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So, another confusion that people get when learning about the so-called “Density matrix” or “Density operator” is that: it is just a representation and we can simply use kets. (if you are reading this and are not familiar with the Dirac notation yet, I suggest you check any intro to bra-ket / Dirac notation, the Wikipedia page, or any short introduction video on quantum computing).

## Definition of a Density matrix

A density matrix is a square matrix that allows us to give a complete description of the quantum state of a physical system. It satisfies the following properties:

• $$\rho \geq 0 \text{ (positive semi-definite)}$$
• $$tr(\rho) = 1$$

It could be used to define the state of a system $$\rho$$ resulting from mixing a system $$\ket{\psi_A}$$ and a system $$\ket{\psi_B}$$. We express it as the following mixture:
$$\rho = \sum_x \mathcal{P}_x \ket{\psi_x}\bra{\psi_x}$$ where $$\ket{\psi_x}$$ is prepared with a probability $$\mathcal{P}_x$$\*

• We call the state a pure state when the matrix $$\rho$$ is of rank 1 (alternatively when $$tr(\rho^2)=1$$). \
• We call it a mixed state when $$tr(\rho^2)<1$$, and it’s made of a mixture of pure states. For a given dimension $$d$$ of the system, the maximally mixed state is obtained when $$tr (\rho^2)= \frac{1}{d}$$

Example
If we have a pure state $$\ket{\psi}$$ prepared with probability 1, we define its density matrix as follows:
$$\rho = \ket{\psi}\bra{\psi}$$.
Now let $$\ket{\psi} = \ket{0}$$, we get the following density matrix:

$$\rho^2 = (\ket{0}\bra{0})^2 =( \begin{bmatrix} 1 \\ 0 \ \end{bmatrix} \times \begin{bmatrix} 1 & 0 \end{bmatrix})^2 = \begin{bmatrix} 1 & 0 \\ 0 & 0 \ \end{bmatrix}$$
$$tr(\rho^2)$$ = 1; This is a pure state.\

## So why do we need Density matrices?

The reason behind this confusion about their usefulness is simply because of the way we commonly introduce this notation. The textbook introduction found in many courses and at many universities goes as explained above by starting with a simple example from a pure state like the one above or with a superposition (a superposition is always a pure state) and then converting it to density matrix. So this gives the impression that it is just another fancy notation, and that we can always start with a ket, and compute the ket-bra. But the fact is that ket notation can be used to build density operators only with states that are represented as kets, to begin with, but there are cases where it does not work the other way around as the initial state can only be a matrix. In fact, if you are given a mixed state, you can’t fully describe the state with only kets.

For instance, let’s take the following example:

Let $$\rho_2 = \begin{bmatrix} 1/2 & 0 \\ 0 & 1/2 \ \end{bmatrix} = 1/2 ( \ket{0}\bra{0} + \ket{1}\bra{1} )$$

Notice here that the 2 conditions from above are met on the trace and positivity, but can we represent this matrix only with kets? The answer is no. It is not possible to find some $$\ket{\psi}$$ such that $$\ket{\psi} \bra{\psi}$$ = the matrix we have

Another important observation is the absence of any off-diagonal elements. In fact, mixed states never have any elements off-diagonal.

There are actually deeper meanings behind this, things that have to do with the “purity” of the state and superpositions. But we are not getting into these details, you may check more adapted learning material or maybe we will get back to this later!

## A little note for the aficionado

There is something known as “quantum state purification” that loosely implies that you can write a pure state out of a density operator. But it is not a 1-1 equivalence. It is impossible to fully transfer the mixedness of a system to a pure state without a loss of part of the system’s description as shown in this article

## So to sum up

• States of the form $$\ket{\psi}$$ are pure states
• Mixed states are initially described as density operators $$\rho$$ and can’t be described with kets only (not without a loss), in physics jargon, a mixed state cannot be described by a wave function, it requires a density matrix for its description.
• Pure states can also be represented with density operators $$\rho = \ket{\psi} \bra{\psi}$$