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I'm trying to understand a deduction but I got stuck at a certain line, which doesn't get into my mind by several reasons.

We are concerning three systems A,B and C. Furthermore there is given a Kraus-operator $ K := \sum_{i=0}^{1} |i>_C \otimes (|0>_A<0|+|1>_A<1|) \otimes (\sqrt{I_i})_B$

where $I_i := \int_{a_i}^{b_i}( dq ~|q><q|)$ are defined in terms of projections on improper eigenstates of the $q$ quadrature.

First of all I try to understand the tensor-product. I think I understand how to build the tensor-product of two matrices, but I don't know how I should interpret the tensor-product between a vector and a matrix. If $|0>_C =\left(\begin{array}{c} 1 \\ 0 \end{array}\right)$ and $|1>_C =\left(\begin{array}{c} 0 \\ 1 \end{array}\right)$, what is $|0>_C \otimes (|0>_A<0|))$ ? My guess would be doing the same as with usual tensor-product $ \left(\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array}\right) .$

But this doesn't make sense, when I proceed in the deduction.

My second queston is how to understand the definition of $I_i$. As far as I know the improper eigenstate of $q$ quadrature is $q \cdot \delta(q-q')$ but I don't see how this is linked with $I_i$.

Thanks in advance for everyone trying to help me understanding this lines!

PS: With quadrature I mean - following the terminology in quantum optics - the real and imaginary part of a point in phase space and the corresponding operators $\hat{q}$ and $\hat{p}$

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Kraus operators often map between Hilbert spaces of different dimension so they are non square matrices. The form $$\vert0\rangle(\vert0\rangle\langle0\vert+\vert1\rangle\langle1\vert)$$ is just a Dirac notation for non square operator that maps $2$ dimensional system into $2^2$ dimensional one.

The rules of tensor products of vectors with matrices are the same and it should read $$ \vert0\rangle\vert0\rangle\langle0\vert= \begin{pmatrix} 1 \\ 0 \end{pmatrix}\otimes \begin{pmatrix} 1 & 0\\ 0 &0 \end{pmatrix}= \begin{pmatrix} 1\begin{pmatrix} 1 & 0\\ 0 &0 \end{pmatrix} \\ 0\begin{pmatrix} 1 & 0\\ 0 &0 \end{pmatrix} \end{pmatrix}= \begin{pmatrix} 1 & 0\\ 0 & 0\\ 0 & 0\\ 0 & 0 \end{pmatrix} $$

Your second question is not clear to me as I'm not sure what you mean by quadrature.

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  • $\begingroup$ Hi and thanks for your answer! Quadrature in this case means the operators $\hat{q} = \frac{1}{\sqrt{2}}( \hat{a}^{\dagger}+\hat{a}) $ and $\hat{p} = \frac{i}{\sqrt{2}}( \hat{a}^{\dagger}-\hat{a}) $ so they stand for the real and complex part of a point in phase-space. I edited my first post and provided a link for further information refering to this terminology $\endgroup$ – pcalc Aug 19 '19 at 6:02
  • $\begingroup$ @pcalc in the definition $I_i$ seems to depend on the mystery integration variables $a_i$ $b_i$. Not sure what they mean but $I_i$ is a projection on the interval in the phase space identified by the integration bounds. $\endgroup$ – oleg Aug 19 '19 at 15:46
  • $\begingroup$ Thanks again for your respond! Yes I also think that they are intended as kind of separating phase space in two intervalls (by setting $b_0 = a_1$ and $a_0 = - \infty$ and $b_1 = \infty$ ). Now my goal is to find some matrix representation of $K$, which means that also this $I_i$ operator has to be made to some kind of matrix (in best case in Fock basis). Now I don't have an idea how I can get this done, because I only know the relation $\int_{-\infty}^{\infty} dq |q><q| = 1$, but due to the eigenstates are deltas, they even aren't part of Hilbertspace, so I don't have a clue how to proceed. $\endgroup$ – pcalc Aug 19 '19 at 17:14

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