Prove that a Bell state is invariant under the single-qubit gate acting on both qubits

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!

• 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. Jun 8 at 20:15

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.