The Density Matrix of a Pure State

Question

It is my understanding that any pure quantum state $|\psi\rangle$ can be represented by the density matrix $|\psi\rangle\langle\psi|$.

It is also my understanding that $|\psi\rangle\langle\psi|$ represents a transformation that projects any vector onto the ray of $|\psi\rangle$, and thus $|\psi\rangle\langle\psi|$ can be identified with the ray of $|\psi\rangle$ itself.

First Question: Is it correct that $|\psi\rangle\langle\psi|$ can be identified with the ray of $|\psi\rangle$?

Second Question: If the answer to the first question is positive, then how is it that $|\psi\rangle\langle\psi|$ can represent the pure state $|\psi\rangle$? I mean, the ray of a vector does not contain information about the magnitude (length) of the vector. Is it simply that we assume that quantum states are normalized and the magnitude of $|\psi\rangle$ is 1 in any case?

Answer 1

Pure states are defined as rays of vectors, although different books choose to give different sorts of emphasis on this. Notice that if $\vert\psi\rangle = \alpha \vert\phi\rangle$ for any constant $\alpha \in \mathbb{C}$, then all of the expectation values, probabilities, and etc associated with $\vert\psi\rangle$ are exactly the same as for $\vert\phi\rangle$. For example, the probability for the transition rate between states $\vert\psi\rangle$ and $\vert\xi\rangle$ (I'll assume the latter to be normalized for convenience) will be \begin{align} \frac{|\langle\xi|\psi\rangle|^2}{\langle\psi|\psi\rangle} &= \frac{|\alpha|^2|\langle\xi|\phi\rangle|^2}{|\alpha|^2 \langle\phi|\phi\rangle}, \\ &= \frac{|\langle\xi|\phi\rangle|^2}{\langle\phi|\phi\rangle}. \end{align}

An expectation value of some observable $\hat{A}$ also satisfies a similar relation, since

\begin{align} \frac{\langle\psi|\hat{A}|\psi\rangle}{\langle\psi|\psi\rangle} &= \frac{|\alpha|^2|\langle\phi|\hat{A}|\phi\rangle}{|\alpha|^2 \langle\phi|\phi\rangle}, \\ &= \frac{\langle\phi|\hat{A}|\phi\rangle}{\langle\phi|\phi\rangle}. \end{align}

In short, yes, $|\psi\rangle\langle\psi|$ can be identified with the ray of $|\psi\rangle$. It can represent the pure state because all pure states are defined up to multiplication by a constant complex number.

Notice the necessity of the constancy: phases that depend on space or time do have dynamical effects since they are part of the wavefunction, and can lead, for example, to the Aharonov–Bohm effect. Furthermore, relative constants between states can also be relevant ($\vert+\rangle = \frac{ 1}{\sqrt{2}}\left(\vert 0 \rangle + \vert 1 \rangle\right)$ and $\vert-\rangle = \frac{ 1}{\sqrt{2}}\left(\vert 0 \rangle - \vert 1 \rangle\right)$ are not the same state, even though the difference appears to be just the phase in front of $\vert 1 \rangle$). Overall phases are irrelevant, but relative phases are relevant.