5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Is this helpful? $\endgroup$ – Charlie Jul 12 at 15:41
  • $\begingroup$ @Charlie I read this before. We know that ket is in vector space and bra in dual space . What is the space of outer product of them that is density matrix? $\endgroup$ – HohO Jul 12 at 15:44
  • 1
    $\begingroup$ it is an element in the space of linear operators from $\mathcal{H}$ to $\mathcal{H}$ (and is psd, has unit trace, etc) $\endgroup$ – 4xion Jul 12 at 16:30
10
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ and |psi><psi| are called the 'pure states' which are the extremal points of the convex set of all density matrices (of the underlying Hlbert space). (so every density matrix is a convex combination of pure states) $\endgroup$ – lalala Jul 13 at 9:32
3
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Is it right that I say $\rho\in V \otimes V_dual$? $\endgroup$ – HohO Jul 12 at 16:29
  • $\begingroup$ Yes, I think you can say that. $\endgroup$ – G. Smith Jul 12 at 16:31
  • $\begingroup$ I corrected my question and you said that is true without any condition? $\endgroup$ – HohO Jul 12 at 16:40
  • 2
    $\begingroup$ It’s not polite to change your question after it has been answered. But at least you have gotten a new answer from Emilio. $\endgroup$ – G. Smith Jul 12 at 16:45
  • $\begingroup$ You’re right . Excuse me. $\endgroup$ – HohO Jul 12 at 16:46
1
$\begingroup$

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 $0$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.