# Projecting $\lvert ++ \rangle$ on Bell Basis

I understand that, projecting $$\lvert 00\rangle$$ on the Bell states would produce $$\lvert\Phi^+\rangle$$. Because,

$$CNOT(H\lvert0\rangle \otimes \lvert0\rangle) = \frac{1}{\sqrt{2}}(\lvert00\rangle + \lvert11)\rangle$$

We can get other Bell states from $$\lvert 01\rangle, \lvert10\rangle, \lvert 11\rangle$$. However, I am having trouble understanding what would happen if I do the same for $$\lvert ++\rangle$$. Like:

$$CNOT(H\lvert+\rangle \otimes \lvert+\rangle) = ?$$

• I'm confused about what you're trying to do. If you're projecting onto $|00\rangle$, then the outcome must be $|00\rangle$ (or an outcome corresponding to one of the other measurement outcomes). This has nothing to do with the unitaries that produce the Bell state. Is it the unitaries you want to apply? Jun 20 '19 at 15:31
• You are right. I think I misunderstood my problem. Projecting those states onto the Bell states is not about applying the unitaries. Jun 20 '19 at 23:48

The Hadamard gate is: $$\frac{1}{\sqrt 2} \left(|0\rangle \langle 0 | + |0\rangle\langle 1| + |1\rangle \langle 0| - |1\rangle \langle 1|\right)$$

And since $$|+\rangle$$ is $$\frac{1}{\sqrt 2}\left(|0\rangle + |1\rangle \right)$$,

you can work out that $$H(|+\rangle) = |0\rangle$$

So, $$CNOT(H|+\rangle \otimes |+\rangle)$$ $$= CNOT(|0\rangle \otimes |+\rangle)$$ $$= |0\rangle \otimes |+\rangle$$

You can also check that $$H^2 = I$$ or that the Hadamard gate is both Unitary and Hermitian. $$H = H^\dagger$$ $$H^\dagger = H^{-1}$$ So, $$H = H^{-1}$$, the Hadamard gate is its own inverse.

What you have done is not projection of $$|00\rangle$$ to get the state $$|\phi^+\rangle$$, but you just applied the unitary that takes the computational basis to the Bell basis.

As you said in the comments, true, if you measure a state in a basis, you will get one of the basis vectors as outcomes with different probabilities. To see that, express the state in hand in the measurement basis.

For ex:

$$|00\rangle = \frac{1}{\sqrt 2} (|\phi^+\rangle + |\phi^-\rangle)$$ so you will get $$|\phi^+\rangle$$ with 50% probability and $$|\phi^-\rangle$$ with 50% probability.

Similarly, on expressing $$|++\rangle$$ in the Bell basis as: $$\frac{1}{\sqrt 2}(|\phi^+\rangle + |\psi^+\rangle)$$ you get each of those states with 50% probability on measuring.

• But, why it's not producing a Bell state? Because, when I measure a state in any basis, the outcome could only be one of the basis states... isn't it? Jun 20 '19 at 9:48

The four Bell states are $$|\Phi_{\pm}\rangle=(|00\rangle\pm|11\rangle)/\sqrt{2}\qquad |\Psi_{\pm}\rangle=(|01\rangle\pm|10\rangle)/\sqrt{2}.$$ So, let's consider what happens then we try and measure in the Bell basis, i.e. project onto one of these four states. If we started with the state $$|00\rangle$$, then we can write it as $$|00\rangle=\frac{1}{\sqrt{2}}(|\Phi_+\rangle+|\Phi_-\rangle).$$ Hence, we would get the answers $$|\Phi_{\pm}\rangle$$ each with probability $$\frac12$$.

Imagine, instead, that your initial state is $$|++\rangle$$. You can write this as $$|++\rangle=\frac{1}{\sqrt{2}}(|\Phi_+\rangle+|\Psi_+\rangle),$$ so when you measure it in the Bell basis, you get the answers $$|\Phi_+\rangle$$ or $$|\Psi_+\rangle$$ with 50:50 probability.