# Simple algebraic explanation for normalizing states

I'm wondering how a set of three 0-state qubits, each prepared identically, like so: When considered together, may produce the fraction: along with their combined states.

This is the entire circuit. Is this an indication of entanglement? Understanding the algebra here will help me to understand more generally how the entanglement occurs. Thanks in advance. The coefficient $$\frac{1}{\sqrt{2^3}}$$ is the normalization factor: if you have a 3-qubit state that is an equal superposition of 8 basis states, its norm still has to be 1; thus the squared amplitude of each basis state has to be $$\frac{1}{8}$$, and the amplitude will be square root of that, i.e. exactly $$\frac{1}{\sqrt{2^3}}$$.

You can also obtain that coefficient from the expansion into a tensor product below.

Speaking of entanglement, these qubits are not actually entangled. You can see that if you try to represent their state as a tensor product of states of 3 individual qubits:

$$\frac{1}{\sqrt{2^3}}(|000\rangle + |001\rangle + ... + |111\rangle) = \frac{1}{\sqrt2}(|0\rangle + |1\rangle) \otimes \frac{1}{\sqrt2}(|0\rangle + |1\rangle) \otimes \frac{1}{\sqrt2}(|0\rangle + |1\rangle)$$

Being able to find such representation is exactly the definition of the qubits not being entangled (in other terms, the state is separable).

• Super helpful, thank you! – neutrino Mar 26 '20 at 13:23
• however, these fractions are not quite the same. – neutrino Mar 26 '20 at 14:06
• Which ones? I might've made a typo somewhere but I don't see it... – Mariia Mykhailova Mar 26 '20 at 23:19