# Definition of density operator/ Matrix

In Sakurai's book, the density operator is defined as $$\rho\equiv \sum_i w_i|i\rangle \langle i|$$ where $$w_i$$ is statistical weight.

Now, I'm reading a book by Parisi, In which it says in Sec 5.1,

We now introduce the density matrix or density operator of the system, denoted as $$\rho$$. This operator is a generalization of the projector $$P_\psi$$. Here we limit ourselves to the discrete case. Let us first discuss the case where the system admits a wave-function description i.e. $$\rho=P_\psi$$.

I don't how they defined $$\rho\equiv|\psi\rangle \langle \psi|$$ What's the benefit of this? It's only the projector. It's also can make sense if $$|\psi$$ is an eigenstate of the system and we are considering pure ensemble so that $$w_i$$ is zero for other states. But here, they are (as it seems) of a general state which can be written in eigenbasis. $$|\psi\rangle =\sum_i c_i|\phi_i\rangle \rightarrow \rho=\sum_{i,j}c_ic_j^*|\phi_j\rangle\langle \phi_i|$$ Can you explain what's the physical interpretation of the above?

Edit: I'm asking, what's the meaning or physical interpretation of the second definition? The first make sense in term of $$\langle A\rangle =\text{Tr}(\rho A)$$ but not the second. It's simply the projection operator. How this become the density operator if system admits a wave-function description?

• Could you elaborate on what exactly your question is? I don't how they defined [...] is, IMO, not clear enough. Further: Can you explain what does this mean? - what are you referring to? Dec 31, 2021 at 13:37
• @Jakob I made few edits, Hope this is clear now. Dec 31, 2021 at 13:41
• Regarding the edit: Even with a pure state of the form $\rho = |\psi\rangle \langle \psi|$, it makes sense to define $\langle A\rangle_{\rho} = \mathrm{Tr} \rho A = \langle \psi|A|\psi\rangle$. Perhaps it would help if you search for a 'more general' definition of density operator. Both 'versions' you give are density operators (by definition). Dec 31, 2021 at 13:42
• @Jakob I'm what's so special about it. I can equivalently call $\rho$ to be the projection operator. Dec 31, 2021 at 13:43
• I have no idea what the question is - the "projector" is just the same as your initial definition with one $w_\psi = 1$ and the other $w_i = 0$. What, exactly, do you want to know? There is no claim here that the two cases are equivalent - the projector is a special case, as your quote explicitly says: "Let us first discuss the case[...]" Dec 31, 2021 at 13:44

"The system admits a wave-function description" means there is just one single state (or "wavefunction") $$\lvert \psi\rangle$$ the system is in with absolute certainty. So in that case, the set of your $$\lvert i\rangle$$ is just the single state $$\lvert \psi\rangle$$ and it occurs with statistical weight $$w_\psi = 1$$. Obviously $$\rho = w_\psi\lvert \psi\rangle\langle \psi\rvert = \lvert \psi\rangle\langle\psi\rvert = P_\psi$$ in this case.