Quantum circuit that puts qubits in "equal" superposition I would like to build a circuit that takes multiple qubit system states and puts them in a superposition with equal amplitudes.
For example:
Let's consider a 4 qubit system. I would like the circuit to start with an input of $\left| 0000 \right>$ and end with the following superposition: $$\frac{1}{\sqrt{4}}\left(\left| 1000 \right> + \left| 1010 \right> + \left| 1001 \right> + \left| 1100 \right>\right).$$
I would like for the choice of states and number of states to be arbitrary. I.e. that I can easily perform the same circuit to generate
$$\frac{1}{\sqrt{3}}\left(\left| 1000 \right> + \left| 1011 \right> + \left| 1101 \right>\right),$$
or
$$\frac{1}{\sqrt{5}}\left(\left| 1000 \right> + \left| 1010 \right> + \left| 1001 \right> + \left| 1100 \right> + \left| 1111 \right>\right).$$
Can anyone help me with this? Any help is appreciated!
EDIT: I have tried creating a matrix representation of such an operator, and I have found one that does what I need it to do, it maps the $\left|00\right>$ to an arbitrary equally-distributed superposition of the basis vectors of my choosing. Here is the matrix I found that acts on 2 qubits to create such a state:
$$
U = \begin{pmatrix}
1/\sqrt{3} & 0 & 0 & 0\\
1/\sqrt{3} & 0 & 0 & 0\\
1/\sqrt{3} & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
\end{pmatrix}.
$$
This gives:
$$
U\left|00\right>= \begin{pmatrix}
1/\sqrt{3} & 0 & 0 & 0\\
1/\sqrt{3} & 0 & 0 & 0\\
1/\sqrt{3} & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
\end{pmatrix}
\begin{pmatrix}
1\\
0\\
0\\
0
\end{pmatrix} =
\frac{1}{\sqrt{3}} \begin{pmatrix}
1\\
1\\
1\\
0
\end{pmatrix} = \frac{1}{\sqrt{3}}\left( \left|00\right> + \left|01\right> + \left|10\right> \right).
$$
The idea is that the first column of the matrix would consist of $N$ non-zero elements, each element contrbuting one basis state to the final superposition, and each element being equal to $\frac{1}{\sqrt{N}}$.
The problem with this matrix, however, is that it is not unitary. Would it be possible to modify this matrix in such a wat that it still has the same result on the $|00>$ state vector, but that it is unitary? This would solve my problem completely.
 A: Just a few hints:

*

*To produce equally distributed superposition containing all combinations of $n$ qubits, simply put Hadamard gate on all qubit.

*Bell states and GHZ states belong among states your are talking about. Here is a circuit how to produce GHZ states (see also article below)

*You might also be interested in W states, i.e. equally distributed superposition of bit strings where only one qubit is in state 1 and others are zero. In article
Efficient quantum algorithms for GHZ and W states, and implementation on the IBM quantum computer an algorithm how to produce such states is described.


EDIT:
Concerning the matrix, it is possible to construct the matrix for example in this way:
$$
\frac{1}{\sqrt{3}}
\begin{pmatrix}
1 & 1 & 0 &-1 \\
1 & -1 & 1 &0 \\
1 & 0 & -1 &1 \\
0 & 1 & 1 &1 \\
\end{pmatrix}
$$
The matrix is unitary (you can check this by a direct multiplication) and  the required state is returned for input $|00\rangle$. You do not have to bother about other inputs because you can always set all input qubits to state $|0\rangle$.
