1
$\begingroup$

Cite from "Geometry of Quantum States: An Introduction to Quantum Entanglement 1st Edition by Ingemar Bengtsson (Author), Karol Zyczkowski (Author)": "Partial transposition applied on a density matrix produces the same spectrum as the transformation of flipping one of both bloch vectors present in its Fano form. Alternatively one may change the signs of all generators $\sigma_i$ of the corresponding group."

Now the problem is, if I want to proof this assertion and thus say

$\rho^{T_B}= \frac{1}{NK} \left [ \mathbb{1}_{NK} + \sum_{i=1}^{N^2-1} \tau_i^A\sigma_i \otimes \mathbb{1}_K + \sum_{j=1}^{K^2-1} \tau_j^B \mathbb{1}_N \otimes \sigma_j^T +\sum_{i,j=1}^{N^2-1} \beta_{ij} \sigma_i\otimes\sigma_j^T \right ]$

this is not equal to flipping the sign of every generator, because:

$\sigma_x^T=\sigma_x, \sigma_z^T=\sigma_z, \sigma_y^T=-\sigma_y$

so only the sign of the y-component changes and I don´t flip every sign.

The book states that:

$\rho^{T_B}= \frac{1}{NK} \left [ \mathbb{1}_{NK} + \sum_{i=1}^{N^2-1} \tau_i^A\sigma_i \otimes \mathbb{1}_K - \sum_{j=1}^{K^2-1} \tau_j^B \mathbb{1}_N \otimes \sigma_j-\sum_{i,j=1}^{N^2-1} \beta_{ij} \sigma_i\otimes\sigma_j \right ]$

so wheres the error?

$\endgroup$
3
$\begingroup$

Partial transposition flips the sign of $\sigma_y$ on the $B$ system. If you now conjugate the resulting matrix with $I\otimes\sigma_y$, you moreover flip the sign of $\sigma_x$ and $\sigma_z$, and thus of all Pauli matrix. On the other hand, this conjugation preserves the spectrum, as claimed.

Note that it is only claimed that the corresponding matrices have the same spectrum, not that the latter is the partial transpose.

EDIT: Indeed, this is exactly what it says in the book right after the equation you cite:

In the two-qubit case, reflection of all three components of the Bloch vector, $\vec\tau^B\mapsto-\vec\tau^B$, is equivalent to changing the sign of its single component $\tau_y^B$ (partial transpose), followed by the $\pi$-rotation along the y-axis.

$\endgroup$
  • $\begingroup$ Thanks, and the spectrum is not changed when applying $\mathbb{1}\otimes\sigma_y$, because the Pauli matrix is unitary? I have one problem left though. I don´t get the same result as in the book, if I just apply $\mathbb{1}\otimes\sigma_y$. Can you provide a short calculation if possible? $\endgroup$ – EpsilonDelta Aug 7 '15 at 13:39
  • $\begingroup$ @EpsilonDelta (i) Yes, because it is unitary and we conjugate with it. (ii) What you have to do is to apply a partial transpose and then conjugate with $I\otimes \sigma_y$. This will map $M\otimes \sigma_i$ to $-M\otimes \sigma_i$ (for any $M$), which is what the book says. (Transpose adds a $-$ to $\sigma_y$, as you noted, and conjugation with $\sigma_y$ adds a $-$ to the other two Paulis.) Does this answer your question? $\endgroup$ – Norbert Schuch Aug 7 '15 at 17:55
  • $\begingroup$ Thanks, one last question: What exactly do you mean by "conjugate with"? $\endgroup$ – EpsilonDelta Aug 8 '15 at 11:37
  • $\begingroup$ @EpsilonDelta By "conjugating A with $U$" I mean mapping $A$ to $UAU^\dagger$. In this case, you would perform a partial transpose and then replace $\rho^{T_B}$ by $(I\otimes\sigma_y) \rho^{T_B} (I\otimes\sigma_y)^\dagger$ (which does not change the spectrum). $\endgroup$ – Norbert Schuch Aug 8 '15 at 12:23
  • 1
    $\begingroup$ Many thanks for your careful explanations Everything looks fine! Yours Karol Życzkowski $\endgroup$ – user111702 Mar 18 '16 at 11:37

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.