ for some $i\neq j$, and numbers $q_1, \ldots, q_n$ with $0< q_i <1$ and $\sum_i q_i =1$, such that $$|\psi\rangle \langle \psi| = \sum_{i=1}^n q_i|\phi_i \rangle \langle \phi_i|$$ and $|\phi_i \rangle \langle \phi_i| \neq |\phi_j\rangle \langle \phi_j|$ for some $i\neq j$?

As far as I understand you already know the following general result.

THEOREM1. Consider an operator $\rho: H \to H$ where $H$ is a complex Hilbert space and $\rho$ is trace class, non-negative and $tr(\rho)=1$. Under these hypotheses, $\rho$ is a pure state if and only if $tr(\rho) = tr(\rho^2)$.

As a consequence, since the operator $\sum_{i=1}^n q_i|\phi_i \rangle \langle \phi_i|$ is trace class, non-negative with unit trace, Q1 may be restated as follows.

Q2. Is it possible to find a set of vectors $\phi_1, \ldots, \phi_n$ with $$n>1\:,$$ $||\phi_i||=1$, possibly $\langle \phi_i|\phi_j \rangle \neq 0$ for some $i\neq j$, and numbers $q_1, \ldots, q_n$ with $0< q_i <1$ and $\sum_i q_i =1$, such that $$tr\left[\left(\sum_{i=1}^n q_i|\phi_i \rangle \langle \phi_i|\right)^2\right] =1$$ and $|\phi_i \rangle \langle \phi_i| \neq |\phi_j\rangle \langle \phi_j|$ for some $i\neq j$?

The answer to Q2 is always negative as soon as $n>1$, and thus

it is not necessary to compute the trace of $\left(\sum_{i=1}^n q_i|\phi_i \rangle \langle \phi_i|\right)^2$, just knowing that $n>1$ is enough to decide that the state $\sum_{i=1}^n q_i|\phi_i \rangle \langle \phi_i|$ is not pure unless $|\phi_i \rangle \langle \phi_i|=|\phi_j \rangle \langle \phi_j|$ for all $i,j$.

The proof is the following. First of all let me introduce the Hilbert-Schmidt scalar product between Hilbert Schmidt operators, and thus trace class operators in particular, $$(\rho|\rho')_{HS} := tr(\rho^*\rho')\:.$$ The associated norm reads $$||\rho||_{HS}= \sqrt{tr(\rho^*\rho)}\:.$$

Theorem1 can equivalently be restated as follows.

THEOREM2. Consider an operator $\rho: H \to H$ where $H$ is a complex Hilbert space and $\rho$ is trace class, non-negative and $tr(\rho)=1$. Under these hypotheses, $\rho$ is a pure state if and only if $||\rho||_{HS}=1$.

Now consider an operator $\rho: H \to H$ of the form $$\rho = \sum_{i=1}^n q_i \rho_i\tag{0}$$ where $\rho_i := |\phi_i \rangle \langle \phi_i|$ with $\phi_i$ and $q_i$ as in Q2. $\rho$ is trace class, non-negative and we want to check if $||\rho||_{HS}=1$ is possible when $n>1$. This condition is equivalent to saying that $\rho$ is pure.

We can always restrict ourselves to deal with a real vector space of trace class operators, since our trace class operators are self-adjoint and the linear combinations we consider are constructed with real (and non-negative) numbers. The scalar product $(\:|\:)_{HS}$ becomes a standard real (symmetric) scalar product in that real subspace.

The crucial observation is that, as it happens in every real vector space equipped with a real scalar product, $$\left|\left|\sum_{i=1}^n x_i\right|\right| \leq \sum_{i=1}^n ||x_i||\tag{1}$$ and "$\leq$" is replaced for "$=$" if and only if $x_i = \alpha_i x$ for some fixed $x$ and non negative numbers $\alpha_i$ where $i=1,\ldots,n$.

In other words, $$\left|\left|\sum_{i=1}^n q_i \rho_i\right|\right|_{HS} \leq \sum_{i=1}^n ||q_i \rho_i||_{HS}\tag{2}$$ and "$\leq$" is replaced for "$=$" if and only if $q_i \rho_i = \alpha_i T$ for some fixed $T$ and non negative numbers $\alpha_i$ where $i=1,\ldots,n$.

Since we know that $$\sum_{i=1}^n ||q_i \rho_i||_{HS}= \sum_{i=1}^n q_i ||\rho_i||_{HS} = \sum_{i=1}^n q_i 1 = \sum_{i=1}^n q_i =1$$ we conclude that If $\rho$ in (0) is pure, then the sign "$\leq$" in (2) is replaced by "$=$", so that $q_i\rho_i = \alpha_i T$ for some fixed operator $T$ and reals $\alpha_i$. Taking the trace of both sides $q_i = \alpha_i tr(T)$ where $tr(T) \neq 0$ because $q_i \neq 0$. Re-defining $T \to \rho_0 := \frac{1}{tr T}T$, we have found that there is a positive trace-class operator $\rho_0$ with unit trace such that $\rho_i= \rho_0$ and furthermore $tr \rho_0^2 = tr \rho_i^2 =1$ so that $\rho_0$ is pure and thus it can be written as $\rho_0 := |\phi_0 \rangle \langle \phi_0|$ for some unit vector $\phi_0$. Summing up, we have obtained that

if $\rho$ in (0) is pure, then $|\phi_i\rangle \langle \phi_i|= |\phi_0 \rangle \langle \phi_0|$ for all $i=1,\ldots, n$.

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  • $\begingroup$ Thank you for your answer, but you assumed that the $ | \Psi_i > $ are orthogonal here : if they are orthogonal then a pure state is also a non pure state only if it is written $ | \Psi_i >< \Psi_i | $. My question was more general with any family of $\Psi_i$. $\endgroup$ – StarBucK Dec 9 '16 at 20:57
  • $\begingroup$ @user3183950 while I totally understand the confusion, note that it does not change the answer above. Every density matrix is diagonal in some orthonormal basis $\big\{|\phi_i\rangle\big\}$, which may have nothing to do with your $\big\{|\psi_i\rangle\big\}$ therefore $\operatorname{Tr}\rho^2 = \operatorname{Tr}\rho = 1$ is both necessary and sufficient. $\endgroup$ – CR Drost Dec 9 '16 at 21:54
  • $\begingroup$ The structure of this answer is a bit hard to parse - it kind of looks like proof by contradiction but then it kind of switches tracks. Maybe make it a bit clearer at the beginning what you're going to show, and how? $\endgroup$ – Emilio Pisanty Dec 10 '16 at 13:28
  • $\begingroup$ @Emilio Pisanty user3183950 I completely changed my answer since I suspected I had completely misunderstood the question. Please let me know if now my answer is more appropriate. $\endgroup$ – Valter Moretti Dec 10 '16 at 16:26
  • $\begingroup$ @Valter Yeah, that's a lot clearer. $\endgroup$ – Emilio Pisanty Dec 10 '16 at 16:38

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