An intuitive explanation of the basis-independency of EPR pair

As usual, we write the EPR pair as $$\frac{1}{\sqrt2}(\left|00\right> + \left|11\right>).$$

A property of the EPR pair is that this definition is basis-independent, which means $$\frac{1}{\sqrt2}(\left|00\right> + \left|11\right>) = \frac{1}{\sqrt2}(\left|uu\right> + \left|vv\right>)$$ as long as $$\left|u\right>$$ and $$\left|v\right>$$ are orthogonal in $$\mathbb{R}^2$$. (As pointed out by @ElioFabri, if we set $$\left|u\right>=\left|0\right>$$ and $$\left|v\right>=i\left|1\right>$$ then the equailty no longer holds. Hence we cannot put them in $$\mathbb{C}^2$$.)

Question: Is there any way to explain this other than an elementary direct computation?

• A property of the EPR pair is that this definition is basis-independent I can't see it's true. If $|u\rangle=|0\rangle$ and $|v\rangle=i\,|1\rangle$ your identity doesn't hold. Feb 21, 2019 at 14:00
• @ElioFabri I think we need orthogonal basis. Feb 21, 2019 at 14:02
• I think we need orthogonal basis. Sure. But $\{|0\rangle, i\,|1\rangle\}$ is orthogonal. Feb 21, 2019 at 14:05
• @ElioFabri You are right. I've modified the OP. Feb 21, 2019 at 14:43
• as long as $|u\rangle$ and $|v\rangle$ are orthogonal in R2. I'm not sure how to interpret that. Do you mean that a real Hilbert space should be used? I would suggest the following. Starting from a base $\{|0\rangle, |1\rangle\}$, in order to keep the EPR pair invariant you may only use new bases obtained from the former through a real orthogonal matrix. Feb 22, 2019 at 13:57

Here's a neat way to see this.

The main element is vectorization. Vectorization of a matrix $$A=\sum A_{i,j}\vert i \rangle\langle j\vert$$ is defined by $$\vert\text{vec}({A})\rangle = \sum A_{i,j}\vert i \rangle\vert j\rangle$$. Essentially, a bra turns into a ket. The cute thing about this is that you've now established a correspondence between matrices and bipartite pure states.

Next, consider what the identity matrix vectorizes to. You have $$\sum_i \vert i \rangle\langle i\vert \rightarrow \sum_i \vert i \rangle\vert i\rangle$$, which is an unnormalized Bell state. Indeed, if you chose to do it in two dimensions and the computational basis, you have your state $$\vert 0\rangle\langle 0\vert + \vert 1\rangle\langle 1\vert \rightarrow \vert\Phi^+\rangle = \vert 00 \rangle + \vert 11\rangle$$ upto normalization. You might be interested in the other Pauli matrices and what they do. I leave it to you to work it out but they simply give you the other Bell states.

But wait, you're free to write the identity as $$\sum_i\vert i \rangle\langle i\vert$$ in any orthonormal basis. So indeed you can have $$i = \{0,1\}$$ or $$i = \{+,-\}$$ or as you wanted to write it, $$i = \{u,v\}$$ for some $$\vert u\rangle \perp \vert v\rangle$$.

In some sense, the intuition is that because the identity matrix vectorizes to the $$\vert\phi^+\rangle$$ state, the freedom in representing the identity using any orthonormal basis carries over to the freedom in representing the vectorized state.

• Elegant! I think there’s only one (maybe trivial) thing I cannot see through. That is, it is not obvious to me that the vectorization is base-independent. Would you please give me some hints? Feb 20, 2019 at 19:38
• I'm not sure if that statement is true. What I can tell you is that there is a vectorization identity $A\otimes B^T\vert\text{vec}(X)\rangle = \vert\text{vec}(AXB)\rangle$. What you are doing when you go from $\vert 00\rangle + \vert 11\rangle$ to $\vert uu\rangle + \vert vv\rangle$ is to apply the unitary $U\otimes U^*$ (note the conjugate!) on $\vert\text{vec}(I)\rangle$. But when you apply the vectorization identity, you get $\vert\text{vec}(UIU^\dagger)\rangle = \vert\text{vec}(I)\rangle$. I'm not sure if there's a more general argument.
– rnva
Feb 20, 2019 at 21:41
• @nr2618 I'm not sure if that statement is true. It can't as vectorization isn't a linear map, turning a bra into a ket. Feb 21, 2019 at 14:01