In quantum mechanics, a pure state of any quantum system is determined by a state $|\psi\rangle$ living in a Hilbert space $\mathcal{H}$. Any Hilbert space has a basis of states and any state can be written as a combination of basis states. The nice thing about Hilbert spaces -among other things- is that there's a notion of inner product which allows us to calculate the distance between states.

However, when we allow statistical mixtures of pure states, a.k.a. mixed states, we stop using states in a Hilbert space in favour of a matrix density. My question is: Is there a space akin to the Hilbert space where mixed states live? Can we calculate the distance between two mixed states like we do with the inner product in a Hilbert space?

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    $\begingroup$ since $\forall\left|\psi\right>\in\mathcal H\exists\rho=\left|\psi\right>\left<\psi\right|\in\mathcal H\otimes\mathcal H^\dagger$, this means that the space for density operators is already fixed once we assert where the basic quantum states live. $\endgroup$ Jul 24, 2023 at 16:31
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    $\begingroup$ I guess that only works for pure states? I agree that the space of pure density matrices should be the same as of states in a Hilbert space, but density matrices can also be mixed, which surely is bigger than a Hilbert space. $\endgroup$ Jul 24, 2023 at 16:38
  • $\begingroup$ Surely one can define a space of density matrices, which also has well-known properties like convexity etc. But you should specify what exactly you mean. Obviously the space of density matrices is not a Hilbert space. $\endgroup$ Jul 24, 2023 at 16:45
  • $\begingroup$ There are ways to define some sort of "distance" between density matrices. As others have suggested, one can view $\rho$ as a vector in $\mathcal{H}\otimes\mathcal{H}^\dagger$, and define the inner product as usual. What you get is the Hilbert-Schmdit product. You can also simply use some form of operator norm of the difference between two density matrices as a measure of the distance. It should be kept in mind that unlike pure states, the "space" of density matrix does not form a vector space. $\endgroup$
    – Meng Cheng
    Jul 24, 2023 at 17:18

3 Answers 3


Let us assume the Hilbert space ${\cal H}$ is generally infinite dimensional. The density matrices (operators) are operators $\rho: {\cal H} \to {\cal H}$ such that

  • they are positive: $\langle \psi|\rho \psi\rangle \geq 0$ for every $\psi\in {\cal H}$ (it easily implies that $\rho$ is selfadjoint and bounded);

  • they are trace class: $\operatorname{tr} (\rho) := \sum_{u\in N} \langle u| \rho u\rangle$ converges in $\mathbb{C}$ and the result does not depend on the chosen Hilbert basis $N\subset {\cal H}$;

  • they are normalized $\operatorname{tr}(\rho)=1$.

Every such $\rho$ can be decomposed as $\rho = \sqrt{\rho}\sqrt{\rho}$ and also, where $U: {\cal H} \to {\cal H}$ is any everywhere defined bounded operator such that $U^\dagger U|_{\operatorname{Ran}(\rho)}= I|_{\operatorname{Ran}(\rho)}$, we can also write $\rho = (U\sqrt{\rho})^\dagger U\sqrt{\rho}$. $U$ is any partial isometry with initial space $\operatorname{Ker}(\rho)^\perp = \operatorname{Ker}(\sqrt{\rho})^\perp$.

The operators $\sqrt{\rho}$ are of Hilbert-Schmidt type and, since the set of Hilbert Schmidt operators is a both side $*$-ideal of $\mathfrak{B}({\cal H})$ (the $\mathrm{C}^*$-algebra of all everywhere defined bounded operators on ${\cal H}$), also $U\sqrt{\rho}$ is Hilbert-Schmidt.

Since the Hilbert-Schmidt operators $T:{\cal H} \to {\cal H}$ satisfy by definition $T^\dagger T$ is trace class, we can conclude (also using the polar decomposition theorem) that

(1) Every Hilbert Schmidt operator $T: {\cal H}\to {\cal H}$ with $\operatorname{tr}(T^\dagger T)\neq 0$ defines a density matrix $\rho_T := \frac{T^\dagger T}{\operatorname{tr}(T^\dagger T)}$

(2) The map $T \to \rho_T$ is such that $\rho_T = \rho_{T'}$ if and only if $UT=U'T'$ for some partial isometries $U$ and $U'$ with initial space $\operatorname{Ker}(\rho)^\perp$

In this sense

mixed states are unit-trace Hilbert-Schmidt operators up to partial isometries.

that is analogous to the fact that

pure states are unit vectors up to phases.

That is not the whole story, since the analogy encompasses the structure of Hilbert space.

In fact, it turns out that the space of Hilbert-Schmidt operators is a Hilbert space in its own right whose inner product is $$\langle A|B\rangle_{HS}:= \operatorname{tr}(A^\dagger B)\:.$$

This fact permits to extend formulas of the pure-state formalism to mixed-state formalism.

If $P_E$ is an orthogonal projector of a spectral measure of some observable $Z$ and a pure state represented by the unit vector $\psi$ is given, $$\langle \psi|P_E \psi\rangle$$ is the probability to measure $E\subset \sigma(Z)$ when the pure state is represented (up to phases) by $\psi$.

The same happens for mixed states. Suppose that $T$ is Hilbert Schmidt with $||T||_{HS}=1$. It defines the density matrix $\rho_T :=T^\dagger T$ and $\rho_T$ defines $T$ up to partial isometries (instead of phases). The probability to measure $E\subset \sigma(Z)$ is now $$\operatorname{tr}(P_E \rho_T) = \operatorname{tr}(P_E T^\dagger T) = \operatorname{tr}((TP_E)^\dagger T) = \langle TP_E|T \rangle_{HS}= \langle T|P_E[T] \rangle_{HS}$$

where I have defined the linear action of the operator $A \in \mathfrak{B}({\cal H})$ on the Hilbert Schmidt operator $T$ as $A[T]:= TA$

As a consequence, for instance the expectation value of $Z$ in the state $\rho_T$ is $$\langle Z\rangle_{\rho_T} = \langle T| Z[T]\rangle_{HS}$$ to be compared with $$\langle Z\rangle_{|\psi\rangle\langle\psi|} = \langle \psi| Z \psi\rangle$$

The analogy does not work at the level of distance between states, because partial isometries have a richer (non-commutative) structure than simple phases.


This question has a very deep answer and cannot be fully covered by a small answer. Therefore, I will leave some references in the end to read more about the structure of the space of density matrices. I will be basing my answer mostly on [1].

Take a Hilbert Space $\mathcal{H}$ of dimension $N$. One can define the space of $\textbf{pure}$ density matrices as $ \{\rho \in End(\mathcal{H}) | \rho^\dagger = \rho, \rho^2 = \rho, Tr(\rho) = 1 \}$. Here the "$End(\mathcal{H})$" means the linear maps that sends vectors on $\mathcal{H}$ to vectors on $\mathcal{H}$. Now one can go from one physical state to the other via a unitary transformation as we want the overall wavefunction to be normalised and the transformations that keep the norm same are unitary. Therefore, on face-value, the space of pure density marices seem to be $U(N)$. However, we are not done yet. To see why, note that without loss of generality, we can always find a basis where the density marix takes the form:

\begin{bmatrix} 1 & 0 & \dots & 0 \\ 0 & 0 & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & 0 \end{bmatrix}

It is clear that if we act on this density matrix with a $U(N)$ marix of the form

\begin{bmatrix} 1 & 0 \\ 0 & A \\ \end{bmatrix}

where $A \in U(N-1)$, the matrix itself won't change. Therefore, the two states related by these transformations are physically the same since they have the same density matrix. To get rid of this degeneracy, we need to mod out $U(N-1)$ from our space. Finally, there is one more degeneracy we have to account for. Note that, any physical state multiplied by a phase is also the same state, i.e $ \vert \psi \rangle \sim e^{i \theta}\vert \psi \rangle $. This phase also adds a degeneracy and therefore we need to remove it as well. Therefore, our final space is

$$ \frac{U(N)}{U(N-1) \times U(1)} = \frac{SU(N)}{U(N-1)}$$

This actually turns out to be the $\textbf{complex projective space}$ $\mathbb{C}P^{N-1} $. One can therefore use the geometry of complex projective spaces to describe the geometry of the space of pure density matrices (e.g Fubini Study Metric). You can read more about it in [2].

The space of mixed state matrices are more general as we remove $\rho^2 = \rho$ condition. Because of this reason, the parametrisation of the physical transformations depend on the eigenvalues of the mixed state density operator and therefore, there is no single space that characterises the space of mixed density operators. However, note that, we can use linear combinations of pure state density operators to write down a mixed density operator. You can read more about this in [1].


[1] Geometry of n-state systems, pure and mixed: M S Byrd et al 2007 J. Phys.: Conf. Ser. 87 012006

[2] Bengtsson, I., & Zyczkowski, K. (2006). Geometry of Quantum States: An Introduction to Quantum Entanglement. Cambridge: Cambridge University Press. doi:10.1017/CBO9780511535048


Density matrices $\rho$ are convex combinations of projections $$ |\psi\rangle\langle\psi|\in \mathcal H\otimes\mathcal H^\dagger\,. $$ To simplify the discussion I assume a finite dimensional $\mathcal H\,.$ A density matrix $$ \rho=\sum_{i=1}^n p_i |\psi_i\rangle\langle\psi_i| $$ represents a pure state if and only if just one of the probabilities $p_i$ is non-zero (and therefore necessarily one). This is equivalent to the Neumann entropy being zero.

  • The Hilbert space $\mathcal H\otimes\mathcal H^\dagger$ is more than large enough to accomodate for all density matrices, in particular for mixed states of a single particle.

  • This is not a contradiction to the known fact that the two-particle pure states live in the similar looking Hilbert space $\mathcal H\otimes\mathcal H\,.$ This space is also way too big since we only need normalized states.

  • When ${\cal H}$ is two-dimensional (we are only interested in spin up and spin down states) then -as we know- the normalized pure states are the surface of the Bloch sphere and the mixed states its interior.


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