# 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? – DaftWullie 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. – Hasan Iqbal 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? – Hasan Iqbal 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.