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:

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

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    $\begingroup$ 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? $\endgroup$
    – DaftWullie
    Commented Jun 20, 2019 at 15:31
  • $\begingroup$ You are right. I think I misunderstood my problem. Projecting those states onto the Bell states is not about applying the unitaries. $\endgroup$ Commented Jun 20, 2019 at 23:48

2 Answers 2


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.

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    $\begingroup$ 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? $\endgroup$ Commented Jun 20, 2019 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.


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