LSTM-Quantum Kernel Method

PennyLane Help
1

I am now trying to replace the VQC circuit with a Quantum Kernel Circuit form circuit on the original QLSTM, but I am not sure whether my input Tensor can be directly split into x1 and x2, and I only want the probability of taking the base state of 0 like this Is it feasible?

 def _circuit_input(self, inputs,weights):
        if inputs.dim() == 1:
            inputs = inputs.unsqueeze(0)
        
       
        if inputs.shape[0] < 2:
            if inputs.shape[1] < 2:
                raise ValueError("Input tensor must have at least 2 elements")
            x1, x2 = torch.chunk(inputs, 2, dim=1)
        else:
            x1, x2 = torch.chunk(inputs, 2, dim=0)

        for wire in self.wires_input:
            qml.Hadamard(wires=wire)
        qml.templates.AngleEmbedding(x1, wires=self.wires_input, rotation='Z')
        qml.templates.AngleEmbedding(x1, wires=self.wires_input, rotation='Y')
        for i in range(len(self.wires_input) - 1):
            qml.CNOT(wires=[self.wires_input[i], self.wires_input[i + 1]])
        
        qml.templates.AngleEmbedding(x1, wires=self.wires_input, rotation='Z')
       
       #Inverse
        qml.templates.AngleEmbedding(x2, wires=self.wires_input, rotation='Z')
        for i in range(len(self.wires_input) - 1, 0, -1):
             qml.CNOT(wires=[self.wires_input[i-1], self.wires_input[i]])
        qml.templates.AngleEmbedding(x2, wires=self.wires_input, rotation='Y')  
        qml.templates.AngleEmbedding(x2, wires=self.wires_input, rotation='Z')
        for wire in self.wires_input:
            qml.Hadamard(wires=wire)
        proj_0 = qml.Projector([0] * len(self.wires_input), wires=self.wires_input)
        return qml.expval(proj_0)
2

Hi @candeer12376 ,

Welcome to the forum!

It’s hard to tell what you’re doing here and whether or not it’s correct. Are you seeing a specific error? If so please provide the full error traceback as well as a minimal (but self-contained) code example so that I can investigate.