# Motivation behind defining mixed states

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?

• The answer to your first question is Yes. May I ask what is a convex combination of arbitrary pure state density operators? Commented Dec 11, 2022 at 6:32
• What exactly is your question? (In particular, it seems that you have more than one question - you should focus on one question per question.) Commented Dec 11, 2022 at 11:23
• 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. Commented Dec 11, 2022 at 11:36

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.