The link here gives a nice description of how partial trace looks in matrix notation. I want a similar explanation for the matrix partial-transposition. How does matrix partial-transposition operation look in the matrix form rather than in Dirac notation?

  • $\begingroup$ Transposition or partial transposition? $\endgroup$ – Norbert Schuch Dec 6 '18 at 17:25
  • $\begingroup$ I meant partial transposition. I just corrected it. $\endgroup$ – Zilch Dec 6 '18 at 17:30
  • $\begingroup$ Which of the linked answers you like? The first can be translated 1-to-1 to the transpose. $\endgroup$ – Norbert Schuch Dec 6 '18 at 19:46
  • $\begingroup$ I like the second one. Where one can write the action of partial tracing as summing to the basis of one party and leaving the other alone (the identity matrix). $\endgroup$ – Zilch Dec 6 '18 at 19:55

An operator $M$ acting on a vector space $V_A \otimes V_B$ can be decomposed as: $M = \sum_{ij} c_{ij} A^{(i)} \otimes B^{(i)}$ with $A^{(i)}$ acting on $V_A$ and $B^{(j)}$ acting on $V_B$. Using matrices to denote $M$, $A^{(i)}$ and $B^{(i)}$: $$ M = \sum_{ij} c_{ij} \left( \begin{array}{ccc} a^{(i)}_{11} B^{(i)} & a^{(i)}_{12} B^{(i)} & \cdots \\ a^{(i)}_{21} B^{(i)} & a^{(i)}_{22} B^{(i)} & \cdots \\ \vdots & \vdots & \ddots \\ \end{array} \right) $$

Then, the partial transpositions are: $$ M^{T_A} = \sum_{ij} c_{ij} \left( \begin{array}{ccc} a^{(i)}_{11} B^{(i)} & a^{(i)}_{21} B^{(i)} & \cdots \\ a^{(i)}_{12} B^{(i)} & a^{(i)}_{22} B^{(i)} & \cdots \\ \vdots & \vdots & \ddots \\ \end{array} \right) \\ M^{T_B} = \sum_{ij} c_{ij} \left( \begin{array}{ccc} a^{(i)}_{11} (B^{(i)})^T & a^{(i)}_{12} (B^{(i)})^T & \cdots \\ a^{(i)}_{21} (B^{(i)})^T & a^{(i)}_{22} (B^{(i)})^T & \cdots \\ \vdots & \vdots & \ddots \\ \end{array} \right) $$


I'm not sure what you mean by "matrix notation". Using brakets, if the matrix $M$ reads $$M=\sum_{ijkl}M_{ik;jl} \lvert i,j\rangle\!\langle k,l\rvert\equiv \sum_{ijkl}M_{ik;jl} \lvert i\rangle\!\langle k\rvert\otimes\lvert j\rangle\!\langle l\rvert,$$ then its partial transpose with respect to the second space is $$M^{T_B}= \sum_{ijkl}M_{ik;jl} \lvert i\rangle\!\langle k\rvert\otimes\lvert l\rangle\!\langle j\rvert.$$

Equivalently, the partial transpose $M^{T_B}$ is that matrix with components $$\big(M^{T_B}\big)_{ij;kl}=M_{ij;lk},$$ where again the first (second) pair of indices refers to the first (second) space.


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