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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?

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  • $\begingroup$ 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. $\endgroup$
    – Elio Fabri
    Feb 21, 2019 at 14:00
  • $\begingroup$ @ElioFabri I think we need orthogonal basis. $\endgroup$
    – Lwins
    Feb 21, 2019 at 14:02
  • $\begingroup$ I think we need orthogonal basis. Sure. But $\{|0\rangle, i\,|1\rangle\}$ is orthogonal. $\endgroup$
    – Elio Fabri
    Feb 21, 2019 at 14:05
  • $\begingroup$ @ElioFabri You are right. I've modified the OP. $\endgroup$
    – Lwins
    Feb 21, 2019 at 14:43
  • $\begingroup$ 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. $\endgroup$
    – Elio Fabri
    Feb 22, 2019 at 13:57

1 Answer 1

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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.

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  • $\begingroup$ 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? $\endgroup$
    – Lwins
    Feb 20, 2019 at 19:38
  • $\begingroup$ 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. $\endgroup$
    – rnva
    Feb 20, 2019 at 21:41
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    $\begingroup$ @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. $\endgroup$
    – Elio Fabri
    Feb 21, 2019 at 14:01

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