# Can you construct a quantum circuit where all qubits are initially in a superposition, but there is still entanglement?

The most basic example of entanglement is when we have 2 qubits, where q0 is in the |+> state and connects to q1 (which is in the |0> state) with a cnot gate: The state is entangled, as the resulting outputs are either 00 or 11 with equal probability, which cannot be replicated with two unconnected qubits. (Statevector is [$$\frac{1}{\sqrt{2}}$$ 0 0 $$\frac{1}{\sqrt{2}}$$]) However, q1 is initially not in a superposition (but |0> instead)

So is there a quantum circuit where all qubits are initially in a superposition, but the state is entangled?

If yes, could you please provide an example? If no, could you explain why? If nobody knows yet, then is there a general consensus on what likely is the case?

Also a side question: Would we consider q1 to be in superposition after the cnot gate? Since it kinda is.

Yes, it is possible to create an entangled state starting from an initial state comprising two disentangled qubits that are both in a superposition of the respective computational basis states on their own. The circuit below provides a concrete example. Here $$\sqrt{X} = \frac{1}{2} \begin{pmatrix} 1+i & 1-i \\ 1-i & 1+i \end{pmatrix}$$, so the initial state of the two qubits before the application of the CNOT is $$\Big[ \frac{1}{2}(1 + i, 1-i)^{T} \Big] \otimes \Big[ \frac{1}{2}(1 + i, 1-i)^{T} \Big] = \frac{1}{2}(i, 1, 1, -i)^{T}$$, which is clearly a product state with both qubits in a superposition of the respective computational basis states. Upon applying the CNOT gate we obtain $$\frac{1}{2}(i, 1, -i, 1)^{T}$$.

We can confirm that this two-qubit state is entangled by defining a bipartition between the two qubits, thereby casting the original $$4$$-dimensional vector into a $$2 \times 2$$ matrix $$\frac{1}{2} \begin{pmatrix} i & 1 \\ -i & 1 \end{pmatrix}$$, where the most significant qubit labels the rows. Performing the singular value decomposition of this matrix, we obtain two nonzero singular values $$s_1 = s_2 = \frac{1}{\sqrt{2}}$$. Hence, the Schmidt rank is $$2$$, so the state is indeed entangled.

Incidentally, if we only consider the Hadamard gate to create the superpositions of $$|0\rangle$$ and $$|1\rangle$$ basis states corresponding to the initial single-qubit states, upon applying the CNOT gate we will never obtain an entangled state. This is made clear by considering the matrix representation of the CNOT gate in the X basis.

• Just to clarify: So an entangled state is not possible if q0 = $(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$ and q1 = $(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$? And a follow up question: since global phases aren't measurable, and $\sqrt{X}$ only has a global phase difference with $R(\frac{\pi}{2})$, does that mean I could replace the $\sqrt{X}$ gates with $R(\frac{\pi}{2})$ and still have an entangled state? Oct 25, 2022 at 1:58
• It is possible to create an entangled state starting from $\frac{1}{\sqrt{2}}(1,1)^{T} \otimes \frac{1}{\sqrt{2}}(1,1)^{T} \equiv |+\rangle \otimes |+\rangle$ if we allow ourselves to use some two-qubit operation other than the CNOT, as the reply by @Mauricio shows. If, however, we impose the application of the CNOT onto the initial disentangled state, then $|+\rangle \otimes |+\rangle$ will not do because the CNOT will leave it unchanged (and thus disentangled). Regarding the latter question, $\sqrt{X} = R_x(\pi/2)$ ignoring the global phase factor, so you could use $R_x(\pi/2)$ instead. Oct 25, 2022 at 10:56

You say, "The states are entangled...". No, the state is entangled. There is one state describing your problem.

I reckon this is perhaps where the problem is in your understanding. Entangled states are by definition not a trivial classical combination of two different states (hence product states).

So is there a quantum circuit where all qubits are initially in a superposition, but they are entangled? Yes ofcourse, just put a H on both instead of just q0. That achieves what you want. Naturally, then the output will look different from 00 and 11 with equal probability.

Would we consider q1 to be in superposition after the cnot gate? Since it kinda is. Yes. You put the CNOT gate exactly for that purpose.

If you only care about the end result it does not matter much if the qubits start entangled or not. You can always add Hadamard gates in order to move them back to zero. Here is the same circuit that you described above with extra gates: Here I just used that $$HZH=X$$. And now both qubit 0 and 1 start in a superposition. (Note that in the $$X$$ basis the states are not in a superposition, so it is basis dependent).