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Suppose I take the Bell state $\Phi^+ = \vert 00\rangle + \vert 11\rangle$.

Is it true that I can choose any basis with basis vectors $\vert b_0\rangle$ and $\vert b_1\rangle$ and reexpress the same Bell state as

$$\Phi^+ = \vert b_0b_0\rangle + \vert b_1b_1\rangle?$$

Now suppose I apply a unitary matrix $U$ on the first qubit whose eigenvalues are $\pm 1$ and eigenvectors are $\vert b_0\rangle$ and $\vert b_1\rangle$. Then I obtain the state

$$(U\otimes I)\Phi^+= \vert b_0b_0\rangle - \vert b_1b_1\rangle$$

Does measuring the first qubit in the $U$ basis automatically put the second qubit in either $\vert b_0\rangle$ and $\vert b_1\rangle$? That is, no matter the choice of $U$, the action of $U$ on one qubit of $\Phi^+$ and subsequent measurement of that qubit in its eigenbasis always collapses the second qubit into one of the eigenvectors of $U$?

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    $\begingroup$ Your first statement is only true so long as there are no complex numbers in the definition of your basis. If you want something similar for all bases, you have to go with perfect anti-correlation instead of perfect correlation, using $|01\rangle-|10\rangle$. $\endgroup$
    – DaftWullie
    Commented May 20 at 14:48

2 Answers 2

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A maximally entangled state like $|00\rangle+|11\rangle$ can be generally written as $$\frac{|00\rangle+|11\rangle}{\sqrt2} = |u_0, \bar u_0\rangle+|u_1,\bar u_1\rangle$$ for any pair of orthonormal states $\{|u_0\rangle,|u_1\rangle\}$, with $|\bar u_i\rangle$ denoting the state whose coefficients are the complex conjugates of those of $|u_i\rangle$. For example, you have $$|00\rangle+|11\rangle = \lvert{+}+\rangle + \lvert{-}{-}\rangle = \lvert LR\rangle + \lvert RL\rangle, $$ where $\lvert L\rangle\equiv \frac1{\sqrt2}(|0\rangle+i|1\rangle)$, $\lvert R\rangle\equiv \frac1{\sqrt2}(|0\rangle -i|1\rangle)$.

This follows from the following general observations:

  1. Any maximally entangled bipartite pure state $|\Psi\rangle$ can be written as $$ |\Psi\rangle = (U\otimes I)|m\rangle = (I\otimes U^T)|m\rangle,$$ for some unitary $U$, and with $|m\rangle\equiv\frac{1}{\sqrt d}\sum_{i=1}^d|i,i\rangle$ denoting the "canonical" maximally entangled state. Explicitly, if $|\Psi\rangle=\frac{1}{\sqrt d}\sum_i |u_i,v_i\rangle$ (any maximally entangled state can be given this form), then the unitary $U=\sum_{ij} |j\rangle\!\langle \bar v_j|u_i\rangle\!\langle i|$ works (note that this is unitary because a product of two unitaries).
  2. It immediately follows that any maximally entangled state $|\Psi\rangle$ can be written as $$|\Psi\rangle = (U\otimes \bar U)|\Psi\rangle$$ for any local unitary $U$. This is the generalisation of the statement above about $|00\rangle+|11\rangle$.

See also this related answer of mine on physics.SE.

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  • $\begingroup$ For the unitary $U = \begin{pmatrix} 1 & 0 \\ 0 & i\end{pmatrix}$, we have $U = U^T$. We also have $U\vert L\rangle = \vert -\rangle$ and $U^T\vert R\rangle = \vert +\rangle$. Doesn't this turn $(U\otimes U^T)\vert\Phi^+\rangle$ into $\vert-+\rangle + \vert +-\rangle$, not $\vert++\rangle + \vert --\rangle$? $\endgroup$
    – Joey
    Commented May 20 at 15:58
  • $\begingroup$ @Joey indeed, you're right. I should have written $\bar U$ instead of $U^T$ in the last equation. This gives $(U\otimes \bar U)|\Psi\rangle=(U^\dagger U\otimes I)|\Psi\rangle=|\Psi\rangle$ as you expect $\endgroup$
    – glS
    Commented May 20 at 16:20
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For your first question: If you pick a change of basis $U$ such that $|b_0\rangle = U|0\rangle$ and $|b_1\rangle = U|1\rangle$, then you are asking for conditions such that $$ (U\otimes U)|\Phi^+\rangle = e^{i\phi}|\Phi^+\rangle \tag{1} $$ (since presumably you don't care about global phase). One nice property of $|\Phi^+\rangle$ is that you may write $$ (U\otimes U)|\Phi^+\rangle = (U^T U\otimes I)|\Phi^+\rangle, \tag{2} $$ and so to satisfy (1) you are equivalently looking for the condition that $U^T U \propto I$. This certainly does not hold in general.

For your second question: Yes, you can think of this as performing a measurement using the POVM $$ \{|b_0\rangle \langle b_0|\otimes I, |b_1\rangle \langle b_1|\otimes I\}. \tag{3} $$ By inspection, the 'collapsed' post-measurement state of the second qubit will be $|b_i\rangle$, conditioned on outcome $b_i$ for the first bit.

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  • $\begingroup$ Could you clarify that even if $U^TU\neq I$, it still is true that applying $U$ to the first qubit and measuring in the $U$-eigenbasis collapses the second state to $\vert b_0\rangle$ or $\vert b_1\rangle$? Does applying $U$ to the first qubit even matter or is the outcome the same in both cases (applying $U$ and then measuring vs just measuring)? $\endgroup$
    – Joey
    Commented May 20 at 15:36

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