# Examples of density operators $\rho=\sum\limits_n p_n|\phi_n\rangle\langle\phi_n|$ in which the states $\{|\phi_n\rangle\}$ are not orthogonal

This thread has seen a *ton* of incorrect statements coming from a number of sides, so it's probably a good idea to set the record straight in a bit more detail, and to provide some more examples of how expressions of this form come up in practice.

So, let's go through a brief rundown of some pertinent points.

**The definition of a density matrix**is just an operator $\rho:\mathcal H \to \mathcal H$ that is self-adjoint and positive semidefinite (and trace class if $\dim(\mathcal H)=\infty$), and whose trace satisfies $$\mathrm{Tr}(\rho)=1.$$ More importantly, this is**all**that's required by the definition. Any operator that satisfies those conditions can legitimately be called a density matrix, period.Because of that,

**all operators that can be expressed in the OP's form**, $$ \rho = \sum_n p_n |\phi_n \rangle\langle \phi_n|, \tag{$*$}$$**are valid density matrices**so long as the component projectors are normalized to $\langle \phi_n|\phi_n \rangle =1$ and the weigths add up to $\sum_n p_n = 1$.**Those two requirements are the**None of the conditions for density-matrix-ness ($\rho^\dagger=\rho$, $\rho\geq 0$, and $\mathrm{Tr}(\rho)=1$) are impacted if the $|\phi_n\rangle$ are not pairwise orthogonal, or if their number exceeds the state space's dimension. That means that it's perfectly fine to take non-orthogonal states in a representation of the form $(*)$.*only*actual requirements.**Explicit examples with non-orthogonal projectors are trivial to construct.**Norbert Shuch's answer contains one example, but if you go looking for them you can build them instantly by just taking*any*collection of unit-normalized vectors weighted by unit-normalized weights $p_n$.To provide one such example explicitly, consider the two-level space $\mathcal H = \mathbb C^2$, and a sequence of $N$ vectors lying equispaced along the equator of its Bloch sphere, giving $$ \rho = \sum_{n=0}^{N-1} p_n |\varphi_n\rangle\langle \varphi_n| \quad \text{for} \quad |\varphi_n\rangle = \frac{1}{\sqrt{2}} \bigg( |0\rangle + e^{i 2\pi n/N} |1\rangle\bigg). \tag{$\star$} $$ Here the weights can be arbitrary so long as $\sum_{n=0}^{N-1} p_n=1$; one obvious choice is $p_n = 1/N$ which gives the maximally-mixed state $\rho = \frac12 \mathbb I$, but there's plenty of other possible choices.

**Representations of the form $(*)$ are not unique.**Suppose, say, that you have some density matrix $\rho$ that you've managed to represent as a sum of normalized projectors in two different ways, say, $$ \rho = \sum_n p_n |\phi_n \rangle\langle \phi_n| = \sum_m q_m |\chi_m \rangle\langle \chi_m|, \tag{$**$}$$ where $\sum_n p_n = 1 = \sum_m q_m$ and $\langle \phi_n|\phi_n \rangle =1=\langle \chi_m|\chi_m \rangle$. Then there are some loose requirements on the two sets of vectors, starting with the fact that $\mathrm{span}\{|\phi_n\rangle\}$ needs to match $\mathrm{span}\{|\chi_m\rangle\}$, but**in general, the layout of the $|\phi_n\rangle$ and the $|\chi_m\rangle$ within that span can be very different**. This is evident in the example $(\star)$ above with equal weights, where $\rho$ is independent of the number $N$ of vectors in your collection, and it can also be represented as $\rho = \tfrac12 \left[ |0\rangle\langle 0| + |1\rangle\langle 1| \right]$.**Representations of the form $(*)$ are interpretations, and little more.**There*is*some physical content in the statement $$ \rho = \sum_n p_n |\phi_n \rangle\langle \phi_n|, \tag{$*$}$$ namely, that you can produce the system state $\rho$ by producing the pure states $|\phi_n\rangle$ with probabilities $p_n$ and then forgetting which pure state you actually produced. However, the operative word there is "can": the fact that that procedure will produce $\rho$ does not say, at all, that it is the*only*possible procedure that will produce that state.**Representations do not imply that the vectors involved are eigenvectors of the resultant density matrix.**That's true*if*the projectors are pairwise orthogonal, but that's not a requirement at all, so it is perfectly possible to construct $\rho$ as a sum of projectors that have nothing to do with the sum's eigenprojectors.It's probably helpful to illustrate this with an explicit example, for clarity. Consider a two-level system that's prepared in a superposition of the form $$ |\theta_\pm\rangle = \cos(\theta/2)|0\rangle \pm \sin(\theta/2)|1\rangle,$$ i.e. an angle $\theta$ down from the north pole of the Bloch sphere, except that each time we flip a fair coin to see which sign of $\theta$ (i.e. which direction on the prime meridian) we take. Then the density matrix reads \begin{align} \rho & = \frac12 \bigg( |\theta_+\rangle\langle\theta_+| +|\theta_-\rangle\langle\theta_-| \bigg) \\ & = \frac12 \bigg( \big(\cos(\theta/2)|0\rangle + \sin(\theta/2)|1\rangle \big) \big(\cos(\theta/2)\langle 0| + \sin(\theta/2)\langle 1| \big) \\ & \qquad + \big(\cos(\theta/2)|0\rangle - \sin(\theta/2)|1\rangle \big) \big(\cos(\theta/2)\langle 0| - \sin(\theta/2)\langle 1| \big) \bigg) %\\ & = \frac12 \bigg( %\big(\cos^2(\theta/2)|0\rangle\langle 0| + \sin(\theta/2)\cos(\theta/2)|1\rangle %\langle 0| + \sin(\theta/2)\cos(\theta/2)|0\rangle \langle 1| + %\sin^2(\theta/2)|1\rangle\langle 1| \big) %\\ & \qquad + %\big(\cos^2(\theta/2)|0\rangle\langle 0| - \sin(\theta/2)\cos(\theta/2)|1\rangle %\langle 0| - \sin(\theta/2)\cos(\theta/2)|0\rangle \langle 1| + %\sin^2(\theta/2)|1\rangle\langle 1| \big) % \bigg) \\ & = \cos^2(\theta/2)|0\rangle\langle 0| + \sin^2(\theta/2)|1\rangle\langle 1| \end{align} because the off-diagonal terms cancel out. In this second representation, we

*do*have orthogonal projectors, so here $|0\rangle$ and $|1\rangle$ are indeed the unique eigenvectors of $\rho$ (unless $\theta=\pi/2$ and $\rho$ is maximally mixed). But that doesn't stop our initial representation, $\rho = \frac12 \left( |\theta_+\rangle\langle\theta_+| +|\theta_-\rangle\langle\theta_-| \right)$, with its non-orthogonal, non-eigenvector components, from also being true.**If a state is built up using non-orthogonal projectors, then it**, and that's perfectly fine. Representations of the form $(*)$ are a dime a dozen if you know where to look. So, you found one that's not the canonical one: great! there's millions where that one came from.*also*has a separate representation in terms of orthogonal projectors**Representations of the form $(*)$ really**If you want to build one yourself, say, for a two-level system, there's a few points that are particularly relevant to the recipe:*are*a dime a dozen.- The Pauli matrices are a basis for all valid density matrices, i.e. if $\rho=\rho^\dagger$ is traceless, then it can be represented as $$ \rho = \tfrac12 \mathbb I + \vec p \cdot \vec \sigma,$$ where $\vec p = (p_x,p_y,p_z)\in \mathbb R^3$ and $\vec \sigma =(\sigma_x, \sigma_y, \sigma_z)$ are the Pauli matrices. (Further, that relationship can be inverted via $\vec p = \mathrm{Tr}(\rho\vec\sigma)$.)
- The positivity condition $\rho\geq 0$ translates into the condition $||\vec p||\leq 1$, i.e. $\vec p$ lives inside the unit ball or its boundary $-$ generally known as the Bloch ball and the Bloch sphere in this context.
- If $|\vec p|=1$, i.e. $\vec p$ is on the Bloch sphere boundary, then $\rho = |\psi\rangle\langle\psi|$ is a pure state, and if you write $|\psi\rangle = \cos(\theta/2) |0\rangle + e^{i\varphi}\sin(\theta/2)|1\rangle$ (which you always can) then $\theta\in [0,\pi]$ and $\varphi\in[0,2\pi)$ are the polar and azimuthal spherical coordinates for $$ \vec p = (\sin(\theta)\cos(\varphi), \sin(\theta)\sin(\varphi), \cos(\theta).$$
- The relationship between $\vec p$ and $\rho$ is linear and bijective.
- If $\rho_1$ and $\rho_2$ are valid density matrices, then any convex combination $$ \rho = q_1 \rho_1 + q_2 \rho_2$$ of the two, with weights adding to $q_1+q_2=1$, is also a valid density matrix.
- Because the relationship between density matrices and Bloch-ball vectors is linear, any convex combination of density matrices translates directly into a convex combination of the corresponding Bloch-ball vectors. Thus, if $ \rho_1 = \tfrac12 \mathbb I + \vec p_1 \cdot \vec \sigma,$ $ \rho_2 = \tfrac12 \mathbb I + \vec p_2 \cdot \vec \sigma,$ and $ \rho = q_1 \rho_1 + q_2 \rho_2$, then $ \vec p= q_1 \vec p_1 + q_2 \vec p_2$ lies on the line that goes from $\vec p_1$ to $\vec p_2$, a fraction $q_1=1-q_2$ of the way in that direction.

So, what does this mean for density-matrix representations? If you have a target density matrix $\rho$ that you want to represent, simply take its Bloch-ball vector $\vec p = \mathrm {Tr}(\rho\vec\sigma)$, and then pick $N$ points $\vec p_n$ on the Bloch sphere itself (the boundary) and weights $q_n$ (normalized to $\sum_n q_n=1$) such that their average $\sum_n q_n \vec p_n=\vec p$ gives you your chosen point. That will then naturally give you a representation of your density matrix as a weighted sum of $N$ pure-state projectors, and you can read off the computational-basis components directly from the spherical coordinates of your chosen extremal points.

Just consider a two-level system and take the three states $|0\rangle$, $|1\rangle$, and $|+\rangle = (|0\rangle+|1\rangle)/\sqrt{2}$. Then, the mixed state $$ \rho = \tfrac13 |0\rangle\langle0| + \tfrac13 |1\rangle\langle1| + \tfrac13 |+\rangle\langle+| $$ is an example for what you are asking for. (Of course, it has also an eigenvalue decomposition where the vectors are orthogonal.)

In case you don't want them to form an (over-complete) basis either, just consider the same example in a three-dimensional space.

A widely used example of the representation of this sort are so-called quasiprobability distributions.

Consider the Harmonic oscillator. You can use the orthonormal coordinate $|x\rangle$, momentum $|p\rangle$ or Fock $|n\rangle$ bases. However there are also nice states, \begin{equation} |\alpha\rangle=e^{\alpha a^\dagger - \alpha^\ast a}|0\rangle \end{equation} known as coherent states. Those are Gaussian wavepackets that are localized both in coordinate and momentum space with $\alpha=\langle x\rangle+i\langle p\rangle$. It's important to stress that the coherent states with different $\alpha$ are not orthogonal.

You can write any density matrix $\rho$ for the Harmonic oscillator as an integral over the phase space, \begin{equation} \rho=\int d^2\alpha\, P(\alpha,\alpha^\ast)|\alpha\rangle\langle\alpha| \end{equation} Obviously from the $\operatorname{Tr}\rho=1$ follows that, \begin{equation} \int d^2\alpha\, P(\alpha,\alpha^\ast)=1 \end{equation} and you often can treat it as a probability distribution in the phase space.

However here comes "quasi" in the "quasiprobability". The function $P(\alpha,\alpha^\ast)$ is allowed to be negative in some regions! The $\rho$ can still be positively defined.

So you can of course consider such representations but remember that some $p_n$ may actually be negative.