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Please help me in building IBM Quantum Experience circuit for: $$ M|0\rangle = \frac{1}{2}(|0\rangle+|1\rangle+|2\rangle+|3\rangle) $$

Edit: Is it possible to make a circuit for a general transformation $M|0\rangle = \frac{1}{\sqrt{m}}\sum_{i=0}^{m-1}|i\rangle$ acting on $\log m$ qubits initialized to $|0\rangle$?

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  • $\begingroup$ Simply put Hadamard gates on two qubits and that's it. $\endgroup$ – Martin Vesely Feb 12 at 19:22
  • $\begingroup$ @MartinVesely can we give circuit for a general transformation $M|0\rangle = \frac{1}{\sqrt{m}}\sum_{i=0}^{m-1}|i\rangle$ by initializing $log~m$ qubits to $|0\rangle$. $\endgroup$ – Adam Levine Feb 13 at 3:10
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Answer to EDIT in question:

It is possible to prepeare a state

$$|\psi\rangle = \frac{1}{\sqrt{m}}\sum_{i=0}^{m-1}|i\rangle,$$

where $m = 2^n$ and $n$ is a number of qubits, by application of operator $\otimes H ^{n} = H_{q_0} \otimes H_{q_1} \otimes \dots \otimes H_{q_{n-1}}$, i.e. Hadamard gate is applied on each qubit involved. As a results, you will get $n=\log{m}$ qubits in superposition described by uniform probability distribution. It is a generalization of construction provided in answer by met927


EDIT: based on comment by Adam Levine: "what if for example $m=5$?"

It that case you have to prepare entangled state and the approach is a little bit more intricated. For $m=5$ you have to prepare state (I switched to more common convention with states expressed in binary numbers instead of decimal):

$$ |\psi\rangle = \frac{1}{\sqrt{5}}(|000\rangle + |001\rangle + |010\rangle+|011\rangle+|100\rangle), $$

i.e. three qubits state where $|101\rangle$, $|110\rangle$ and $|111\rangle$ have zero probability.

To prepare an arbitrary state, you can employ approach presented in this article: Transformation of quantum states using uniformly controlled rotations.

Based on the article, a circuit preparing state $|\psi\rangle$ is this:

Circuit

Here you can see results provided by IBM Q simulator:

enter image description here

Apparently, the circuit prepared desired state with uniformly distributed values from $|000\rangle$ to $|100\rangle$.


EDIT 2: here is a code in QASM of the circuit above:

OPENQASM 2.0;
include "qelib1.inc";

qreg q[3];
creg c[3];

ry(0.927) q[0];

ry(pi/4) q[1];
cx q[0],q[1];
ry(pi/4) q[1];
cx q[0],q[1];

ry(pi/2) q[2];
cx q[1],q[2];
ry(-pi/4) q[2];
cx q[0],q[2];
ry(pi/4) q[2];
cx q[1],q[2];
cx q[0],q[2];

measure q[0] -> c[2];
measure q[1] -> c[1];
measure q[2] -> c[0];
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    $\begingroup$ If $m \neq 2^{n}$ for example $m=5$, what should be the approach? $\endgroup$ – Adam Levine Feb 13 at 5:49
  • $\begingroup$ @AdamLevine: Good question, see expanded answer. $\endgroup$ – Martin Vesely Feb 13 at 7:16
  • $\begingroup$ Please share code for the above circuit $\endgroup$ – Adam Levine Feb 13 at 9:38
  • $\begingroup$ @AdamLevine: Please find it in the answer. $\endgroup$ – Martin Vesely Feb 13 at 11:12
  • $\begingroup$ Dear Martin Vesely, following the paper - " Transformation of quantum states using uniformly controlled rotations." did you solve it by pen and paper? In paper, $|a\rangle \mapsto |e_{1}\rangle$ is given (this is not applicable for our transformation since it is already in $|e_{1}\rangle$ form), but how do I convert $|e_{1}\rangle \mapsto |b\rangle$ (where $|b\rangle$ is the state we require)? $\endgroup$ – Adam Levine Feb 14 at 15:17
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The circuit you have described is formed of 2 qubits, both in an equal superposition. This can be achieved by applying the H gate to both qubits, as this puts each qubit into superposition and takes us up to 4 possible states. In the IBM Quantum Experience, this circuit would look like 2 qubit superposition

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