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.

Overall, my intution is that one bit tells us how the qubits pair with each other, the other bit tells us the relative phase, and these are the only possible degrees of freedom in an orthonormal basis of two qubits where all basis states are maximally entangled.

  • $\begingroup$ Hi! I am wondering if you could expand on why the 2nd term has to take on the form $|\bar{0},\bar{y}\rangle=|1,\bar{y}\rangle$? $\endgroup$
    – Alexia.
    Aug 4 '21 at 16:15
  • $\begingroup$ 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) $\endgroup$ Aug 4 '21 at 17:57

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