# How do you write the Hadamard operator on two qubits in braket notation?

I understand how to write the Hadamard operator on one qubit in braket notation using $$\newcommand\bra[1]{\left\langle#1\right|}\newcommand\ket[1]{\left|#1\right\rangle} H=\sum_{i,j} \bra{w_{j}}H\ket{v_{i}} \ket{w_{j}} \bra{v_{i}}$$ where $$H_{ij}= \sum_{i,j} \bra{w_{j}}H\ket{v_{i}}$$

From my understanding, $$\ket{w_{i}}=\ket{v_{i}}=\ket{0}$$ and $$\ket{w_{j}}=\ket{v_{j}}=\ket{1}$$

But I don't understand how to translate the matrix into braket notation when we range over $$i$$ and $$j$$ in $${0,1,2,3}$$. Does the new basis become $$\ket{00}, \ket{01}, \ket{10}, \ket{11}$$?

Hadamard in Dirac notation is $$H=\frac{1}{\sqrt{2}}(|0\rangle\langle 0|+|0\rangle\langle 1|+|1\rangle\langle 0|-|1\rangle\langle 1|).$$ For two qubits, you take the tensor product $$H\otimes H$$. You get $$\frac12\left(|00\rangle\langle 00|+|00\rangle\langle 01|+|00\rangle\langle 10|+|00\rangle\langle 11| +|01\rangle\langle 00|-|01\rangle\langle 01|+|01\rangle\langle 10|-|01\rangle\langle 11| +|10\rangle\langle 00|+|10\rangle\langle 01|-|10\rangle\langle 10|-|10\rangle\langle 11| +|11\rangle\langle 00|-|11\rangle\langle 01|-|11\rangle\langle 10|+|11\rangle\langle 11|\right)$$ (essentially, you're looking for combinations $$|1\rangle\langle 1|$$ to put a negative sign on). If I were to write this as a formula, I'd have $$H=\frac{1}{\sqrt{2}}\sum_{x,y\in\{0,1\}}(-1)^{x\cdot y}|x\rangle\langle y|$$ and $$H\otimes H=\frac{1}{2}\sum_{x_1,x_2,y_1,y_2\in\{0,1\}}(-1)^{x_1\cdot y_1}(-1)^{x_2\cdot y_2}|x_1x_2\rangle\langle y_1y_2|.$$