# CNOT's control qubit preceded by Hadamard: Why is sqrt also applied to target qubit?

I'm going through the Quantum computing for the very curious (Matuschak & Nielsen) tutorial. The example shows a $$\operatorname{CNOT}$$ gate where the input of the control bit is preceded by a Hadamard gate

The output of the $$H$$ gate (@ point 2) is $$\frac{|0\rangle+|1\rangle}{\sqrt{2}}$$. So far so good.

The tutorial then writes:

For the two qubits it doesn't affect the second qubit at all, and so it takes $$|00\rangle$$ to $$\frac{|00\rangle + |10\rangle}{\sqrt 2}$$

The end result is: $$\frac{|00\rangle + |11\rangle}{\sqrt 2}$$

(I think) I understand:

• the first qubit of $$|00\rangle$$ is for the control bit and the second for the target qubit
• a $$|10\rangle$$ would output a $$|11\rangle$$ in a $$\operatorname{CNOT}$$ gate

However, the piece I don't understand is why is the $$\frac{1}{\sqrt{2}}$$ also part of the target qubit. At point A, the input of the target qubit, remains unchanged a $$|0\rangle$$, no? Why does it suddenly have a $$\frac{1}{\sqrt 2}$$?

• quantum states have norm $1$ May 5 '19 at 17:33
• Hi, tokosh. Welcome to Quantum Computing SE! It is preferable that you use MathJax to typeset your posts. I've edited the question on your behalf. May 6 '19 at 5:09
• @SanchayanDutta Thanks. I tried to use tex, but somehow I couldn't make it display correctly. May 6 '19 at 5:13

You should forget about amplitudes as only applying to single qubits. Amplitudes apply to the states of an entire system.

Mathematically, this comes from a property of the tensor product: $$(\alpha|0\rangle)\otimes(\beta|0\rangle)=(\beta|0\rangle)\otimes(\alpha|0\rangle)=(\alpha\beta|0\rangle)\otimes |0\rangle=|0\rangle\otimes(\alpha\beta|0\rangle)=\alpha\beta(|0\rangle\otimes |0\rangle)$$ we can move numbers around the tensor product arbitrarily.

Physically, what does this mean? Well, if I know my first qubit has a probability amplitude $$\alpha$$ for being in state $$|0\rangle$$, and the second qubit is guaranteed to be in the state $$|0\rangle$$, then what is the probability amplitude for the two qubits together to be in the state $$|00\rangle$$? It's $$\alpha$$. Even if you're not yet comfortable with the maths of probability amplitudes, go back to probabilities and check my claim is consistent with what you know: if the first qubit is $$|0\rangle$$ with probability $$|\alpha|^2$$, and the second is guaranteed to be $$|0\rangle$$, what that the probability that the first is $$|0\rangle$$ and the second is $$|0\rangle$$? $$|\alpha|^2$$.

In general, you can do any computation without worrying about the norm because you cannot actually measure the norm of a quantum state.

In the simplest superposition $$a |0\rangle + b |1\rangle$$, you know that the probability to measure $$|0\rangle$$ (resp. $$|1\rangle$$) is proportional to $$|a|^2$$ (resp. $$|b|^2$$). Therefore, as the total must be equal to 1, if we want to directly consider the squared norms as probabilities, it implies $$|a|^2 + |b|^2 = 1$$.

As a convention, it is often supposed that the norm of a quantum state is $$1$$. Some authors don't bother to write it though.

Similarly, the quantum gates are unitary matrices. The Hadamard gate $$H$$ from your question verifies indeed $$HH^* = I$$. A consequence is that its eigenvalues are all on the unit circle, therefore it also won't change the norm of a quantum state (it is an isometry for the Euclidean norm).

A related question is whether the phase factor (a multiplicative factor of norm 1, see Wikipedia) matters. That is, are $$|0\rangle$$ and $$e^{i\pi/3}|0\rangle$$ different?

The answer (see here for more details) is that you don't really care about the phase because all experimental values are averages obtained as real eigenvalues of an Hermitial operator, hence the phase is ignored. However, the relative phases between states matter, like in $$|0\rangle + e^{i\pi/3}|1\rangle$$.

• I'm afraid I still don't get it. What would happen if there was another H gate in the target gate (between A and cross-circle)? What would the state be before the CNOT? May 6 '19 at 2:20