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?

  • $\begingroup$ 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? $\endgroup$ Dec 31, 2021 at 13:37
  • $\begingroup$ @Jakob I made few edits, Hope this is clear now. $\endgroup$ Dec 31, 2021 at 13:41
  • $\begingroup$ 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). $\endgroup$ Dec 31, 2021 at 13:42
  • $\begingroup$ @Jakob I'm what's so special about it. I can equivalently call $\rho$ to be the projection operator. $\endgroup$ Dec 31, 2021 at 13:43
  • $\begingroup$ 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[...]" $\endgroup$
    – ACuriousMind
    Dec 31, 2021 at 13:44

1 Answer 1


"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.

  • $\begingroup$ Why is not measurement in 2 steps, lets take the singlet state : initial state : pure singlet state, intermediate state : mixed state with statistical weights given by qm born rule. End state : pure state chosen among the possible states ? $\endgroup$
    – Cretin2
    Jan 17 at 7:12
  • $\begingroup$ In fact they give the same average and probabilities, only powers of the density matrices were different $\endgroup$
    – Cretin2
    Jan 17 at 10:00

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