# Why do Bell states have all real coefficients?

Canonically, the four Bell or EPR states for 2-qubit systems are given by:

$$|\Phi^{\pm}\rangle = \frac{1}{\sqrt{2}} \left( |00\rangle \pm |11\rangle\right)$$

$$|\psi^{\pm}\rangle = \frac{1}{\sqrt{2}} \left( |01\rangle \pm |10\rangle\right)$$.

I'm wondering why is it that we never mention states such as $$|\psi\rangle = \frac{1}{\sqrt{2}} \left( |00\rangle + i|11\rangle\right)$$? That state is also entangled, and it is maximally entangled (if you take either partial trace, you get the identity matrix, thus maximally mixed). Is there a reason why we don't call those "Bell states", or is it simply the case that there is no unique way of defining them?

I would call $$|\psi\rangle = \frac{1}{\sqrt 2} (|00\rangle + i|11\rangle)$$ also a Bell state due to the properties that you have already listed. There is nothing really special of $$|\Phi^+\rangle$$ over $$|\psi\rangle$$.

Though, it is easier to calculate with real coefficients and the four standard Bell states do form a orthonormal basis of the 2-qubit Hilbert space. (But you can also construct an ONB with maximally entangled states with non-real coefficients, of course.)

It's because we can get away with it. It's always easier to deal with only real coefficients.

For dimension $$d$$ there is a very simple and convenient construction of generalized Bell states: $$|\Psi_{ij}\rangle = (I \otimes X^iZ^j)|\phi^+\rangle,$$ where $$X,Z$$ are the $$d$$-dimensional shift and clock operators, and $$|\phi^+\rangle = \frac1{\sqrt d}\sum_{i=0}^{d-1}|ii\rangle$$. Note that they always form an orthonormal basis and are maximally entangled, but for $$d \ge 3$$ they have complex coefficients.

First I thought this was necessarily so, but as Craig Gidney pointed out one can build a real Bell basis for power of 2 dimensions simply by taking tensor products (and permuting subsystems) of the $$d=2$$ case. It is unclear, however, whether it exists in all dimensions.

Any maximally entangled state can be written in the form $$(I \otimes U) |\phi^+\rangle$$ for some unitary $$U$$, and the inner product of two states in this form is given by $$\operatorname{tr}(V^\dagger U)/d$$ (if they are built with unitaries $$U,V$$). Therefore finding a real Bell basis in dimension $$d$$ reduces to finding a set of $$d^2$$ real unitaries $$\{U_i\}$$ such that $$\operatorname{tr}(U_i^\dagger U_j) = d \delta_{ij}$$.

• Don't power of 2 values of d have real-only bases via tensor products of the d=2 basis? Commented Jul 22 at 14:36
• You're right. And also I managed to brute force a real Bell basis for $d=3$. I'll edit the answer. Thanks for the remark. Commented Jul 22 at 15:35
• You should include the brute forced basis for d=3. See also en.wikipedia.org/wiki/Gell-Mann_matrices Commented Jul 22 at 16:04
• It was mistaken. Commented Jul 22 at 17:16