I know that in order to make a two quit entangled state, this quantum circuit is used: Hadamard + CNOT gate

But I was wondering if there are any other gate combinations which also create entangled two quit states.

If there are other quantum circuits which achieve this, I was wondering if they would also create any one of the four bell states. Thanks in advance.

Cross-posted on physics.SE


The circuit you gave certainly generated an entangled state. In particular, it generates the state $|\psi\rangle = \dfrac{|00\rangle + |11\rangle}{\sqrt{2}}$ . But this is just one particular entangled state, there are many more.

For instance, the following circuit

enter image description here

prepares the state $|\psi \rangle = \dfrac{|01\rangle + |10\rangle}{\sqrt{2}} $ which is another Bell state.

Note that if instead of using the Hadamard gate, $H$, I have replaced it with a $RY$ rotation then I would still generate an entangled state.

enter image description here

Since the above circuit will generate a state in the form $|\psi \rangle = \dfrac{ \alpha|01\rangle + \beta|10\rangle}{\sqrt{2}} $ where $|\alpha|^2 + |\beta|^2| = 1$. Also, note that at $\theta = \pi/2$, you get back the first circuit.

You can also use the $RXX$ gate, for example:

enter image description here

This will generate the state $|\psi \rangle = \dfrac{|00\rangle - i|11\rangle}{\sqrt{2}}$ which is also an entangled state. This gate has the matrix representation in the computational basis as $RXX = \begin{pmatrix} \cos(\theta) & 0 & 0 & -i\sin(\theta) \\ 0 & \cos(\theta) & -i\sin(\theta) & 0 \\ 0 & -i\sin(\theta) & \cos(\theta) & 0 \\ -i\sin(\theta) & 0 & 0 & \cos(\theta) \end{pmatrix} $.

  • $\begingroup$ Ok thank you. Is there a way mathematically, such as maybe taking the transpose conjugate of a circuit, to find out which of the four bell states two qubits are in? $\endgroup$ – Will Mar 3 at 22:49

Yes, there are many other quantum circuits that prepare entangled states. In fact, in certain sense that can be made precise, almost all possible quantum circuits on two qubits produce entangled states. This corresponds to the fact that almost all pure two-qubit states are entangled. It is the product states that are rare (more precisely: they form a measure zero subset of all pure states).

That said, there are circuits that do not produce any entanglement. For example, any cirucit that lacks two-qubit gates or any circuit that consists of single-qubit gates and SWAP gates has this property. Also, running any circuit in reverse destroys the entanglement created by running it forward.

Remark on comparing circuits: Generally speaking, almost any question of the form "Circuit $C$ produces states with property $P$. Are there other circuits that do this?" will have an affirmative answer, because we can always make changes to a circuit by adding or moving around gates without affecting the unitary operation it applies to its input. This is analogous to how we can modify a computer program by adding or moving around instructions without affecting its output. For example, in the circuit in the question we can replace CNOT with CZ sandwiched between two Hadamards on the target qubit or we can add two consecutive X gates anywhere in the circuit.

Example circuit which prepares the Bell state $(|00\rangle + |11\rangle)/\sqrt{2}$ without using the CNOT gate is the following

$$ (S \otimes X) \circ \sqrt{\text{SWAP}} \circ (S \otimes X) $$

where the gates are as described in the Wikipedia article about quantum gates.

  • $\begingroup$ Ok thank you that makes a lot of sense. So if I was to apply what you said to Bell states, is there a way to differentiate the many quantum circuits that can make any entangled state vs. the ones which generate one of the four bell states? $\endgroup$ – Will Mar 3 at 22:44
  • $\begingroup$ Yes, you can compute the circuits' unitary matrices and compare the first column. If the circuits are given as experiments then you can use quantum state tomography on the outputs. Note that this requires a large number of circuit executions. $\endgroup$ – Adam Zalcman Mar 4 at 16:56

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