I am a physics undergraduate. I just read through Sakurai's section 3.4 (3rd ed) on Density Operators and Pure vs. Mixed states.

Are pure states as kets generalized to mixed states as operators in order to retain all the properties we know and love of states as kets while affording the ability to introduce relative population weights of such states that do not interfere with such ket properties (like interference)?

For example, if we tried to stick with the ket state formalism, then we wouldn't be able to distinguish relative population weights from complex amplitudes and thus between a more "classical" probability from the "quantum" probability?

Additionally, would we write a general density operator as

$$\rho = \sum_j w_j\left(\sum_{\textbf{i}_n} c_{\textbf{i}_n}|\textbf{i}_n\rangle \langle \textbf{i}_n |\right)$$

where $| \textbf{i}_n \rangle = |i_1\rangle \otimes |i_2 \rangle \otimes ...|i_n\rangle$ and $\sum_i w_i = 1$?

In other words, would we write an arbitrary general density matrix as a convex combination of arbitrary pure state density operators?

  • $\begingroup$ The answer to your first question is Yes. May I ask what is a convex combination of arbitrary pure state density operators? $\endgroup$
    – user35952
    Commented Dec 11, 2022 at 6:32
  • $\begingroup$ What exactly is your question? (In particular, it seems that you have more than one question - you should focus on one question per question.) $\endgroup$ Commented Dec 11, 2022 at 11:23
  • 1
    $\begingroup$ I would say the reason why we "generalize" to mixed states is that we must do so: If you consider a pure quantum state on a system with two parts A and B, where B might be huge - the experimental apparatus, lab, universe, or whatever else - then the correct description for your experiment on the system A alone is precisely a "mixed state", or density matrix. The other option is to carry around the whole pure description of A+B, which has many more parameters and a lot of entirely useless information. $\endgroup$ Commented Dec 11, 2022 at 11:36

2 Answers 2


Yes, for everything. Mixed states introduce classical probability aspects to the description, while the kets only deal with "quantum probability".

Yes, it is a convex combination. In fact, notice all states $\rho$ can be written in the form $$\rho = \lambda \rho_1 + (1 - \lambda) \rho_2,$$ where $\rho_i$ are some other states and $\lambda \in [0,1]$. Other authors define a pure state to be a state where one always has either $\lambda = 0$ or $\lambda = 1$, i.e., it only admits trivial convex combinations. Mixed states are then any state that isn't pure.

A pedagogical discussion of these themes in some more mathematical settings can be found in arXiv: 1909.06232 [math-ph]. I believe it can be quite useful to understand the concept of pure and mixed states and, as far as I remember, it already takes this point of view of convex combinations from the start. It is also a nice introduction to the algebraic approach to quantum mechanics.


Density operators are an intuitive way to describe a lack of information within the reference system (RS).

The lack of information can be seen as an entanglement of the RS with an extra one, e.g. the environment. As the RS information is entangled with the environment, these two are not separable.

Once understood that, the next step comes from an interest in characterizing the lack of information. You need mixed states to do that properly.


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