Dissipative learning of a quantum classifier

Abstract

The expectation that quantum computation might bring performance advantages in machine learning algorithms motivates the work on the quantum versions of artificial neural networks. In this study, we analyse the learning dynamics of a quantum classifier model that works as an open quantum system which is an alternative to the standard quantum circuit model. According to the obtained results, the model can be successfully trained with a gradient descent (GD)-based algorithm. The fact that these optimisation processes have been obtained with continuous dynamics, shows promise for the development of a differentiable activation function for the classifier model.

Appendix A. Derivation of the cost function

In this section, we present the mathematical justifications for numerical calculations in the text. First, we substitute $\nu =g$ in eq. (19)

$$\begin{aligned} \delta {g}_i=-\eta \frac{\partial C}{\partial {g}_i} \end{aligned}$$

(A.1)

and obtain the expression for cost function by taking the partial derivative with respect to the coupling constant g.

$$\begin{aligned} \frac{\partial C}{\partial {g}_i}=(\langle \sigma _z^0\rangle _{\textrm{des}}^{ss}-\langle \sigma _z^0\rangle _{\textrm{act}}^{ss})\bigg (-\frac{\partial \langle \sigma _z^0\rangle _{\textrm{act}}^{ss}}{\partial {g}_i}\bigg ). \end{aligned}$$

(A.2)

In our current example, we have two information reservoirs corresponding to specific magnetisations. Therefore, the actual steady-state magnetisation (eq. (13)) reads as

$$\begin{aligned} A=\langle \sigma _z^0\rangle _{\textrm{act}}^{ss}=\frac{g_{1}^{2}\langle \sigma _z^1\rangle +g_{2}^{2}\langle \sigma _z^2\rangle }{g_{1}^{2}+g_{2}^{2}}. \end{aligned}$$

(A.3)

According to the recipe to derive the cost function, the partial derivatives with respect to $g_1$ and $g_2$ separately are obtained as

$$\begin{aligned}&\frac{\partial A}{\partial {g}_1}=\frac{2g_{1}\langle \sigma _z^1\rangle (g_{1}^{2}+g_{2}^{2})-2g_{1}(g_{1}^{2}\langle \sigma _z^1\rangle +g_{2}^{2}\langle \sigma _z^2\rangle )}{(g_{1}^{2}+g_{2}^{2})^{2}}\nonumber \\&\frac{\partial A}{\partial {g}_2}=\frac{2g_{2}\langle \sigma _z^2\rangle (g_{1}^{2}+g_{2}^{2})-2g_{2}(g_{1}^{2}\langle \sigma _z^1\rangle +g_{2}^{2}\langle \sigma _z^2\rangle )}{(g_{1}^{2}+g_{2}^{2})^{2}}. \end{aligned}$$

(A.4)

In our example, the desired magnetisation $\langle \sigma _z^0\rangle _{\textrm{des}}^{ss}=0.4$ is a constant value in the cost function. Using the expression (A.1), (A.2), (A.3) and (A.4), eq. (18) in expressed as follows:

$$\begin{aligned}&(g_1)_{k+1}=(g_1)_{k}+\delta (g_1)_{k}\nonumber \\&(g_2)_{k+1}=(g_2)_{k}+\delta (g_2)_{k}. \end{aligned}$$

(A.5)

Next, we substitute $\nu =\theta$ in eq. (19) as

$$\begin{aligned} \delta {\theta }_i=-\eta \frac{\partial C}{\partial {\theta }_i}. \end{aligned}$$

(A.6)

Regarding eq. (15), one can easily see that the magnetisation of the ith reservoir is $\langle \sigma _z\rangle _i=\langle \sigma _{i}^+\sigma _{i}^-\rangle - \langle \sigma _{i}^-\sigma _{i}^+\rangle$. Therefore, azimuth parameter-dependent expression of the magnetisation can be easily written as $\langle \sigma _z\rangle _i=\cos \theta _{i}$.

Equation (A.7) is obtained when we take the partial derivative of the cost function with respect to $\theta$.

$$\begin{aligned} \frac{\partial C}{\partial {\theta }_i}=(\langle \sigma _z^0\rangle _{\textrm{des}}^{ss}-\langle \sigma _z^0\rangle _{\textrm{act}}^{ss})\bigg (-\frac{\partial \langle \sigma _z^0\rangle _{\textrm{act}}^{ss}}{\partial {\theta }_i}\bigg ). \end{aligned}$$

(A.7)

In our current example, we have two information reservoirs corresponding to specific magnetisations. Therefore, the actual steady-state magnetisation (eq. (13)) reads as

$$\begin{aligned} A=\langle \sigma _z^0\rangle _{\textrm{act}}^{ss}=\frac{g_{1}^{2}\cos \theta _{1}+g_{2}^{2}\cos \theta _{2}}{g_{1}^{2}+g_{2}^{2}}. \end{aligned}$$

(A.8)

According to the recipe to derive the cost function, the partial derivatives with respect to $\theta _{1}$ and $\theta _{2}$ separately are obtained as

$$\begin{aligned}&\frac{\partial A}{\partial {\theta }_1}=-\frac{g_{1}^{2}\sin \theta _{1}}{g_{1}^{2}+g_{2}^{2}}\nonumber \\&\frac{\partial A}{\partial {\theta }_2}=-\frac{g_{2}^{2}\sin \theta _{2}}{g_{1}^{2}+g_{2}^{2}}. \end{aligned}$$

(A.9)

In our example, the desired magnetisation $\langle \sigma _z^0\rangle _{\textrm{des}}^{ss}=0$ a constant value in the cost function. Using the expression (A.6), (A.7), (A.8) and (A.9), eq. (A.7) is expressed as follows,

$$\begin{aligned}&\mathbf {(\theta )}_{k+1}=\mathbf {(\theta _1)}_{k}+\delta \mathbf {(\theta _1)}_{k}\nonumber \\&\mathbf {(\theta _2)}_{k+1}=\mathbf {(\theta _2)}_{k}+\delta \mathbf {(\theta _2)}_{k}. \end{aligned}$$

(A.10)

Let us edit eq. (18) for $\nu =\phi$

$$\begin{aligned} \delta {\phi }_i=-\eta \frac{\partial C}{\partial {\phi }_i}. \end{aligned}$$

(A.11)

Equation (A.12) is obtained when we take the partial derivative of the cost function with respect to $\phi$.

$$\begin{aligned} \frac{\partial C}{\partial {\phi }_i}=(\langle \sigma _y^0\rangle _{\textrm{des}}^{ss}-\langle \sigma _y^0\rangle _{\textrm{act}}^{ss})\bigg (-\frac{\partial \langle \sigma _y^0\rangle _{\textrm{act}}^{ss}}{\partial {\phi }_i}\bigg ). \end{aligned}$$

(A.12)

In our current example, we have two information reservoirs corresponding to specific magnetisations. Therefore, the actual steady-state magnetisation (eq. (13)) by using eq. (15) reads as

$$\begin{aligned}{} & {} A=\langle \sigma _y^0\rangle _{\textrm{act}}^{ss}\nonumber \\ {}{} & {} \quad =-r\tau \frac{g_{1}^{3}\sin \theta _{1}\cos \theta _{1}\cos \phi _{1}+g_{1}g_{2}^{2}\sin \theta _{1}\cos \theta _{2}\cos \phi _{1}+g_{1}^{2}g_{2}\cos \theta _{1}\sin \theta _{2}\cos \phi _{2}+g_{2}^{3}\sin \theta _{2}\cos \theta _{2}\cos \phi _{2}}{g_{1}^{2}+g_{2}^{2}}.\nonumber \\ \end{aligned}$$

(A.13)

According to the recipe to derive the cost function, the partial derivatives with respect to $\phi _{1}$ and $\phi _{2}$ separately are obtained as

$$\begin{aligned}&\frac{\partial A}{\partial {\phi }_1}=r\tau \frac{g_{1}^{3}\sin \theta _{1}\cos \theta _{1}\sin \phi _{1}+g_{1}g_{2}^{2}\sin \theta _{1}\cos \theta _{2}\sin \phi _{1}}{g_{1}^{2}+g_{2}^{2}}\nonumber \\&\frac{\partial A}{\partial {\phi }_2}=r\tau \frac{g_{1}^{2}g_{2}\cos \theta _{1}\sin \theta _{2}\sin \phi _{2}+g_{2}^{3}\sin \theta _{2}\cos \theta _{2}\sin \phi _{2}}{g_{1}^{2}+g_{2}^{2}}. \end{aligned}$$

(A.14)

In our example, the desired magnetisation is $\langle \sigma _y^0\rangle _{\textrm{des}}^{ss}=0$ a constant value in the cost function. Using the expression (A.6), (A.12), (A.13) and (A.14)

$$\begin{aligned}&\mathbf {(\phi )}_{k+1}=\mathbf {(\phi _1)}_{k}+\delta \mathbf {(\phi _1)}_{k}\nonumber \\&\mathbf {(\phi _2)}_{k+1}=\mathbf {(\phi _2)}_{k}+\delta \mathbf {(\phi _2)}_{k}. \end{aligned}$$

(A.15)