I have a Bell state ${\Psi}^{-}= \frac{1}{\sqrt2} (|01\rangle - |10\rangle).$

How can I prove that this state is invariant (up to a global phase), when doing the same unitary $U$ on each qubit?

That is, how can I show that, for all $2\times 2$ unitaries $U$, we have:

$$(U\otimes{U})|{\Psi}^{-}\rangle = e^{i\theta}|{\Psi}^{-}\rangle?$$

Actually I don’t really understand if $U$ is generic, or it’s some particular operation. Is there some data missing?

I know $U$ acts only one one qubit but the last expression acts on two qubits.

I’m really lost.

Any help would be appreciated. Thanks!

  • 4
    $\begingroup$ Welcome to QCSE. This sounds like homework/coursework; you did a reasonable job of explaining the question and where you're getting stuck I think but the more generalized you can make it, the better it will be. $U$ is a generic unitary that only acts on one qubit, but $U\otimes U$ - that is, $U$ tensored with itself - will act on two qubits. You're asked to prove that $\vert\Psi^-\rangle$ is an eigenstate of $U\otimes U$ for any $U$. You might want to experiment with different $U$, such as any of the Pauli gates or the Hadamard gate. $\endgroup$ Jun 8, 2021 at 20:15

2 Answers 2


You can find $$U|0\rangle \otimes U|1\rangle$$ and $$U|1\rangle \otimes U|0\rangle$$ in the standard basis, where $$|0\rangle \equiv \begin{bmatrix} 1\\0 \end{bmatrix},\\ |1\rangle \equiv \begin{bmatrix} 0\\1 \end{bmatrix},$$ $$|01\rangle = |0\rangle \otimes |1\rangle,\\ |01\rangle = |1\rangle \otimes |0\rangle.$$ You know that a 2x2 unitary $U$ should take this form: $$ U = e^{i\phi/2}\begin{bmatrix}e^{i\phi_1}\cos{\theta} & e^{i\phi_2}\sin{\theta}\\ -e^{-i\phi_2}\sin{\theta} & e^{-i\phi_1}\cos{\theta} \end{bmatrix} $$ (see https://en.wikipedia.org/wiki/Unitary_matrix). Then,

  • $ U|0\rangle =\\ =e^{i\phi/2}\begin{bmatrix}e^{i\phi_1}\cos{\theta} & e^{i\phi_2}\sin{\theta}\\ -e^{-i\phi_2}\sin{\theta} & e^{-i\phi_1}\cos{\theta} \end{bmatrix}\begin{bmatrix} 1\\0 \end{bmatrix}\\ = e^{i\phi/2}\begin{bmatrix}e^{i\phi_1}\cos{\theta}\\-e^{-i\phi_2}\sin{\theta}\end{bmatrix} ,$
  • $ U|1\rangle =\\ =e^{i\phi/2}\begin{bmatrix}e^{i\phi_1}\cos{\theta} & e^{i\phi_2}\sin{\theta}\\ -e^{-i\phi_2}\sin{\theta} & e^{-i\phi_1}\cos{\theta} \end{bmatrix}\begin{bmatrix} 0\\1 \end{bmatrix}\\ =e^{i\phi/2}\begin{bmatrix} e^{i\phi_2}\sin{\theta}\\ e^{-i\phi_1}\cos{\theta} \end{bmatrix}.$

Using this,

  • $U|0\rangle \otimes U|1\rangle =\\ = e^{i\phi/2}\begin{bmatrix}e^{i\phi_1}\cos{\theta}\\-e^{-i\phi_2}\sin{\theta}\end{bmatrix} \otimes e^{i\phi/2}\begin{bmatrix} e^{i\phi_2}\sin{\theta}\\ e^{-i\phi_1}\cos{\theta} \end{bmatrix}\\ = e^{i\phi}\begin{bmatrix} e^{i\phi_1}\cos{\theta}e^{i\phi_2}\sin{\theta}\\ e^{i\phi_1}\cos{\theta}e^{-i\phi_1}\cos{\theta}\\ -e^{-i\phi_2}\sin{\theta}e^{i\phi_2}\sin{\theta}\\ -e^{-i\phi_2}\sin{\theta}e^{-i\phi_1}\cos{\theta} \end{bmatrix}\\ = e^{i\phi}\begin{bmatrix} e^{i(\phi_1+\phi_2)}\cos{\theta}\sin{\theta}\\ \cos^2{\theta}\\ -\sin^2{\theta}\\ -e^{-i(\phi_2+\phi_1)}\sin{\theta}\cos{\theta} \end{bmatrix},$

  • $U|1\rangle \otimes U|0\rangle =\\ e^{i\phi/2}\begin{bmatrix} e^{i\phi_2}\sin{\theta}\\ e^{-i\phi_1}\cos{\theta} \end{bmatrix} \otimes e^{i\phi/2}\begin{bmatrix}e^{i\phi_1}\cos{\theta}\\-e^{-i\phi_2}\sin{\theta}\end{bmatrix}\\ =e^{i\phi} \begin{bmatrix} e^{i\phi_2}\sin{\theta}e^{i\phi_1}\cos{\theta}\\ -e^{i\phi_2}\sin{\theta}e^{-i\phi_2}\sin{\theta}\\ e^{-i\phi_1}\cos{\theta}e^{i\phi_1}\cos{\theta}\\ -e^{-i\phi_1}\cos{\theta}e^{-i\phi_2}\sin{\theta} \end{bmatrix}\\ =e^{i\phi} \begin{bmatrix} e^{i(\phi_1+\phi_2)}\cos{\theta}\sin{\theta}\\ -\sin^2{\theta}\\\cos^2{\theta}\\-e^{-i(\phi_1+\phi_2)}\cos{\theta}\sin{\theta}\end{bmatrix}. $ Then, $$U|0\rangle \otimes U|1\rangle - U|1\rangle \otimes U|0\rangle\\ =e^{i\phi} \begin{bmatrix} 0\\ \cos^2{\theta}+\sin^2{\theta}\\ -\sin^2{\theta}-\cos^2{\theta}\\0\end{bmatrix}\\ =e^{i\phi} \begin{bmatrix} 0\\1\\ -1\\0\end{bmatrix}. $$ I didn't include the normalization constant of $\frac{1}{\sqrt{2}}$ in these calculations, but clearly, applying $U \otimes U$ to this Bell state results in the same state multiplied by $e^{i\phi}$.


Any unitary matrix can be written as $SU(2)$ matrix discarding global phase. The general form for 2x2 $SU(2)$ is $\begin{pmatrix}a & b\\-b^*&a^*\end{pmatrix}$, where $a$ and $b$ are two complex number satisfying $|a|^2+|b|^2=1$, star stands for complex conjugate.

Question becomes whether $\langle \Psi|U\otimes U|\Psi\rangle$. Replace $|\Psi\rangle = \frac{1}{\sqrt2} (|01\rangle - |10\rangle)$ into $\langle \Psi|U\otimes U|\Psi\rangle$, and change $\langle 0|U|0\rangle = a, \langle 0|U|1\rangle=b,\langle 1|U|0\rangle=-b^*,\langle 1|U|1\rangle=a^*$, we can get $\langle \Psi|U\otimes U|\Psi\rangle=1$ .

Your question is also refer to that the maximal entangled state remain maximal entangled with the operation of LOCC.


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