What quantum gates are needed to get the state $|01rangle+|10rangle$ from $|00rangle$?

Question

I was wondering if I start with two qubits in the state
$$|00rangle$$
If it's possible to apply gates to get it to the state
$$frac{|01rangle + |10rangle}{sqrt{2}}$$
I have tried applying the Hadamard Gate, Controlled X etc, But I couldn't make this state. So I'm curious if it's possible and I am just missing something very obvious.

Bruno De Souza Leão · Accepted Answer

Applying Hadamard + CNOT takes you from $|00rangle$ to $dfrac{|00rangle + |11rangle}{sqrt{2}}$. Now, just apply the single-qubit $X$ Pauli operator (which swaps $|0rangle$ with $|1rangle$ and vice-versa) to either one of the two qubits, and you get the target state $dfrac{|01rangle + |10rangle}{sqrt{2}}$.

Correct answer by Bruno De Souza Leão on August 4, 2021

Karol · Answer

Let's consider a space of four states ${|00rangle, |01rangle, |10rangle, |11rangle}$. The question is whether exist any operator for which,
$$
left(begin{array}{cccc}
a_{11} & a_{12} & a_{13} & a_{14} 
a_{21} & a_{22} & a_{23} & a_{24} 
a_{31} & a_{32} & a_{33} & a_{34} 
a_{41} & a_{42} & a_{43} & a_{44} 
end{array}right) 
left(
begin{array}{c}
1  0  0  0
end{array}
right) = frac{1}{sqrt{2}} 
left(
begin{array}{c}
0  1  0  1
end{array}
right)
$$
Of course one can invent such an operator, e.g.
$$
hat{a} = frac{1}{sqrt{2}}left(begin{array}{cccc}
0 & a_{12} & a_{13} & a_{14} 
1 & a_{22} & a_{23} & a_{24} 
0 & a_{32} & a_{33} & a_{34} 
1 & a_{42} & a_{43} & a_{44} 
end{array}right),
$$
but I do not know if it corresponds to any known quantum gate. Remaining $a_{ij}$ can be arbitrary.

What quantum gates are needed to get the state $|01rangle+|10rangle$ from $|00rangle$?

2 Answers

Add your own answers!

Ask a Question