Space of density matrix

Question

If we say that the $|\psi\rangle$ is in the Hilbert space what does the space of density matrix?

Is this phrase true?

$|\psi\rangle \langle\psi| \in V \otimes V_\mathrm{dual}$

$V$ is vector space.

Answer 1

Is this phrase true?

$|\psi\rangle \langle\psi| \in V \otimes V_\mathrm{dual}$

Yes, this is a correct way to see things, but the more usual view is to note that $$ V \otimes V_\mathrm{dual} \cong \mathrm{End}(V), \tag{$*$} $$ i.e., the tensor space $V \otimes V_\mathrm{dual}$ is canonically isomorphic to the vector space of endomorphisms in $V$, i.e. to the space of linear operators $\rho: V\to V$.

It's important to note that the isomorphism $(*)$ is only strictly valid in finite dimension, and that in infinite dimensionality you need to be careful with what you allow and what you don't. Thus, in infinite dimensionality, density matrices are normally required to be trace-class, positive semi-definite, self-adjoint linear operators over the system's Hilbert space. But, ultimately, the isomorphism $(*)$ is still morally true, though.

Answer 2

The density matrix is a representation in a particular basis of a linear operator on the Hilbert space called the density operator. This operator lives in the space of all linear operators on the Hilbert space.

NOTE: The OP asked an additional question after I answered.

Answer 3

So here's a bit more on this to complement @EmilioPisanty 's answer.

If $\vert\psi\rangle$ is in some irreducible representation $\lambda$, then $\vert\psi\rangle\langle \psi\vert$ lives in $\lambda\otimes \lambda^*$ where $\lambda^*$ is the representation conjugate to $\lambda$.

Thus for instance, if $\vert\psi\rangle$ is a state of angular momentum $J$ (in a finite dimensional irrep of dimension $(2J+1)$, $\vert\psi\rangle\langle \psi\vert$ lives in the tensor product space $J\otimes J$ (since $J^*$ is the same as $J$) and this tensor product decomposes as a direct sum of angular momentum subspaces: \begin{align} J\otimes J = \displaystyle{\oplus_{L=0}^{2J}} L \, . \end{align} As a consequence, an operator of the form $\vert JM\rangle \langle J M'|$ with $[\hat J_0,\vert JM\rangle \langle J M'|=(M-M')\vert JM\rangle \langle J M'|$ can be expressed as a sum of tensor operators $T^L_{M-M'}$ with $L$ in the range above (and obviously $L\ge |M-M'|$ as a secondary condition.)

In general $\lambda^*$ is not equivalent to $\lambda$ so that, for instance in the case of $SU(3)$, for $\vert\psi\rangle$ a state in the irrep $(p,q)$, then \begin{align} \vert\psi\rangle\langle \psi\vert =(p,q)\otimes (q,p) \end{align} is a direct sum of $SU(3)$ irrep, and this direct sum will contain multiple copies of some irreps unless $p=0$ or $q=0$. This is the more general case.

If $\vert\psi\rangle$ belongs to an infinite-dimensional representation - say to the $SU(1,1)$ irrep $k=1/4$ (in one limit of a positive discrete series) which contains the even h.o. states $\vert 0\rangle, \vert 2\rangle, \ldots \vert 2r\rangle,\ldots$, then $\langle\psi\vert$ is in the corresponding negative series, and (for instance) $\vert 2r\rangle\langle 2r\vert$ lives in \begin{align} \textstyle\frac{1}{4}\otimes\left(\frac{1}{4}\right)^* \end{align} This tensor product does not decompose in a direct sum of positive discrete series but as an integral over representations in the continuous series.

The easy way to see this is that the diagonal $\mathfrak{su}(1,1)$ operator $\hat K_0=\frac{1}{2}\hat H$ has eigenvalue $k+m>0$ on any state $\vert k,m\rangle$ or operator $\hat T^k_m$ transforming in a positive discrete series, but \begin{align} [\hat K_0,\vert 2r\rangle\langle 2r\vert]=0 \end{align} showing that $\vert 2r\rangle\langle 2r\vert$ cannot have any piece in a discrete (positive or negative) series. Indeed only continuous series accommodate eigenvalues $0$.