# Why can every Bell state be written as $|\beta_{xy}\rangle=\frac1{\sqrt2}(|0,y\rangle + (-1)^x|1,\bar y\rangle)$?

In Nielsen and Chuang, there's the following paragraph:

The mnemonic notation $$|\beta_{00}\rangle, |\beta_{01}\rangle, |\beta_{10}\rangle, |\beta_{11}\rangle$$ may be understood via the equations $$|\beta_{xy}\rangle \equiv \frac{|0,y\rangle + (-1)^x|1,\bar y\rangle}{\sqrt{2}},$$ where $$\bar y$$ is the negation of $$y$$.

I can tell that this is true by substituting the value of x and y into the equation, however, I am wondering if there's any intuition for why this equation holds true for all bell states?

There are four Bell states, which is true because they form a basis of the two-qubit Hilbert space. Four options can be parametrized by two bits of information, so we may as well label those two bits by $$x$$ and $$y$$.
Since all of the basis states are entangled, and there are only two degrees of freedom per qubit, the first qubit must be $$|0\rangle$$ in one branch of the superposition and $$|1\rangle$$ in the other branch; the same holds true for the second qubit. By simply labelling the first term to have the first qubit being $$|0\rangle$$, we can call the first term $$|0,y\rangle$$ without loss of generality. We know immediately that the second term must take the form $$|\bar{0},\bar{y}\rangle=|1,\bar{y}\rangle$$, where $$(\bar{0},\bar{1})=(1,0)$$. That takes care of the intution behind $$y$$.
The only other degree of freedom left to us is the relative phase, which must be parametrized by the other bit $$x$$. We need the basis to be orthonormal, so the only possible phases that work are $$\pm 1$$. This is what the final bit does.
• Hi! I am wondering if you could expand on why the 2nd term has to take on the form $|\bar{0},\bar{y}\rangle=|1,\bar{y}\rangle$? Aug 4 '21 at 16:15
• If you try any other state, there won't be maximal entanglement. Notably, if you try using $|\bar{0},y\rangle$ or $|0,\bar{y}\rangle$, the state will be separable. If you try something like $\left(\alpha|{0}\rangle+\beta|\bar{0}\right)\otimes|\bar{y}\rangle$, for example, you'll find maximal entanglement when $\alpha=0$ for any measure of entanglement (eg the purity of the reduced density matrix) Aug 4 '21 at 17:57