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Note: Cross-posted on Physics SE.


I am reading a research article based on quantum image watermarking (PDF here). The authors have defined some unitary transforms for the watermarking schemes, which is where I am having great difficulty in understanding. The operators defined as: $$U_1=I^{\otimes 3q-1}\otimes U\otimes |yx\rangle\langle yx|+I^{\otimes3q}\otimes \left(\sum_{j=0}^{2^n-1}\sum_{i=0, ji\neq yx}^{2^n-1}|ji\rangle\langle ji|\right)$$This is the operator where $q=8$, $n=8$, $U=\begin{bmatrix} 0&1\\ 1&0\end{bmatrix}$. This is an operator acting on $$|I(\theta)\rangle=\dfrac{1}{2^n}\left(\sum_{j=0}^{2^n-1}\sum_{i=0}^{2^n-1}|C(j,i)\rangle\otimes |ji\rangle \right)$$ Can somebody explain the operator and how does this act on $I(\theta)$. I have a sound linear algebra background but as soon as I see these quantum notations I have a hard time understanding them.

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$I^{\otimes 23} = I\otimes I\otimes I\otimes \cdots \otimes I$ (containing 23 identity operators, each presumably being $2\times 2$)

The $\otimes$ operator is just the Left Kronecker Product.

Assuming that $y$ and $x$ represent qubits, then $|yx\rangle\langle yx|$ is some $4\times 4$ matrix, which can be calculated as the outer product of the column vector $|yx\rangle$ and the row vector which is its transpose: $\langle yx|$ (this is called Dirac notation).

The first of two summands in your operator is, therefore, a $2^{26} \times 2^{26}$ matrix which you can calculate by applying the Left Kronecker Product to 3 matrices (the $2^{23} \times 2^{23}$ identity matrix, the $U$ operator, and the 4x4 matrix $|yx\rangle\langle yx|$.

The information I've given you above should also be sufficient for you to now figure out on your own how to get the matrix for the 2nd term of the summand.


As for the 2nd part of your question, which is about how to "act on $I(\theta)$" with the operator $U_1$. It just means the same thing (the $I(\theta)$ will be some $2^{26} \times 1$ column vector in Dirac notation, and it can be converted to matrix notation again using the Kronecker product, which is usually taught for matrices, but if you think of column vectors as matrices where the number of columns is 1, then you can apply the Kronecker product to vectors with no problems).

This is about all I can tell you without knowing more about the paper itself (such as what $C(i,j)$ means and what $x$ and $y$ mean. The article you referred to is not free.

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  • $\begingroup$ Okay, and i have also attached the research article so that you can comprehend a bit more of what my problem is $\endgroup$ – fff Mar 20 at 11:50
  • $\begingroup$ The second summand will be $(2^{2n})-1$ matrices of dimension $2\times 2$? But with a + sign between them, do I just add them? $\endgroup$ – fff Mar 20 at 12:32
  • $\begingroup$ Looks like $x,y$ are $n$-bit strings, so the actual size of the matrix $U_1$ is $2^{3q-1}\cdot 2\cdot 2^{2n} \times 2^{3q-1}\cdot 2\cdot 2^{2n}$. $\endgroup$ – Danylo Y Mar 20 at 20:03
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    $\begingroup$ @Upstart Every bit-string corresponds to a single basis vector. There are $2^k$ of $k$-bit strings, so the dimension is $2^k$. $\endgroup$ – Danylo Y Mar 21 at 11:54
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    $\begingroup$ Yes, they explicitly say that $2n + 3q$ qubits are used to store the image. And $k$ qubits form a $2^k$-dimensional space. $\endgroup$ – Danylo Y Mar 21 at 14:09
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You can think of a pixel state as a combination of color ($3q$ qubits) and position ($2n$ qubits). Authors represent the whole image as a superposition of pixels. This is a lossy image encoding, since measurement in standard basis will give the value of 1 random pixel.

Operator $𝑈_1$ just flips the least significant qubit of color encoding for a particular pixel at a given $x,y$ location. This is basically a multiply controlled $X$ operation, an extended version of the simple CNOT gate.


EDIT: after some research I must add, that authors of the original encoding scheme of this type (called NEQR, with grayscale encoding instead of RGB, see here) claim that you can "retrieve digital images from quantum images accurately".

From my point of view this is cheating, because to do this retrieval you need a lot of copies of the image state. The minimum number of copies must be at least the number of pixels in the image, but in general you need much bigger number of copies, infinite in the worst case.

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