A quantum mechanics-based framework for image processing and its application to image segmentation

Abstract

Quantum mechanics provides the physical laws governing microscopic systems. A novel and generic framework based on quantum mechanics for image processing is proposed in this paper. The basic idea is to map each image element to a quantum system. This enables the utilization of the quantum mechanics powerful theory in solving image processing problems. The initial states of the image elements are evolved to the final states, controlled by an external force derived from the image features. The final states can be designed to correspond to the class of the element providing solutions to image segmentation, object recognition, and image classification problems. In this work, the formulation of the framework for a single-object segmentation problem is developed. The proposed algorithm based on this framework consists of four major steps. The first step is designing and estimating the operator that controls the evolution process from image features. The states associated with the pixels of the image are initialized in the second step. In the third step, the system is evolved. Finally, a measurement is performed to determine the output. The presented algorithm is tested on noiseless and noisy synthetic images as well as natural images. The average of the obtained results is 98.5 % for sensitivity and 99.7 % for specificity. A comparison with other segmentation algorithms is performed showing the superior performance of the proposed method. The application of the introduced quantum-based framework to image segmentation demonstrates high efficiency in handling different types of images. Moreover, it can be extended to multi-object segmentation and utilized in other applications in the fields of signal and image processing.

Appendix: Additional proofs

1.1 Measurement operator

Let the general state vector of a two-state quantum system is

$$\begin{aligned} |\psi \rangle =\alpha |0\rangle +\beta |1\rangle \end{aligned}$$

(25)

Working in computational basis where

$$\begin{aligned} |0\rangle =\begin{pmatrix}1 \\ 0\end{pmatrix} , |1\rangle =\begin{pmatrix}0 \\ 1\end{pmatrix} \end{aligned}$$

(26)

Then

$$\begin{aligned} |\psi \rangle =\begin{pmatrix}\alpha \\ \beta \end{pmatrix} \end{aligned}$$

(27)

The Hermitian conjugate (complex conjugate transpose) of the state vector is

$$\begin{aligned} \langle \psi |=\alpha ^{\dagger }\langle 0|+\beta ^{\dagger }\langle 1| \end{aligned}$$

(28)

Choosing the same basis, we get:

$$\begin{aligned} \langle \psi |=\begin{pmatrix}\alpha ^{\dagger }&\beta ^{\dagger }\end{pmatrix} \end{aligned}$$

(29)

The inner product of two states $|\phi \rangle$, $|\psi \rangle$ is calculated by multiplying $\langle \psi |$ and $|\phi \rangle$. The short-hand notation for this operation is $\langle \psi |\phi \rangle$. Now, taking the inner product of the vectors $|\psi \rangle$ and $|1\rangle$:

$$\begin{aligned} \langle 1 | \psi \rangle =\langle 1|\left( \alpha |0\rangle +\beta |1\rangle \right) =\alpha \langle 1|0\rangle +\beta \langle 1|1\rangle \end{aligned}$$

(30)

Since the states $|0\rangle$ and $|1\rangle$ are orthonormal as they form the basis of the Hilbert space of the system, therefore: $\langle 1|0\rangle =0$ and $\langle 1|1\rangle =1$. Thus,

$$\begin{aligned} \langle 1 | \psi \rangle =\beta \end{aligned}$$

(31)

Taking the Hermitian conjugate of the last equation, we get

$$\begin{aligned} \langle \psi | 1\rangle =\beta ^{\dagger } \end{aligned}$$

(32)

From the postulates of quantum mechanics, the probability of finding the system in state $|1\rangle$ after measurement is given by

$$\begin{aligned} p(1)=\langle \psi |M^{\dagger } M|\psi \rangle \end{aligned}$$

(33)

where M is the measurement operator for the state $|1\rangle$, taken to be

$$\begin{aligned} M=|1\rangle \langle 1| \end{aligned}$$

(34)

We can prove that this is the correct form of the operator by substituting in the mentioned postulate and using the previous results:

$$\begin{aligned} p(1)=\langle \psi |(|1\rangle \langle 1|)(|1\rangle \langle 1|)|\psi \rangle =\langle \psi | 1\rangle \langle 1 | 1\rangle \langle 1 | \psi \rangle = \beta ^{\dagger } 1 \beta =|\beta |^2 \end{aligned}$$

(35)

Which agrees with the physical interpretation of $\beta$. This proves that this is the correct form of the measurement operator. This operator will be needed in the next section.

1.2 Hamiltonian design

Assume now we apply the operator U on the state $|\psi \rangle$ to evolve it, where

$$\begin{aligned} U=e^{-\frac{i}{\hbar } \int _0^t{H(\tau )\hbox {d}\tau }}\end{aligned}$$

(36)

The probability of being in state $|1\rangle$ after measurement is:

$$\begin{aligned} p(1)=\langle \psi |U^{\dagger }M^{\dagger }M U |\psi \rangle \end{aligned}$$

(37)

but

$$\begin{aligned} M^{\dagger } M=|1\rangle \langle 1 | 1\rangle \langle 1|=|1\rangle \langle 1| \end{aligned}$$

(38)

So

$$\begin{aligned} p(1)=\langle \psi |U^{\dagger }|1\rangle \langle 1| U |\psi \rangle \end{aligned}$$

(39)

Now, in our design we took the Hamiltonian to be in the form of

$$\begin{aligned} H = i \hbar g(t) \begin{pmatrix} 0 &{} -1 \\ 1 &{} 0 \end{pmatrix} \end{aligned}$$

(40)

so as to make sure that we are in cases 1 or 2 where we can find a closed-form solution. This chosen form ensures the commutation of the operator at different time instants. We now begin calculating the matrix U. Given that the eigen decomposition of the matrix:

$$\begin{aligned} \mathbf {S}=\begin{pmatrix} 0 &{} -1 \\ 1 &{} 0 \end{pmatrix} \end{aligned}$$

(41)

is given by:

$$\begin{aligned} \mathbf {S}=\mathbf {Q} \mathbf {D} \mathbf {Q}^{\dagger }=\frac{1}{\sqrt{2}} \begin{pmatrix}1 &{} 1 \\ -i &{} i \end{pmatrix} \begin{pmatrix} i &{} 0\\ 0 &{}-i \end{pmatrix} \frac{1}{\sqrt{2}}\begin{pmatrix}1 &{} i \\ 1 &{} -i \end{pmatrix} \end{aligned}$$

(42)

Then, we can calculate the matrix:

$$\begin{aligned} U=f(\mathbf {S})=\mathbf {Q} f(\mathbf {D}) \mathbf {Q}^{\dagger } \end{aligned}$$

(43)

where

$$\begin{aligned} \begin{aligned}f(\mathbf {S})&=\exp \left( -\frac{i}{\hbar }\int _0^t{H(\tau )\hbox {d}\tau }\right) \\&=\exp \left( -\frac{i}{\hbar }i \hbar \mathbf {S} \int _0^t{g(\tau )\hbox {d}\tau }\right) \\&=\exp \left( \mathbf {S}\tilde{g}(t)\right) \end{aligned} \end{aligned}$$

(44)

So,

$$\begin{aligned} f(\mathbf {D})=\begin{pmatrix}e^{i\tilde{g}(t)} &{} 0\\ 0 &{} e^{-i\tilde{g}(t)}\end{pmatrix}\end{aligned}$$

(45)

Therefore,

$$\begin{aligned} U=\frac{1}{2} \begin{pmatrix}1 &{} 1 \\ -i &{} i \end{pmatrix} \begin{pmatrix}e^{i\tilde{g}(t)} &{} 0\\ 0 &{} e^{-i\tilde{g}(t)}\end{pmatrix}\begin{pmatrix}1 &{} i \\ 1 &{} -i \end{pmatrix} \end{aligned}$$

(46)

In our work, we chose the initial state $|\psi \rangle =|0\rangle$. Return back to calculate the probabilities.

$$\begin{aligned} \langle 1| U |\psi \rangle =\langle 1| U |0\rangle \end{aligned}$$

(47)

So,

$$\begin{aligned} \begin{aligned}\langle 1| U |0\rangle&=\begin{pmatrix}0&1\end{pmatrix}\frac{1}{2} \begin{pmatrix}1 &{} 1 \\ -i &{} i \end{pmatrix} \begin{pmatrix}e^{i\tilde{g}(t)} &{} 0\\ 0 &{} e^{-i\tilde{g}(t)}\end{pmatrix}\begin{pmatrix}1 &{} i \\ 1 &{} -i \end{pmatrix}\begin{pmatrix}1\\ 0\end{pmatrix} \\&=\frac{1}{2}\begin{pmatrix}-i&i\end{pmatrix} \begin{pmatrix}e^{i\tilde{g}(t)} &{} 0\\ 0 &{} e^{-i\tilde{g}(t)}\end{pmatrix} \begin{pmatrix}1\\ 1\end{pmatrix} \\&=\frac{1}{2}\begin{pmatrix}-i e^{i\tilde{g}(t)}&i e^{-i\tilde{g}(t)} \end{pmatrix}\begin{pmatrix}1\\ 1\end{pmatrix}\\ {}&=\sin (\tilde{g}(t)) \end{aligned} \end{aligned}$$

(48)

Finally,

$$\begin{aligned} \begin{aligned}p(1)&=\langle \psi |U^{\dagger }|1\rangle \langle 1| U |\psi \rangle \\&=\langle 0|U^{\dagger }|1\rangle \langle 1| U |0\rangle \\&={\sin (\tilde{g}(t))}^{\dagger }\sin (\tilde{g}(t))\\ {}&=\sin ^2(\tilde{g}(t))\end{aligned} \end{aligned}$$

(49)

Now we see that if

$$\begin{aligned} g(\tau )=c \end{aligned}$$

(50)

then

$$\begin{aligned} \tilde{g}(t)=\int _0^t{g(\tau )\hbox {d}\tau }=c t \end{aligned}$$

(51)

and we get

$$\begin{aligned} p(1)=\sin ^2(c t) \end{aligned}$$

(52)

which has an oscillatory behavior. This would require to choose a time instant to perform the measurement. On the other hand, if we choose

$$\begin{aligned} g(\tau )=c e^{-\tau } \end{aligned}$$

(53)

then

$$\begin{aligned} \tilde{g}(t)=\int _0^t{g(\tau )\hbox {d}\tau }=c -c e^{-t} \end{aligned}$$

(54)

and we get

$$\begin{aligned} p(1)=\sin ^2\left( c -c e^{-t}\right) \end{aligned}$$

(55)

Clearly, if we take the limit as $t \rightarrow \infty$

$$\begin{aligned} p(1)=\sin ^2(c) \end{aligned}$$

(56)

which is an exact value, that does not depend on time. If we choose $c=2\pi$ then $p(1)=0$ which means the final state is $|0\rangle$ and if we choose $c=\frac{\pi }{2}$ then $p(1)=1$ which means the final state is $|1\rangle$. This proves that the chosen form of the Hamiltonian can take the initial state $|0\rangle$ to either of the two states $|0\rangle$ or $|1\rangle$ depending on the value of c. if we choose c carefully to depend on the feature vector, we can reach the final states that correspond to the classification of the pixel. The function g(t) can be any function such that the integral

$$\begin{aligned} \int _0^\infty {g(\tau )d\tau } \end{aligned}$$

(57)

exists. Satisfying this condition obviates us from choosing a time instant to measure, since we let the system evolve for sufficiently long time before measuring.