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

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

Let us introduce the notation $\newcommand{\tr}{\operatorname{tr}}P_i\equiv \lvert\psi_i\rangle\!\langle\psi_i\rvert$. Note that each $P_i$ is a (normal) projector with unit trace.

The question is then equivalent to the following:

when can a convex combination of (normal) trace-$1$ projectors $P_i$ be a trace-$1$ projector?

The answer is that this is the case if and only if the projections are all equal (or in other words, it is never true, except for the trivial cases).

To show this, suppose $\rho=\sum_i p_i P_i$ with $p_i>0, \sum_i p_i=1$.

A normalised state $\rho$ is pure if and only if $\rho^2=\rho$, and if and only if $\tr(\rho^2)=\tr(\rho)$. We have $$ \rho^2 = \sum_i p_i^2 P_i + \sum_{i\neq j}p_i p_j P_i P_j.$$ and thus $$\tr(\rho^2) = \sum_i p_i^2 + 2 \sum_{i<j}p_i p_j\tr(P_i P_j)\le \left(\sum_i p_i\right)^2=1,$$ where the inequality becomes an identity if and only if $\tr(P_i P_j)=1$ for all $i,j$, that is, if and only if $P_i=P_j$.

Another way to prove this is to pass by the eigendecomposition of $\rho$.

If $\rho$ is a trace-$1$ projection, then there is some vector $v$ such that $\rho v=v$. This would imply $\sum_i p_i P_i v=v$, and thus $\sum_i p_i \langle v,P_i v\rangle=1$. But $\langle v,P_i v\rangle\in[0,1]$ for all $i$, and thus the only way for such convex combination to equal $1$ is that all terms do, i.e. $\langle v,P_i v\rangle=1$ for all $i$, and thus $P_i=vv^\dagger$ for all $i$.

Yet another argument is to observe that a state is pure if and only its rank is $1$, and if and only if its support is one-dimensional. For a convex combination (more generally, a sum with positive coefficients) of rank-$1$ projections to have one-dimensional support, each component must also have the same (one-dimensional) support, hence $P_i=P_j$.

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