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I would like to simulate the 4 "Bell States" on the IBM composer?

How can I best implement those 4 Bell states using the existing set of gates ?

Here below you see the definition of the 4 Bell states.

The 4 bell states

The first bell state can be easily implemented on the composer through a Hadamard gate and a CNOT gate as follows:

first Bell state

but how can I implement the other 3 bell states ?

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Remember that the Hadamard (H) gate maps $|0\rangle$ to $\frac{|0\rangle+|1\rangle}{\sqrt{2}}$ and $|1\rangle$ to $\frac{|0\rangle-|1\rangle}{\sqrt{2}}$, while the CNOT gate has the following conversion table:

enter image description here

So, you can use the same circuit:

enter image description here

but begin in the states $|01\rangle$, $|10\rangle$ and $|11\rangle$ to get the other three Bell states (you can easily convert to these states from $|00\rangle$ using the Pauli-X gate).

enter image description here

Source: Todd Brun's lecture notes, Part-11

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One of the possible solutions:

$|\Phi^+\rangle = \textrm{CNOT} \cdot H_1 |00 \rangle$
$|\Phi^-\rangle = Z_1 |\Phi^+\rangle = Z_1 \cdot \textrm{CNOT} \cdot H_1 |00 \rangle$
$|\Psi^+\rangle = X_2 |\Phi^+\rangle = X_2 \cdot \textrm{CNOT} \cdot H_1 |00 \rangle$
$|\Psi^-\rangle = Z_1 |\Psi^+\rangle = Z_1 \cdot X_2 \cdot \textrm{CNOT} \cdot H_1 |00 \rangle$

Where $Z_i$ and $X_i$ act on the $i^\textrm{th}$ qubit.


So here is $|\Phi^-\rangle$:

enter image description here

Here is $|\Psi^+\rangle$:

enter image description here

And here is $|\Psi^-\rangle$:

enter image description here


Another solution is:

$|\Phi^+\rangle = \textrm{CNOT} \cdot H_1 |00 \rangle$
$|\Phi^-\rangle = X_1 \cdot \textrm{CNOT} \cdot H_1 |00 \rangle$
$|\Psi^+\rangle = X_2 \cdot \textrm{CNOT} \cdot H_1 |00 \rangle$
$|\Psi^-\rangle = X_1 X_2\cdot \textrm{CNOT} \cdot H_1 |00 \rangle$

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  • $\begingroup$ Are you assuming that we can only begin in the state $|00\rangle$? Is that a necessity for the IBM Q composer? Asking because I don't have experience with the IBM Q composer. $\endgroup$ – Sanchayan Dutta Jun 7 '18 at 15:45
  • $\begingroup$ @Blue, yes IBM Q composer seems always to begin in state $|00\rangle$ but it can be easily converted to e.g. state $|01\rangle$ by using an $X$ gate. $\endgroup$ – JanVdA Jun 7 '18 at 16:01
  • $\begingroup$ I've included the solution that starts with states different from $|00\rangle$, but do I have to put the diagrams in there too or is what I have good enough for you? $\endgroup$ – user1271772 Jun 7 '18 at 16:11

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