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It crossed my mind when reviewing density matrices that if you were looking at a composite system consisting of three subsystems, (indexed by three quantum numbers: $<i,j,k|\rho|i,j,k>$) then the density matrix would be three dimensional. The populations would then live along the diagonal of a cube. Does this ever show up in quantum mechanics? Can a three dimensional density matrix always be recast as two dimensional by grouping two subsystems as one?

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  • $\begingroup$ To clarify, I meant 3 dimensional as in $n\times n\times n$, not a $3\times 3$ matrix. $\endgroup$
    – aRockStr
    Oct 4, 2019 at 4:26

2 Answers 2

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The density matrix has the dimension of the Hilbert space. Thus, for states with $\ell=1$ the density matrix would be $3\times 3$, irrespective of the number of constituents.

The easiest way to see this is that elements in $\rho$ are of the from $\vert i\rangle\langle j\vert$ so clearly if $i,j$ run over a basis set of size $n$ then $\rho$ is an $n\times n$ matrix.

It may be possible to organize the constituents so they are invariant under some transformations but the density matrix of $3$ spin-1/2 particles is still $8\times 8$.

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Consider three spin-1/2 systems: $|\theta\rangle,|\phi\rangle,|\chi\rangle$, each a linear superposition of the states $\{|\uparrow\rangle,|\downarrow\rangle\}$. We can represent each system as a vector:

$|\theta\rangle=\begin{pmatrix}\theta_{1} \\\theta_{2} \\ \end{pmatrix},|\phi\rangle=\begin{pmatrix}\phi_{1} \\\phi_{2} \\ \end{pmatrix},|\chi\rangle=\begin{pmatrix}\chi_{1} \\\chi_{2} \\ \end{pmatrix}$.

A general state $|\psi\rangle$ of two of the subsystems, say $|\theta\rangle$ and $|\phi\rangle$, is the tensor product $|\theta\rangle$ and $|\phi\rangle$ which can also be represented by a vector:

$|\psi\rangle=|\theta\rangle\otimes|\phi\rangle=|\theta\phi\rangle=\begin{pmatrix}\theta_1\phi_1 \\\theta_1\phi_2 \\ \theta_2\phi_1 \\\theta_2\phi_2\end{pmatrix}\equiv\begin{pmatrix}\psi_{11}\\\psi_{12}\\\psi_{21}\\\psi_{22}\\\end{pmatrix},$

or equivalently

$|\psi\rangle=\sum_{ab}\psi_{ab}|ab\rangle=\psi_{11}|11\rangle+\psi_{12}|12\rangle+\psi_{21}|21\rangle+\psi_{22}|22\rangle.$

The density matrix is then defined to be

$\rho=|\psi\rangle\langle\psi|=\sum_{abcd}\psi_{ab}\psi_{cd}^*|ab\rangle\langle cd|,$

or

$\rho = \begin{pmatrix}\psi_{11}\psi_{11}^* & \psi_{11}\psi_{12}^* &\psi_{11}\psi_{21}^* & \psi_{11}\psi_{22}^* \\ \psi_{12}\psi_{11}^* & \psi_{12}\psi_{12}^* &\psi_{12}\psi_{21}^* & \psi_{12}\psi_{22}^* \\ \psi_{21}\psi_{11}^* & \psi_{21}\psi_{12}^* &\psi_{21}\psi_{21}^* & \psi_{21}\psi_{22}^* \\ \psi_{22}\psi_{11}^* & \psi_{22}\psi_{12}^* &\psi_{22}\psi_{21}^* & \psi_{22}\psi_{22}^*\end{pmatrix}.$

A general state of the three subsystems is tensor product of the three subsystems

$|\psi\rangle=|\theta\rangle\otimes|\phi\rangle\otimes|\chi\rangle=|\theta\rangle\otimes(|\phi\rangle\otimes|\chi\rangle)=|\theta\rangle\otimes(|\phi\chi\rangle)=|\theta\phi\chi\rangle,$

or

$|\psi\rangle=|\theta\phi\chi\rangle=\begin{pmatrix}\theta_1\phi_1\chi_1 \\ \theta_1\phi_1\chi_2 \\\theta_1\phi_2\chi_1 \\ \theta_1\phi_2\chi_2 \\ \theta_2\phi_1\chi_1 \\\theta_2\phi_1\chi_2 \\\theta_2\phi_2\chi_1\\\theta_2\phi_2\chi_2 \end{pmatrix}\equiv\begin{pmatrix}\psi_{111}\\\psi_{112}\\\psi_{121}\\\psi_{122}\\\psi_{211}\\\psi_{212}\\\psi_{221}\\\psi_{222}\\\end{pmatrix}.$

The density matrix is

$\rho=|\psi\rangle\langle\psi|=\sum_{abcdef}\psi_{abc}\psi_{def}^*|abc\rangle\langle def|.$

In comparison to the 2-subsystem case, we see that we will still have a two-dimensional matrix, however the matrix is $8\times 8$.

I guess my confusion was that I thought the number of indices corresponded to the number of dimensions of the density matrix. However, the density matrix is simply the outer product of two vectors, which will always be two-dimensional. It seems that the size of the density matrix will be $2^n\times2^m,$ where $n,m$ are the dimensions of the Hilbert spaces of the subsystems.

Source: American Journal of Physics 77, 244 (2009); doi: 10.1119/1.3043847

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